AoNArt of Nanoelectronics
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03 · Advanced

Miller effect

Derive the Miller transformation from the current through a two-node impedance rather than memorizing a multiplied-capacitance rule. Show that an impedance Z between nodes v1 and v2=Av·v1 can be replaced, for that transfer relation, by Zin=Z/(1−Av) and Zout=Z/(1−1/Av). Specialize to Cgd/Cμ, explain why inverting gain creates a large apparent input capacitance, why a follower can bootstrap the same capacitor, and where the approximation fails when Av is frequency dependent.

220–280 minAdvanced
In this lesson
  • Derive the Miller input equivalent directly from i1=(v1−v2)/Z.
  • Derive the Miller output equivalent directly from i2=(v2−v1)/Z.
  • State the exact conditions under which the two-node element replacement preserves the terminal currents for a specified voltage ratio Av.
  • Specialize the theorem to a capacitor and obtain Cin,M=C(1−Av) and Cout,M=C(1−1/Av).
  • Explain why a large negative gain multiplies the apparent input capacitance while a gain near +1 bootstraps it.
  • Apply the result to Cgd in a common-source stage and Cμ in a common-emitter stage.
  • Derive a first-order input pole including source resistance, input resistance and Miller-multiplied capacitance.
  • Explain why Miller splitting does not remove the physical feed-forward path or the RHP zero derived in Chapter 02.
  • Explain why a frequency-dependent Av(s) makes the exact Miller equivalents frequency dependent rather than simple constant capacitors.
  • Use the theorem as a controlled reduction after, not before, identifying the real input/output nodes and the gain between them.
Starting question

Chapter 02 showed that Cgd can create both loading and feed-forward behavior. The Miller effect is a controlled way to reinterpret a two-node impedance when the voltage ratio between its terminals is known. The theorem is algebraic; the familiar “capacitance multiplication” is only one special case.

03.1Start from the bridge current

Figure 3.1 — A Miller capacitor is first a bridge between two nodes

The one question is: what current does this capacitor actually demand from the input node?

v1Cv2v2 = A_v · v1i1 = sC(v1 − v2)i2 = sC(v2 − v1)

The capacitor is not connected to ground: it sees v1−v2. Any equivalent input capacitance therefore depends on the voltage ratio Av=v2/v1 created by the rest of the circuit.

Let an impedance Z connect nodes v1 and v2. Define Av=v2/v1 for the circuit condition being analyzed. The current leaving node 1 through the bridge is

i1=(v1−v2)/Z
=v1(1−Av)/Z

Therefore a grounded impedance connected from node 1 to the reference node would draw the same current if

Zin,M=Z/(1−Av)

Nothing has been assumed about capacitors yet.

03.2The output equivalent follows from the other terminal current

At node 2, define current leaving node 2 through the same bridge:

i2=(v2−v1)/Z
=v2(1−1/Av)/Z

so the output-side grounded equivalent is

Zout,M=Z/(1−1/Av)

Figure 3.2 — Miller equivalent for a specified Av

The one question is: which grounded elements reproduce the same terminal currents?

v1C(1−Av)v2C(1−1/Av)two grounded equivalents replace one bridging capacitor only for the declared node ratio

The replacement is exact for the voltage ratio Av used in the transformation. If Av varies with frequency, the exact equivalent 'capacitances' are themselves frequency dependent.

03.3Specialize to a capacitor

For Z=1/(sC), the equivalent admittances remain proportional to s when Av is treated as frequency independent over the range of interest. Thus

Cin,M=C(1−Av)
Cout,M=C(1−1/Av)

If Av is negative, 1−Av=1+|Av| and the apparent input capacitance can be much larger than the physical bridge capacitance.

Example: Av=−10 and C=0.20 pF give Cin,M=2.20 pF and Cout,M=0.22 pF.

03.4Why an inverting stage suffers strong input loading

For a common-source or common-emitter stage in its midband region, a useful first approximation is

Av≈−gmRo,eff

A bridge capacitance Cgd or Cμ then contributes approximately

Cbridge→in≈C(1+gmRo,eff)

The multiplication is not extra physical charge-storage area. It is the extra current demanded because the far plate moves in the opposite direction with amplified amplitude.

03.5Worked common-source input-pole estimate

Take Av≈−20, Cgs=0.80 pF and Cgd=0.20 pF. The Miller input capacitance due to Cgd is

Cgd,M,in=0.20 pF·(1−(−20))=4.20 pF
Cin,total≈0.80 pF+4.20 pF=5.00 pF

If the resistance seen by the input node after independent-source zeroing is 20 kΩ,

fp,in≈1/[2π·20 kΩ·5.00 pF]≈1.59 MHz

Ignoring Miller and using only Cgs+Cgd=1.00 pF would predict about 7.96 MHz — a factor-five error in this simplified example.

03.6Bootstrapping is the opposite voltage-motion effect

If the far node follows the input in phase, Av is positive and close to unity. Then

Cin,M=C(1−Av)≪C

The capacitor sees only a small differential voltage, so the input source supplies little charging current. This is the principle of capacitive bootstrapping.

Do not over-interpret the output Miller equivalent as a passive capacitor in this case: for 0<Av<1, 1−1/Av is negative. The active circuit relation is what makes the algebraic equivalent possible.

03.7If Av depends on frequency, the exact Miller equivalent does too

The exact theorem may use Av(s):

Zin,M(s)=Z(s)/[1−Av(s)]
Zout,M(s)=Z(s)/[1−1/Av(s)]

For a capacitor, the equivalent admittance is sC[1−Av(s)]. If Av(s) has poles and phase shift, that expression is not generally the admittance of one constant capacitor. The simple capacitance multiplication therefore uses a gain approximation valid over a restricted frequency range.

03.8Miller splitting does not erase the RHP zero

Chapter 02 derived a feed-forward RHP zero from the original Cgd branch. Replacing Cgd by two constant grounded capacitors using a midband Av is useful for estimating loading and a dominant pole, but that approximation no longer represents the full direct feed-forward current over all frequencies.

Therefore use the original bridge circuit for exact pole/zero analysis; use the simple Miller split when a nearly constant local gain makes it a controlled approximation.

03.9Common-emitter and common-source are the same topology lesson

In a BJT common-emitter stage, Cμ bridges base and collector. In a MOS common-source stage, Cgd bridges gate and drain. The device physics setting the capacitance values differs, but the circuit-theory transformation is identical once the two node voltages are defined.

Cμ,in≈Cμ(1−Av,bc)
Cgd,in≈Cgd(1−Av,gd)

03.10Mastery exercises

  1. Starting from i1=(v1−v2)/Z, derive Zin,M without using a memorized theorem.
  2. For Av=−30 and Cμ=0.10 pF, find the input and output Miller capacitances.
  3. For Av=+0.95 and C=2 pF, find the apparent input capacitance and explain the physical meaning.
  4. A common-source stage has Cgs=0.7 pF, Cgd=0.15 pF, Av≈−15 and Rin,seen=12 kΩ. Estimate its Miller-dominated input pole.
  5. Why can the simple Miller capacitance model miss the RHP zero?
  6. What changes if Av is replaced by Av(s)?
Solutions

1. i1=v1(1−Av)/Z=v1/Zin,M, hence Zin,M=Z/(1−Av).

2. Cin=0.10·31=3.10 pF; Cout=0.10·(1+1/30)≈0.1033 pF.

3. Cin=2 pF·0.05=0.10 pF. The two capacitor plates move almost together, so little differential voltage and little input charging current remain.

4. Cin≈0.7+0.15·16=3.10 pF; fp≈1/[2π·12 kΩ·3.10 pF]≈4.28 MHz.

5. A constant-gain Miller split removes the explicit feed-forward branch from the reduced circuit and is intended for loading/pole estimation, not exact zero recovery.

6. The equivalent admittances become frequency dependent: sC[1−Av(s)] and sC[1−1/Av(s)].

Chapter 04 uses this result to show why a cascode suppresses the voltage swing across the input-output capacitance and therefore suppresses Miller multiplication.

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