Frequency Response of Amplifiers

Frequency response defines an amplifier's usable bandwidth and how gain rolls off — vital for choosing devices and understanding the gain-bandwidth trade-off. GATE EC tests corner frequencies, Miller effect and Bode behaviour (1–2 marks).

Key results

• Low-frequency rolloff: coupling/bypass capacitors set lower cutoff f_L (each C-R gives a pole).
• High-frequency rolloff: device/junction capacitances (C_π, C_μ, C_gs, C_gd) set upper cutoff f_H.
• Bandwidth BW = f_H − f_L ≈ f_H (since f_H >> f_L).
• Miller effect: a feedback capacitance C_f between input and output appears at the input as C_f(1+|A_v|) — drastically lowers f_H in inverting stages.
• Gain-Bandwidth Product (GBW) ≈ constant: raising gain lowers bandwidth proportionally.

Region	Limiting element	Slope
Low freq	coupling/bypass C	+20 dB/decade rise
Midband	flat gain	0
High freq	junction/parasitic C	−20 dB/decade fall

Exam Tricks & Tips

• 🎯 Miller capacitance = C_gd(1+|A_v|) (or C_μ) — the inverting gain multiplies the feedback cap, killing bandwidth.
• 🎯 GBW ≈ constant: for a single-pole amplifier, gain × bandwidth is fixed — trade one for the other.
• 🎯 Each RC gives a pole at f = 1/(2πRC); the largest low-frequency RC sets f_L, the smallest high-frequency RC sets f_H.
• 🎯 Cascode and common-gate stages avoid the Miller effect → wide bandwidth.
• 🎯 −3 dB points (half-power) define the band edges; gain rolls off ±20 dB/decade per pole.
• ❌ Common mistake: ignoring the (1+|A_v|) Miller multiplication when estimating f_H of an inverting stage.

Expected exam pattern

1–2 marks: compute f_L/f_H, apply Miller effect to find input capacitance/f_H, or use GBW to trade gain for bandwidth.

Quick recap

f_L from coupling/bypass caps, f_H from device caps; BW≈f_H. Miller: C_in=C_f(1+|A_v|) shrinks f_H in inverting stages. GBW≈constant. Cascode dodges Miller. Poles at 1/(2πRC), −20 dB/decade each.

Cover the answer, recall, then check. 11 cards on amplifier frequency response for GATE EC.

Q1. What sets the low-frequency cutoff of an amplifier?
A1. Coupling and bypass capacitors — each forms an RC high-pass pole.

Q2. What sets the high-frequency cutoff?
A2. Internal/parasitic capacitances (C_π, C_μ for BJT; C_gs, C_gd for MOSFET).

Q3. State the Miller effect.
A3. A capacitance C_f bridging input and output of an inverting stage appears at the input as C_f(1 + |A_v|).

Q4. Why does Miller effect reduce bandwidth?
A4. The multiplied input capacitance forms a larger input RC, lowering the high-frequency pole f_H.

Q5. State the gain-bandwidth product property.
A5. For a single-pole amplifier, gain × bandwidth ≈ constant; increasing gain proportionally reduces bandwidth.

Q6. How is the −3 dB (corner) frequency of an RC found?
A6. f = 1/(2πRC).

Q7. Approximate bandwidth in terms of f_L and f_H.
A7. BW = f_H − f_L ≈ f_H (since f_H >> f_L).

Q8. Which configurations avoid the Miller effect?
A8. Common-gate/common-base and cascode stages (no inverting gain across the feedback cap).

Q9. What is the rolloff slope per pole?
A9. −20 dB/decade (−6 dB/octave).

Q10. In midband, what limits the gain?
A10. Nothing capacitive — coupling caps are shorts and device caps are opens, so gain is flat (maximum).

Q11. If a single-pole amp has GBW = 10 MHz, what bandwidth at a gain of 100?
A11. BW = 10 MHz / 100 = 100 kHz.

No amplifier has infinite bandwidth — coupling capacitors kill the low frequencies and device/parasitic capacitances kill the high frequencies. Frequency response describes this passband, and the Miller effect explains why high-gain stages are surprisingly slow. GATE tests corner frequencies, the Miller multiplication, and the gain-bandwidth product.

Core concept: Coupling/bypass capacitors set the low-frequency roll-off; device and load capacitances (amplified by the Miller effect) set the high-frequency roll-off; the midband gain sits between the two corner frequencies.

The theory

Beginner — the three bands

• Low frequency: coupling and bypass capacitors have high impedance, attenuating the signal. Each sets a pole fL = 1/(2πRC) using the resistance it sees.
• Midband: all coupling caps are shorts, all device caps are opens → flat, maximum gain.
• High frequency: device capacitances (Cπ, Cµ for BJT; Cgs, Cgd for MOSFET) shunt the signal, giving high-frequency poles fH = 1/(2πRC).
  Bandwidth = fH − fL ≈ fH (since fH ≫ fL).

Intermediate — the Miller effect

A capacitance Cgd (or Cµ) bridging input and output of an inverting amplifier of gain −Av appears at the input as:
CMiller = Cgd·(1 + |Av|).
So a small feedback capacitance is multiplied by the gain — the dominant high-frequency limiter in common-source/common-emitter stages. This is why high-gain stages have low bandwidth, and why cascode/common-gate stages (no Miller multiplication) are faster.

Advanced — gain-bandwidth product and fT

• Gain-bandwidth product (GBW): for a single-pole amplifier, gain × bandwidth = constant. Trading gain for bandwidth is the fundamental compromise.
• Transition frequency fT: where the transistor's short-circuit current gain falls to 1; fT = gm/(2π(Cπ + Cµ)) for a BJT, gm/(2π(Cgs + Cgd)) for a MOSFET — the intrinsic speed limit of the device.
• Dominant-pole approximation: the lowest fH pole dominates the bandwidth; find the node with the largest R·C product.

Worked example

A common-source amp has |Av| = 20, Cgd = 1 pF, Cgs = 5 pF, and drives a source resistance Rsig = 10 kΩ. Estimate the input-side high-frequency pole.
Miller capacitance: CM = Cgd(1 + |Av|) = 1 pF × 21 = 21 pF.
Total input capacitance ≈ Cgs + CM = 5 + 21 = 26 pF.
fH = 1/(2π·Rsig·Cin) = 1/(2π × 10⁴ × 26×10⁻¹²) = 1/(1.63×10⁻⁶) ≈ 612 kHz.
Without Miller, Cin ≈ 6 pF would give ~2.65 MHz — the Miller effect cut the bandwidth ~4×.

GATE relevance

1–2 mark NAT/MCQ: low/high corner frequency from R and C, Miller-multiplied input capacitance, gain-bandwidth trade-off, dominant-pole bandwidth, and fT. Cascode's bandwidth advantage is a favourite conceptual question.

Exam tricks and shortcuts

• Each capacitor gives a pole f = 1/(2πRC); use the resistance that capacitor "sees."
• Miller: bridging cap seen at input = Cgd(1 + |Av|) — the high-gain bandwidth killer.
• GBW ≈ constant for single-pole; cascode/CG avoids Miller → wider band.
• Mnemonic: "High gain buys low bandwidth" (Miller + GBW).

Key formulas

• Lower/upper cutoff: f_L, f_H at −3 dB; bandwidth BW = f_H − f_L.
• Gain–bandwidth product: GBW = A_v × BW = constant.
• Pole from RC: f = 1/(2πRC).
• Miller effect: C_in = C(1 + |A_v|) (multiplies feedback capacitance).
• Roll-off: −20 dB/decade per pole.

High-frequency gain falls due to device capacitances (Miller effect); the gain–bandwidth product is fixed.

Worked Example

Problem: An amplifier has a midband voltage gain of A₀ = 100 and a single dominant (high-frequency) pole at f_p = 20 kHz. Find the gain-bandwidth product, the −3 dB bandwidth, and the gain magnitude at 200 kHz.

Solution:
For a single-pole amplifier the magnitude response is
|A(f)| = A₀ / √(1 + (f/f_p)²).

−3 dB bandwidth: the gain falls to A₀/√2 exactly at the pole, so
f_(−3dB) = f_p = 20 kHz.

Gain-bandwidth product (constant for a single-pole system):
GBW = A₀ × f_p = 100 × 20 kHz = 2 MHz.

Gain at 200 kHz. Since 200 kHz ≫ f_p, use the exact formula with f/f_p = 200/20 = 10:
|A| = 100 / √(1 + 10²) = 100 / √101 = 100/10.05 ≈ 9.95 ≈ 10.
(Equivalently, well above the pole, |A| ≈ GBW/f = 2 MHz / 200 kHz = 10.)

In decibels this is 20·log₁₀(10) = 20 dB, i.e. the gain has dropped 20 dB per decade from midband.

Answer: GBW = 2 MHz, −3 dB bandwidth = 20 kHz, and |A(200 kHz)| ≈ 10 (20 dB).