I'm having some trouble understanding the effect that charge injection has on capacitors - more specifically, when there is unequal charge on a capacitor.

Consider the following circuit: enter image description here

My understanding is as follows

  • Assume that Vin is at a constant 1V. Initially, the CK is high and the MOSFET is on with a channel present. Let's also assume that the Ron is neglidble and there is enough time for the capacitor to settle at a final voltage. The capacitor has a voltage of 1V across it and thus Vout = 1V
  • The CK goes low just as in the image. The free electrons present in the channel need to go somewhere now since the channel is no longer present as the MOSFET is off now. Some of these electrons go to the source Vin and are absorbed there by an ideal voltage source and some of these electrons go to the top plate of the capacitor.
  • Now, here comes the confusion. Most books say that the 1V on the capacitor will decrease slightly to let's say 0.96V and that can be proven by the Q=CV. I want a more intuitive understanding. My understanding is that these "injected" electrons come to the top plate of the capacitor which I assume is the + plate. These new electrons recombine with positve charges there resulting in a smaller number of positive charges on the top plate and hence the bottom negative plate loses some electrons too as there is not enough attraction for them. Thus capacitor voltage decreases.

Q1: Is my understanding correct?

I don't think it's correct (particularly the part about how the electron recombines - I feel that's only relevant to semiconductor physics) because of the following section in Razavi's book which suggests how to cancel this effect. Consider the following circuit:

enter image description here

Now, if we use my same logic of the electron cancelling the positive charge on the top plate. Then in this case, as soon as M1 goes low, charge q1 is injected to the top positive plate of capacitor, recombining with a "hole" and thus when M2 goes high, there is nothing for it to absorb since that initial charge injected by M1 has already decreased the capacitor voltage. What will M2 even do at this point?

Does my argument seem reasonable?

  • \$\begingroup\$ If you don't get a good answer here, look up Jim Todsen or Bonnie Baker. They won't be where I remember them, but they know this stuff well. Jim is an IC designer. Bonnie, when I knew her, was more on the field applications side but doing a lot of writing on these topics. The ACF2101 is the only part I know of where they took as much care to balance the charge injection and removal when switching an integrator on and off. Those are the two folks I've spoken with (more with Jim, though.) I don't want to confuse you with my picture; better to get it from folks who really can give you all of it. \$\endgroup\$
    – jonk
    Commented Feb 21, 2019 at 1:05
  • 1
    \$\begingroup\$ Recombination is not a thing on a metal-plate capacitor. Recombination is a phenomenon that happens in a semiconductor that can have both electrons and holes, and metal doesn't have holes in that sense. So perhaps your confusion stems from not understanding how conduction works. \$\endgroup\$
    – TimWescott
    Commented Feb 21, 2019 at 1:55

1 Answer 1


I'd say it's a lot simpler if you just stick with the standard electrical engineering convention of voltage, current, and charge instead of trying to think in terms of electrons (and/or "holes"). That's how the standard electrical engineering curriculum teaches it. So we just build up from simple fundamentals:

  • Electric current is just the flow of charge from point A to point B.
  • A capacitor builds up voltage as it accumulates charge, as defined by Q=CV.
  • A MOSFET has a small amount of parasitic capacitance between all terminals( Reference)

Showing the above circuit with the parasitic capacitance added:


simulate this circuit – Schematic created using CircuitLab

  • We start with Vin at a constant 1V and CK high (let's say 5V). The MOSFET M1 is ON, so Cds is shorted out and both Cgd and Cgs have +4V across them. CH has +1V across it, and Vout is +1V as well.
  • When CK goes low (0V) then M1 turns off. We can ignore Cgd, since there is no way for it to source or sink current to CH. We can also ignore Cds, since it is both a very small capacitance and a very small voltage across the capacitor (0.00 - 0.04V). But Cgs is the important one. Pulling CK down to 0V means we must introduce a voltage of -1V across Cgs, versus the +4V it had before.

That's the key point. The capacitor starts out with Q1 = Cgs*4V and then must consume enough charge to re-stabilize at Q2 = Cgs*(-1V), so it draws a total charge of (Q2-Q1) out of CH. Since CH is much larger than Cgs, this 5V change on Cgs corresponds to a 0.04V change across CH.

In the circuit shown by Razavi, there is another parasitic capacitance introduced which has a complementary voltage across it, so any excess charge introduced by M1 would be consumed by M2, and vice versa, leaving the net charge on CH unchanged.

  • \$\begingroup\$ Although the intuition is correct, linear parasitic capacitances are not a good representation of charge injection in a MOSFET during a switching transition. As the channel collapses or forms, the lowest impedance node will be receiving the majority of the charge. \$\endgroup\$ Commented Feb 23, 2019 at 12:54
  • \$\begingroup\$ Ì believe this is referring to clock feedthrough (or capacitve feedthrough) which is a seperate phenomenon to what I'm referring to here. I'm concerned about what happens to electrons physically present in the channel of the MOSFET when the CK goes low. It has to go somewhere. Correct me if I'm wrong. I may be getting confused myself. \$\endgroup\$ Commented Mar 2, 2019 at 17:57
  • \$\begingroup\$ @AlfroJang80 if I'm understanding your question and comment correctly (that's a 50/50 shot), then no, it's the same thing. Unfortunately keeping track of the electrons is the reverse of standard EE terminology (where "current" flows from positive to negative) but let's give it a shot. When we start CH has +1V, so the bottom plate has -1V worth of electrons more than the top plate. When CK goes low it injects some electrons. Those settle in to the top plate, balancing out the effect of some -- but not all -- of the electrons on the bottom plate. The net effect is the same. \$\endgroup\$
    – Mr. Snrub
    Commented Mar 2, 2019 at 20:05
  • \$\begingroup\$ @Mr.Snrub Ah. That makes a lot more sense. So this balancing out of electrons essentially doesn't mean recombination. It's just adjusting the net amount of electrons difference on the two plates and hence the voltage. I think this question is more to do with my lack of capacitor understanding. Thank you very much! \$\endgroup\$ Commented Mar 2, 2019 at 21:57

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