Feedback amplifiers
An amplifier that distorts 7.7 % drops to 1.2 % with a loop gain of 10. Feedback trades surplus gain for linearity.
01The loop gain
A feedback amplifier takes a sample of its output, multiplies it by a factor \( \beta \) and subtracts it from the input. If the amplifier without feedback has gain \( A \), the closed-loop gain is:
Feedback trades gain for other qualities. The most direct trade is desensitivity: a relative variation of \( A \) reaches the feedback gain divided by \( 1 + T \):
The multistage amplifiers of Applied Electronics I have large but imprecise gains; feedback turns them into small, exact gains.
02The four topologies
The sample of the output can be a voltage (shunt connection across the output) or a current (series connection), and the feedback signal can be subtracted as a voltage (in series with the input) or as a current (in parallel). Each combination stabilizes a different transfer function and modifies the impedances by the factor \( 1 + T \):
| Topology (input–output) | Stabilizes | Input | Output |
|---|---|---|---|
| Series–shunt | Voltage gain | \( R_i(1+T) \) | \( R_o/(1+T) \) |
| Shunt–series | Current gain | \( R_i/(1+T) \) | \( R_o(1+T) \) |
| Series–series | Transconductance | \( R_i(1+T) \) | \( R_o(1+T) \) |
| Shunt–shunt | Transresistance | \( R_i/(1+T) \) | \( R_o/(1+T) \) |
The rule is simple: a series connection multiplies the impedance by \( 1 + T \) and a shunt connection divides it. The noninverting amplifier with an op amp is series–shunt: very high input impedance and almost zero output impedance. The common-emitter stage with an unbypassed emitter resistor is series–series, and with a collector-to-base resistor, shunt–shunt.
In real circuits the feedback network loads the amplifier. The rigorous analysis first separates the amplifier loaded by the network, calculates \( A \) and \( \beta \) of that version, and only then applies the formulas. Another, more direct way is to calculate \( T \) by breaking the loop at a point and to apply Blackman's theorem for the impedances.
03Distortion, noise and bandwidth
Feedback continuously compares the output with the input: if the amplifier distorts, the error appears at the amplifier input and is corrected. The distortion components generated inside the loop end up divided, approximately, by \( 1 + T \), when compared at equal output amplitude. The same happens with bandwidth: with a dominant pole at \( f_H \),
An amplifier of gain \( A \) with an asymmetric hyperbolic-tangent transfer function, which saturates at +7 V and −13 V. With and without feedback, the input is adjusted to obtain the same output amplitude, and the waveforms (above) and the harmonics (below, in dB relative to the fundamental) are compared. The feedback output is calculated by solving the loop at each instant.
There is something that feedback does not improve: the noise that enters along with the signal, or that the first stage generates. That noise is amplified just like the signal. The signal-to-noise ratio is decided at the input stage, before the loop.
04In the lab
Build a two-stage transistor amplifier, and measure its gain without feedback and with a series-voltage feedback network. Swap the transistor for another with a different β and compare the variations.
Measure the input and output impedances with and without feedback and verify the factor \( 1 + T \).
With a spectrum analyzer or the FFT of an oscilloscope, measure the harmonic distortion with and without feedback, adjusting the input for the same output.
05Common mistakes
- Comparing distortion at equal input. With feedback the output is much smaller; the valid comparison is at equal output.
- Forgetting the loading of the feedback network. Without including it, \( A \) and \( \beta \) come out wrong and so do the impedances.
- Expecting feedback to lower the input noise. It only reduces what is generated inside the loop, after the input.
- Confusing negative feedback with positive. If the sign is reversed, \( 1 + T \) can vanish and the circuit oscillates.
06Self-assessment
An amplifier with gain 2000 is fed back with β = 0.05. What gain does it have?
\( 2000/(1 + 100) \approx 19.8 \).
In the same circuit, if the amplifier gain drops to 1000, how much does the feedback gain change?
It becomes \( 1000/51 = 19.6 \): a 1 % change for a 50 % one.
An amplifier with a 1 kΩ output resistance is fed back in series–shunt with T = 99. What output resistance does it have?
10 Ω.
An amplifier with gain 10⁴ has a pole at 100 Hz. What bandwidth does it have with gain 100?
About 10 kHz: the gain–bandwidth product, 1 MHz, is conserved.
07Further reading
- Adel S. Sedra and Kenneth C. Smith. Microelectronic Circuits. 8th ed., Oxford University Press, 2020. The four topologies with the method of separating the amplifier loaded by the network.
- Paul R. Gray, Paul J. Hurst, Stephen H. Lewis and Robert G. Meyer. Analysis and Design of Analog Integrated Circuits. 5th ed., Wiley, 2009. Feedback with Blackman's theorem and loop-gain analysis.
- Jacob Millman and Arvin Grabel. Microelectronics. 6th ed. (Spanish edition, “Microelectrónica”), Hispano Europea, 1993. A classic treatment of feedback amplifiers, available in Spanish.