Global negative feedback refers to the "feeding back" of a small amount of signal from a later part of the circuit to an earlier part, usually from a tap on the output transformer back to the phase inverter. The use of global negative feedback does several things: it flattens and extends the frequency response, it reduces distortion generated in the stages encompassed by the feedback loop, and it reduces the effective output impedance of the amplifier, which increases the damping factor, producing a "tighter" tone.
While this may sound like a great idea, it is not without drawbacks for guitar amplification purposes. All of these things affect the tone in some manner. Negative feedback can make an amp sound more "solid-state" by reducing the interaction between the speakers and the amplifier. It can also make the transition from clean to distortion more abrupt, as the negative feedback loop fails when there is insufficient forward loop gain to maintain the loop, which occurs when the output hits clipping. Whether or not to use feedback is a subjective matter, left to the taste of the designer. The design process is relatively simple.
How it works
The amount of voltage fed back determines the amount of gain reduction and the amount of distortion reduction, as well as the effective output impedance. The more voltage fed back, the less distortion, the lower the effective output impedance, the higher the damping factor, and the lower the gain of the stages enclosed by the feedback loop.
Typically, in a guitar amp, somewhere around 6-10dB of feedback is used. If you have 6dB of feedback, for instance, and it takes 2V at the phase inverter input to achieve output clipping, if you removed the feedback, it would only take 1V at the phase inverter input to achieve output clipping. In other words, there is a voltage gain reduction of 6dB, or a factor of two, in the stages enclosed by the feedback loop. This is achieved by feeding back a certain percentage of the output voltage to an earlier point in the circuit, the phase inverter. The more voltage fed back, the more the voltage gain reduction, as mentioned previously. There is a danger in using too much global negative feedback, however, as the amplifier may oscillate due to phase shifts in the transformer and other circuits. The amount of feedback also determines the amount of range of the presence control, as it attenuates the voltage being fed back at higher frequencies, which in turn boosts those frequencies at the output.
The series feedback resistor, in conjunction with the resistor to ground, determines the amount of voltage being fed back. If you want to feed back more voltage, you make the series resistor smaller, or the shunt resistor larger, or you use a higher impedance tap on the output transformer.
The actual resistor values used in the feedback attenuator aren't that important, as their ratio determines the amount of feedback. The shunt resistor value is usually fixed by the phase inverter design requirements, and the series resistor is then sized according to the desired amount of feedback, given the voltage available at the output. Note that Marshall typically uses 100K/5K attenuator, while Fender uses a 820ohms/100ohms. You can get the same attenuation from a 10K/500ohm pair as you would from a 100K/5K pair. In addition, if you were using a 100K/5K attenuator running from the 16 ohm tap, you would get roughly the same amount of feedback if you used a 47K/5K attenuator running from the 4 ohm tap. Note that the tap voltage is not linear with respect to the impedance, it varies linearly with the square root of the impedance. That is, the voltage on the 8 ohm tap is not half the voltage on the 16 ohm tap, rather, the voltage on the 4 ohm tap is half the voltage on the 16 ohm tap. It helps if you think of the equation for power: P = V^2/R. If you have 100W into 16 ohms, the voltage is V = sqrt(100*16) = 40V RMS. If you have 100W into 8 ohms, the voltage is V = sqrt(100*8) = 28.28V RMS. If you have 100W into 4 ohms, the voltage is V = sqrt(100*4) = 20V RMS.
When designing an amplifier from scratch, it doesn't really matter which transformer tap you use. What you are trying to do is obtain a certain amount of negative feedback, which is determined by the ratio of the two feedback resistors (the series feedback resistor from the output and the shunt feedback resistor that goes to ground, which form an attenuator). The tap simply determines the initial amount of voltage applied to the feedback network, which is then adjusted to achieve the desired amount of signal reduction, or negative feedback.
The design process
Following is a schematic of a typical amplifier output stage:
While this circuit looks complicated at first glance, it can be broken down into the simple block diagram of a non-inverting feedback amplifier as shown below:
The block labeled "A" is the open loop forward gain path. This is the gain of the output stage if the feedback loop is removed, i.e. if Rf is disconnected. Rf and Ri are the series and shunt feedback resistors, respectively. Ri is the value of the presence control pot in the above schematic diagram, or 5K. Rf is the 100K series feedback resistor. Ro is the internal output impedance of the amplifier.
It can be shown1 that the equation for the closed loop gain of this amplifier is:Acl = A*(Ri+Rf) / (Ri + Rf + Ro + Ri*A)
which simplifies to the following equation if the output resistance, Ro, is very small in comparison to the feedback resistance values, as is usually the case in guitar amplifiers:Acl = A / (1 + A* Ri / (Ri + Rf))
Calculating the closed-loop gain and feedback factor
First you must calculate (or measure) the open-loop forward gain of the output stage portion where feedback is to be applied. For instance, in a "standard" Marshall type amp configuration as shown above, the input is applied to the first input of the phase inverter, and the feedback is applied to the second input. The output transformer primary taps are arranged so the output voltage is in phase with the first input. This means that the feedback voltage will subtract from the overall gain if added into the second input, producing negative feedback, which is what we want. If the transformer primary taps are reversed, the feedback will be positive, resulting in uncontrollable oscillation.
For the circuit shown above, the forward gain, G, to the 16 ohm output, when loaded with 16 ohms, is around 41, depending on the plate voltage, the output transformer, the tubes, etc.
Second the feedback factor must be calculated. For the circuit shown above, the feedback factor is:
H = 5K/(100K + 5K) = 0.048
Now the closed-loop gain can be calculated using the gain formula:
Acl = 41/(1 + 41 * 0.048) = 13.8
This corresponds to a gain reduction of 13.8/41 = 0.337, or in dB: 20*log(41/13.8) = -9.5 dB.
Calculating the required feedback resistor value for a given feedback factor
This process can be used in reverse, to determine the required value of feedback resistor for a given amount of negative feedback.
For example, if -6 dB of negative feedback is desired, and a 5K resistor (or presence pot) is used for Ri, the value of the feedback resistor can be determined by rearranging the closed-loop gain equation as follows:Rf = -Ri* (Acl + Acl* G - G) / (Acl - G)
First, the required closed loop gain, Acl, is determined by the required -6dB of feedback as follows:Acl = log-1(-6dB/20) * G = 0.5 * 41 = 20.5
Note: the equation dB = 20 * log (V1/V2) was rearranged to solve for (V1/V2) in this example. Thus, (V1/V2) = log-1(dB/20). This, multiplied by the open-loop gain, gives the required closed-loop gain for the desired amount of feedback in dB.
Next, Rf is calculated as follows:Rf = -5K * (20.5 + 20.5 * 41 - 41)/(20.5 - 41) = 200K
The new feedback factor, H, is:H = 5K/(200K + 5K) = 0.024
Using a different impedance tap on the output transformer
The above design assumed the feedback voltage was taken from the 16 ohm tap. What if the feedback is moved to the 4 ohm tap? What will the resulting feedback factor be, and how can it be returned to the original desired feedback factor of -6dB?
Since the tap voltage varies with the square root of the impedance tap (as explained above), for a 1V input to the phase inverter, which results in a 15.5V output on the 16 ohm tap, the output voltage on the 4 ohm tap will be 7.75V as shown in the equation below:V(4 ohm) = V(16 ohm) * sqrt(4/16) = 20.5V * sqrt(0.25) = 20.5V * 0.5 = 10.25 volts.
Since the output is still taken from the 16 ohm tap, but the feedback has been moved to the 4 ohm tap, the open-loop gain remains the same, at 31. However, the feedback factor is now attenuated by a factor of two, because the amount of voltage being fed back through the attenuator is now half the original value when the feedback was taken off of the 16 ohm tap.
The new feedback factor, H, is as follows:H = H2 * 5K / (5K + 200K) = 0.012
where H2 = 7.75/15.5 = 0.5
The new closed-loop gain (to the 16 ohm output tap) will be:Acl = 41/(1 + 41 * 0.012) = 27.5
This corresponds to a gain reduction of 27.5/41 = 0.67, or in dB: 20*log(27.5/41) = -3.5dB.
In order to get the feedback factor back to -6dB, the series feedback resistor will have to be recalculated to compensate for the attenuation of two. If the original closed-loop gain equation is modified to include the new attenuation factor, H2, the new equation for closed-loop gain will be:Acl = G / (1 + G * H2 * Ri / (Ri + Rf))
Therefore, the new value of Rf, to achieve the original desired Acl of 20.5, can be calculated as follows:
Rf = -Ri* (Acl + Acl*G*H2 - G) / (Acl - G)soRf = -5K* (20.5 + 20.5* 41*0.5 - 41) / (20.5 - 41) = 97.5KAnother way to look at this is that the required voltage at the feedback summing point must remain the same, but only half the original voltage is present on the input of the divider. The new value of Rf can be calculated by determining the original voltage, and the resulting current through the feedback resistor, and calculating a new value of feedback resistor by subtracting the summing voltage from the new input voltage, and dividing by the required current.
For example, if 20.5V was originally present, the voltage at the summing point is 20.5*5K/(5K+200K) = 0.317V. This results in a current of 20.5V-0.317V/200K = 100uA. The new voltage is 10.25V, so the required resistor is calculated as 10.25V-0.5V/100uA = 97.5K.
Using the formula for input impedance, along with the originally calculated value of 13.8, and assuming a grid resistor, Rg, of 1Meg, feedback resistor, Rf, of 100k, and input resistor, Ri, of 5K, the input impedance would be:Zin = Rg / (1V - Acl*Ri/(Ri+Rf))
= 1 Meg/(1V- 13.8*5K/(5K+100K)
= 2.9 Meg
Effective Output Impedance
Using the formula for output impedance, along with the originally calculated open-loop gain of 41, and assuming a feedback resistor, Rf, of 100k, and an input resistor, Ri, of 5K, and an internal output impedance of 16 ohms, the closed-loop effective output impedance would be:
Zout = ((Ri + Rf) * Ro) / (Ri + Rf + Ro + Ri*A)
= (5K + 100K) * 16 / (5K + 100K + 16 + 5K*41)
= 5.2 ohms
The Effect of Changes in Load Impedance
Note that in a tube amp, the load impedance greatly affects the open-loop gain, because the internal plate resistance of the typical pentodes is very high, so the effective output impedance would be rather large if you didn't have a load connected. The impedance seen looking into the output would be equal to the effective plate resistance of the tubes divided by the impedance ratio of the tubes. When a load is connected, it reflects back an impedance equal to its value multiplied by the impedance ratio of the transformer. This means that the effective internal output impedance is equal to the output load in parallel with the tube plate resistance reflected to the secondary. It is still fairly close to the load resistance, because the plate resistance of a typical pentode is quite large.
What this all means is that the open-loop gain is going to change when a different load impedance is connected to the same tap. This change in open-loop gain changes the effective output impedance and the overall closed-loop gain of the amplifier.
For example, in the above example, the open-loop gain on the 16 ohm tap with a 16 ohm load is 41. This leads to a closed-loop gain of 13.8 and an effective internal output impedance of 5.2 ohms. There is no additional attenuation due to the load resistance, because it figures into the internal output resistance and the open-loop gain. It is effectively "inside" the amplifier block at that point.
Now, if the transformer tap isn't changed, and an 8 ohm load is connected to the 16 ohm tap, the open-loop gain will drop to around 21. This will result in a new closed-loop gain of:Acl = 21/(1 + 21 * 0.048) = 10.5
and a new effective output impedance of:Zout = ((Ri + Rf) * Ro) / (Ri + Rf + Ro + Ri*A)
= (5K + 100K) * 8 / (5K + 100K + 8 + 5K*21)
= 4 ohms
Once again, there is no additional attenuation, because it is taken into account with the change in open-loop gain.
Note that the feedback factor doesn't change, but the change in closed loop gain is due instead to the change in open-loop gain, which was rather small to begin with.
The actual measured values will be off by a small amount if the internal amplifier output resistance, Ro, was ignored. This can be compensated by simply adjusting the value of the feedback resistor slightly from the calculated value to achieve the desired gain reduction. In the case of a guitar amplifier, the output resistance is usually very small in comparison to the feedback resistor values, so it can be ignored.
In order to test and verify the feedback calculations and the final design. the amplifier is constructed and a known input voltage, for example, 1V p-p is applied to the phase inverter input. The output voltage with the feedback resistor both in and out of circuit is measured, which gives the open-loop and the closed-loop voltage gain, which can then be compared to the calculated values.
Appendix A: The math behind the non-inverting global feedback amplifier:
1Feedback principles and circuit analysis
In order to analyze the circuit to determine the equations to calculate gain, input and output impedances, and frequency response, the circuit must first be converted to block diagram form. The basic top-level block diagram of the above circuit is shown below:
Ri is the input resistance, Rf is the feedback resistance, and Ro is the internal output resistance of the amplifier stage being used. The block labeled "A" represents the open-loop gain of the amplifier being used.
This top-level block must be converted to a full block diagram detailing the entire feedback system. This is done by deriving equations for the forward paths and feedback paths of the top-level block, and determining the overall transfer function of the system.
First, the system is broken down into the separate forward gain and feedback attenuation paths. The non-inverting amplifier is simpler with respect to the inverting amplifier, in that it only has one forward path.
Forward path:The forward path has a gain element, A, which amplifies the difference, or error voltage, between the input and the attenuated feedback path, to produce a voltage at the output of amplifier A, so the transfer function of this block would be simply:G = A
Feedback path:There is a single distinct feedback attenuation path, but the output is taken from the junction of two attenuators in this path. The first attenuator is from the output of amplifier A to the junction of Ri, Rf, and the inverting input of amplifier A. This transfer function can be derived using the voltage divider rule as follows:Ve = Vo*(Ri + Rf)/(R i+ Rf + Ro)so the transfer function of this block would be:H1 = Ve1/Vi = (Ri + Rf) / (Ri + Rf + Ro)(Note: Ve1 is the component of the error voltage at the inverting input to the amplifier A.)
The second attenuation path is from Vout to the junction of Ri, Rf, and the input of amplifier A. This transfer function can also be derived by using the voltage divider rule as follows:Ve2 = Vo * Ri/(R i + Rf)
so the transfer function of this block would be:H2 = Ve2/Vo = Ri / (Ri + Rf)
As a check, the combined transfer function of the feedback path should be equal to H1*H2, orH = (Ri+Rf)/(Ri+Rf+Ro) * Ri/(Ri+Rf) = Ri/(Ri+Rf+Ro)
This gives the correct attenuation from the output the gain block to the inverting input, so it is correct.
The completed block diagram:
The completed block diagram showing all forward and feedback paths is shown below.
Where:G = A
H1 = (Ri + Rf) / (Ri + Rf + Ro)
H2 = Ri / (Ri + Rf)Now that the overall block diagram is complete, the gain equations and the overall transfer function for the system can be derived. Note the labeling of critical nodes: R is the input signal, C is the output signal, E is the error signal, and D is the amplifier output signal. In order to derive the output equations and impedance equations, it is necessary to determine these intermediate point equations as well. The derivations are as follows:
E = R - D* (H1*H2)or
D = E * G
= [R - D*(H1*H2] * G
= R*G - D*H1*H2*GD = R*G/(1+H1*H2*G)The transfer function of the output, C, is then:C = D*H1Substituting for G, H1, and H2:
= R*G*H1/(1+H1*H2*G)C = (R* A*(Ri + Rf) / (Ri + Rf + Ro)) / ((1 + (Ri + Rf) / (Ri + Rf + Ro) * (Ri / (Ri + Rf)) * G))simplifying:C = R* A*(Ri+Rf) / (Ri + Rf + Ro + Ri*A)This is the final equation for the output voltage, C, for a given input voltage, R.
The transfer function for this amplifier (Vout/Vin) is then C/R, or:Vout/Vin = C/R = A*(Ri+Rf) / (Ri + Rf + Ro + Ri*A)
so the closed loop gain is:Acl = A*(Ri+Rf) / (Ri + Rf + Ro + Ri*A)
If Ro, the internal amplifier output resistance, is small enough in comparison to the feedback resistance and the load resistance, it can be ignored in order to simplify the calculations, without too much error. The transfer function of this circuit can then be described using the standard feedback formulas as follows:Vout/Vin = G / (1+GH)
where G = the forward gain of the amplifier
H = the feedback factor
In the above block diagram, if Ro is zero, the feedback factor is the attenuation of the feedback divider resistors, or:H = Ri / (Ri + Rf)
This simplifies to the following equation:
Acl = G/(1+GH)
or:Acl = A / (1 + A* Ri / (Ri + Rf))
which can be rewritten as:Acl = A* (Ri+Rf) ./ (Ri + Rf + Ri * A)
which can be seen as the same result if Ro is replaced by zero in the original closed-loop gain formula.
Following is a block diagram of the new simplified transfer function:
The input impedance can be determined from the the top-level block diagram shown below and the original output stage circuit diagram show above:
Since the non-inverting input ideally draws no current, the input impedance is essentially infinite for an ideal amplifier. In the case of the actual phase inverter used in a guitar amplifier as shown above, the input impedance depends on the amount of feedback used and the size of the input grid resistor, Rg. The voltage at the top of the grid resistor is the input voltage, and the voltage at the bottom of the grid resistor is the error voltage at the feedback divider point. Therefore, the current that flows in the grid resistor is the difference between those two voltages divided by the grid resistance. The input impedance can be determined by assuming a 1V input test signal, and dividing this 1V by the calculated current.
Zin = 1V/Iin = 1V/((1V-E)/Rg)) = Rg/(1V - E)
Where E is the resulting voltage at the inverting input of the amplifier A when the 1V test signal is applied to the input, and is calculated as follows:E = Vout * Ri/(Ri + Rf) = Acl*Ri / (Ri+Rf)
Therefore, the input impedance would be:Zin = Rg / (1V - Acl*Ri/(Ri+Rf))
It can be shown that if the open-loop gain is infinite, the input impedance will also go to infinity, because the closed loop gain will be equal to (1 + Rf/Ri), which will result in division by zero in the above input impedance equation, indicating a Zin of infinity.
The output impedance can be determined from the above derived equations and the original top-level block diagram.
In order to determine the output impedance, the input must be grounded, and a test voltage of 1V applied to the output. The resulting current is calculated, and the output impedance is equal to the 1 volt test input divided by this current. Looking at the top level block diagram, the current flow resulting from this 1V test input would split into two paths as shown below:
The first current, I1, would be equal to:I1 = (1V - E) / Rf
Where E is the resulting voltage at the inverting input of the amplifier A when the 1V test signal is applied to the output, and is calculated as follows:E = 1V * Ri/(Ri + Rf)
The second current, I2, would be equal to:I2 = (1V - E*(-A))/Ro = (1V + E*A)/Ro
The total current, I, would be the sum of I1 and I2 as below:I = I1 + I2
orI = (1V - E) / Rf + (1V + E*A)/Ro
The resulting output impedance would be:
Zout = 1V/I
= 1V / [(1V - E) / Rf + (1V + E*A)/Ro]
substituting for E and simplifying:Zout = ((Ri + Rf) * Ro) / (Ri + Rf + Ro + Ri*A)
if Ro can be ignored and assumed zero, the equation simplifies to:Zout = 0
The three main equations used in the design of a single-stage negative feedback amplifier are the gain equation, the input impedance equation, and the output impedance equation. Following are the simplified equations, and the resulting equations when Ro can be ignored:
- Acl = A*(Ri+Rf) / (Ri + Rf + Ro + Ri*A)
- Acl (neglecting Ro) = A / (1 + A* Ri / (Ri + Rf))
- Zin = Rg / (1V - Acl*Ri/(Ri+Rf))
- Zin, if closed loop gain is high enough, is essentially infinite
- Zout = ((Ri + Rf) * Ro) / (Ri + Rf + Ro + Ri*A)
- Zout (neglecting Ro) = 0
Note that A is always a positive quantity for negative feedback.
Copyright © 2000, Randall Aiken. May not be reproduced in any form without written approval from Aiken Amplification.