**Designing Single-Stage Inverting Feedback Amplifiers**

**General **

The single-stage inverting feedback amplifier is a useful circuit for guitar amplifier design. The main features of the amplifier are phase inversion, very high linearity, gain which is relatively independent of tube parameters (more on this later), extremely low output impedance, which makes it ideal for driving tone stacks or effects loops, very wide and flat frequency response, and a fixed input impedance approximately equal to the value of the input resistor, which can be relatively low or high.

**The inverting feedback amplifier configuration **

The amplifier works on the principle of local negative feedback Following is a schematic diagram of a typical single-stage inverting feedback amplifier:

The amplifier is built around a standard common-cathode stage, composed of V1A, Rp, Rk, and Ck. These components determine the open-loop gain of the amplifier. Rl is the load resistor, which is most commonly the input resistance of the next stage. Ci is the input coupling capacitor, which is used to block any DC voltage from the previous stage from affecting the grid bias of the inverting amplifier stage, as well as control the lower -3dB frequency point of the stage. Co is the output coupling capacitor, which is used to block the DC voltage at the plate of the amplifier from the grid circuit and the output. Components Ri and Rf are the feedback components which determine the overall gain, input, and output impedance of the amplifier.

Note that for calculation purposes involving a driving stage with a non-zero output impedance, Ri is actually the value of the input resistor plus the output impedance of the previous stage. This can lead to drastic differences between the calculated value and the measured value if it isn't taken into account. Another way to look at it is that the output of the previous stage is attenuated by the voltage divider formed by it's own internal output impedance and the input impedance of the inverting feedback amplifier. It is usually simpler to add the output impedance of the previous stage to the value of Ri, and treat the input voltage as the unloaded output voltage of the previous stage. Either way, the resulting composite stage gain is the same.

**Designing a single-stage feedback amplifier **

It can be shown^{1}that the equations for the gain, input impedance, and output impedance for the single-stage feedback amplifier are:

- Acl = Vout/Vin = (Ra + A*Rf) / (Ri + Rf + Ra - Ri*A)

- Zin = (Ri * A - Ri - Rf - Ra)/(A-1)

- Zout = (Ri + Rf) * [Ra / (Ri + Rf + Ra - Ri*A)]

Where:Acl = closed loop gainFor an ideal inverting feedback amplifier, with zero internal amplifier output resistance, these equations simplify to:

A = open loop gain

Ri = input resistance

Rf = feedback resistance

Ra = internal output resistance of the amplification stageNote that for negative feedback, A must be a negative quantity. If A is positive, the feedback will be positive.

- Acl = A*Rf / (Ri + Rf - Ri*A)

- Zin = (Ri * A - Ri - Rf)/(A-1)

- Zout = 0

Note that for negative feedback, A must be a negative quantity. If A is positive, the feedback will be positive.

For an ideal inverting feedback amplifier with both zero internal amplifier output resistance and infinite gain, these equations further simplify to:

- Acl = -Rf/Ri

- Zin = Ri

- Zout = 0
These equations show that the output voltage is dependent upon the feedback elements, the output impedance of the amplifier, and the gain of the amplifier stage. The output impedance can be ignored, and the second or third set of equations can be used if the gain of the amplifier is large enough and the output impedance is small in comparison to the feedback amplifier resistor values.

In order to design a negative feedback amplifier, the forward open-loop gain path must first be designed. For this example, a 12AX7 is chosen, because it has the highest gain of the common preamp tubes.

Assuming a 12AX7 is chosen, and a gain of -2 is desired, with a lower -3dB frequency of 50Hz, and an input impedance of 100k. Following are the calculations for the gain and output impedance:

Amplifier design:

- Some typical 12AX7 numbers:

plate resistance: ra = 62.5K

amplification factor: mu = 100

- A plate load resistance of 100K and a cathode resistance of 820 ohms are chosen to provide good gain and linearity in the forward path. In order to maximize the gain and minimize the output impedance, the cathode must be fully-bypassed at the frequencies of operation. The midband gain for the 12AX7 with a fully-bypassed cathode is:

Av = -(mu*Rl)/(Rl+ra)

= -(100*100K)/(100K+62.5K)

= -61.5

Where:

Av = the open-loop gain of the amplifier

Rl = the load resistor (470K in this example)

ra = the internal plate resistance (62.5K for a typical 12AX7)

mu = the mu of the tube (100 for a typical 12AX7)

Note that the negative sign indicates phase inversion at the plate, relative to the grid input.

- Next, the output impedance of the stage is calculated. Since the internal plate resistance is effectively in parallel with the plate load resistor, the output impedance (for the signal taken off the plate) will be:

Ra = ra || Rl

= 62.5K || 100K

= 38.5K

Note: the symbol "||" means "in parallel with". Resistors in parallel add in reciprocal, i.e. 1/Rt = 1/R1 + 1/R2.

- Next, the resistance seen looking into the cathode must be calculated, as it is needed to determine the value of bypass capacitor needed.

The resistance seen looking into the cathode ( with Rk unbypassed) is:

Rk' = (Rl+ra)/(mu+1)

= (100K + 62.5K)/(101)

= 1.6K

Therefore, the total cathode resistance is the parallel combination of the cathode resistance, Rk', and the cathode resistor, Rk, as below:

Rcathode = Rk' || Rk

= 1.6K || 820

= 542 ohms

- Next, the value of the cathode bypass capacitor must be calculated. Since the gain of the amplifier must be high at the chosen lower cutoff frequency of 50Hz, the minimum capacitor value is chosen using the following equation for the upper shelving frequency breakpoint of the amplifier.

f = 1/(2*PI*R*C)

solving for C:

C= 1/(2*PI*R*f)

The minimum value of bypass capacitor is therefore:

C= 1/(2*PI*542*50Hz) = 5.8uF

The amplifier gain will be down around -3dB at this point, corresponding to a gain decrease of 0.707 times 61.5, or 43.5. However, in order to maximize gain and keep the phase shift associated with the cathode bypass capacitor to a minimum, the capacitor value should be increased to around twice this value, or 10uF. If the capacitor is made too large, however, the stage will suffer from poor transient response, as the cathode voltage will slowly move up and down, causing the plate current, and thus, the plate voltage, to also shift up and down. This can result in an audible distortion and nonlinearities in response to a large transient input. For this reason, the capacitor should be made no larger than necessary.This bypass capacitor can be eliminated entirely, if the tube has sufficient gain to provide the necessary gain margin for the feedback loop to provide the desired gain and input/output impedances. Removing the capacitor in this case will result in a gain of 31.9 and an output impedance of 68.2K for the open-loop amplifier.

Now that the open-loop gain element has been characterized, the rest of the feedback amplifier can be designed, and the closed-loop gain can be determined.

Feedback element design:

- For the circuit chosen, the open-loop gain and output impedance values were calculated above to be:

A = - 61.5

Ra = 38.5K

If the exact required gain of -2 is needed, the gain equation can be solved for Rf, and Ri is chosen to give the desired minimum input impedance. However, for this example, it is useful to assume ideal amplifier characteristics to choose the value of Rf, so the amount of error can be seen when using the actual equations.

Since the input impedance is desired to be 100k, the value of Ri is chosen to be 100k, because the ideal amplifier equations show the input impedance equal to Ri, and the non-ideal amplifier shows that the input resistance will be greater than Ri, but never less than Ri.

Therefore:

Ri = 100k

- Next the feedback resistor is calculated, based upon the ideal amplifier equations and the required gain.

The ideal gain equation is as follows:

Vout = -Vin*(Rf/Ri)

Solving for Rf:

Rf = -(Vout/Vin)*Ri

Since a gain of -2 is desired, Rf is calculated as:

Rf = -(-2)*Ri = 2*100K = 200K

- Now all parameters are known, so the actual amplifier performance can be determined, and the resulting gain and impedances can be compared to the ideal values.

A = -61.5

Ra = 38.5K

Ri = 100K

Rf = 200K

The voltage gain can be calculated as follows:

Vout = Vin* (Ra + A*Rf) / (Ri + Rf + Ra - Ri*A)

= Vin * (38.5K - 61.5*200K) / (100K + 200K + 38.5K - 100K*(-61.5))

= Vin * (-1.89)

Therefore, the closed-loop gain (or Acl) of the amplifier is Vout/Vin, or -1.89. This differs from our desired ideal gain of -2 by 5.5%, which is close enough for most applications.If the 820 ohm cathode resistor is unbypassed, the equations are as follows:

ra'(unbypassed Rk) = ra + (mu + 1)*Rk

= 62.5K + (101)*820

= 145.3KRa = 145.3K || 100K = 59.23K

Av(unbypassed Rk) = -(mu*Rp)/(Rp+ra')

= (100*100K)/(100K+145.3K)

= 40.8This gives the following parameters:

A = -40.8 (negative to indicate inverting amplifier)

Ra = 59.2K

Ri = 100K

Rf = 200K

The new voltage gain would be:

Vout = Vin* (Ra + A*Rf) / (Ri + Rf + Ra - Ri*A)

= Vin * (59.2K - 40.8*200K) / (100K + 200K + 59.2K - 100K*(-40.8))

= Vin * (-1.83)

Therefore, the closed-loop gain of the amplifier is Vout/Vin, or -1.83. This differs from our desired ideal gain of -2 by 8.5%, which is probably still close enough for most applications.A better way to approach the design is to solve the closed-loop gain equation for Rf, and use the given parameters to solve for it. This will provide the exact value of Rf to achieve the desired gain.

Acl = Vout/Vin = (Ra + A*Rf)

(Ri+Rf+Ra -Ri*A)Solving for Rf:

Rf = ((Ri+Ra-(Ri*A))*Acl)-Ra

(A-Acl)Now that we have an equation for Rf, we can plug in the values from above to get the required feedback resistor for an exact closed-loop gain of -2:

Fully-bypassed cathode:

Acl = -2 (negative because it is inverting)

A = -61.5 (negative because it is inverting)

Ra = 38.5K

Ri = 100KRf =( (100K+38.5K-(100K*(-61.5))*(-2)) -38.5K = 212K

-61.5-(-2)

Unbypassed cathode:

Acl = -2

A = -40.8

Ra = 59.2K

Ri = 100KRf = ((100K+59.2K-(100K*(-40.8))*(-2)) -59.2K = 220K

-40.8-(-2)The input impedance for the bypassed cathode stage can be calculated as follows:

Zin = (Ri * A - Ri - Rf - Ra)/(A-1)

= (100K * (-61.5) - 100K - 200K - 38.5K) / (-61.5 - 1)

= 103.8K

Therefore, the input impedance of the amplifier is 103.8K. This differs from the desired ideal input impedance of 100k by 3.8%, which is close enough for most applications.If the cathode resistor is unbypassed, the input impedance would be:

Zin = (Ri * A - Ri - Rf - Ra)/(A-1)

= (100K * (-40.8) - 100K - 200K - 68.2K) / (-40.8 - 1)

= 108.2K

Therefore, the input impedance of the amplifier is 108.2K. This differs from the desired ideal input impedance of 100k by 8.2%, which is still close enough for most applications.

The output impedance can be calculated as follows:

Zout = (Ri + Rf) * [Ra / (Ri + Rf + Ra - Ri*A)]

= (100K + 200K) * [38.5K/(100K + 200K + 38.5K - 100K*(-61.5)]

= 1.78K

Therefore, the output impedance of the amplifier is 1.78K. This differs from the desired ideal output impedance of zero by a large amount, but is quite low in comparison to most vacuum tube amplifier stages (even cathode followers), so it should be close enough for most applications, as vacuum-tube circuitry typically works into rather large impedances.If the cathode resistor is unbypassed, the new output impedance would be:

Zout = (Ri + Rf) * [Ra / (Ri + Rf + Ra - Ri*A)]

= (100K + 200K) * [59.2K/(100K + 200K + 59.2K - 100K*(-40.8)]

= 4K

This is a 2.25 times increase in output impedance, but it is still very low compared to the typical output impedance of a common cathode stage.

- Next the frequency response components must be calculated.

Since the output coupling capacitor in the above circuit is inside the feedback loop, it is only marginally important in the actual output frequency response, as any rolloff due to it's capacitive reactance will tend to be taken out by the feedback, as long as there is enough excess gain at the frequencies of interest. However, if the reactance is large compared to the output impedance, it will introduce a frequency-dependent increase in output impedance, which will affect the above gain equations. This increasing output impedance will both increase the gain and the output impedance of the feedback network, so too small a capacitor will either peak the response at low frequencies, or roll off the gain at low frequencies, depending upon the actual output impedance and the capacitor value in relation to the feedback resistor. In addition, if the feedback resistance is too low in comparison to the reactance of the capacitor, the frequency response will also be adversely affected.

For these reasons, it is best to choose the output coupling cap reactance based upon the feedback resistance value. It should be around ten times the value required to achieve the desired frequency response, and can be calculated as follows:

Co = 10 * 1/(2*pi*f*Rf) = 10 * 1/(2*pi*50*200K) = 0.16uF (choose 0.15uF as the nearest standard value)

Once again, this capacitor should not be made any larger than necessary, or the transient response of the stage will be poor.The main frequency determining component is the input capacitor, Ci. It, in conjunction with the input impedance, Zi, determines the lower -3dB rolloff point of the amplifier, and can be calculated as follows:

Ci = 1/(2*pi*f*Zi) = 1/(2*pi*50*103.8K) = 0.03uF (choose 0.033uF as the nearest standard value)

Note that large bandwidths cannot be obtained with high closed-loop gains. This is because the gain-bandwidth product of the amplifier is fixed, so any increases in gain will result in a reduction in bandwidth. By the same token, a reduction in gain will result in an increase in bandwidth (however, the low frequency response will remain controlled by Ci). Because of this, the amplifier can have a very high upper -3dB point (around 450kHz with the above circuit!), and it may be desirable to limit the frequency response somewhat. This can be accomplished by placing a capacitor in parallel with the feedback resistor, Rf. Assuming an upper -3dB limit of 20kHz is desired, The capacitor value is calculated as follows.

Ci = 1/(2*pi*f*Rf) = 1/(2*pi*20kHz*200K) = 39pF

The finished circuit:

The final circuit is shown below:

Note that the tube gets its grid DC bias reference from Rf and Rl. If the sum of the feedback resistor and the load resistor is very large, a 1Meg resistor should be added from the grid of the tube to ground, to keep the grid bias stable. This will not affect the gain, because the grid node is a "virtual ground" and has very little voltage on it. The voltage can be calculated as the error voltage E as follows (assuming an input voltage of 1V):

E = Vin*(Rf + Ra)/[(Ri + Rf + Ra) - Ri*A] = 0.037V or 37mV

Compared to the input voltage of 1V, this 37mV is very small. Because this is a virtual ground input, it can be used as a summing junction. As can be seen by the error voltage equation, the larger the open-loop gain, A, the smaller this error voltage will be, and the better the isolation between inputs. If another input is connected through a 0.033uF capacitor and a 100K resistor, the output voltage will be the sum of the two inputs multiplied by the closed-loop gain of the amplifier. This circuit makes a very good adder with a fixed input impedance, very little crosstalk between channels, and a very low output impedance. Multiple inputs can be summed into this junction.Following is an example of a two-input summing junction amplifier based on the above design:

The output will be equal to Vin1 * Acl + Vin2 * Acl, or (Vin1 + Vin2)*Acl, where Acl is the above calculated closed-loop gain.

Because the output impedance of the above circuit is so low, it can drive loads down to around 10 times it's output impedance with very little attenuation. The above circuit, with an output impedance of 1.78K can drive a 10K load with only a 1dB loss in signal level over the 100K load shown.

Circuit variations:

There are several variations on the above circuit, varying in where the coupling capacitor is placed. Following are a few of them, along with comments on the circuit differences:

The above circuit moves the output coupling capacitor outside the feedback loop. This requires an additional grid bias resistor, Rg, to provide the proper grid bias reference for the tube. This circuit now has two low-frequency control points: Ci/Ri and Co/Rl. The feedback coupling capacitor, Cc, should be calculated in the same manner as Co in the previous example, because it is dependent upon the value of the feedback resistor, Rf.Note that the input coupling capacitor, Ci, can be eliminated if the driving circuit has no DC component. This circuit cannot drive as low load impedance as the previous circuit unless the output coupling capacitor, Co, is made larger. For example, in order to obtain a 50Hz lower -3dB point when driving a 10K load with this circuit, the coupling capacitor must be:

Co = 1/(2*pi*f*Rf) = 1/(2*pi*50*10K) = 0.32uF (choose 0.33uF as the nearest standard value)

This also means that the source impedance varies with frequency, which the first circuit does not exhibit, except at very low frequencies.

The above circuit moves the output coupling capacitor outside the feedback loop, and moves the feedback coupling capacitor to the grid circuit. This requires an additional grid bias resistor, Rg, to provide the proper grid bias reference for the tube. This circuit now also has two low-frequency control points: Ci/Ri and Co/Rl. The feedback coupling capacitor, Cc, should be calculated in the same manner as Co in the previous example, because it is dependent upon the value of the feedback resistor, Rf.Note that the input coupling capacitor, Ci, can be eliminated only if the driving circuit has the same DC level as this circuit. This circuit is used in some older Fender amplifiers, such as the 6G6 Bassman where it is DC-coupled to the plate of the previous stage. In the case of the Bassman, Ri is not present, and is equal to the effective output resistance of the previous DC-coupled stage. Also, Ck is omitted in the Bassman, likely to improve the transient response at the expense of loop gain.

**Conclusions **

Aside from the potential transient response problems, the single-stage negative feedback amplifier is ideal for use in places where a flat, uncolored, controlled frequency response is desired, and where the low output impedance and controlled input impedance is useful.

**Appendix A: The math behind the single-stage inverting feedback amplifier: **

^{1}**Feedback 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 Ra 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. By superposition, if an input signal is applied to Vin, it will produce an output voltage at Vout that is developed from two distinct paths, one through Ri, A, and Ra, and one through Ri and Rf. The situation gets complicated because the output, Vout, is effectively taken from a tap on the feedback divider resistance composed of Rf + Ra.

Forward paths:The forward attenuation of the first path before the amplifier is determined by the voltage divider rule, and is as follows:Ve1 = Vi*(Rf + Ra) / (Ri + Rf + Ra)so the transfer function of this block would be:G1 = Ve1 / Vi = (Rf + Ra) / (Ri + Rf + Ra)(Note: Ve1 is the component of the error voltage at the input to the amplifier A.)The first path has another forward gain element, A, which amplifies the error voltage Ve1, to produce a voltage at the output of amplifier A, so the transfer function of this block would be:

G2 = AThis second path also has a forward gain path formed by the attenuation of Ri, Rf, and Ra. The forward gain of this path is determined by the voltage divider rule as follows:Ve2 = Vi * Ra / (Ri + Rf + Ra)so the transfer function of this block would be:G3 = Ve2 / Vi = Ra / (Ri + Rf + Ra)(Note: Ve2 is the component of the error voltage at the input to the amplifier A.)

Feedback paths:There are two distinct feedback attenuation paths. The first is from the output of amplifier A to the junction of Ri, Rf, and the input of amplifier A. This transfer function can be derived using the voltage divider rule as follows:Ve3 = Vo*(Ri + Rf)/(R i+ Rf + Ra)so the transfer function of this block would be:H1 = Ve3/Vi = (Ri + Rf) / (Ri + Rf + Ra)(Note: Ve3 is the component of the error voltage at the input to the amplifier A.)The second feedback 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:

Ve4 = Vo * Ri/(R i + Rf)so the transfer function of this block would be:H2 = Ve4/Vo = Ri / (Ri + Rf)

The completed block diagram:The completed block diagram showing all forward and feedback paths is shown below.

Where:G1 = (Rf + Ra) / (Ri + Rf + Ra)

G2 = A

G3 = Ra / (Ri + Rf + Ra)

H1 = (Ri + Rf) / (Ri + Rf + Ra)

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*G1) + D* (H1*H2)orD = E * G2

= [(R*G1) + D*(H1*H2] * G2

= R*G1*G2 + D*H1*H2*G2D = R*G1*G2/(1-H1*H2*G2)This leads to the overall transfer function for the output, C, by substituting the equations for D, H1, and G3 into the equation for C:

C = D*H1 + R*G3C = R*G1*G2/(1-H1*H2*G2) * H1 + R*G3or

= R*G1*G2*H1/(1-H1*H2*G2) + R*G3

= R* [G1*G2*H1/(1 - H1*H2*G2) + G3]C = R* [G1*G2*H1/(1 - H1*H2*G2) + G3]Substituting for G1, G2, G3, H1, and H2:C = R* [[(Rf + Ra) / (Ri + Rf + Ra)] * A * [(Ri + Rf) / (Ri + Rf + Ra)] ] / [1 -[ [(Ri + Rf) / (Ri + Rf + Ra)] * [Ri / (Ri + Rf)] * A) + [Ra / (Ri + Rf + Ra)]]simplifying:C = R* (Ra + A*Rf) / (Ri + Rf + Ra - Ra*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 = (Ra + A*Rf) / (Ri + Rf + Ra - Ra*A)so the closed loop gain is:Acl = (Ra + A*Rf) / (Ri + Rf + Ra - Ra*A)Input impedance:The input impedance can be determined from the above derived equations and the top-level block diagram shown below:

If a voltage is applied to the input, the current that will flow in input resistor Ri is equal to the input voltage R minus the error voltage E, divided by the resistance, Ri (as shown in the top level block diagram). This means that the equivalent input impedance is equal to the input voltage R divided by the resulting current as follows:

Zin = R / [(R-E)/Ri]Note that E is the error voltage that results from an input voltage R, and can be calculated as follows:E = (R*G1) + D* (H1*H2)substituting for D:E = (R*G1) + [R*G1*G2/(1-H1*H2*G2)]* (H1*H2)substituting for G1, G2, H1, and H2:E = (R*[(Rf + Ra) / (Ri + Rf + Ra)]) + [R*[(Rf + Ra) / (Ri + Rf + Ra)]*A/(1-[(Ri + Rf) / (Ri + Rf + Ra)]*[Ri / (Ri + Rf)]*A)]* ([(Ri + Rf) / (Ri + Rf + Ra)]*[Ri / (Ri + Rf)])simplifying:E = R*(Rf + Ra)/[(Ri + Rf + Ra) - Ri*A]substituting back into the equation for Zin:Zin = R/[(R - R*(Rf + Ra)/[(Ri + Rf + Ra) - Ri*A)]/Ri]this simplifies toZin = (Ri * A - Ri - Rf - Ra)/(A-1)Output impedance: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 - E2) / RfWhere E2 is the resulting voltage at the input of the amplifier A when the 1V test signal is applied to the output (it is not the same value of E for the above derived gain equations), and is calculated as follows:E2 = 1V * Ri/(Ri + Rf)The second current, I2, would be equal to:I2 = (1V - E2*A)/RaThe total current, I, would be the sum of I1 and I2 as below:I = I1 + I2orI = (1V - E2) / Rf + (1V - E2*A)/RaThe resulting output impedance would be:simplifying:

Zout = 1V/I

= 1V / [(1V - E2) / Rf + (1V - E2*A)/Ra]

= 1V / [(1V - 1V * Ri/(Ri + Rf))/Rf + (1V - (1V * Ri/(Ri + Rf))*A/Ra]Zout = (Ri + Rf) * [Ra / (Ri+Rf+Ra-Ri*A)]Summary: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:

- Acl = (Ra + A*Rf) / (Ri + Rf + Ra - Ri*A)

- Zin = (Ri * A - Ri - Rf - Ra)/(A-1)

- Zout = (Ri + Rf) * [Ra / (Ri+Rf+Ra-Ri*A)]
Note that for negative feedback, A must be a negative quantity. If A is positive, the feedback will be positive.

Copyright © 2000-2015, Randall Aiken. May not be reproduced in any form without written approval from Aiken Amplification.

*Revised 03/18/15 *