Roles of the coupling and bypass capacitors in a CE transistor

Question

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.

Answer 1

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).

Answer 2

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.

Answer 3

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).

Answer 4

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.

Answer 5

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.