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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.

https://i.imgur.com/1UVHs8H.png

gripen
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    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. – Neil_UK Oct 08 '20 at 21:16
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    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. – jonk Oct 09 '20 at 04:53
  • @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... – Circuit fantasist Oct 10 '20 at 18:00
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    @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. – jonk Oct 10 '20 at 18:45
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    @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. – jonk Oct 10 '20 at 18:46
  • @Circuitfantasist If you have one of your answers which you feel is particularly maligned but also is intuitive (to you) and gives correct results, feel free to point it out to me. I may be able to give a try there. I probably can afford ***one*** such attempt. – jonk Oct 10 '20 at 19:03
  • @jonk, I also think that here are many paths to the truth and we have the freedom to choose our loved one. Most people choose well-trodden paths from someone else; others have the courage to make new ones themselves. Regarding your last comment, I would like you to see my last question about the charge amplifier - https://electronics.stackexchange.com/questions/525548/does-a-charge-amplifier-really-amplify-the-charge-if-not-what-does-it-amplif. Tomorrow I will replace the pictures with more beautiful. But it would be interesting for me to see your opinion about this situation... – Circuit fantasist Oct 10 '20 at 19:17
  • ... where they don't give me the opportunity to ask my question and answer it (you will understand what I am speaking about if you browse the comments in all the four questions). I have distant memories of such hostility in Wikipedia, now it surprised me and found me unprepared here... – Circuit fantasist Oct 10 '20 at 19:17
  • @Circuitfantasist There's nothing wrong with attempting to find new ways to view the world. Certainly, Einstein's initial viewpoint regarding the special theory of relativity didn't immediately get a good reception -- quite the opposite. His view is more complex than Newton's. There's no escaping that fact. But it's more correct, too, in some circumstances. Over time, what takes place is that people know ***where*** to apply the simpler view and ***where*** to apply the more complex one. There are also different ***levels*** of view. Lumped vs distributed, for example. – jonk Oct 10 '20 at 19:25

4 Answers4

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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).

LvW
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  • So the role of C3 is to make the emitter voltage "stiff"? – Circuit fantasist Oct 10 '20 at 15:45
  • 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. – LvW Oct 10 '20 at 15:50
  • So, more precisely speaking, the emitter voltage is "stiff" regarding the AC variations and "soft" regarding the DC variations? – Circuit fantasist Oct 10 '20 at 15:53
  • Without C3 the resistor Re causes current-controlled voltage feedback. Loop gain is Gloop=-gm*Re. – LvW Oct 10 '20 at 15:58
  • 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. – Circuit fantasist Oct 10 '20 at 16:44
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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.
Graham Stevenson
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    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. – LvW Oct 10 '20 at 09:07
  • 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 ! – Graham Stevenson Oct 10 '20 at 11:38
  • 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) – G36 Oct 10 '20 at 13:29
  • 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 – Graham Stevenson Oct 10 '20 at 17:07
  • @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?' – Circuit fantasist Oct 10 '20 at 17:09
  • Yes, worth an explanation I agree. – Graham Stevenson Oct 10 '20 at 17:11
  • @Graham Stevenson, Let's try then to find the explanation. I see two reasons for now - one obvious and another not so... – Circuit fantasist Oct 10 '20 at 18:04
  • @GrahamStevenson But 517Hz is a "zero" frequency. From this frequency (517Hz) the voltage gain starts to increases from Rc/(Re + 1/gm) ≈ Rc/Re ≈ 6 V/V towards Rc*gm ≈ 440 V/V and the circuit will reach 440 V/V at the pole frequency Fp ≈ 0.16/(1/gm * C3) ≈ 38kHz if Ic ≈ 3mA – G36 Oct 10 '20 at 22:55
  • Pardon ? I agree with your 'in band' gain calculation (440, assuming 3mA Ic) but I can't make sense of your turnover frequencies. A more normal value of C3 for AF would be say, 1000uF and it's throwing me. Gain will reduce to 6 at 'DC' yes. OK I see what you're saying, The elaborate equations prevented me from seeing it. If you increase C3 to a more realistic 1000uF (for audio use), then the LF turnover is OK. – Graham Stevenson Oct 10 '20 at 23:36
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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.

Generalization

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.

Analogies

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).

Circuit fantasist
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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.

Eloy Calatrava
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