I was reading the chapter of Microelectronic Circuits (Sedra Smith) about negative feedback systems, and it shows an example of Miller frequency compensation with the capacitance Cf shown in the following scheme (a common emitter amplifier which is part of a cascade of amplifiers):

enter image description here

It shows how that capacitance moves a pole in the s plane, and this allows us to make the closed loop total system stable:

enter image description here

I have a basic doubt: if you see Cf, it is in the same place of the BJT common - emitter parasitic capacitance (or, in case of MOSFETS, gate - drain capacitance), i.e. a capacitance between output and input.

But sometimes I have read that these parasitic capacitances are terrible regarding stability, and that problem was important also for old vacuum tubes. In "the design of CMOS radiofrequency electronic circuits" (Lee) for instance it is shown as Cgd of a MOS amplifier may determine an input admittance with negative real part, which means that the system is not stable.

So I do not understand why I read both that a Miller feedback capacitance may be useful to get frequency compensation and to get the stability, and also that a parasitic capacitance included exactly in the same position in the circuit may provoke unstability.

  • \$\begingroup\$ We simply use a Miller effect to our advantage and we place a dominant-pole capacitor between a base-collector junction in the CE stage. Thanks to this we can use a smaller capacitor size (because the effective capacitance at the base is increased due to the Miller effect) and we get a stable amplifier because of the gain roll-off at −20dB per decade rate. \$\endgroup\$
    – G36
    Commented Feb 2, 2020 at 9:17
  • \$\begingroup\$ Remember the classical frequency compensation method for opamps: They are unity gain stable (100% feedback) because the first pole is shifted to a very low frequency (10...200 Hz). And the price we have to pay for universal stability: Smaller closed-loop bandwidth. \$\endgroup\$
    – LvW
    Commented Feb 2, 2020 at 10:15
  • \$\begingroup\$ "for instance it is shown as Cgd of a MOS amplifier may determine an input admittance with negative real part, which means that the system is not stable" Could you leave the actual text for that? Or perhaps a reference to check further. \$\endgroup\$
    – edmz
    Commented Feb 2, 2020 at 17:43

2 Answers 2


An unstable closed loop system can be made stable in two ways: -

  1. Lowering the open loop gain or,
  2. Ensuring that the open loop phase change does not reach 180 degrees before the open loop gain drops below unity.

Clearly, for people who make opamps, they like to sell you something that has a massive DC open loop gain so, if the overall gain is dropped (to fix instabilities when running closed loop) then the device will look unattractive in applications that require ultimate DC precision.

That’s the back story for opamps but, it applies to any control loop; good positional accuracy is usually a strong desire whereas speed of response might be seen as secondary.

So, step in the miller capacitor. It is used extensively in opamps to ensure that the overall phase change from DC to high frequencies doesn’t reach 180 degrees by the time the open loop gain has fallen below unity.

But sometimes I have read that these parasitic capacitances are terrible regarding stability

If you have a control system that is stable but has a phase margin of only a few degrees then adding any more phase angle without seriously reducing gain, will usually result in instability. However, if the phase change is added such that it dominates the existing poles of the system, then stability can be regained.

It is this last part that works in an opamp. Without the miller compensation, the opamp might undergo massive phase changes at really high frequencies (and well below the point where gain has dropped to unity) and, if high levels of negative feedback were applied it would become an oscillator.

So, the miller capacitor is used to dominate those areas and force the gain to fall below unity before the phase change can reach 180 degrees. But, if your miller capacitor circuit is only slightly significant at too high a frequency, it can cause instability to an otherwise stable system.

  • \$\begingroup\$ @Kinka-Byo do you need more clarification here? \$\endgroup\$
    – Andy aka
    Commented Feb 3, 2020 at 9:19
  • \$\begingroup\$ clear, thank you very much! \$\endgroup\$
    – Kinka-Byo
    Commented Feb 6, 2020 at 7:17

The circuit analysis you described indeed assumes that the presence of the feedback capacitor does not cause any oscillations and the circuit is absolutely stable (defined below). If the circuit is stable, then phase margin can be used to measure the relative stability. But usually resistively loaded circuits are always stable and the test for absolute stability is unnecessary. More details are summarized below.
In the following, I will be considering a MOS device to keep the math simple but similar arguments can be applied to BJT's.

Stability: Definition

The stability of a circuit can be determined in two ways: absolute stability and relative stability. These are defined as follows:

  • A circuit is absolutely stable if all of its poles lie in the left half s-plane.
  • The relative stability (of absolutely stable circuit) measures how close the circuit is from being unstable. It is calculated via the phase and gain margins.

From hereon, the stability of a circuit is implies the absolute stability.
An open-loop system is always stable as the real circuit components cannot produce right half plane poles. Thus, stability is always discussed in context of a feedback system. An unstable system with right half plane poles can be created by a positive feedback system. The output voltage of such system would show a exponential growing response implying an output impedance with negative real part (negative resistance).
Thus, presence of negative resistance implies right half plane poles and unstable system.

Stability: Common Source (CS) Amplifier

A common source amplifier without \$C_{gd}\$ is an open-loop or a unilateral system and hence is always stable. Addition of the \$C_{gd}\$, however, creates a feedback loop which can become unstable under certain conditions. As explained above, both the input impedance and the output impedance should have positive resistance for a stable amplifier. The stability of the circuit now depends on the loading conditions.
- For a resistively loaded CS amplifier (as you show in the figure), the input impedance is given by: $$Z_{in} = \frac{R_L}{1+g_m R_L} - j\frac{1}{\omega C_f(1+g_mR_L)}$$ The input resistance is always positive and the circuit is thus stable with this load element.
- For an inductively loaded CS amplifier, the real part of input impedance is given by: $$Re\{Z_{in}\} = \frac{(\omega^2LC_f-1)g_mL}{(1+\omega^2g_m^2L^2)C_f}$$ Clearly, for \$\omega < \frac{1}{\sqrt{LC_f}}\$, the input resistance is negative giving rise to unstable system which would oscillate. But it would be stable at higher frequencies.
The CS transistor is thus a conditionally stable system whose stability depends on the output load.
In fact, at low frequencies the CS amplifier is always conditionally stable as it has high power gain. As frequency goes up, the power gain goes down and after a certain knee frequency, the amplifier is unconditionally stable. Beyond the \$f_{max}\$ the power gain becomes less than unity and the transistor ceases to be an active element.

General Two Port Stability
In general, a two port network is unconditionally stable if both the input impedance and output impedance have a positive resistance (real part) for all types of source and load impedances.
In terms of reflection coefficient, it implies: \$|\Gamma_{in}|, |\Gamma_{out}| < 1\$. From this, it can be shown that the amplifier is unconditionally stable if: $$k = \frac{1 - |S_{11}|^2 - |S_{22}|^2 + |\Delta|^2}{2|S_{12}S_{21}|} > 1, |\Delta| < 1$$ Here, \$\Delta = S_{11}S_{22} - S_{12}S_{21}\$.
The factor \$k\$ is known as Rollet's stability-factor.
Usually, we don't need general purpose circuits which are stable for all kinds of loading conditions. And the k-factor only needs to be calculated for the particular load we are interested to drive.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.