Note: Some of the questions on this section of the website reflect my opinions, and they are just that - my opinions, and not necessarily to be taken as the gospel truth. Everyone has different ideas of what it takes to build a good sounding, high-quality amplifier. Some manufacturers will insist only on original parts, cloth wire, etc. while others use only the best quality components. Some insist only on aluminum chassis while others say steel is the only correct choice. Because tone is such a subjective thing, there can be no single "right" answer to some of these questions. Use your ears and formulate your own opinions.
Q:What is the best type of resistor to use - metal film, carbon film, or carbon composition?
A: It depends on what you are trying to accomplish. If you want the absolute lowest noise, metal film is the only way to go. However, some people prefer the tone of carbon composition resistors, which they say are subjectively "warmer". Unfortunately, carbon composition resistors have notoriously poor drift and noise characteristics, and are responsible for most of the "crackling" and "frying" sounds heard in most old Fenders and other type amps. Carbon film resistors are a good compromise between the low noise and excellent stability of metal film and the subjective "warmth" of carbon composition. Larger wattage resistors are quieter than smaller ones, due to the geometry of the part, so you should always use the highest power rating you can find, particularly in carbon composition resistors. In addition, most lower power resistors, such as 1/2W and 1W units, are only rated for a maximum of 250V to 350V. I recommend using only 2W 750V resistors for plate load resistors and other locations where large voltage swings or high turn-on voltage surges are expected. The first stage of the amplifier and other low-level stages, such as reverb recovery amps, are the most critical for noise, because the signal level is lowest at that point, so proper attention to the type of resistor used will have the most impact on overall amplifier noise performance.
Q:What kind of resistors do you recommend for power supply decoupling use?
A: I recommend 5W (or a higher power rating if needed) 1% 750V metal oxide resistors for this application. They have higher surge/overload capability than the film types, excellent long-term stability, and flameproof construction. Note that lower-power metal oxides, such as 1W and 2W are usually only rated for 350V and should be avoided for this application. Even though the voltage across the resistor in normal operation is only 50 volts or so, it can be much higher at turn on, or if a tube fails. Another good type to use is the Ohmite OX/OY series ceramic composition resistors. These don't have the tight temperature coefficient or tolerance (typically 10%) that metal oxides do, but they are designed for heavy surge use.
Q:What about screen grid resistors?
A: I recommend 5W 1% 750V metal oxide flameproof resistors for this application. The flameproof "cement box" resistors are also good, but are usually only available in 5% and 10%, and must be rated for at least 7W or 10W in order to get an appropriate voltage rating, because the 5W and lower units are usually only rated for 350V. In normal operation, the screen grid resistor only has a few volts dropped across it, so a 350V resistor is overkill. However, in the event of an internal tube short, the voltage drop can be much higher, up to the level of the supply, and the current through the screen resistor will increase dramatically, increasing the voltage drop across it. Normally, the fuse will blow fairly quickly, but sometimes not before the screen resistor is toasted. There are two possible failure modes, over-voltage and over-dissipation due to the increased current from the short. If the voltage rating is too low, the resistor element will flash over and either burn up or carbon track and change value, leading to eventual failure down the road. If the voltage rating is made higher than the supply voltage, there is only one possible failure mode of the resistor, over-dissipation. The metal oxide flameproofs or OX/OY ceramics can usually take a higher-than-rated surge for a short time before they burn up. If the fuse is appropriately rated, it will blow before this over-dissipation can harm the resistor. A quick fuse and tube swap and the amp is back in business, instead of having to replace the screen resistors.
Q:What kind of potentiometers do you recommend?
A: I prefer to use sealed RV4 mil-style 2W potentiometers. They are available in both hot-molded carbon and conductive plastic elements. I originally used the conductive plastic Clarostat units, but their inconsistent torque and feel caused me to switch to the hot-molded carbon types from PEC, which I now use exclusively. They are extremely well-built units, even better looking than the Clarostat, but more expensive. These RV4 type pots are more expensive than other types, but their sealed construction, reliability, and precision feel makes them worth the extra cost. UPDATE: I have to admit I was wrong on this issue. After having quite a few amps in the field for going on 18 years now, I am starting to see a few complaints of noisy pots from people who own my amps. Normally, I would just tell them to open the amp up and spray some contact cleaner in the pots. Unfortunately, you can't do this with a sealed pot, so you have no choice but to replace the pot. No pots are going to last forever, so you may as well use a cheaper pot like the Alpha or CTS, that can be cleaned or replaced for a fraction of the cost of one PEC pot. The Alphas have a nice smooth torque feel to them. For dual-ganged pots in PPIMV uses, the PEC still has the best matching between sections, so I would still recommend using them, unless you can find another brand with good matching.
Q:Why do you recommend polypropylene film/foil capacitors?
A: Because they are the best, in terms of temperature stability, and, subjectively, tone. The interior of a guitar amplifier chassis tends to get very hot, particularly in a combo, where the tubes are hanging upside down beneath the chassis and the heat rises upward. Polyester and metallized film capacitors have very poor temperature characteristics compared to polypropylene film/foil, which means the amplifier tone will change as the amplifier heats up. A 5% or 10% drift in value can produce a very noticeable shift in tone, depending upon how the capacitor is used in the circuit, and all caps will drift in the same direction, which is determined by the characteristics of the dielectric material, so the effect is multiplied. Also, polypropylene capacitors are far better in terms of dissipation factor and dielectric absorption, which some people say makes them sound better. I use them because they have the best characteristics, even though they cost quite a bit more than the cheaper polyester caps. The long-term temperature stability and quality construction makes them worth it, in my opinion. The most commonly used coupling cap, the Sprague Orange Drop, is available in polypropylene film/foil, but only in a radial-lead package. These are fine for PC board amplifiers, but some manufacturers use them for point-to-point boards by bending the leads out at the body of the capacitor. This usually leads to cracking of the epoxy seal around the lead entry points, which can allow moisture into the capacitor, leading to drift and potential failure. I use only axial-leaded capacitors, because they are designed for point-to-point style mounting, and the epoxy end seals do not crack when they are mounted on the board.
Q:What type capacitor do you recommend for small values in the pF range, such as in tone controls?Q:I've heard that it is best to run electrolytic capacitors as close as possible to their rated voltage. Is this true?
A: Silver mica or polystyrene film capacitors are the best choices. They have the best temperature stability and most transparent tone. However, the polystyrene film caps in the pF range usually have very tiny leads, which may not be robust enough to stand up to the rigors of road use. There are very few capacitor manufacturers making low-pF value film capacitors in other dielectrics, such as polypropylene or polyester, and they are nearly impossible to find. Ceramic capacitors should be avoided, as they are usually horribly microphonic, have very high odd-order harmonic distortion (the "bad, or harsh-sounding" kind), typically mostly 3rd and 5th and 7th harmonics, and have poor temperature stability. If ceramics must be used, the COG/NPO type has the best temperature stability and much, much lower harmonic distortion - stay away from X7R and Z5U types, and ceramic disk capacitors, in particular. Silver mica capacitors aren't without their problems, however, as they seem to have a high failure rate in the field, generally in the form of leaking DC. If used in a tone stack, for instance, they can make all the pots sound "scratchy" due to the presence of a DC voltage across them from a leaky silver mica treble capacitor.
A: No. Electrolytic capacitor life is not infinite. It is controlled by a number of factors, but mainly the temperature at which it operates, which is set by two things: ambient temperature and internal heating. Ambient temperature is the biggest factor, so if you are running a capacitor in a hot tube amp, be sure to use one with extended temperature rating if possible. The second factor, internal heating, depends on the ripple current in the capacitor and it's internal resistance.
Capacitor life is directly proportional to the ratio of rated voltage/operating voltage. If you run a 500V cap that has a specified lifetime of 10,000 hours at 250V, it will instead last 20,000 hours. If you run it at 500V as specified, however, it will only last 10,000 hours, which is what it is rated for. So, as you see, there is no benefit to running a capacitor right at it's rated voltage, it is better to run them conservatively with respect to voltage and temperature for maximum lifespan.
Film capacitors and ceramic capacitors are even more dramatic - their lifespan is proportional to the rated/operating voltage raised to the 7th power and 3rd power, respectively, so a film cap rated at 10,000 hours at 500V run at 250v will last 1,280,000 hours, and a ceramic cap under the same conditions will last 80,000 hours, while an electrolytic will last 20,000 hours as I mentioned previously.
Go here for more info and a handy web calculator for capacitor lifespan: http://www.illinoiscapacitor.com/techcenter/lifecalculators.asp
Q: What kind of transformers do you recommend?
A: I have my transformers custom-made for the amplifier they are to be used in. I typically use Heyboer or TMI transformers. For most general purpose home-building, Hammond transformers are readily available and reasonably priced.
Q: Do I need to replace my driver tube every time I replace my power tubes?
A: No. This is an "old wive's tale" propagated by the tube vendors who want to sell you more tubes. The "driver" tube, or phase inverter as it is more appropriately known, is under no more stress than any of the other preamp tubes. The output stage of a normal guitar amplifier is AC-coupled, class AB1 or class A1. The "1" suffix indicates that the tubes do not draw grid current during any portion of the input cycle. A "2" suffix indicates grid current during some portion of the cycle, as in class AB2. Since almost all guitar amps are class AB1 or A1 (it takes a DC-coupled cathode-follower or interstage transformer to be able to drive the power tubes into the positive grid region), the phase inverter does not have to supply any current to the grids (it can't really source current anyway - it only sinks current through the plate load resistor, which is quite large, and inherently limits the plate current to a few mA). The AC coupling (capacitor coupling) between the phase inverter and the output tubes precludes any flow of grid current anyway. The output tube grids will merely act as a grid clamp, shifting the bias downward as the output is driven harder. This in no way stresses the phase inverter tube. In addition, the plate load resistors and the bias current in a typical phase inverter are identical to the preamp stage values, indicating that they are operating in the same area, dissipation-wise, so they cannot be "wearing out" any faster. In fact, the reverb driver tube on a typical amplifier is dissipating more power than the phase inverter, and should be replaced more often, if anything. Don't be misled by the higher plate voltage on the phase inverter either, because the cathode is usually sitting somewhere around 30V - 100V above ground, which lowers the plate-to-cathode voltage by that amount. The plate-to-cathode voltage is what determines the power dissipation of the tube, not the absolute plate voltage.
Q: Why do you use point-to-point instead of PCB? Because it sounds better?
A: No. There is no difference in tone between a properly designed PC board and a properly layed out turret board or other point-to-point construction technique. Note that I used the term "properly designed". There is a world of difference in the tone of an improper PCB layout and a proper one, just as there is between a proper PTP layout and an improper one. If high-impedance traces are run too close to other parts of the circuit, unwanted coupling can occur, leading to oscillations in the worst case, or just odd tone caused by frequency cancellations. This is not a problem that is unique to PC boards, as some "gurus" would have you believe. If you run a wire on a PTP layout too close to another wire or component, you will have the same problems. Don't believe the hype about "tiny capacitances" and "co-planar" traces robbing high end. PC boards are used in the gigahertz range, and are perfectly adequate for the audio frequency range. The problem comes in when an amp manufacturer uses a layout guy who doesn't know anything about proper layout of high-impedance circuitry, and when cheap, single-sided PC boards with non-plated-through holes are used. Now, why don't I use PC boards? Simple. A turret board provides better mechanical reliability and ease of maintenance or modification. I have seen far too many PC boards with pads and traces ripped up after one or two component changes. A turret board allows both a good mechanical connection and a good solder connection, and components are easy to remove and replace many times without any damage to the board. Turret boards can even incorporate ground planes or circuit traces, without the downside of worrying about traces and pads lifting up during repeated removal and replacement of parts. I don't use eyelet boards, because they don't allow a good mechanical connection; instead, they depend solely on the solder for mechanical strength at the connection. In addition, solder tends to blob up on the underside of the board, potentially leading to loose solder balls inside the amp, or accidental short circuits. Will I ever use PC boards? Probably at some point, especially on the more complex amplifier designs, mainly for consistency of manufacture or when the order volume outruns our small manufacturing capability, but you can bet that they will be better and more serviceable than any PC board you've ever seen in a guitar amplifier. I have been laying out PC boards for many, many years in my previous audio/video/medical engineering jobs, some of them being very complex multi-layer boards with traces/spaces as fine as 5 mils, some even using surface-mount components on both sides of the board, operating at frequencies from DC to the high MHz region, so I know all the ins and outs of proper PC board design and manufacture, and won't be at the mercy of an outside contract layout guy.
Q: Can I change the preamp cathode or plate resistors to change the gain or tone of my amp?
A: Yes, but you should change the plate resistor value when you change the cathode resistor value, in order to compensate the quiescent plate voltage shift, to rebias the tube near the center of the plate voltage swing, or to the place it was originally biased. If you don't change the plate resistor, the change in the static DC bias point may cause the tube to clip very asymmetrically, and the headroom will be lower, which may or may not be a good thing, depending on your needs. The two go hand-in-hand. If you change the cathode resistor, the plate resistor should (ideally) be changed, and if you change the plate resistor, the cathode resistor should be changed, unless you are designing for an asymmetric bias point. At any rate, it is always a good idea to check the operating conditions of the tube circuit on a scope to make sure things aren't getting out of hand. People tend to view preamp tubes differently from power tubes, but in reality, they are the same animal, just on a different scale. Preamp tubes need to be biased, just as power amp tubes do, and there are ranges that are more optimum than others. This is why it is not a fair test to just swap different tube types and compare tones. A 12AX7 will require different plate and cathode resistors than a 12AT7, for instance, for optimum bias point and tone.
Q: Can I substitute a 12AT7/12AU7/etc. for a 12AX7?
A: Yes and no. While they are the same pinout, the tubes are very different with regards to internal plate resistance, bias current, gain, etc. It won't usually hurt anything to substitute them, but it is not a fair tonal comparison if you don't also change the cathode and plate resistors to optimally bias the tube. A straight swap without regard for circuit operating conditions will lead to incorrect conclusions regarding the characteristic tone of the tube.
Q: How can I find out if the hum in my amp is caused by the filament wiring?
A: Disconnect the 6.3V AC wires from the power transformer, and temporarily connect in a 6V lantern battery. If the hum is caused by filament wiring, or by induction from the filament inside the tube, it will go away with the battery connected. If the hum is caused by something else, such as a ground loop, it will not go away when the filaments are run on DC.
Q: How can I get rid of the hum if it is caused by the filaments?
A: The best way to get rid of the hum is to generate a DC filament supply, but this isn't always practical, because of the high dissipation caused by the large current draw of the filaments. First, make sure the filament wires are run as twisted pairs, then make sure they aren't routed too close to sensitive grid wires or next to coupling capacitors. If that doesn't do it, try "elevating" the center tap of the filament winding (or junction of the two filament reference resistors) to a potential above the cathode voltage of the tubes, typically between 10V - 50V or so. You can generate this voltage with a couple of resistors arranged as a voltage divider from the plate supply. Be sure to bypass the junction of the resistors to ground with a suitable filter capacitor, or you may inject some buzz or noise into the amplifier from the power supply.
Q: What if the hum is not caused by the filaments?
A: If the amp still hums when a battery is substituted for the filament supply, the cause is most likely one of four things: either the AC wiring (plate supply, filament supply, or mains input) is running too close to a sensitive stage, the transformers themselves are inducing hum into sensitive stages, there is too much residual ripple on the main power or bias supplies, or there is ground loop hum caused by a poor grounding scheme in the amplifier. The true cause has to be determined by a process of elimination. Sometimes it is advantageous to pull preamp tubes, one by one, starting at the input, to determine if the hum is getting in to a particular stage.
Q: One intriguing thing about some amps is the 1K resistor used for the screen supply filter instead of an inductor. Does it really work like a small choke? How much inductance would such a resistor really have? Would this inductance be enough to significantly reduce power supply ripple (compared to e.g. a metal film resistor of the same resistance)?
A: The inductance of the resistor is negligible at the frequencies of interest in the power supply. For example, using a typical inductance of 0.04mH for a 1K wirewound resistor, you would get an inductive reactance of 0.03 ohms at 120hz. This, in conjunction with the usual 50uF filter cap, which has a reactance of 26.53 ohms at 120Hz, would give an attenuation of around -0.01dB at 120Hz (the power supply ripple is twice the 60Hz mains frequency because it is full-wave rectified).
By contrast, the 1K resistor itself, which has a reactance of 1K at 120Hz, gives an attenuation of around -32dB at 120Hz in conjunction with the 50uF capacitor, which would reduce the residual 120Hz hum by a factor of around 39 times.
Okay, you might ask, why use a choke? Well, a 6H choke would have a typical resistance of of only 100 ohms or so, so it would drop less voltage than the 1K resistor, for DC. However, for the 120Hz AC ripple riding on the DC voltage, the inductive reactance would make it look like a whopping 4.5k resistor. This 4.5K, along with the 50uF cap, would result in an attenuation of -44dB at 120Hz (ignoring the effect of the 100 ohm choke resistance, which would be negligible compared to the 4.5K reactance of the inductor, but it does help to dampen the overshoot at the corner frequency of 9.2Hz). So, by using a 6H choke, you get 12dB more attenuation of hum (a factor of 4 times less ripple) along with less voltage drop to the screens (more power output) and less preamp hum.
The only thing a large resistor buys you is screen supply sag, which you may or may not like. You can always do both - use a choke and add a largish series resistor for better ripple rejection and screen supply sag.
Q: Why do some amps connect the center tap of the filament winding, or the junction of the two resistors off the filament string, to the cathodes of the output tubes?
A: In a cathode-biased amplifier, the cathode is at a positive voltage, somewhere around 10-40V with respect to ground. If you elevate the filament "reference" above the potential of the cathode by connecting the center tap to this point, you can effectively reduce the amount of hum coupled into the tube. This is because the filament is now positive with respect to the cathode, so the cathode doesn't attract electrons (i.e. hum) from the filament. This is a very inexpensive and easy method of reducing the hum in an amplifier without having to go to a DC filament supply. In a fixed bias amplifier, the output tube cathodes are usually at ground potential, so you have to add a voltage divider from the plate supply to generate the elevated filament reference. You can experiment with the voltage level to determine the value that best minimizes the hum. Be sure to bypass the junction of the resistors to ground with a suitable filter capacitor, or you may inject some buzz or noise into the amplifier from the power supply.
Q: Are cathode followers gain stages?
A: Yes. However, the problem is that people associate "gain stages" with, well...gain. The cathode follower has a maximum theoretical gain of unity, and typically a gain of around 0.5 to 0.7 or so. To us engineering types, a gain of less than unity is still called a gain (that is, unless it's called a loss or an attenuation, in which case the attenuation is the reciprocal of the gain), so a stage with a gain of unity still has a gain - a gain of 1. The cathode follower *is* an amplifier stage, but not a voltage amplifier in the typical sense. It is used as a "buffer" amplifier, which means it has a high input impedance and a low output impedance. This means it does not appreciably load the previous stage it is connected to, and the very low output impedance allows it to drive low impedance loads without much signal voltage loss. For example, if the previous stage had an output impedance of 100K, and you tried to connect a stage with a 10K input impedance to it, you would only get 9% of your original signal when you connected the second stage, because of the voltage divider formed by the 100K output impedance and the 10K input impedance. If you insert a cathode follower with a 1K output impedance and a 1Meg input impedance, the 10K stage can be driven with 90% of the original signal, because there is now effectively a 1K:10K voltage divider instead of a 100K:10K voltage divider. The cathode follower is basically an active "impedance transformer", in this sense. The reason the cathode follower is used in driving a tone stack is not only because the tone control network impedance is relatively low in comparison to the output impedance of the previous stage, so it would cause a loss of gain, but more importantly, the tone stack is a filter network that is designed to ideally be driven from a zero source impedance to achieve it's proper frequency response. The cathode follower provides a very low source impedance that allows the tone stack to work as designed. If the tone stack is driven from too large a source impedance, not only will there be a loss of gain, but there will be a different frequency response to the network, typically quite a few dB loss of the highs. The cathode follower prevents this loss, allowing the tone stack to retain more of it's theoretical frequency response.
Q: Does cathode biasing mean class A?
A: Absolutely not. You can have a cathode-biased class AB or class A amplifier, just as you can have a fixed-bias class A or class AB amplifier.
Q: Is it true that the only true class A amplifier is a single-ended amplifier?
A: No. You can design a true class A single-ended or push-pull amplifier. The presence of a phase inverter tube does not automatically mean the amplifier is class AB, just as the presence of a cathode resistor doesn't automatically mean the amplifier is class A. It is all a function of where the output tubes are biased, and under what voltage/impedance conditions they are operating. In fact, unless you know the plate voltage, plate bias current, and output transformer reflected impedance, you can't tell the class of an amplifier just by looking at the schematic. A push-pull class A amplifier differs from a single-ended class A amplifier tonally in that it cancels even-order harmonics generated in the output stage (but passes through even harmonics generated in the preamp stages, of course). It also has inherent power supply rejection for lower hum and noise levels than a single-ended class A amplifier. Typically, a push-pull class A amplifier will clip rather symmetrically, while a single-ended class A amplifier usually clips asymmetrically, most often rounded on one side while hard-clipped on the other. While both amps are indeed true class A amplifiers, their tones are dramatically different. This further illustrates the fact that there is really no such thing as a "class A" tone.
Q: Are class AB amplifiers actually running in class A at lower volumes, as is commonly claimed?
A: No. They are operating in conditions similar to class A, but they are not actually "class A at lower volumes". Now, what are the differences, you might ask? Well, for one, the class AB amplifier is biased in a more non-linear portion of the characteristic curves, which means it has more harmonic distortion than a true class A amplifier, even when running "clean". Also, the efficiency will be greater than is theoretically possible with a class A amplifier at these levels. There is a very real difference in tone and operating conditions between a true class A 10W amplifier running at say, 1W, and a 10W class AB amplifier running at 1W. Same output level, same overall power level,but a different class of operation, different amount of distortion, different efficiency, and a different tone, even though neither one of them is in cutoff for any portion of the output cycle at that low level. This is due to the bias point differences and load line differences. The differences become even more apparent when the amplifiers are run at their full undistorted output power. The true class A amplifier will have no crossover distortion, while the class AB amplifier will. The average plate current for the true class A amplifier will not change, or will change very little, from idle to full output power, while the average plate current in a class AB amplifier will increase dramatically. This will lead to "sag" in the power supply that doesn't exist in the true class A amplifier, which again results in a tonal change.
Q:Take a single EL34. The RCA manual says the maximum plate dissipation is 25 watts. Put the plate voltage at 400V. Set the bias so that the current is 60mA (24 watts dissipation). Put the plate voltage at 300V. Set the bias so that the current is 80mA. Both give you a dissipation of 24 watts. Both seem to be within operating parameters for this tube, but I'm guessing the tube is going to have a different effect on the signal. Is there a basic, overview explanation on how the tube is going to operate under these two conditions?
A: You cannot bias an amp at any arbitrary voltage for maximum dissipation, and expect to end up with a properly operating class A amp. While the amplifier is biased at idle to a point not exceeding the plate dissipation, which seems okay on the surface, this is only half the story.
When you put a signal into the amp, the plate dissipation changes from the idle value. It can either go up, down, or stay the same, depending on the idle bias point. This is because the plate dissipation is dependent on the average signal at the plate, along with the plate voltage, idle bias current, and plate load impedance.
If the tube is biased so that it just hits plate current cutoff/saturation on top and bottom at the same time, the tube is said to be operating at the limit of class A operation. The average current draw at full signal will be the same as the average current draw at idle with no signal applied. This is because the average value of the unclipped, full power output sine wave is zero - there is an equal area above the center as there is below the center of the waveform. This "ideal" symmetrical bias point can only be achieved at one plate voltage, if the constraint of biasing to exactly max dissipation at idle is applied. It is interesting to note that the plate dissipation will actually drop at full power, compared to idle, because the plate dissipation is equal to the DC input power minus the output signal power, which can be at most equal to 1/2 the idle DC input power. Essentially, the part of the DC input power that is not passed on to the load must be dissipated as heat in the plate of the tube. In a true class A amplifier at idle, no power is developed in the load, so all the power is dissipated in the tube. At full output power (unclipped sine wave), half the DC input power is used to produce the AC output signal, the other half is dissipated as heat in the tube.
Now, if you bias that same tube to max dissipation at a significantly higher voltage/lower current point, it will again be okay at idle, because it isn't exceeding the max dissipation rating for the tube. However, when a signal is applied, the tube will reach saturation before it reaches cutoff on the other side, so the average area above the zero bias line will be larger than the average area below the zero bias line, even when the tube is not clipped on top or bottom. This results in a net average increase in plate dissipation over the no-signal bias point. The tube, which was perfectly happy at idle, may now exceeding the max plate dissipation with a signal applied, and the plates may start to glow red when you start playing, and go back to normal when you stop playing. If you were to use cathode bias instead of fixed bias, the increase in average plate current will cause an increase in cathode voltage, which will act to reduce the plate current and counteract the increase in dissipation, at the expense of a shift in operating point.
If you were to bias the tube to a lower idle dissipation, say 70% of the max dissipation, the increase in plate dissipation that occurs at max signal is now offset by the lower average idle dissipation, so the tube is again operating in a safe area at all times. In this case, however, you no longer have a true class A amplifier, because the plate current will go into cutoff at some point before it hits saturation on the other side, because the bias is offset relative to the available grid swing. The tube has now been biased to class AB operation. This may be fine for a single-ended amp, provided that you don't mind a bit of asymmetrical clipping when you drive the amp hard, which creates a predominant even-order harmonic structure. In the case of a push-pull amp, the other tube "takes over" when the first one goes into cutoff, so the output is not clipped, even though the plate current of the first tube has completely shut off for a portion of time.
As mentioned earlier, there is a bias shift that can occur when the grid signal is AC-coupled into the tube through the typical coupling caps present in most guitar amplifiers. When the grid drive voltage exceeds the cathode voltage, the grid voltage is "clamped" at the peak to a point near the cathode voltage. As the signal is increased, the peak stays at the same point, but the "center" of the grid drive signal is shifted downward. This changes the duty-cycle of the plate waveform, and therefore changes the average dissipation in the output tube, in a manner dependent on the magnitude of the applied signal.
You can use a load-line superimposed on the characteristics of the tube to determine the max plate voltage, idle current, and load impedance that will allow you to bias the tube to the optimum class A operating point. The aforementioned EL34 typically cannot be run at a plate voltage much higher than 250V without exceeding the "true" limiting class A operating range at reasonable distortion levels. You can run a "flatter" load line, i.e. higher impedance load, and colder bias up to a certain point to try and eke out as much symmetry as you can if you are using higher plate voltages. The downside to this higher impedance load is increased harmonic distortion and nonlinearity at "clean" outputs, and higher plate voltage swings that may exceed tube maximum ratings or output transformer insulation ratings. For a better explanation of this, see this paper on biasing -The Last Word On Biasing.
Q: How far can the load line in a class AB amplifier go above the maximum dissipation limit?Q:Any time one tube or set of tubes is said to be "pushing" and the other "pulling" inherently, the amp is not operating Class A, right?
A: You can allow excursions into the "no-man's land" above the max dissipation curve *if* (a) you don't stay there very long and (b) the average dissipation is less than the max plate dissipation. In the real world, a guitar signal generally runs up and down the load line all over the place and doesn't stay at any one dissipation for any particular amount of time, except at idle or full power if you are getting feedback. If the load line crosses over into the max dissipation area at idle, you're screwed. If the load line crosses over in the middle somewhere, there will be a period of time where the signal goes above the max dissipation, and a corresponding period of time where it goes below the max dissipation. In class AB, the time above the max dissipation curve will always be less than the time below the curve, because the tube is in cutoff for some portion of the cycle (the load line goes below the zero current axis at higher plate swings). This allows you to use a load line that goes above the max dissipation curve for some amount of time. Calculating the max distance you can go above the limit is not straightforward, because the voltage and current functions are no longer sinusoidal, they are clipped sinusoids, or clipped square waves.
There are three main variables that control static and active dissipation: (1) plate voltage (2) plate current and (3) reflected load impedance. Once the amp is built, the only one you usually have control over is plate current, so if the tube plates are glowing red at any point in the operation, you have no choice but to lower the plate current if the average dissipation if too high.
In an "ideal" 11W dissipation tube, for example, if you are running 250V and an 875 ohm (3.5K p-p) load, you can't run a bias of more than 32mA without exceeding the dissipation at any point on the load line with either sine wave or square wave excitation. The idle dissipation at this point will be 8W. The worst-case squarewave dissipation of 11W will occur at around 110V peak above the idle point of 250V, or a peak-to-peak swing of 220V around the idle 250V point.
Not true. In a true class A push-pull amplifier, one tube is indeed "pushing" and the other tube is "pulling", but neither reaches cutoff at any point up to the unclipped full output level.
The difference between a class A push-pull output stage and a class B push-pull output stage is primarily the bias point. The class A output stage is biased in the center of the linear portion of the transfer characteristics, and the class B output stage is biased at cutoff.
In a class A amp at idle, with no signal present, each tube is drawing (ideally) an amount of current halfway between cutoff and saturation. For example purposes, let's assume this is 100mA, and assume the plate voltage is 250V. The tube is therefore idling at 250V*100mA = 25W. Now, if a signal is applied to each of the two output tubes from the phase inverter, one tube's current will increase and the other tube's current will decrease by the same amount (because the phase inverter generates two drive signals that are 180 degrees out of phase).Let's assume that the input signal amplitude is enough to drive the first tube to 150mA - in this case the second tube is now at 50mA, because it decreased by the same amount as the increase of the other tube. Taken to the limit, if the input voltage is enough to drive the first tube to 199mA, the second tube will be driven to 1mA, which is right at the limit of cutoff. If this is the maximum clean output of the amp, and the saturation point occurs at 200mA, the amp is operating in perfect class A, because neither tube has ever hit cutoff or saturation up to the maximum clean output of the amp. The output transformer sums these two complete, unclipped, out-of-phase sine waves to generate a sine wave of twice the equivalent level of one side (this is how you get twice the output power of a single tube running class A single-ended).
If you average the current draw over the full sine wave, the increases and decreases cancel each other out and the average current is the same as the idle value, or 100mA. Therefore, there is no change in the current drawn from the supply. The plate dissipation in a class A amplifier output tube goes down with applied signal, because as the current goes up, the voltage goes down, and as the voltage goes up, the current goes down (inverting) and the power dissipated in the tube is the product of the voltage multiplied by the current. At the extreme, when the current is zero and the voltage is at max, the instantaneous dissipation in the tube is zero, and at the other extreme, when the voltage is zero and the current is at max, the instantaneous dissipation is also zero. Since the average plate dissipation in the tube decreases, the tube can be biased at the max rated power at idle.
In a class B amp, since the tubes are biased at 0mA, one tube will increase to a maximum as a signal is applied, but the other tube will stay in cutoff until the second half of the input cycle. Then that tube will increase in current while the other tube stays in cutoff. These two "halves" will sum together in the output transformer to make a full cycle. Since the tubes aren't perfectly linear in the region near cutoff, there is a type of distortion introduced called "crossover distortion", caused when the two tubes "hand off" operation to each other. Because the tubes are in cutoff for half of each cycle, the average plate dissipation is much lower than it is for class A, so more output power can be obtained from the class B configuration.
Class AB operation is obtained by biasing to a point in between class B and class A. for a portion of both half cycles of the input sine wave, both tubes are conducting current in opposition as in class A, but one tube hits cutoff before the other tube hits saturation. The other tube's current keeps increasing up to a maximum, and back down to the "changeover" point, which is above the zero current point. The maximum output power is less than in class B, but more than in class A. The distortion is less than in class B, but more than in class A. Most push-pull guitar amps run in this class of operation.
In a class AB or class B amp, the power dissipation will increase or decrease depending on operating conditions and applied signal level. This is why they have to be biased at a lower dissipation than the max rating of the tube, typically 70% max. If you bias them at the rated dissipation, they will be fine at idle, but they will run too hot when you start playing.
Q: I'm still confused about this "class A" stuff - how about a more general explanation?
A: The general idea is this: Picture a sine wave going positive above zero volts and then negative below zero volts. The way the tube is biased and the circuit is set up determines when the tube (or tubes) is on. It gets a bit shaky when you are dealing with push-pull, which uses tubes in pairs (or quartets, or more, always in multiples of two), or single-ended, which uses one tube (or more, but they are in parallel with the first one, not in push-pull).
Class A: All tubes are "on" for the entire sine wave, in either single-ended or push-pull. In single-ended class A, one tube (or two or more in parallel) amplifies the whole sine wave. In class A push-pull, one tube is "pushing", while the other tube is "pulling", but they are both on for the entire waveform. In push-pull class A, one tube amplifies the sine wave, the other amplifies an upside-down version of it (180 degrees out of phase), and the two complete signals are added together in the output transformer to produce a new sine wave of higher power than you could get with a single tube in the single-ended configuration. This class of operation is characterized by high-linearity and low distortion, which makes it ideal for audiophile-type gear, and very low efficiency (DC power in vs. AC signal out), which means you don't get a lot of watts for your money. All single-ended output stage guitar amps operate in class A mode.
Class B: This class is usually used for push-pull only, although you could have a single-ended class B amp, but it wouldn't sound very good for audio. In class B push-pull, on the top half of the sine wave, one tube is on ("pushing"), while the other tube is off ("idle"). Then, on the bottom half of the sine wave, the other tube is on ("pulling") while the first tube is off ("idle"). Essentially, one tube amplifies the top portion of the sine wave, while the other tube amplifies the bottom portion. The two "halves" are added together in the output transformer to make a new, completely-whole sine wave, so there is no clipping caused by the amplification of only half the wave by each tube. This mode is characterized by very high efficiency, so you get more power than you can in class A, and a type of distortion known as "crossover distortion", which is caused by the non-linear overlapping of the turn-off and turn-on times of the tubes.
Class AB: This class is sort of in-between class A and class B, where the tubes are biased higher than class B, but not as high as class A, so one tube still turns off at some point in the sine wave, but not at the exact zero crossing. This allows more "overlap" in the two halves, so there is little or no crossover distortion, but the efficiency is still much higher than class A, so you get more watts for your buck. Most push-pull guitar amps operate in class AB mode.
There is some contention in the amp tech community about how to define the class of an amplifier. The class designation has to be made at the full, unclipped rated output power of the amplifer, otherwise, it is meaningless. Or, as stated in the RCA receiving tube manual: "The classification depends primarily upon the fraction of input cycle during which plate current is expected to flow under rated full-load conditions". The key phrase being "under reted full-load condtions", which is a requirement for amplifier classifications to be meaningful..
Q: "Aiken claims that if the bias is not at half the saturation current, the amplifier is not true class A. In reality, with a small enough signal, Class A operation can be achieved with virtually any bias between cutoff and saturation. Aiken is relying not on the class definitions themselves, but on the interpretation of an anonymous author of an obscure Army document."
A: Let me refer you to a highly-regarded textbook on the subject, Eastman's "Fundamentals of Vacuum Tubes", 1941, page 332, which is not at all an "obscure" document:
Notice that is says:
"The class AB amplifier operates with a bias somewhere between that required for class A and that required for class B amplification." Earlier in the book, Eastman details the different biasing conditions for class A and class B amplifiers.
And one more, from the also highly-regarded Frederick Terman's "Radio Engineering", Third Edition, 1947:
Notice it says:
"The Class AB amplifier is a push-pull system in which the grid bias is adjusted to a value intermediate between that which would be used for Class A power amplification, and that which would be appropriate for Class B amplification".
These two references clearly indicates that there is a "proper" biasing point when designing a class A, class B, or class AB amplifier, that which puts the tube in the most linear portion of its operating range.
And lastly, look at the bottom of the page, where it says:
"...and have to a considerable extent displaced the true Class A push-pull amplifier".
That, along with the two previous references regarding specific bias points for each class, rebuts the statement quoted above, and supports my statements that there is indeed such a thing as a "true" class A amplifier. If you don't believe Eastman and Terman, then I can't offer any further help.
Oh, and both of these support what is stated in that "obscure" Army document referenced in the quotation, which, by the way, for completeness of references, is TM11-670, "Special Purpose Oscillators and Amplifiers", Department of the Army, July 1952, which is available in it's entirety here. I highly recommend you read it for a good understanding of the subject:
Q: You recommend biasing the Intruder 30 watt head to 45mV - 50mV on the dual external pots which corresponds to 45 - 50 mA per side. This is loaded with a pair of EL34's. Is this specific to his head and others like this?
A: Yes, that is specific to the Intruder (and Invader) 30W head, which runs very close to true class A push-pull, at around 300-320VDC on the plates.
You can't just set the bias of a particular tube to "xx mA", without taking into account the plate voltage the amp is running at, and in some cases, the design of the phase inverter.
Higher plate voltages generally require lower bias currents to keep the tube running in the safe dissipation area. This usually means the amp is running in class AB or, if the plate voltage is extremely high, class B, with the tubes biased at or near cutoff because of the high plate voltage involved. If the amp is running class A, or very near, you can bias it much hotter, as is the case in the 30W Intruder.
The other thing to take into consideration is the phase inverter output voltage swing. As you adjust the DC bias voltage, you also are adjusting the baseline negative DC voltage level of the drive signal to the grids of the output tubes. If you set the bias for less current, what you are actually doing is raising this negative DC baseline voltage up towards ground.
As you raise this voltage, you are effectively limiting the headroom of the signal, because it will clip when the peak voltage hits zero volts. If the phase inverter is designed for a lot of gain/output voltage swing, it will distort very early and possibly go into "blocking" distortion because of the clamping action at the grids of the output tubes when the signal hits zero volts on the peak. This can sound "mushy", so you must bias the amp colder to compensate. In the case of the 30W Intruder, the phase inverter output voltage swing is designed to be lower than a 50W amplifier of similar design.
The bottom line is this: you can't just arbitrarily say that any tube should be biased at some particular range of currents, because it depends on the circuit design of the amplifier. The manufacturer should provide the specified biasing procedure for the tubes suitable for use in each particular amplifier.
Q:A class A amplifier's tubes are is working at full power all the time, whether it is sitting at idle or playing full blast, right? That's why the tubes don't last as long?
A: No. Contrary to common misconceptions that are quoted in all the magazines, a true class A amplifier is not running at full power all the time, whether at idle or full output. The plate dissipation of the output tube (or tubes) is usually highest at idle (they are typically biased to max dissipation, or even slightly higher in some guitar amps), and the plate dissipation decreases as it starts delivering power to the load. In other words, the power that is dissipated as heat in the tubes at idle is passed on to the load at full power. If the output is a pure sine wave, the dissipation splits evenly between the tube and the load. For example, if an EL84 is biased at 12W idle in true class A, it will put out 6W into the load at full power (assuming no other losses, for this example), and the tube will dissipate 6W at full power. As you can see, the tube's actually run cooler at full power than the do at idle in a true class A amplifier. In addition, if the output is a square wave instead of a sine wave, the tube's dissipation will be zero at full power, and the load will receive the entire 12W (ideally speaking, ignoring other losses and assuming a purely resistive load). This is because the plate voltage and plate current in an output tube are 180 degrees out of phase, so when the plate voltage is at max, the plate current is at zero, and vice-versa, so the instantaneous power, which is the product of the voltage and current, is zero at all times.
Q:Class A watts are different from class AB watts - they're louder, right?
A: Absolutely not. This is a common misconception put forth by people who don't know what the hell they are talking about. Watts are watts, and they are completely independent of the class of operation of an amplifier. The proper way to measure and amplifier's maximum output is at the onset of clipping into a purely resistive load. Apparent loudness is a different subject entirely, and depends on a lot of factors.
Q:On a Marshall the bias feeds the two grid resistors 150k for 6550 and 220k for EL34, why do they change this?
A: The 6550 has a lower maximum grid #1 circuit resistance specification than the EL34. Actually, it is spec'd for 50K max in fixed bias, 250K max in cathode bias, but is typically run at 100K in fixed bias. The EL34is spec'd for 700K max in cathode bias and 500K max in fixed bias. Too high a grid resistance can cause the tube to fail because there is a small amount of gas in the tube which can cause grid current to flow. This grid current will cause a voltage drop across the grid resistor that will increase the bias current in the tube and in turn, increase the plate current. This can eventually end up in a runaway condition and the tube will redplate and croak.
Q:Is there any way to turn a push-pull amplifier into a single-ended amplifier, like pulling all the tubes on one side?
A: The output transformer in a single-ended amplifier must be gapped to prevent DC offset in the primary from saturating the core. A push-pull output transformer is not gapped, so if you try removing the tubes on one side, there will be an offset DC current that will saturate the core and cause a loss of low end (or worse!). The easy way to do this is to leave both sets of tubes in, but remove the signal going to one side, but leave it biased the same current as the other side. This will cause the primary DC currents to cancel out, so you don't get transformer saturation. This will effectively result in single-ended operation of the push-pull output stage. Note that this will only work if you bias the amp into class A, because you will only be running one side, and it will clip if biased at class AB or class B. You will also have to reduce the signal level going to the output tubes compared to the level at class AB, otherwise it will clip too soon. The best candidate for this type of mode switching is a true class A push-pull amplifier, because then you can easily switch from push-pull to single-ended simply by removing the signal drive to one of the sides.
Q: How can I measure the output power of my amplifier if I don't have an oscilloscope, just with a signal generator and a meter?
A. In order to get an accurate measurement of output power, you should use a purely resistive load. A reactive load, like a speaker, will have an impedance that varies with frequency. Tube amps are not constant-voltage devices, so the output voltage will vary with changes in load impedance, which will cause you to get an inaccurate measurement of amplifier output power if you assume a constant resistance load equal to the speaker impedance.
Also, current and voltage are not in phase in a reactive load. The current waveform either leads or lags the voltage waveform. To get a truly accurate measurement of power into a reactive load, you have to multiply the voltage by the current, and then multiply this result by the cosine of the phase angle between the voltage and the current. This is why you should use a purely resistive load for power measurements - the voltage and current are in phase in a purely resistive load, so the cosine of the phase angle between the two (zero degrees) is equal to one, so you can ignore it and just multiply the voltage by the current to get the real power.
Also note that the RMS power in a square wave is exactly twice the RMS power of a sine wave of equal amplitude. This is because the RMS value of the sine wave is equal to 1/sqrt(2), or 0.707 times the peak, while the RMS value of the square wave is equal to the peak value. Since power is voltage squared divided by resistance, the sine wave power equation has a factor of (1/sqrt(2))^2, or 1/2, while the square wave power has a factor of 1, so the square wave power is twice that of the sine wave power. If you have an RMS reading meter, and can't tell where the clip point is, you may be reading square wave power and trick yourself into thinking your amp puts out more power than it really does. It can sometimes be difficult to tell the onset of clipping in a tube amp, it is usually much smoother than the onset of clipping in a solid-state amp.
Having said all this, if you are just looking for a ballpark figure, a meter and a load resistor will do fine, if you have the amp set just prior to where you start to hear distortion when a speaker is connected. Sub a load resistor equal to the speaker impedance, measure the RMS AC voltage, square it, and divide by the measured value of the load resistor. Be *sure* your meter reads in true RMS, or you will get an incorrect result. Use a 400Hz signal as the input source, because this is where most speakers have their "flattest", or nominal, impedance.
Q: What about effects loops degrades the signal so much that you chose not to include them as a stock feature in your amps, despite their usefulness? Is it the loop itself, or what you put in the loop? Is there no way to "true bypass" them so those of us who are willing to give up a little bit of tonal purity to have that feature can do so, while those who want the pure tone can just bypass the loop?
A: Since effects loops are usually best placed as far back in the signal chain as possible, and they work on relatively low signal levels, there can be a noise penalty as well as tonal change when adding them. In general, time-based effects, such as reverb, chorus, and echo need to be added after distortion to sound "natural". This is impossible to do in an amplifier designed to get it's distortion from the output stage, unless you use a wet/dry two amp setup or run the amp into a load, then to effects, and then to another clean amp to get your stage volume. This is a lot of equipment to carry around, although some guys do it. Amps that get their distortion exclusively from the preamp section will function much better with an effects loop, because the output stage is designed to run clean, so the effects can be used to better advantage.
Putting the effects as late as possible in the signal chain of a non-master amp is at best a compromise. In my Invader amp, for example, the reverb signal is taken off after the cathode follower/tone stack and fed back into the "other" side of the phase inverter. In this way, I can get the reverb after the preamp stage/cathode follower distortion, which helps a lot, but it is still not as good as it would be if I could put the reverb after the output transformer distortion.
Typical effects take either guitar level signals (very low level) or line level signals, either consumer (-10dBV or 0.316V RMS) or professional (+4dBu or 1.23V RMS). In order to avoid overloading the input of these devices with the signal levels in a typical tube guitar amp (which can range from a few volts p-p to a few hundred volts p-p, depending on where you are taking the signal from), you must attenuate the signal going to the effects device and then amplify the return signal back up the the original level. This attenuation/amplification process usually increases the noise floor of the signal (especially if you are using a tube recovery stage), so the amplifier will be noisier. A parallel effects loop can minimize this effect, unless you run it full wet.
Another factor to consider is the placement of the loop. Since the return signal can be a very low-level signal, bringing it back to a later stage in the amp near the large amplitude signals and output/power transformers can cause unwanted coupling if the circuitry is not well shielded.
In addition to the tonal and noise impact, a "proper" effects loop must have a buffered output, usually meaning a low-impedance drive from a cathode follower or <shudder> opamp or MOSFET source follower, (although there are other ways to get low impedance drive, including transformers and other methods I use in my amps). To implement this using tubes requires 1/2 of a 12AX7. It will take another 1/2 of a 12AX7 to bring the return signal back up to tube amp levels (or perhaps two sections, depending on where the send signal came from in the circuit). A single 12AX7 is not a large price to pay for a serial effects loop, but things get a bit more complex if you want to do a good parallel loop with a proper mix control and send/return level controls or switchable -10/+4 compensation. In order to get low-impedance cathode-follower drive and a good high-impedance return, enough gain, and a good low-noise parallel, continuously-variable mix circuit, it generally takes two 12AX7's, and quite a few passive components. This is a lot of stuff to add to the amp, which will increase the labor and cost of the amp. However, if done properly, it will not appreciably degrade the tone of the amp, other than the signal-to-noise ratio degradation. The more complex parallel loop has the advantage of not passing the dry signal through the effect, so the tone is not affected too much.
Some manufacturers elect to use an unbuffered loop stuck between the preamp and the phase inverter input, but this is at best a very poor, but cheap, implementation.
The best lower-cost compromise would be a series loop with a switchable bypass that took it out of the circuit when not being used. Most designs use a switching jack to bypass the loop when no cable is plugged in, but these are notorious for getting dirty contacts and becoming intermittent. A real loop bypass switch is the preferred method.
Q: Some time ago I purchased an amp from a well-known boutique amplifier maker (whose name I won't mention here). This expensive amp has "new/old" switch on the back. The manual explains that the "new" mode corresponds to the pentode mode and the "old" to the triode one. The trouble is that in the "old" triode mode the amp produces ghost- notes that are out of pitch! Hearing out-of-pitch after-sounds really confuses my playing. The manufacturer says "that's normal" and points to the Matchless brand that "does the same"... I'd be curious to hear Randall's opinion on this subject.
A: In general, triode mode is much more sensitive to insufficiently-filtered power supplies (which is what causes most "ghost-noting") than pentode mode. This is because in pentode mode, the screen grid supply is usually taken off from a point after a filter choke/filter capacitor. The plate supply is usually taken off from the first filter cap, which has a lot more ripple voltage on it. This is okay in pentode mode, because, in a pentode, the screen grid voltage has far more control over the plate current than the plate element does, so the ripple/hum/noise on the plate supply doesn't get into the output. Also, there is some inherent power supply rejection due to the push-pull arrangement.
When you switch to triode, mode, what you are doing is moving the screen grid connection from the nice, quiet, filtered point and connecting it to the plate of the output tube. Now, there is no relatively stabilized screen supply to keep the plate current quiet, and the triode-mode tube will respond to the under-filtered supply.
The tube acts as a "mixer" in this case, where the 120Hz (or 100Hz if you are on the other side of the pond) full-wave rectified ripple voltage mixes with the note you are playing, and generates sum and difference frequencies that are not harmonically related to the note. These "out-of-tune" frequencies sound like a second note following your playing, and can get really annoying. This is closely related to something called "intermodulation distortion" which occurs in all tube amps, regardless of pentode/triode mode, as a byproduct of the overdrive/distortion generation itself, but not at that great a level.
The solution to your triode mode problem is more filtering in the power supply. Some pentode amps will also exhibit this problem if they have insufficient filtering in the power supply section, or in the screen supply. As mentioned previously, there will be some intermodulation distortion present in all amps, but you can usually easily distinguish between it and "ghost-noting" caused by a poorly-filtered supply. Improperly designed bias supplies can also cause this problem, again due to lack of filtering resulting in too much ripple voltage on the supply. In addition, "ghost-noting" can be caused by the speakers themselves. Some brands/models are worse than others.
Q:I have an amplifier that has a "class A" switch on the back that connects the screen grids to the plates of the output tubes. Is this really class A?
A: No. That is a clever marketing ploy to rename the half-power switch and make you think you are getting a "class A" amp, which must be better than class AB, right?
Triode mode, in addition to reducing the output power, results in higher damping factor of the output than pentode mode. Damping factor is basically the ratio of the amplifier's load impedance to its own internal output impedance. If the damping factor is very high, the amplifier more tightly controls the speaker movement. This also means the speaker impedance peaks and dips have less influence on the amplifier's output frequency response. This is one reason hi-fi guys prefer triode-connected pentodes or pure triodes - they have to use less negative feedback to control the damping factor and achieve flat frequency response.
Triode mode, with its higher damping factor, makes the amplifier less responsive to speaker changes, so it sounds subjectively a bit less "tubey", and also lowers the power output. Unfortunately, flat response and high damping factor makes guitar amps sound sterile and "solid-state", so too much negative feedback or too high a damping factor is not a good thing. If triode mode is used, it is usually best to also disconnect the negative feedback when in this mode. Unfortunately, if you do this, you will also lose your presence control if there is one.
Triode mode can also sometimes result in more hum if the power supply is not adequately filtered, because there is no longer a choke and large capacitor on the screens to provide a filtered, low-ripple DC supply. Instead, the screens are tied to the plate, so you lose all the inherent power supply rejection provided by the screen grid.
Triode mode is NOT "class A". Stupid marketeers...or perhaps diabolically clever marketing geniuses?
Q:What is "swirl"?
A: "Swirl"is a dynamically-changing, slightly "phasey" sound as a note or chord decays, which is common to some tube amps. Typically, "swirl" is caused by a midrange "dip" or varying duty-cycle change in a clipped square wave that changes position as the note decays, giving a sort of mild phase shifter effect.
What happens is that first the phase inverter or output stage clips and produces a flat square wave. As the note decays, the signal level decreases, and the midrange frequencies start getting "unclipped" (either by the fact that their frequency band level is lower, or by phase cancellations due to the unequal phase shift with respect to frequency caused by tone controls and other RC phase shifts that occur in a gain stage) and show up as a "dip" in the top of the square wave, which will move back and forth along the top as the fundamental and other harmonics shift the operating point. Even if the clipping ratio isn't extreme enough to show the "dip" on the scope, the duty-cycle of the square wave will usually be dynamically changing as well.
Since the preamp stages are all AC-coupled to each other, the operating point shifts as the signal gets smaller, due to slight "blocking" distortion, where the gain stage clamps the top peak to a point slightly above it's cathode voltage, while allowing the wave to still increase in the negative direction. As the signal decays, it shifts upward and changes the duty-cycle of the clipping. It is this ever-changing shifting of the operating point that causes the "swirl" effect. The trick to good "swirl" is in the correct staging of the gain and frequency breakpoints of each gain stage in the amp, particularly in the phase inverter and output stage.
A similar effect can be caused by too much drive from the phase inverter to the output tubes. As the note decays, a riding "buzz" can be heard coming in and out. This is crossover distortion aggravated by too much signal swing to the output tube grids. Reducing the signal levels at the output of the phase inverter will cure this.
Another cause of a "swirly" sound is a parasitic oscillation that is riding on the output signal, causing intermodulation distortion.
Q:I'm working on scaling a Baxandall tone stack to be driven from a speaker output and I'm wondering what the input impedance should be in this scenario?
A: Tone stacks in general, like many filter networks, should be driven from an (ideally) zero source impedance and into an (ideally) infinite load impedance. In practical terms, the load impedance should be at least 10 times the characteristic impedance of the filter network, and the source impedance driving it should be as small as possible. By driving/loading your tone stack (filter network) in this manner, you will make the frequency response as close as possible to the theoretical ideal it was designed for, and the corner frequencies of the controls will occur where they are supposed to, and there will be minimum loss through the network. The output of an amplifier, at 4, 8, or even 16 ohms is, for all practical purposes as would relate to a tone stack, an zero ohm source, so you have that part covered.
Now, the only thing you have to worry about is the load impedance. This is where you need to look in order to figure out how to scale your Bax stack. What is the input impedance of the circuit your tone stack is driving into? If it is the grid of a preamp tube, with a 1Meg resistor to ground, you might make the assumption that the input impedance is 1Meg (actually, it varies with frequency, and drops lower as the frequency increases). However, if you are driving into the cathode of a tube, the input impedance may be on the order of only a few hundred ohms.
A general "rule of thumb", is that the load impedance must be at least 10 times the characteristic impedance in order to be considered negligible, because you will get only a 10% loss through the circuit. It is even better if you have a load that is 100 times greater. The downside of higher impedances is greater susceptibility to noise and interference from neighboring circuitry.
The last thing you need to know is the characteristic impedance of the network you are inserting into the path. This can be rather difficult to determine in some cases, because it is frequency-dependent. You can usually guess at it by assuming the capacitors are short circuits at particular frequencies and noting the largest resistance to ground or in series with the signal, and use that as your characteristic impedance.
For example, if you have a standard Bax stack with a couple of 1Meg pots, you can assume the impedance plot of the network looks like a maximum of around 1Meg at some frequencies. This would mean you could drive it from a source impedance of 100K minimum, preferably 10K or less (ideally zero) and into a load of 10Meg minimum, or ideally 100Meg or so, but again, you'd run into noise/interference problems. Assuming you were trying to drive into a 100K load, you'd want to scale your Bax stack to an impedance of 10K or lower. Assuming you were trying to drive a 10K load, you'd want to scale your Bax stack to an impedance of 1K or lower.
Q: Why are resistive attenuators harder on an amp than an inductive attenuator or speaker load?
A: A resistive load is not any worse than a speaker load or reactive attenuator load. In fact, the opposite is true - a purely resistive load dissipates all the amp's power in the load, while a reactive load varies the dissipation between the load and the output tubes, depending on the phase angle of the reactive component of the load. The worst case load for an amplifier is a purely inductive load, with a phase shift of 90 degrees between the supplied voltage and current. In this case, when the voltage across the load is zero, the current is maximum, which means that the output device now has maximum voltage across it at maximum current, which results in maximum dissipation. In effect, the load gets no power while the output devices are cooking!
In addition, a reactive load has a very high impedance at the low frequency resonant point (typically up to 4 or 5 times the nominal 400Hz impedance) and a rising impedance at higher frequencies that can go up to many times the nominal impedance. These high, reactive impedances can cause very high, frequency-dependent voltages on the reflected primary impedance of the output transformer, which can cause arcing. The problem is much worse if the impedance is mismatched. The main factor in amps blowing up from attenuator use is not the fact that most attenuators are resistive, it is the fact that the amp is run full-out all the time, something that would not normally happen because the amp would be too damned loud to use that way.
Another factor in amplifier damage from attenuator use is when the amp is run into an attenuator that is not properly impedance matched to the amplifier. Contrary to some attenuator manufacturer's claims, there is no way to make an "automatic" impedance matching attenuator, or a "one-size-fit's all" product. Their products are single-impedance attenuators marketed as "safe" for all load impedances. Lower than normal impedances cause higher than normal currents in output tubes, and higher than normal impedances increase the risk of arcing in tubes, sockets, and output transformers. I believe the "resistive load is bad" argument has the same origins as some of the other common misconceptions in the amp world. Look on any amp designers or tech's workbench, and you will find a purely resistive dummy load for amp design and testing. All Aiken amplifiers are designed and tested flat out at full power into a purely resistive load for long periods of time, in addition to being tested using reactive loads and actual speakers.
Q: Why do you test your amps on a purely resistive load when they are not going to be run on a purely resistive load?I don't know anything about amp building but I do know about engineering, and when I do a test on something I will try to duplicate the exact environment in which it will be run in the real world.The first is that it is the only way to easily see the true power output of the amp. If you run into a reactive load, you can't easily see how much power the amp is putting out, because you have to take into account the frequency-dependent impedance of the reactive load, and there is a cosine of the phase angle to be taken into consideration to calculate the real power. If you run into a purely resistive load, the voltage and current are in phase, so the power you see on the scope is what is really being put out by the amp (well, you have to do some basic math with the voltage and the load resistance, but there is no phase angle or frequency-varying impedance to take into account).
A: Now, that is an excellent question! There are several reasons to use a resistive load.
The second reason is that reactive loads cause all kinds of overshoots to appear on the edges of the square waves when you drive the output stage into clipping, and you can't tell whether they are actual overshoots that you don't want, which could point to a problem in the amp, or just normal reactive load artifacts that you don't necessarily need to worry about.
The third reason is that you need a load impedance that is flat with respect to frequency in order to run frequency/phase response plots of the amp with a network analyzer to check the design of various stages or the entire amp, to see how much effect the tone controls have, what frequency the breakpoints are at, etc. A reactive load will skew the results because of the frequency-dependent impedance magnitude changes.
Once the amp is happy running into the normal resistive bench dummy load, it is then hooked up to a reactive load for the final stability testing. I like the Marshall Power Brake for this because it is designed with no limiting of the high-frequency rise on the impedance (which is probably why they have a reputation for sounding buzzy), and can really aggravate an amp that is on the edge of oscillating. You might not catch it into your bench test speaker, but it might oscillate into, say, a 4x12 cab. The PowerBrake is a pretty rough load for most amps, so it is a good test.
After the amp is checked out to make sure all is working as designed, then you test it into a speaker, using the most important pieces of test gear of all - your guitar and your ears!
Q: I have a Dr. Z AirBrake attenuator and am ordering an Ultimate Attenuator. Are these safe to use with my Sabre?
A: The Airbrake is a modified L-pad attenuator which presents a widely varying load to the amp, depending on the impedance of the cabinet connected to it and the attenuation setting (contrary to their marketing claims, there is no such thing as a "one-size-fits-all" attenuator that properly matches all impedances - this is in reality an 8 ohm unit being marketed as "universal", because it presents a "close enough" load in most cases, although it doesn't really match that well even at 8 ohms for higher attenuation levels - up there it is closer to a 16 ohm load no matter what impedance cabinet you have connected). For example, at -3dB attenuation, the impedance is 11 ohms for a 16 ohm cab, 8 ohms for an 8 ohm cab, and around 6 ohms for a 4 ohm cab. At -12dB, the impedance is 15 ohms for a 16 ohm cab, 14 ohms for an 8 ohm cab, and 13 ohms for a 4 ohm cab. The more you attenuate, the more this type of attenuator approaches a 25 ohm load. Here is a link to the impedance load you can expect to see as a function of the speaker load and attenuation setting. Also note that the attenuation steps are only accurate for an 8 ohm speaker load. If the load is 16 ohms, the attenuation steps are 1/2, i.e. at -12dB you are really only -6dB down, and at 4 ohms they are twice, i.e. at -12dB you are really -24dB down.
Be sure to set the Sabre to 16 ohms when using the Dr. Z AirBrake with either a 16, 8, or 4 ohm cabinet at high levels of attenuation, but don't forget that at bypass or at one or two clicks down, you'll need to switch back to 16, 8, or 4 ohms to match the cabinet being used. Running the amp at 16 ohms into an 8 ohm or 4 ohm load at high power will accelerate tube wear and could cause damage, so be sure to match the load whenever possible. Higher than normal impedances can cause output transformer damage or tube arcing.
Beware that the Ultimate Attenuator is a 30 ohm load (or so I've been told, I've never had one in to test, but again, there is no way to "automatically" properly match all impedances as claimed). This 30 ohm load is not a proper load for the Sabre's output transformer, so use it at your own risk - your warranty will not cover any damage resulting from this upwardly mismatched load using this device, and I have had reports from Sabre customers of "flashing" inside the tube near the base when used with this device, and have seen reports of oscillations and output transformer failure from users of other brands of amps. Be sure to set the impedance selector for 16 ohms when using this device with any cabinet impedance.
Of the two, the AirBrake is the safer unit to use in my opinion (and it sounds good, too), but I think both of them should be better designed to correctly match impedances.
Q: Can you elaborate a bit more about L-pads as an attenuator?
A: There are two ways to make an L-type attenuator:
(1) Match the impedance of the L-pad in the direction of the series arm. This is the "traditional" L-pad configuration. In this case, the input impedance stays constant, and the output impedance gets lower as you increase the attenuation, down to a theoretical minimum of zero ohms. This configuration sounds like crap for a guitar amp, because the decreasing output impedance increases the damping factor and removes all "tubeyness" from the tone at lower volumes, because the amp no longer reacts to the variations in speaker impedance.
(2) Match the impedance of the L-pad in the direction of the shunt arm. In this case, the input impedance also stays constant, but the output impedance gets higher as you increase the attenuation. This is great for guitar amps, up to a point. The increasing output impedance lowers the damping factor, which enhances the interaction between the amp and the speaker, giving a natural bass and treble boost as you increase the attenuation (sort of a "built-in" Fletcher-Munson effect compensation!). The problem is that it gets to be too much, and you end up with too much bass and treble boost and "hangover effects" as it was called in the old days of audio, and you get a flubby, fizzy tone at high levels of attenuation. This can be alleviated by limiting the maximum output impedance with an additional shunt resistor at the output. I used this method of attenuation on my second-generation Invader and Tomcat amplifiers, which sounded pretty good all the way down to whisper levels with the VAR control. It was designed to correctly match the 4, 8, or 16 ohm settings of the rear-panel impedance switch. The first generation pf the Invader used a reactive load, and the second generation was purely resistive, but I used a modified L-type matched in the direction of the shunt arm. It sounded a bit different than the reactive attenuator version, but I think I preferred it, as it was a bit smoother.
The Airbrake is basically a type 2 L-pad, but it is made with a fixed resistor in the shunt arm instead of the tapped/variable resistor normally required to keep the input impedance constant. This was likely done to save money and labor, because you don't need a tapped resistor and you don't have to wire it to another gang on the rotary switch. Because of this, the input impedance does not stay constant with variations in attenuation, as you can see in the graph shown here.
You can also design a "T-type" attenuator that matches the impedances in both directions. This will give a more constant tone with changes in attenuation, without the extra increase in bass and treble as the attenuation increases. While this will give a more "accurate" tone, as can be shown by recording and playing back at a normalized volume, it may not be as pleasing to the ear because it doesn't automatically compensate for the loss of bass and treble with volume reduction as the L-type does, but it also won't get as flubby and fizzy, either. A properly-compensated type 2 L-attenuator is probably the best sounding, but it takes a bit more circuitry to tailor the output impedance for best tone at all volumes..
Q: Is it normal for a blown fuse to shut the amp down completely?
A. An amp typically has two fuses - one for the "mains" or AC supply, and a second one for the "HT" or high-voltage section of the amplifier. Some amps use only the mains fuse, and others have several fuses, including some that fuse the output tubes individually, etc. It is perfectly normal for a blown mains fuse to shut the entire amp down. That's what it's job is - to shut the amp off to protect it in the event of a short circuit. Sometimes a fuse will blow for no apparent reason (called "nuisance" fuse blowing). An amplifier typically has large filter capacitors that can draw quite a bit of inrush current when the amp is first turned on and they are fully discharged, and this can occasionally cause a fuse to blow, particularly if the fuse is old. If your amp blows a fuse, you should replace it with one of the same type and rating. If it still blows the fuse, the amp needs further servicing.
What I typically tell customers to do first is to pull out all the power tubes and rectifier tube (if there is one), and put in a new fuse. Turn the amp on, and if the pilot light comes on, the power transformer and wiring are fine. If the fuse blows, you probably have blown rectifier diodes (if there is no tube rectifier in the amp). If the amp had a tube rectifier, put it back in and turn the amp on again. If the fuse blows, it was a bad rectifier tube. If not, the rectifier tube is probably okay. Then install the power tubes. If the fuse blows, one of the output tubes is bad, and the entire set should be replaced. If the fuse doesn't blow, and the amp works fine, it was likely just a nuisance fuse blowing, and the amp should be monitored carefully for the first hour or so of playing just to make sure everything is okay. Never keep feeding fuses to an amp that is blowing them, you'll likely cause permanent damage. Stop after the second one blows, unless you've isolated the cause as I've described it above.
Q: I understand you can put a 50V reverse-biased zener diode in the center-tap of a power transformer and lower the output voltage by 50V. Exactly how does this work?Another way to look at it is this: it makes the center-tap negative by 50V during the capacitor charging pulse. Since this is negative with respect to ground (and the output voltage), it "subtracts" 50V from the output voltage. Either way, the "real" total voltage differential between B+ and the center-tap remains the same, however, since we have level-shifted the center-tap negatively by 50V, the output voltage is lowered by 50V. The drawback is that you don't get something for nothing. In this case, the price you pay is in power dissipation. The voltage dropped across the zener multiplied by the current through the zener results in a power loss that is dissipated in the zener as heat.
A. A zener diode is simply a diode that has a designed-in "breakdown" reverse voltage that is fairly tightly controlled. In the forward direction, it is a "normal" diode, with a voltage drop of around 0.7V. If you drive current through it backwards, it will drop a voltage equal to the zener voltage. This makes it useful as a level-shifter. Note that "regular" diodes also have a breakdown voltage, and will conduct in the reverse direction if you exceed that voltage. If you put a 50V reverse zener in the center-tap <>of the power transformer, it will conduct on the filter capacitor charging pulses and make the center-tap <>more negative than ground by 50V. Since the main rectifiers can't conduct until the voltage is higher than the voltage stored on the filter capacitor, conduction will be delayed until the voltage is effectively 50V higher than it was before you put the zener in, which in turn reduces the output voltage by 50V.
Q: I have seen a circuit where two 1N4148 diodes are connected in reverse-parallel in series with the signal between the coupling capacitor from the tube's plate.and the voltage divider feeding the next stage. What does this circuit do? Is is a clipping circuit?
A: No, it is not a clipping circuit. That is what is commonly called a "coring" circuit. It is used for noise reduction.
Here's how it works: the reverse-paralleled diodes will not conduct in either direction until the signal exceeds their turn-on threshold, which is typically 0.7V. This "cores out" the center of the waveform, so low-level signals can't pass through the diode network. Since the noise is a constant, low-level signal, it gets blocked, but the higher-level audio signal passes through. It must be used in a high-enough signal level point where the 0.7V drop is not too significant with respect to the audio signal level at that point.
The drawback to this circuit is that it introduces what is effectively crossover distortion, because it chops the center out of the waveform and connects the upper and lower parts back together. If you put a sine wave into the circuit, you'll see it has a "notch" at the zero crossing. This can be audible, typically as a higher-frequency "buzz" accompanying the note, more noticeable as the note decays. Capacitive filtering can improve the tone by removing the upper harmonics. This is typically done with a small value capacitor across the voltage divider resistors following the coring stage.
Q: While eyeballing a schematic of an amp i noticed something different. Instead of the usualy 4.7k resistor to ground coming off from the tail 10k, there was a 10k instead. So a 10k tail and the 10k to ground in place of the 4.7k you usually see. I had been messing with the NFB lately and i realized i never played with that value. So i tried a 10k, then finally i just put a 25k pot there. the more resistance the softer the sound much like if you install a pot in place of the NFB resistor and turn it to smaller resistance, thoigh in this case that tone happens as you go higher in resistance. Can someone try and explain to me in laymans terms as much as possible whats going one there?
A: The amount of negative feedback is controlled by a voltage divider consisting of the series resistor, commonly called the "feedback" resistor", and the shunt resistor, either a fixed resistor or the resistance of the presence control pot.
The ratio of the resistors in this voltage divider determine the amount of voltage "fed back", and thus, the amount of gain reduction, bandwidth increase, noise reduction, and output impedance reduction. The more feedback voltage you apply, the "tighter" the amp gets.
There's a surprise for you - I'll bet you thought the 16 ohm tap had an output impedance of 16 ohms, right? It actually changes, depending on the amount of global negative feedback you apply. This doesn't mean you connect a different load impedance when it changes, it simply means the effective output impedance changes depending on how much feedback you apply.
The ratio of the internal "virtual" output impedance to the speaker load impedance determines the damping ratio, which in turn determines how much the amp reacts to the varying impedance characteristics of the speaker. The higher the damping ratio, the "tighter" the sound.
Therefore, a feedback resistor of 100k and a shunt resistor of 4.7k is roughly the same amount of feedback you would get if you used a 200k and 10k. If, however, you left the 100K alone and changed the 4.7k to a 10k, you would then have more feedback voltage. You can vary either the feedback resistor, the shunt resistor, or both. Normally, you leave the shunt resistor alone, because it also controls the headroom and balance of the PI, along with the 10K "tail" resistor.
Q: My amplifier is picking up radio stations - what causes this, and how can I fix it?
A: RFI (radio frequency interference) is usually caused by bad cables, bad solder joints, long unshielded wires from the input jack to the first tube and/or bad ground, no "grid stopper" resistor on the grid of the first preamp tube, or no tube shields on the tubes. In some cases, RFI can get in on the AC mains. Possible fixes:
- First, unplug the cable to see if the RFI goes away. If so, it is likely the cable or the guitar picking it up, or lack of grid stopper resistor on the first tube.
- Try pulling out the preamp tubes and putting them back in a few times - they can have corrosion that is causing rectification of RF, and the wiping action of removing/replacing the tubes can help clean them in a pinch.
- If it has isolated input jacks, put a 0.01uF cap from the *ground* side of the jack directly to the chassis at the closest point, such as on a ring terminal mounted behind a switch, or a on a pot or something. This will kill RFI deader'n hell in most cases. Ceramic disks from Radio Shack will work fine, or any kind of film capacitor. Sometimes you can test this to see if it is the cause by sticking a wire or paper clip on the metal body of the guitar cable where it plugs into the amp and touch it to the front panel on a bare metal spot.
- If it has no grid stoppers, put a 10K-68K resistor on the grid pin of the first tube, right at the socket. You can also use a ferrite bead installed right at the grid.
- Resolder the joints on the sockets, jacks, and components in the input stage.
- If it has no tube shields, add them.
- Install an internal AC power filter, such as those made by Corcom (available from Digi-key). You must use a filter appropriately sized for the mains voltage and current draw of the amplifier.
- Install a clamp-on ferrite core on the AC cord right at the point where it is plugged into the amp.
- If there isn't one, install a metal shield plate on the cabinet underneath the open side of the chassis. The best material for the shield is aluminum gutter flashing (available in rolls from Home Depot or Lowe's). It can easily be cut into shape with scissors and stapled into the bottom of the cabinet with one of those heavy-duty construction staplers also sold at home-improvement stores (use the 10mm staples, not the 8mm ones).
Q: Can you speak a bit about grounding rules?
A: Grounding isn't rocket science, but it helps to plan ahead. Following are a few things that usually work best for me:
- Make sure the center tap of the HT goes directly to the first filter cap ground. There is a lot of high-current noise traveling in this wire. Do not connect this point to the chassis, unless it is the only ground point (see item 4 below).
- Locate the remaining preamp filter caps near their respective stages, and connect their ground to the local node grounds of the stage they bypass. Also run a buss wire along the grounds of each the filter caps back to the main filter cap. This is a "modified star/buss" approach where each stage has it's own local star ground and each ties into the global buss.
- Make sure you don't route the choke wires near any sensitive preamp nodes - there are also lots of current spikes traveling in these wires, and they will radiate or couple into adjacent wires.
- Whether you use a star ground, buss ground, or a modified version of both, connect the entire ground system to only one point in the chassis, preferably right at the input jack, using a PEM stud or secure bolt with lockwasher/loctite. Do not connect the ground system to both the input jack and back at the main power supply ground node or you're asking for ground loop hum. If you are running a buss rail, tie it to the chassis right at the input jack (preferred) or right at the main filter cap, but never both. Do NOT use the input jack nut/solder lug as the ground connection, always use a bolt or PEM fastener that stays securely connected to the chassis.
- If you are using isolated input jacks, run a 0.01uF cap from the ground/sleeve connection directly to the chassis with as short a leads as possible. If you are using non-isolated jacks, and you've connected the power supply ground buss/star to ground right near the jack, you're okay, but you may want to add the capacitor anyway in case the connection gets oxidized or corroded.
- Never depend on the input jack ground connection through the nut and sleeve lockwasher. Always run a separate ground wire directly from the input jack ground to the bottom of the cathode resistor of the first tube it goes to.
- Never ground the pots/switches/jacks to the power supply ground rail or buss. Treat every amplification stage as if it were a differential amplifier that amplifies the difference between it's grid and it's cathode (or the bottom of the cathode resistor). If you have a volume pot that goes to the grid of a tube, run a wire from the ground of that pot to the bottom of the cathode resistor for that tube. Then run a wire from that point down to the supply buss ground or star ground.
- Always reference circuit grounds to the later stage that is receiving the signal, not the driving stage's ground. For example, in the volume pot example above, reference the pot ground to the stage that the wiper is connected to, not the stage that drives the top side of the pot. If you have a tone stack circuit, reference the ground of the tone stack to the stage following the tone stack, not the stage driving it.
- Always connect the AC mains safety ground (green wire) directly to the chassis, close to the point where it enters, with enough slack in the wire to maintain the ground wire integrity if the other wires get pulled out by accident.
Q: Does negative feedback reduce harmonics and distortion?
A: Negative feedback will reduce distortion in the stage that the feedback is wrapped around. Basically, it tries to make the output "look" like the input, so if any harmonic distortion (odd or even) is added by the stage, the feedback will try to remove it.
Negative voltage feedback does the following things:
(1) Reduces distortion created in the stage(s) feedback is applied around. Note that it will not reduce distortion added in previous stages.
(2) Flattens frequency response (removes peaks and dips and extends the bandwidth on both high and low end, resulting in a flatter overall response).
(3) Reduces output impedance - this causes less interaction with the peaks and dips in the speaker's impedance with respect to frequency, causing the amp to react less to the speaker.
Note that feedback only works to do these things if the stage is not clipping and there is enough excess gain for it to work with. Once the stage hits clipping, all bets are off. Basically, feedback trades off excess gain for flatter response and less distortion and a more "ideal" output impedance. This may or may not be a good thing, depending on your perspective and tonal tastes. In general, too much feedback can make an amp sound "solid-state", while too little can make an amp "boomy" or "loose" feeling, especially if the output stage is improperly designed.
Q: What works best as a master volume: pre Phase inverter, or post phase inverter? What are the tradeoffs in using either?
A: The post-PI master volume is best for amps that are single-channel affairs, like a non-master Marshall. It allows you to get a relatively decent "cranked output stage" tone at low levels, as the phase inverter distorts in a manner that is similar to the output tube distortion. This is because it is a differential amplifier, which has "graceful" overload characteristics, because when one side hogs all the current from the "tail", the other side has to cave in at the same time, so you get a clipping that is reminiscent of a push-pull output stage clipping.
The drawback of a post phase inverter master volume is that your tone controls won't have very much effect, because you are clipping the signal after the tone controls, so increasing the level of any of the frequency bands will not increase the level of the output because it is already clipped. Also, if you have an effects loop, your time-based effects, like delay and reverb, will be muddy and indistinct if you drive the amp hard, because you are distorting both the original signal and all the repeats and reflections of the echo and reverb, so it doesn't sound like natural reverb, which does not distort all the secondary echos.
Another drawback of the post-phase inverter master volume is that it reduces the effectiveness of a presence or depth control. This is because the gain control is inside the feedback loop, and decreasing the forward gain of a feedback amplifier reduces the effect of the feedback, which the presence and depth controls depend on in order to work. You can simultaneously increase the amount of feedback to counteract this, but it is limited to a small amount of range because there just isn't enough forward gain to work with once you get down low on the signal level.
The pre-PI master allows you to put tone controls and an effects loop after the master, so your tone controls will have much wider range, since you can now increase and decrease frequency bands because you aren't clipping the signal afterward and limiting the range. Since the effects loop comes after the distortion, the reverbs and delays sound more natural, as you are delaying and reverbing the distortion, not distortion the delay and reverb.
The pre-PI master also has the advantage of making it easier to implement channel-switching, because you can switch between two masters, going into one effects loop. In addition, since the output stage and phase inverter don't have a gain control stuck in the middle, the presence and depth controls have the full forward gain of the output stage to work with, so you can have wide ranging high and low end boosts from these controls.
The downside of the pre-PI master is that now all of your distortion has to come from the preamp stages, and some amps don't have a good-sounding preamp distortion, as the designer lets too much high end through and it sounds like a box o' bees, or they constrain the bandwidth too much and it sounds like Carlos Santana in a garbage can, or they let too much low end through and it sounds like an amplified fart.
Q: I measured my power transformer before wiring up the tubes, and the voltages are a lot higher than they claim. For example, the 6.3V winding is reading 7.5V! What is wrong?
A: Open-circuit voltages are always higher than loaded secondary windings.
Transformers are rated for a specific output voltage at the specified rated current draw. The deviation from this voltage is called the "regulation", or the percent change in voltage from no load to full load.
This is why you have to be careful when spec'ing out a transformer - you might think, "I want to allow for a lot of extra current to add reliability or to have extra margin in case I want to add more tubes later...", but if you don't also spec a tighter voltage regulation, you will end up with a higher secondary voltage than you expected, because you aren't loading the transformer at its rated load.
The downside to a tighter voltage regulation spec is that the cost and size/weight of the transformer will increase. This is because the wire gauges on the secondaries will have to be larger for less voltage drop between no-load and full-load conditions.
Q: Okay, then, how do I determine what my voltages will be if I don't load my transformer at the full rated current load?
A: If you purchase a transformer that is rated at, say, 6.3V 3A, and you load it at 3A, you will get 6.3V. If, however, you load it at only 1A, you will get something else.
The only way to know what that voltage will be is to find the spec for voltage regulation (if they provide one) or take the no-load voltage and calculate it out based on the full-load voltage at the specified current. It will be a linear function (because it is a resistive voltage drop due to winding resistance, ignoring core losses), so you can interpolate to find the in-between points.
For example, if you have a 6.3V 3A rated transformer, and you want to load it at 1A, and you've measured the no-load voltage to be 7.5V, you will have a "droop" of 7.5-6.3 = 1.2V from no load (0A) to full load (3A).
Since you are loading at only 1A, the output voltage will be 7.5V - (1.2V/3A)*1A = 7.1V.
If, instead, you loaded it at 2A, you'd get: 7.5V - (1.2V/3A)*2A = 6.7V.
If you loaded it at full load, you would get 7.5V - (1.2V/3A)*3A = 6.3V, as expected.
(Notice how the "A" cancels out in the equation and you are left with V, indicating the correct units in the equation).
For filament windings, you can simply add up the filament currents of the tubes. Most 12xxx tubes are wired for 6.3V and draw 300mA, so if you have, say, 4 12AX7s, you would have a current draw of 1.2A.
There is usually only one 5V rectifier tube, so you can just use it's filament current draw directly in the equation.
Q: Attempting to simultaneously operate a 4-ohm and a 8-ohm speaker from the 4-ohm and 8-ohm secondary taps of an output transformer...not a good idea---but, why?
A: If you run a 4 ohm speaker on the 4 ohm tap, it will reflect back, say, 8K to the primary. Now, if you also connect an 8 ohm speaker to the 8 ohm tap, it will also reflect back as 8K to the primary, so the tubes now are seeing a 4K load instead of the 8K they require. That's why it is a bad idea. Now, if you connected an 8 ohm speaker to the 4 ohm tap, and a 16 ohm speaker to the 8 ohm tap, the reflected impedance would be correct.
Q: What is the purpose of the 1Meg resistor on the input jack?
A: The 1Meg resistor is there to provide a ground reference for the grid. The tube works by developing a bias voltage across the cathode resistor with respect to the grid, which must be at or near ground, DC-wise, or it won't work.
Since the tube's control grid is a very high impedance, much higher than the value of the resistor, the resistor value also sets the input impedance of the amp. The input impedance is not extremely critical, as long as it is high enough, because preamp stages are voltage amplifier stages, not power amplifier stages, and you don't match impedances like you do with a power output stage, where you are striving for maximum power transfer. In a voltage amplifier stage, the goal is to make the input impedance high for maximum voltage transfer, and to avoid loading the output of the driving stage, which, in this case, is the guitar pickup. A typical rule of thumb is that the load impedance should be at least ten times the value of the driving source impedance in order to be negligible.
Now, do you need a resistor there with a guitar amp? Not always, because a "standard" guitar pickup is a coil of wire with one end grounded. The pickup coil will then become the DC ground return for the amplifier's input stage, and it is in parallel with the 1Meg resistor, so a resistor is not always necessary.
However, it is a very good idea to add it, because you don't know what might be plugged into the input. The driving source may be AC-coupled though a capacitor from a stompbox output, which would result in the grid floating.
What will happen with a floating grid? Will it cause the tube to melt down? No. The typical 100K plate load resistor will limit the maximum current to the plate to around 2.5 mA at most, and the resulting voltage drop across the resistor would limit the plate voltage to around 50V at most, so there is not enough power dissipation to cause any damage. The worst case that would happen would be the resistor would burn up if it weren't rated for high enough power. In addition, the tube's grid is such a high impedance, it will sit there and "float" for awhile, so you may not even get much current.
Having said this, the best approach is to add a blocking capacitor in series with the input, with the 1Meg grid reference resistor after it. This will provide a stable DC biasing point for the tube no matter what is plugged into the input, even a floating cable or an AC-coupled stompbox, or an effects unit with a high DC offset at the output.
This capacitor is a necessity on a grid-leak biased amp, like the Magnatone T-32, because the tube wouldn't be able to develop enough grid-leak bias voltage at the grid when connected to a guitar amp pickup's resistance without the DC-blocking capacitor. As mentioned, it is also a good idea on a cathode-biased amp, and is used in some commercial amplifiers, like the Peavey Classic 30, for instance. It will also keep the guitar volume pot from making scratchy noises when turned up and down if there is a bit of grid current in the first preamp tube, which will develop a small DC voltage at the grid that can result in these noises.
Q: A lot of amps have effects loops. Which ones are buffered and what does that mean?
A: "Buffering" means the fx loop circuit has an amplifier with a low-impedance output to drive cables and effects inputs without signal loss. There is also usually an accompanying buffer on the return side to receive the signal from the effects unit. Essentially, the "buffer" isolates the effect from the amplifier circuitry.
A typical rack-mount line-level type effect may have an input impedance of around 10K ohms, so if you drive it with the typically high-impedances found in guitar amplifier stages, you'll lose signal level, and may get unwanted clipping/distortion.
In addition, a high impedance is far more susceptible to high-frequency rolloff due to cable capacitance, so a low-impedance drive is best to preserve frequency response.
A buffer is typically an active amplifier stage, such as a cathode follower (tube), emitter follower (BJT), or source follower (FET), but can also be an opamp (IC), or even a transformer with a step-down ratio to transform the high impedance of the tube circuit to a low impedance. The transformer gives the added benefit of isolation to prevent ground loops.
There is also typically a buffer amplifier on the return side as well. This can also take the form of a discrete tube, transistor, or FET, or an opamp or another transformer. The return buffer will have a high input impedance for maximum signal transfer without loading down the effect output.
In generic terms, a "buffer" is an amplifier that has a high input impedance and a low output impedance. This allows it to drive low impedance and/or high capacitance loads without losses. It may or may not have voltage gain. Typically, the effects send buffer will have a gain of less than one, as the signal from the circuit is usually much larger than the effects input requires. The return buffer, on the other hand, will usually have a gain of greater than one, as it has to make up the difference in signal level between the output of the effects unit and the signal level inside the amplifier.
Q: I'm not 100% clear on what fully bypassed means. Is it when the cathode is connected directly to ground or when there's any cap paralleling the cathode resistor?
A: Fully-bypassed means the value of the bypass capacitor is large enough to completely bypass the cathode at all frequencies being amplified. For example, a Marshall plexi has a 330uF cap bypassing an 820 ohm resistor. The "stock" Fender preamp stages have 25uF caps bypassing 1.5K resistors. Both of these are large enough to fully bypass the cathode resistor at all frequencies of interest to guitars.
Partially-bypassed means the value of the bypass capacitor is not large enough to completely bypass the cathode at all frequencies. For example, the bright channel of a Marshall has a 2.7K cathode resistor bypassed by a 0.68uf capacitor. This value is not large enough to completely bypass the cathode at all frequencies, so you end up with more gain at midrange and high frequencies. The partially-bypassed stage forms a "shelving" highpass filter. Changing the value of the capacitor shifts the center point of the "shelf", and changing the value of the cathode resistor changes the ratio of the max/min gain of the shelving response.
Unbypassed means no cathode bypass capacitor at all. In this case, the gain of the stage is greatly reduced, because of the local negative feedback formed by the AC signal current flowing through the cathode resistor, which generates a voltage at the cathode that subtracts from the gain of the stage. The overall gain is set by the ratio of the plate load resistor in conjunction with the internal plate resistance of the tube and the value of the cathode resistor.
You can get a rough idea of whether or not the capacitor value is large enough to fully-bypass the cathode by comparing it's capacitive reactance to the value of the cathode resistor at the lowest frequency being amplified.
For example, the low E string is around 80Hz, so if you use the capacitive reactance formula, Xc = 1/(2*pi*f*C), you will find that a 330uF cap has a reactance of around 6 ohms at 80Hz, a 25uF cap has a reactance of around 80 ohms at 80Hz, and a 0.68uF capacitor has a reactance of 2.9K ohms. As you can see, 6 ohms is so much lower than the 820 ohms in the Marshall circuit, and 80 ohms is way lower than the 1.5K ohms in the Fender circuit that they will dominate and fully-bypass the stage. However, the 2.9K reactance of the 0.68uF cap is almost equal to the 2.7K of the Marshall bright channel cathode resistor, so it will not fully bypass the stage, and a shelving response will result.
Q: I was wondering if you can answer a question regarding balanced triodes? What would you consider more important a characteristic for determining balance between two triodes, transconductance or amplification factor? Or is there something else that is even more important?
A: Short answer: both have to be matched for true balancing.
Long answer: There are three tube factors: mu, gm, and rp. Since mu = gm*rp, if rp is equal for the two tubes, it doesn't matter if you match for mu or gm, you will get the same result (the same gain in circuit). However, if rp is not equal, you will get very different results. You can calculate rp as: rp = mu/gm, so if you measure both mu and gm you can calculate rp. Gain in circuit is equal to : A = mu*RL/(RL+rp).
For example, if rp=62.5K for both tubes, you will get the following gains in circuit for two tubes with vastly different mu and gm, because the ratio is the same to get 62.5k for rp:
1: mu = 100, gm = 1600, rp=62.5K -> gain in circuit with 100K load resistor = 61.5
2: mu = 50, gm = 800, rp=62.5K -> gain in circuit with 100K load resistor = 61.5
So, if you want two tubes to give the same gain in circuit, you need to match for the ratio of mu/gm to be equal. The ratio in the above example is 2, because 100/50 = 1600/800 = 2.
However, that would only give the same gain if the load resistance, RL, were constant with respect to frequency. Since this is not the case in most tube circuits, because they drive capacitively-coupled loads or tone stacks, the gain will vary with frequency, because RL is actually the parallel combination of the load resistor and the input impedance of the following circuit.
So, to be truly matched in circuit, you would have to match for both mu and gm to be the same, which would mean rp would also be the same for both tubes.
Q: Why is it, when discussing circuits in internet forums, that people with no technical background always seem to claim that they can hear so much better than someone with an electronics background? :)
A: That's the "golden ears" strategy, a time-honored internet arguing technique used with limited success by those who are usually "technically challenged", i.e. "internet armchair engineers", who cannot produce supporting data to back up their side of the argument. These people tend to rely heavily on tenuous, unprovable concepts, such as "magic" and "luck"to describe circuits, rather than anything concrete, like mathematics, scope plots, or plain old reason and common-sense.
The laws of "golden ears" are a bit complex. Here they are, at least as far as I have been able to ascertain:
(1) "You are not allowed to make any mention of golden ears until you are losing an argument and have no way to save face."
(2) "He who first proclaims to have golden ears is the only one in the argument who can truly have golden ears." The opponent, therefore, must, by the rules, have tin ears, since there can only be one golden-eared person per argument. Now, everyone else on the board may have golden ears, but only if they choose not to perform the test in question that may disprove the theory. These people may be brought up as often as possible, until that point where they actually perform said test and obtain results not in accordance with the original golden-eared proponent's position.
(3) "Golden ears can only exist on a person who has no formal education in the subject matter." The minute a person receives an engineering degree, his ears automatically turn to tin, and he cannot be trusted with even the simplest test of tone or frequency response, no matter how long he or she has been in the field of interest.
(4) "Once you have proclaimed golden-ear status, you can no longer answer any questions, nor put forth anything of a technical nature, because you may slip up and show that you really have no clue as to what you are talking about." At this point, your golden ears have become tarnished, and everyone will see the emperor has no clothes, so this situation should be avoided at all costs. Should this happen, you have no choice but to either slink away and hope nobody notices and soon forgets about you, or loudly proclaim victory in the vain hope that someone will believe you.
I'm sure there are more rules, I'll post them as I learn them.
Q: You claim purple tolex sounds best. What about green and other colors?
A: Here is the standard tonal color hierarchy:
Purple: the most high tone, a royal color indeed, to be played only by those secure in their manhood.
Red: can sometimes sound as good as purple, and can be played by anyone, even those not secure in their manhood.
Pink: to be played only by girls, and those secure in someone else's manhood.
Orange: a rare tone, not often found. Ideally should be elephant grain, although these are seldom heard in the wild, except in old 70's Marshalls. Orange levant is a close substitute, and it is available nowadays. Rough-grain orange "nubtex" is below black in terms of tone, and should be avoided, unless your amp company name is "Orange". If your amp company name is "Orange", you should consider changing your tolex to elephant grain, to make it more readily available to those of us who want to use it without having to custom-order five 50-yard rolls.
Green (dark blackish green only, not that lime crap): Plays well with gold-paneled amps, imparts an "elegance" to the tone, and can be played in all types of bands, including wedding gigs and jazz clubs, so it is a versatile color. This color coordinates well with older players who are feeling their oats and want to step out a bit from basic black, and make a fashion statement without appearing to make a fashion statement, and for those afraid to stand out too much by playing a brighter color.
Fawn: a very seldom used color, it can sound very good, but again, only on an amp with a gold faceplate.
White: what can you say about white? It's white. White has no color, but it still sounds better than black, so there.
Black: your basic tone, the standard by which all others are judged.
Q: Who are your favorite guitarists?
A: Michael Schenker is my all-time favorite guitarist, especially on his albums with UFO. The first album I ever bought as a teenage kid was UFO's "Force It", mainly because of the picture on the back cover showing Michael Schenker with a Gibson Flying V and a bunch of Marshall stacks. That was the coolest thing I had ever seen in all my life at the time! The second album I ever bought was UFO's "Phenomenon". I wore the grooves out on those two albums, trying to learn the songs. Schenker had the most amazing tone and feel in his playing, and it made a huge impact on me. I hope someday to be able to meet him. Second would be Gary Moore - he is another master of tone and phrasing in a rock/blues context. I never tire of listening to his playing. Sadly, he is no longer with us.. In other styles, Shawn Colvin is my all-time favorite acoustic guitarist. She has an amazing voice and incredible mastery of alternate tunings and phrasings. Other guitarists/groups I like to listen to are Jeff Beck, Van Halen, Tom Petty, Bryan Adams, SRV, Hellecasters, early Rod Stewart/Faces, Aerosmith, Judas Priest, and of course, AC/DC (great Marshall tones there!). I'm a closet 80's hair band fanatic, and I've always had a place in my heart for Joan Jett and Lita Ford, too! There is very little "new" music I can stand to listen to. It all sounds the same to me - mushy, overly-distorted guitars with no dynamics or tone and Cookie Monster vocals. I must be getting old - that's what my father said about the music I like!. :) I also think modern country music has some of the best guitar tones and playing nowadays. For country music, I really like Keith Urban, Brad Paisley, Marty Stuart, Miranda Lambert, and Vince Gill.
Q: Are the Vox AC-30 and other clones really Class A?
A: Read this article and decide for yourself - Is the Vox AC-30 really class A?
Q: Is there any advantage to using solder with a 2% silver content?
A: Yes. Silver solder keeps werewolves away from your amp.
Q: Is solid-state really as evil as you say?
A: Yes. Update: I'm becoming rather fond of the new breed of high-voltage depletion-mode MOSFETs, so I may change my mind about this...
Note: this section is periodically updated when new information is added or an error is found.
Copyright © 1999-2017, Randall Aiken. May not be reproduced in any form without written approval from Aiken Amplification.