I'm looking at some transistor radio circuits and the book I'm following makes the following statement with regard to AC gain: "By raising the DC voltage at the collector, the internal collector-base capacitances of the transistor are reduced".

Here's a circuit to help understand what's going on:

enter image description here

Here the author has killed the DC gain by placing an inductor in parallel with R2 whilst leaving the AC gain intact: collector reactance (L1||R2) / emitter reactance (C3||R3).

I have a reasonable grasp of the effect of Miller Capacitance on an inverting amplifier, where the inverted output acts negatively on the input. What I don't understand is why increasing the collector voltage acts to reduce collector-base capacitance.

Does the increase in current flowing through the collector-emitter junction have anything to do with it?

  • 3
    \$\begingroup\$ What happens to the capacitance of a diode (PN junction) in reverse mode as a function of the (reverse) voltage ? This effect is used in varicap diodes but any PN junction exhibits this behaviour. \$\endgroup\$ Commented Nov 5, 2015 at 13:07
  • \$\begingroup\$ I have no idea. But, if you turn your question into a fully fledged answer, you'll have made my day. \$\endgroup\$
    – Buck8pe
    Commented Nov 5, 2015 at 13:18
  • \$\begingroup\$ Open your favorite textbook on semiconductor physics and look it up then ! If you want to be able to work with analog circuits, this is a thing you need to know and understand. It's not hard. \$\endgroup\$ Commented Nov 5, 2015 at 13:19
  • \$\begingroup\$ Is this the same as the "Miller Effect", or is that only on FETs? I've never quite understood it either. \$\endgroup\$
    – pjc50
    Commented Nov 5, 2015 at 13:45
  • 1
    \$\begingroup\$ Thanks for the hint @FakeMoustache, a quick search for varactors gave me the intuition I needed. \$\endgroup\$
    – Buck8pe
    Commented Nov 5, 2015 at 13:49

1 Answer 1


As @FakeMoustache hinted in a comment to your question, the explanation lies in the behavior of a reverse-biased PN junction, because that's what Q1's collector-base junction is in your circuit.

From a macroscopic point of view any reverse-biased PN junction acts like a parallel-plate capacitor whose capacitance (called transition capacitance \$C_T\$) depends inversely on the reverse voltage \$V_R\$. The relationship is not linear, but it is approximately:

$$ C_T = K \dfrac{1}{\sqrt{V_0 + V_R}} $$

where \$V_0\$ is the voltage gap created by the junction and \$K\$ is a constant.


Struggling to remember the exact form of the formula (there are half a dozen of ways of writing down that relationship, depending on which physical parameters of the junction you want to emphasize) I found a more intuitive formula in this Google book:

$$ C_T = \dfrac{C_0}{(1 + V_R)^n} $$

Note: That formula has an error in it (dimensional analysis debunks it). Probably \$V_R\$ is meant to be the relative voltage with respect to some reference. I guess the correct formula should be: $$ C_T = \dfrac{C_0}{\left(1 + \dfrac{V_R}{V_0}\right)^n} $$

where \$C_0\$ is the capacitance when no bias is applied and \$n\$ depends on how the junction is doped: \$n = \frac 1 2\$ for step-graded junctions, whereas \$n=\frac 1 3 \$ for linearly-graded junctions.

Another interesting article on the subject (tougher semiconductor physics stuff) explains how to derive that relationship (in yet another form!).

  • 1
    \$\begingroup\$ Yep, @FakeMoustache put me on the right track and for those interested the images in this link (en.wikipedia.org/wiki/Varicap) give you a fair idea of what's happening. \$\endgroup\$
    – Buck8pe
    Commented Nov 5, 2015 at 13:52
  • \$\begingroup\$ Exactly :-) Ct goes down when Vr goes up ! The increased reverse voltage causes the size of the depletion layer to expand, this decreases the capacitance. Visualize it, now remember forever :-) \$\endgroup\$ Commented Nov 5, 2015 at 13:53

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.