I wonder what the purpose of the coupling capacitors and the bypass capacitors are and the type of effect each capacitor has on the circuit.

As I've understood it, you use coupling capacitors to block DC and avoid interference (what does that really mean?) and the bypass capacitor is to have a short-circuit in the emitter of the transistor (and what does this really mean?).

If I vary the values of C1 and C2, what kind of effect does it have on the bandwidth of the amplified signal?

What happens if we have a large or a small bypass capacitor (C3), what effect does it have?

So in essence, what does the three capacitors do in the circuit, i.e. what low pass and high pass effect does it have?

I have an LTspice schematic that we can get a visualization from.


  • 1
    \$\begingroup\$ Usually, there will be a small resistor in series with C3. Usually, C2 is followed by some sort of load that defines a resistance to ground. The answers to most of your questions will be undefined without these components. \$\endgroup\$
    – Neil_UK
    Oct 8, 2020 at 21:16
  • 1
    \$\begingroup\$ gripen, Just imagine that these capacitors develop a voltage across them that is stable (doesn't vary much.) Just the right amount of voltage so that they can attach to an input source or an output destination without interfering with either DC operating point. For example, if the C2 output ties to another amplifier stage with a different DC operating point, then C2 will charge up to just the right voltage so that the quiescent point of the current stage's collector voltage mates up well with the quiescent point of the next stage's base voltage. If that's 3V diff. then the cap will have 3V. \$\endgroup\$
    – jonk
    Oct 9, 2020 at 4:53
  • \$\begingroup\$ @jonk, Nice intuitive explanations... but I thought you were just a math genius... I imagine a chain of cascaded stages connected by self-adjusting "shock absorbers" that transfer the "movements" from a stage to stage... Even more interesting is to picture the charge movement from a charged capacitor to another empty capacitor. Have you visited my last four questions about this phenomenon (all they closed)? Do you care about their fate or you just watch indifferently how they are methodically destroyed? I need support in this difficult time... \$\endgroup\$ Oct 10, 2020 at 18:00
  • 4
    \$\begingroup\$ @Circuitfantasist I've found your explanations to be unintuitive. They are also very long and very detailed. It takes substantial effort on my part to work out all of the details of where we part company and (from my perspective) you wind up choosing what appears to be the more difficult mental approaches I could imagine. After trying to wade through a few where I could see how much work it would take to turn them back right-side-upwards where they'd be intuitive (to me), I've given up on the idea. I'm not saying you are wrong. I'm just saying your viewpoints are very far removed from mine. \$\endgroup\$
    – jonk
    Oct 10, 2020 at 18:45
  • 1
    \$\begingroup\$ @Circuitfantasist This is very much to me like the difference between the Platonic view of the planets and the Copernican view. They both work (perhaps.) They differ somewhat in results, here and there, but not so much you can say one is right and one is wrong. But one is MUCH simpler. (To me.) But a Platonicist would argue oppositely. I just don't have the time to engage the debate, is all. \$\endgroup\$
    – jonk
    Oct 10, 2020 at 18:46

5 Answers 5


The role of C3:

It is the main purpose of C3 to restrict the negative feedback effect (caused by Re) to DC and very low frequencies (below the desired operating frequencies).

This feedback effect (for DC) is very inportant because it makes the DC operating point less sensitive to parts tolerances and variations of the transistors B-Value (B=Ic/Ib).

As another effect, negative feedback reduces the gain value - and if somebody does not want such a reduction, the feedback effect must be cancelled for operating frequencies (bypassing Re with a capacitor C3).

  • \$\begingroup\$ So the role of C3 is to make the emitter voltage "stiff"? \$\endgroup\$ Oct 10, 2020 at 15:45
  • \$\begingroup\$ No - it is the emitter resistor Re which produces negative feedback for DC - thereby reducing the influence of the unknown values of Ib (base current) and Vbe. (Both values are to be estimated during the design process). The capacitor C3 removes this feedback effect for operating frequencies. Without C3 the gain would be smaller but more insensitive to tolerances of BJT parameters. \$\endgroup\$
    – LvW
    Oct 10, 2020 at 15:50
  • \$\begingroup\$ So, more precisely speaking, the emitter voltage is "stiff" regarding the AC variations and "soft" regarding the DC variations? \$\endgroup\$ Oct 10, 2020 at 15:53
  • \$\begingroup\$ Without C3 the resistor Re causes current-controlled voltage feedback. Loop gain is Gloop=-gm*Re. \$\endgroup\$
    – LvW
    Oct 10, 2020 at 15:58
  • \$\begingroup\$ Interesting... As though, two voltage sources - the emitter follower output and the capacitor, are connected to each other. Initially, the first charges the second... but then the second interferes with the first when it tries to change its output voltage quickly (AC). In other words, the capacitor is an "ungrateful" voltage source:) BTW a similar interaction can be seen in a differential pair where two voltage sources (emitter followers) are connected in parallel. As a result, the emitter voltage is "stiff" at differential mode and "soft" at common-mode. \$\endgroup\$ Oct 10, 2020 at 16:44

Yes, the coupling capacitors block DC and pass AC (simple analysis). They have no effect on 'blocking interference'. Perhaps you would be kind enough to advise us where you got such an idea.

If you vary the values of C1 and C2 the low frequency response of the circuit will be affected. The first order (6dB/octave or 20dB/decade) HPF formed by the illustrated value of C1 and (R1 in parallel with R2) has a corner frequency of 20Hz (-3dB). That would be considered unacceptable today for even moderately quality audio. Increase to 4.7uF or 10uF at no serious cost and the HPF drops to 4 or 2 Hz with attendant improvement in phase response.

Likewise for C2 taking into account the load impedance of the following circuit

The same conditions apply to C3 and Re regarding their time constant and the effect on frequency response as to C1.R and C2.R . The value shown has a corner frequency in combination with Re of 517Hz ! A value of 220 or 470uF would be more suitable for AF use.

Note that Re *(more usually called RE) is included to provide stability of the operational point by introducing (shunt derived - series applied) negative feedback. Without a capacitor to bypass it, the gain would be reduced to approximately Rc/Re.

  • Re is more usually reserved for the dynamic impedance at the emitter, which is 27 ohms @ 1mA.
  • 1
    \$\begingroup\$ Quote: "...corner frequency in combination with Re of 515Hz". This statement is not correct. The total (effective) resistance which determines the lower corner frequency is Re in parallel with the input resistance at the emitter node (re=1/gm). For a DC current of app. 1mA this results in re=1/gm=25 Ohm. The corner frequency will be app at 13 kHz (for many applications to large). Therefore, the Capacitor C3 should be enlarged. \$\endgroup\$
    – LvW
    Oct 10, 2020 at 9:07
  • \$\begingroup\$ The circuit is drawn oddly. I missed it too at first. Re is in parallel with C3. Actually it's 517Hz, a typo. Mea Culpa ! \$\endgroup\$ Oct 10, 2020 at 11:38
  • \$\begingroup\$ The LvW is right, the pole frequency will be equal to fp = 0.16/( 1/gm||RE * C3) and the zero frequancy is at Fz = 0.16/(RE * C3) \$\endgroup\$
    – G36
    Oct 10, 2020 at 13:29
  • \$\begingroup\$ Look at the circuit very carefully. C3 and Re share the same node at the emitter of Q1. Their other ends are connected to GND. They're in parallel. C3 is the 'bypass capacitor' for Re. I agree that Re is normally reserved for the transistor's emitter dynamic resistance but in this case it's in fact what is often called RE. So Fc = 1/2.pi.Re.C3 = 1/2*3.142*615*0.5^-6 = 517 \$\endgroup\$ Oct 10, 2020 at 17:07
  • \$\begingroup\$ @Graham Stevenson, Nice explanation... I would only ask you, 'Why the bypass capacitor is "in combination with a series resistor"? What problem does the resistor solve?' \$\endgroup\$ Oct 10, 2020 at 17:09

I think the OP needs a simple, clear and intuitive explanation of the two electrical arrangements - a charged capacitor connected in series (coupling capacitor) and in parallel (bypass capacitor) to another voltage source.

In such cases, I even suggest to think of the charged capacitor of as a "rechargeable battery"... as a simpler electrical analogy. It is well charged and keeps up its constant voltage.

Coupling capacitors

In the OP's circuit, the input coupling capacitor C3 is charged to the bias voltage set by the R1-R2 voltage divider (I hope OP has some idea what it is)... and is connected in series to the AC input voltage source. So, its (bias) voltage adds to ("shifts up") the bipolar input voltage variations... what the transistor needs.

Similarly, the output coupling capacitor C2 is charged to the output bias (quiescent) voltage... and is connected in series to the output collector voltage. But here, its voltage is subtracted from ("shifts down") the output collector voltage.

So, in both cases, charged "coupling" capacitors (with a constant voltage) are connected in series to voltage sources with varying (AC) voltage. The only difference is their polarity.

Bypass capacitors

In the OP's circuit, the blocking capacitor C3 is connected in parallel to the emitter resistor thus "copying" the voltage across it (I will not discuss what this voltage is). So, it keeps this (its) voltage constant when it tries to change. Thus it fixes the emitter voltage.


So, both coupling and blocking capacitors are the same - a charged capacitor acting as a constant voltage source. But in the first case it is connected in series while in the second - in parallel to another voltage source.

And both coupling and blocking capacitors do the same - they keep the voltage across themselves constant. Only, in the first case, they transfer the voltage variations while, in the second case, they "kill" them.


A shock absorber is a very good mechanical analogy of the capacitor:

  • When acting as a "coupling capacitor", it transfers the movement (e.g., of a spring) of the one end to the other.

  • When acting as a "bypass capacitor" (connected "in parallel" to a spring), it blocks the movements of its ends relative to each other (turns into a hard rod).


The role of C1 is to avoid the DC component of V2 to enter in the circuit. If DC component enters the transistor will change its operating point, called Q-point (from quiescent) and dangerously can enter in a saturation mode. In saturation mode the transistor does not amplify, just acts as a switch. Role of C2 is already explained above but has the same role, although in the circuit you plotted is useless as one of its ends is open circuit. Role of C3 is already explained above.

So, capacitors in here are to block DC for the reason explained, nothing to do with interference.


Ely said it quite well but here is the same thing but expanded. C1, for example, helps to remove any DC offset in the input signal (eg from a microphone etc) and ensures that only the AC component is amplified. If you didn't do that, you'd upset the delicate biasing at the base of the transistor. Note that the biasing ensures that the base is at or around 0.6V which is the threshold at which the transistor starts to respond. If we exposed the base to the full DC coming from the input, which might be at 2v say, that would immediately saturate the transistor and it would be able to amplify anything. The same applies to C2.


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.