02 · Advanced

Poles and zeros in amplifiers

Move from device capacitances to the frequency response of a complete biased stage. Derive poles from the natural-response denominator rather than assigning one pole per capacitor, derive zeros from feed-forward terms in the transfer numerator, solve a separated two-node amplifier exactly, introduce open-circuit time constants as an approximation tool, and show how Cgd/Cμ can create a right-half-plane zero even before the Miller approximation is invoked.

230–290 minAdvanced
In this lesson
  • Define amplifier poles as roots of the small-signal network characteristic denominator and zeros as roots of the selected transfer-function numerator.
  • Explain why the number and location of poles are properties of the coupled network rather than a one-capacitor/one-pole lookup rule.
  • Derive a two-node unilateral common-source response with separate input and output RC poles.
  • Relate well-separated first-order poles to Bode magnitude/phase behavior while retaining the exact denominator for quantitative work.
  • Derive the resistance seen by a capacitor with independent sources zeroed and the other high-frequency capacitors opened.
  • Use open-circuit time constants to estimate a dominant high-frequency pole and state when the estimate is not reliable.
  • Derive the simple common-source transfer with Cgd feed-forward and identify the right-half-plane zero at ωz≈gm/Cgd under the stated reduced-model assumptions.
  • Explain the different phase effects of left-half-plane and right-half-plane zeros.
  • Distinguish pole/zero cancellation in an algebraic transfer from robust removal of an internal dynamic mode.
  • Apply a workflow: biased small-signal circuit → nodal equations → determinant/numerator → exact or approximate poles/zeros → Bode interpretation → approximation checks.
Starting question

Chapter 01 identified capacitances. Chapter 02 asks the harder question: what dynamic modes do those capacitances create once they interact with the actual network impedances? A capacitor does not own a pole. Poles belong to the coupled linearized circuit.

02.1Poles come from the natural-response denominator

After linearization, write the small-signal nodal equations in the Laplace domain as

Y(s)·v(s)=b(s)u(s)

For an internally stable finite-dimensional lumped network, the characteristic denominator is generated by det Y(s). The natural modes satisfy

det Y(s)=0

The transfer from one declared input to one declared output can be written N(s)/D(s); its poles are roots of D(s) after any exact algebraic cancellations. The important modeling habit is to derive D(s) from the network, not to attach a guessed pole to every capacitor.

02.2A two-node unilateral amplifier factorizes cleanly

Figure 2.1 — Two nodes, two time constants

The one question is: how do input and output RC networks become two poles when reverse coupling is neglected?

vsRsigvgRGCingm·vgRoutCoutvoRin = Rsig || RG after the independent source is zeroed

The source drives the gate through Rsig while RG and Cin load the input node. The output node contains Rout, Cout and the gm·vg controlled source. In this reduced unilateral model the two nodes factor; a real stage with Cgd/Cμ couples them.

Neglect reverse coupling from output to input. Let Rin=Rsig||RG be the resistance seen by Cin after the independent source is zeroed. Then

vg/vs=[RG/(Rsig+RG)]·1/(1+sτin)
τin=(Rsig||RG)Cin

At the output,

vo/vg=−gmRout/(1+sτout)
τout=RoutCout

so the source-to-output response is the product of two first-order factors. This exact factorization belongs to the reduced unilateral circuit, not automatically to a real transistor with Cgd or Cμ.

02.3Well-separated poles recover the familiar Bode picture

H(s)=H0/[(1+s/ωp1)(1+s/ωp2)]

If ωp1≪ωp2, the first pole dominates the initial bandwidth loss: magnitude falls at approximately −20 dB/dec after ωp1, then −40 dB/dec after ωp2, and each real pole contributes up to −90° phase lag. If the poles are close, the exact product must be evaluated; the label “dominant pole” is then weak or meaningless.

02.4Open-circuit time constants estimate the first denominator coefficient

For each high-frequency capacitor Ci, zero the independent sources, open the other high-frequency capacitors, and compute the resistance Ri⁰ seen by Ci in the remaining linearized network. Define

τi0=Ri0Ci
Στi0

The sum of zero-value/open-circuit time constants gives the first-order denominator coefficient for the high-frequency capacitive network. When one real pole dominates, a widely used estimate is

ωH≈1/Στi0

This is an approximation tool, not an exact rule for arbitrary clustered poles, complex poles, important zeros or strong feedback. If accuracy matters, solve the nodal determinant or evaluate the full transfer.

02.5Zeros come from the transfer numerator

A zero is not a natural mode. It is a frequency at which different paths from the chosen input to the chosen output cancel. Therefore zeros depend on the declared transfer. A feed-forward path can create a zero even when it adds no new energy-storage state of its own.

H(s)=N(s)/D(s)
N(sz)=0 ⇒ H(sz)=0

Different outputs of the same physical circuit can have different zeros while sharing the same internal pole set.

02.6Cgd feed-forward creates a RHP zero in a simple common-source model

Figure 2.2 — A feed-forward path can create a RHP zero

The one question is: what does Cgd do when an ideal source fixes the gate voltage?

viCgdgm·viRoCovosource and body are AC-grounded in this reduced common-source model

With vi imposed, Cgd injects a direct input-to-output current in parallel with the transconductance path. In this reduced model the two contributions cancel at s=+gm/Cgd, so the zero lies in the right-half plane.

Assume an ideal voltage source fixes vi, the source/body are AC-grounded, Ro is the total resistive load, Co is the output capacitance excluding Cgd, and ro has already been absorbed into Ro. KCL at the drain gives

vo[1/Ro+s(Co+Cgd)]+vi[gm−sCgd]=0

Hence

vo/vi=−gmRo·(1−s/ωz)/(1+s/ωp)
ωp=1/[Ro(Co+Cgd)]
ωz=gm/Cgd

Because the zero factor is 1−s/ωz, its zero is at s=+ωz: a right-half-plane zero. It has the same magnitude factor as a left-half-plane zero but contributes phase lag rather than phase lead.

02.7Worked pole/zero estimate

Let gm=5 mS, Ro=10 kΩ, Co=1.0 pF and Cgd=0.10 pF. Then

fp=1/[2π·10 kΩ·1.10 pF]≈14.5 MHz
fz,RHP=5 mS/[2π·0.10 pF]≈7.96 GHz
Av0=−gmRo=−50

The zero is far above the first pole in this numerical example, so the stage first behaves like a one-pole low-pass. That ordering is not universal; shrinking capacitance, changing gm or loading Ro moves the characteristic frequencies differently.

02.8Pole-zero cancellation is fragile

If a numerator zero equals a denominator pole algebraically, that input-output transfer can show a cancellation. But transistor parameters, capacitances, process variation, temperature and loading move poles and zeros. Near-cancellation can therefore create hidden slow or lightly damped behavior that reappears under another output or perturbation.

Do not equate “the plotted transfer looks cancelled” with “the physical state disappeared.”

02.9Workflow for amplifier frequency response

  1. Find the DC operating point and verify device regions.
  2. Build the midband small-signal conductance model.
  3. Add only the charge-storage elements needed over the frequency range of interest.
  4. Include source, bias, load and package/measurement impedances that are electrically relevant.
  5. Write nodal equations before simplifying coupled capacitances.
  6. Extract exact poles/zeros or use a controlled approximation such as open-circuit time constants.
  7. Compare the reduced result with the full transfer and check the frequency range where the approximation holds.

02.10Mastery exercises

  1. For Rsig=20 kΩ, RG=80 kΩ and Cin=5 pF, find the input pole of Figure 2.1.
  2. For Rout=10 kΩ and Cout=2 pF, find the output pole. Which pole dominates?
  3. Show directly from the drain KCL of Figure 2.2 that the numerator is gm−sCgd.
  4. With gm=4 mS and Cgd=0.2 pF, find the RHP-zero frequency.
  5. Explain why a right-half-plane zero adds phase lag even though its magnitude asymptote rises at +20 dB/dec.
  6. State one condition under which 1/ΣRi⁰Ci is a poor bandwidth estimate.
Solutions

1. Rin=20 kΩ||80 kΩ=16 kΩ; fp,in=1/[2π·16 kΩ·5 pF]≈1.99 MHz.

2. fp,out=1/[2π·10 kΩ·2 pF]≈7.96 MHz, so the input pole dominates this reduced two-pole example.

3. KCL gives vo[1/Ro+s(Co+Cgd)]+vi[gm−sCgd]=0.

4. fz=gm/(2πCgd)=4 mS/(2π·0.2 pF)≈3.18 GHz.

5. The RHP factor is 1−jω/ωz; its argument is negative while its magnitude equals that of 1+jω/ωz.

6. Examples: clustered/complex poles, an important nearby zero, strong frequency-dependent feedback, or no clearly dominant pole.

Chapter 03 will now return to the input-output capacitance and derive the Miller transformation rather than treating it as a memorized “capacitance multiplication” slogan.