Frequency response of amplifiers

Why it matters

An instrument amplifier must pass every frequency in the measured signal with the same gain: a vibration channel needs a flat response to tens of kHz, a thermocouple channel needs a response down to DC. Coupling capacitors cut off the low end; transistor and stray capacitances cut off the high end. Knowing where those corners fall, and which capacitor sets each one, lets you design the bandwidth instead of discovering it on the bench.

Key ideas

Three bands. A capacitively coupled amplifier has a low-frequency region (gain falls because coupling and bypass capacitors no longer act as shorts), a mid-band region (all large capacitors are shorts, all small device capacitances are open; gain A_M is flat) and a high-frequency region (device and stray capacitances start to short the signal). A DC-coupled amplifier (such as an op-amp) has no low-frequency roll-off.

Corner (−3 dB) frequencies. The lower and upper cut-off frequencies f_L and f_H are where the gain falls to A_M/√2 (70.7 %), i.e. 3 dB below mid-band and the output power halves. Bandwidth BW = f_H − f_L ≈ f_H when f_H ≫ f_L.

Single-pole response. Each RC pair contributes one pole. A single pole gives a 20 dB/decade (6 dB/octave) roll-off and 45° of extra phase at the corner, reaching 90° far beyond it. On a Bode plot the magnitude is approximated by straight lines that meet at the corner.

Short-circuit and open-circuit time constants. For f_L, take each coupling or bypass capacitor in turn with the others shorted and find the resistance it sees; f_L ≈ Σ 1/(2π·R_i·C_i). For f_H, take each device capacitance with the others open; f_H ≈ 1/(2π·Σ R_i·C_i). If one term clearly dominates, that capacitor sets the corner (the "dominant pole").

The emitter bypass capacitor sees a very small resistance (R_E in parallel with about r_e plus the source resistance divided by β + 1), so it usually sets f_L unless made very large.

Miller effect. A capacitance C_F bridging an inverting gain −K (such as C_μ from base to collector of a CE stage) appears at the input as C_F·(1 + K). With K ≈ 100 a 2 pF C_μ looks like 200 pF, which usually dominates f_H. Cascode (CE + CB) and follower stages avoid this because the bridged gain is small.

Transition frequency. f_T = g_m/(2π·(C_π + C_μ)) is the frequency at which the BJT's short-circuit current gain falls to 1. It is a figure of merit for the device. The β cut-off frequency f_β = f_T/β.

Gain-bandwidth product. For a single-pole amplifier A_M·f_H is constant, so trading gain for bandwidth is possible with feedback (an op-amp with GBW = 1 MHz configured for a gain of 10 has f_H ≈ 100 kHz).

Cascading. Connecting n identical stages multiplies the gains but shrinks the bandwidth: f_H(n) = f_H·√(2^(1/n) − 1).

Formulas

f_c = 1 / (2π·R·C) — corner of a single RC pole (Hz); R the resistance seen by the capacitor (Ω), C in F.

A(f) = A_M / √(1 + (f/f_H)²) — single-pole high-frequency response; A(f) = A_M / √(1 + (f_L/f)²) — single-pole low-frequency response.

Gain (dB) = 20·log₁₀(A) — voltage gain in decibels.

f_L ≈ Σ 1/(2π·R_i·C_i) — short-circuit time-constant estimate; f_H ≈ 1 / (2π·Σ R_i·C_i) — open-circuit time-constant estimate.

C_M = C_F·(1 + K) — Miller input capacitance; K = magnitude of the inverting gain across C_F.

f_T = g_m / (2π·(C_π + C_μ)), f_β = f_T / β.

GBW = A_M·f_H — constant for a single-pole amplifier.

f_H(n) = f_H·√(2^(1/n) − 1), f_L(n) = f_L / √(2^(1/n) − 1) — n identical non-interacting stages.

Worked examples

Example 1 (standard: low-frequency corners). A CE stage has R_sig = 1 kΩ, R_in = 2.22 kΩ, R_C = 5 kΩ and R_L = 5 kΩ. The input coupling capacitor is 10 µF and the output coupling capacitor is 1 µF (emitter bypass very large). Find the two low-frequency corners and f_L.

1. Input capacitor sees R_sig + R_in = 3.22 kΩ: f₁ = 1/(2π × 3220 × 10×10⁻⁶) = 4.94 Hz.
2. Output capacitor sees R_C + R_L = 10 kΩ: f₂ = 1/(2π × 10 000 × 1×10⁻⁶) = 15.9 Hz.
3. f₂ is more than three times f₁, so it dominates: f_L ≈ 15.9 Hz (the sum rule gives about 21 Hz as an upper estimate).

Answer: f₁ ≈ 4.9 Hz, f₂ ≈ 15.9 Hz, f_L ≈ 16 Hz (dominated by the output capacitor).

Example 2 (GATE level: Miller effect). A CE stage has g_m = 40 mS, r_π = 2.5 kΩ, C_π = 10 pF, C_μ = 2 pF and a total collector load R'_L = 2.5 kΩ. R_sig = 1 kΩ and R_B = 20 kΩ. Estimate f_H.

1. Mid-band gain from base to collector: K = g_m·R'_L = 0.040 × 2500 = 100.
2. C_M = C_μ(1 + K) = 2 pF × 101 = 202 pF.
3. Total input capacitance: C_in = C_π + C_M = 212 pF.
4. Resistance seen by C_in: R_sig ∥ R_B ∥ r_π = 1 kΩ ∥ 20 kΩ ∥ 2.5 kΩ = 690 Ω.
5. f_H = 1/(2π·R·C_in) = 1/(2π × 690 × 212×10⁻¹²) = 1.09 MHz.

Answer: f_H ≈ 1.09 MHz. Without the Miller multiplication (C_in = 12 pF) the corner would be near 19 MHz: the 2 pF C_μ costs most of the bandwidth.

Common mistakes

• Using the wrong resistance in f = 1/(2πRC); it must be the Thevenin resistance seen by that capacitor with the sources zeroed.
• Forgetting the (1 + K) Miller factor, or applying it to a non-inverting follower.
• Adding cut-off frequencies of cascaded stages instead of applying the √(2^(1/n) − 1) shrinkage.
• Confusing −3 dB with half the gain; −3 dB is 0.707 of the voltage gain (half the power), −6 dB is half the voltage gain.
• Using 20 dB/decade for a two-pole region (it is 40 dB/decade).

For GATE IN

Expect questions on −3 dB frequencies from given R and C, Bode magnitude at a given frequency, gain at a decade beyond the corner, Miller capacitance, f_T and the gain-bandwidth trade-off, and the bandwidth of cascaded identical stages. Practise identifying which capacitor dominates and reading slopes off a Bode plot.

Quick check

1. A single-pole amplifier has A_M = 60 dB and f_H = 10 kHz. What is the gain at 100 kHz (in dB)?
2. A 1 pF capacitor bridges an inverting gain of −49. What is the Miller input capacitance?
3. Two identical stages each have f_H = 1 MHz. What is the overall f_H?
4. An op-amp has GBW = 3 MHz. What is f_H at a closed-loop gain of 30?

Answers: 1. About 40 dB. 2. 50 pF. 3. About 0.64 MHz. 4. 100 kHz.