Feedback amplifiers and their properties

Why it matters

Transistor gain varies from device to device and drifts with temperature, yet an instrument amplifier must have a gain you can trust to 0.1 %. Negative feedback is the trick that makes this possible: it trades surplus raw gain for a closed-loop gain set by precision resistors, and at the same time widens bandwidth, reduces distortion and lets you choose whether the amplifier looks like a high- or low-impedance source and load. Every op-amp circuit in the following topics is a feedback amplifier.

Key ideas

The basic loop. A basic amplifier of gain A drives the output x_o. A feedback network samples the output and returns x_f = β·x_o, which is subtracted from the input x_s at a mixing point, so the amplifier sees the error x_i = x_s − x_f. Solving gives the closed-loop gain A_f = A/(1 + Aβ). The quantity Aβ is the loop gain and (1 + Aβ) is the desensitivity factor D.

When Aβ ≫ 1 the closed-loop gain becomes A_f ≈ 1/β: it depends only on the feedback network (usually resistors), not on the transistors.

Benefits of negative feedback (each by the factor D):

• Gain desensitivity: dA_f/A_f = (1/D)·dA/A.
• Bandwidth: for a single-pole amplifier the upper cut-off rises to f_H·D and the lower cut-off falls to f_L/D, so gain × bandwidth is unchanged.
• Distortion: nonlinear distortion generated inside the loop falls by D (for the same output level).
• Noise or interference injected inside the loop (after the first stage) is reduced relative to the signal; noise entering at the input with the signal is not.
• Impedances change by D in a direction set by the topology.

Four topologies. Name them by what is sampled at the output and how the feedback is mixed at the input.

• Voltage-series (series-shunt): samples output voltage, mixes in series with the input voltage. Voltage amplifier. R_in rises by D, R_out falls by D. Example: non-inverting op-amp amplifier, emitter follower.
• Current-series (series-series): samples output current, mixes a voltage in series. Transconductance amplifier. R_in rises, R_out rises. Example: CE stage with unbypassed R_E.
• Voltage-shunt (shunt-shunt): samples output voltage, mixes a current at the input node. Transresistance amplifier. R_in falls, R_out falls. Example: inverting op-amp amplifier, collector-to-base bias resistor.
• Current-shunt (shunt-series): samples output current, mixes a current. Current amplifier. R_in falls, R_out rises. Rule of thumb: series mixing raises input resistance, shunt mixing lowers it; voltage sampling lowers output resistance, current sampling raises it.

Positive feedback (1 + Aβ < 1) increases gain; when Aβ = −1, i.e. loop gain of magnitude 1 with the right phase, the circuit oscillates (Barkhausen criterion). Comparators with hysteresis and oscillators use positive feedback deliberately.

Stability. In a real amplifier A has several poles, so its phase shifts with frequency. If the loop phase reaches −180° while |Aβ| is still above 1, negative feedback turns positive and the amplifier oscillates. Phase margin (how far the phase is from −180° where |Aβ| = 1) of about 45°–60° is a common design target; more feedback (larger β) makes this harder, which is why op-amps are frequency-compensated.

Formulas

A_f = A / (1 + A·β) — closed-loop gain; A open-loop gain, β feedback factor (both dimensionless for a voltage amplifier).

A_f ≈ 1/β when A·β ≫ 1.

dA_f/A_f = (1/(1 + A·β))·(dA/A) — gain sensitivity.

f_Hf = f_H·(1 + A·β), f_Lf = f_L / (1 + A·β) — single-pole amplifier.

R_if = R_i·(1 + A·β) (series mixing), R_if = R_i / (1 + A·β) (shunt mixing).

R_of = R_o / (1 + A·β) (voltage sampling), R_of = R_o·(1 + A·β) (current sampling). Here A is the open-loop gain measured with the appropriate load (and the output short-circuit or open-circuit value where textbooks specify).

D_f = D / (1 + A·β) — distortion with feedback.

Worked examples

Example 1 (standard: desensitivity). An amplifier has A = 1000 and β = 0.01. Find A_f. If A falls by 20 % because of ageing, by how much does A_f fall?

1. A_f = A/(1 + Aβ) = 1000/(1 + 10) = 90.9.
2. New A = 800: A_f = 800/(1 + 8) = 88.9.
3. Change = (90.9 − 88.9)/90.9 = 2.2 %. (The differential formula gives 20 %/11 = 1.8 %, accurate only for small changes.)

Answer: A_f = 90.9; a 20 % fall in A gives only about a 2.2 % fall in A_f.

Example 2 (GATE level: voltage-series amplifier). A voltage amplifier has A = 10⁴, R_i = 10 kΩ, R_o = 100 Ω and f_H = 1 kHz (single pole). Voltage-series feedback with β = 0.02 is applied. Find A_f, R_if, R_of and f_Hf.

1. Desensitivity: D = 1 + Aβ = 1 + 10⁴ × 0.02 = 201.
2. A_f = A/D = 10⁴/201 = 49.75 (close to 1/β = 50).
3. Series mixing: R_if = R_i·D = 10 kΩ × 201 = 2.01 MΩ.
4. Voltage sampling: R_of = R_o/D = 100/201 = 0.498 Ω.
5. f_Hf = f_H·D = 1 kHz × 201 = 201 kHz.

Answer: A_f ≈ 49.8, R_if ≈ 2.01 MΩ, R_of ≈ 0.50 Ω, f_Hf ≈ 201 kHz. Gain × bandwidth = 49.75 × 201 kHz ≈ 10 MHz, the same as 10⁴ × 1 kHz.

Common mistakes

• Using A_f = A/(1 − Aβ) with negative-feedback sign conventions mixed up; decide the sign once.
• Applying R_in × D to a shunt-mixing circuit (it should be divided).
• Assuming feedback reduces noise that enters with the input signal; it does not.
• Forgetting that the bandwidth improvement holds exactly only for a single-pole amplifier.
• Using the differential sensitivity formula for large changes in A.
• Thinking more feedback always makes an amplifier more stable; with several poles it can cause oscillation.

For GATE IN

Expect numericals on A_f, β from given open- and closed-loop gains, the percentage change in A_f for a change in A, bandwidth extension, and input or output resistance with feedback. Also expect topology identification from a circuit (which quantity is sampled, how it is mixed) and loop-gain or phase-margin reasoning. Practise the four topology rules until they are automatic.

Quick check

1. A = 200 and A_f = 20. What is β?
2. A = 10⁵ and β = 0.1. What is A_f to three significant figures?
3. Which topology lowers both input and output resistance?
4. A single-pole amplifier has GBW = 2 MHz. With feedback giving A_f = 20, what is f_H?

Answers: 1. 0.045. 2. 10.0. 3. Voltage-shunt (shunt-shunt). 4. 100 kHz.