How Much Does the Miller Effect Cut a Common-Emitter Amplifier's Bandwidth?

Studyloaf Team•October 10, 2026•Last verified October 10, 2026•8 min read

In the example worked below, by a factor of about 20: the 2 pF base-collector capacitance Cμ is multiplied by (1 + gmR'L) = 97.7 and appears at the input as 195.4 pF. That drags the upper cutoff frequency from 22.1 MHz, where it would be if Cμ were zero, down to about 1.07 MHz by the Miller estimate (1.04 MHz exactly).

What you'll learn

  • Why a small capacitor between an inverting amplifier's input and output looks large from the input
  • How to replace Cμ with an input and an output capacitance, step by step
  • A worked common-emitter example with gm = 38.7 mA/V, R'L = 2.5 kΩ and Cμ = 2 pF
  • How close the Miller estimate gets to the exact cutoff frequency
  • Why trading gain for bandwidth works so directly in this stage

Why does a 2 pF capacitor act like 195 pF?

Take an inverting stage with voltage gain A = vo ÷ vπ from the base node to the collector, where A is negative. A capacitor C connected from input to output carries a current sC(vπ − vo) = sCvπ(1 − A). From the input node this current looks exactly like the current of a capacitor C(1 − A) connected to ground. With A = −gmR'L the factor 1 − A becomes 1 + gmR'L, so the larger the gain, the larger the capacitance the source sees. In a common-emitter stage that capacitor is Cμ, the base-collector junction capacitance, and the input capacitance it creates is called the Miller capacitance, CM = Cμ(1 + gmR'L). It adds to Cπ, the base-emitter capacitance, which is already across the input.

What happens on the output side?

The same current seen from the collector is sCvo(1 − 1/A), so the output side gets a capacitance C(1 − 1/A). For a large inverting gain this is only slightly more than C itself: 2.02 pF in the example. It sets a much higher-frequency pole than the input side, so the input pole dominates the bandwidth.

Worked example: a common-emitter stage at 1 mA

High-frequency hybrid-pi model of the common-emitter stageThe source vs drives Rs into the base node. From the base node, r pi and C pi go to ground, and C mu bridges from the base node to the collector node. At the collector a dependent current source gm times v pi and the load R prime L go to ground. Values: Rs 1 kilohm, r pi 2.59 kilohms, C pi 10 picofarads, C mu 2 picofarads, gm 38.7 milliamps per volt, R prime L 2.5 kilohms.vsRsrπCπCμgm·vπR'LB'voRs = 1 kΩ, rπ = 2.59 kΩ, Cπ = 10 pFCμ = 2 pF, gm = 38.7 mA/V, R'L = 2.5 kΩ
The hybrid-pi model at high frequency. Assumptions: the bias resistors are much larger than rπ and are ignored, coupling and bypass capacitors are short circuits at these frequencies, and the transistor's output resistance is neglected.
  • Bias: IC = 1 mA, β = 100, VT = 25.85 mV (300 K). So gm = IC ÷ VT = 38.68 mA/V and rπ = β ÷ gm = 2.585 kΩ.
  • Load: RC = 5 kΩ in parallel with RL = 5 kΩ gives R'L = 2.5 kΩ.
  • Gain from base to collector: A = −gmR'L = −38.68 mA/V × 2.5 kΩ = −96.7.
  • Midband gain from the source, including the divider at the input: −96.7 × 2.585 ÷ (1 + 2.585) = −69.7, which is 36.9 dB.
  • Miller capacitance: CM = Cμ(1 + gmR'L) = 2 pF × 97.7 = 195.4 pF. Total input capacitance: Cin = Cπ + CM = 10 + 195.4 = 205.4 pF.
  • Resistance seen by Cin: Rs in parallel with rπ = 1 kΩ ∥ 2.585 kΩ = 721 Ω.
  • Upper cutoff: fH ≈ 1 ÷ (2π × 721 Ω × 205.4 pF) = 1.07 MHz. With Cμ removed it would be 1 ÷ (2π × 721 Ω × 10 pF) = 22.1 MHz, about 20.5 times higher.
The input side after the Miller approximationC mu is replaced by a Miller capacitance CM equal to C mu times 1 plus gm R prime L, which is 195.4 picofarads, in parallel with C pi of 10 picofarads at the base node. The total 205.4 picofarads sees Rs in parallel with r pi, 721 ohms, so the upper cutoff is about 1.07 megahertz. On the output side a capacitance of 2.02 picofarads appears across R prime L.vsRsrπCπCMCM = Cμ(1 + gm·R'L) = 195.4 pFCin = Cπ + CM = 205.4 pFOutput side: 2.02 pF across R'LfH ≈ 1 ÷ (2π × 721 Ω × 205.4 pF) = 1.07 MHz
After the Miller approximation the input side is a single RC: 721 Ω (Rs ∥ rπ) driving Cπ + CM = 205.4 pF.

How accurate is the Miller estimate?

Computed gain against frequency, with and without C muExact magnitude response of the hybrid-pi model from 10 kilohertz to 100 megahertz. The midband gain is 36.9 decibels. With C mu of 2 picofarads the gain is 3 decibels down at 1.04 megahertz; with C mu set to zero it would stay up until 22.1 megahertz.10k100k1M10M100M010203040frequency (Hz)gain (dB)Cμ = 2 pFCμ = 01.04 MHz22.1 MHz
The exact magnitude response of the hybrid-pi model, computed by nodal analysis at 161 frequencies. The −3 dB point is 1.04 MHz with Cμ = 2 pF and 22.1 MHz with Cμ = 0.

Solving the full two-node circuit at each frequency puts the true −3 dB point at 1.04 MHz, so the Miller estimate of 1.07 MHz is about 3% high. The error comes from using the midband gain to size CM: near the cutoff the collector voltage has started to fall and lag, so the real multiplication is a little different, and the output-side capacitance and the feed-forward path through Cμ also shift the answer slightly. A second shortcut, the open-circuit time-constant method, sums each capacitor's time constant: 721 Ω × 10 pF = 7.2 ns for Cπ, and (721 × 97.7 + 2500) Ω × 2 pF = 145.9 ns for Cμ. That gives 1 ÷ (2π × 153.1 ns) = 1.04 MHz, within 0.2% of the exact result here. For a GATE answer either is acceptable unless the question names a method.

Why does lowering the gain buy bandwidth?

Because CM grows with the gain itself, the stage trades the two almost one for one. Halve R'L to 1.25 kΩ and the gain from base to collector halves to −48.4, while CM falls to 2 pF × 49.4 = 98.7 pF and the Miller estimate of fH roughly doubles to 2.03 MHz. The same idea explains the fixes for the Miller effect: a cascode keeps the common-emitter transistor's collector at a nearly fixed voltage so its own gain is close to −1, and a common-base stage has no capacitor bridging an inverting input and output at all. It is also the effect that dominant-pole (Miller) compensation in op-amps uses on purpose.

Common mistakes

  • Multiplying by |A| instead of (1 + |A|). The factor is 1 − A with A negative, so 97.7, not 96.7.
  • Using the gain from the source (−69.7) to size CM. The voltage across Cμ depends on the gain from base to collector (−96.7).
  • Forgetting Cπ, or forgetting that Cin sees Rs in parallel with rπ rather than Rs alone.
  • Applying the same multiplication to a non-inverting stage. With a gain near +1 the factor 1 − A is near zero and the capacitance almost vanishes.
  • Treating the Miller estimate as exact. It slightly overestimates fH here; a stated method in the question decides which answer is expected.

Check your understanding

  • Q1. In the same stage, Cμ is 1 pF instead of 2 pF. What is CM, and roughly what is fH? Answer: CM = 1 pF × 97.7 = 97.7 pF, so fH ≈ 1 ÷ (2π × 721 Ω × 107.7 pF) = 2.05 MHz.
  • Q2. R'L is reduced to 1.25 kΩ. What are the new base-to-collector gain and Miller estimate of fH? Answer: A = −38.68 × 1.25 = −48.4, CM = 98.7 pF and fH ≈ 1 ÷ (2π × 721 × 108.7 pF) = 2.03 MHz.
  • Q3. What capacitance does Cμ add to the output side in the original example? Answer: Cμ(1 − 1/A) = 2 pF × (1 + 1/96.7) = 2.02 pF.

Related

Official source

Last verified

The GATE 2027 EC syllabus lists the frequency response of BJT and MOSFET amplifiers, and dominant-pole (Miller) compensation, under Section 4, Analog Circuits (checked against the official PDF from IIT Madras on 10 October 2026). Every number here, including the exact frequency response, was recomputed in code.

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