# Capacitors in Amplifiers

I am reading about single stage CE Amplifiers and I don't understand how the capacitors work here.

1) Coupling Capacitors:

"The coupling capacitor is used to isolates the d.c. of one stage from the next stage and allows the a.c. signal only."

As per my understanding, when an a.c. input is given to a capacitor, it superposes with the d.c bias voltage and we get something like this:

This is again only an alternating signal. There are no d.c. components. So, why do we need the capacitor? I found an answer on the internet that said that the capacitor removes the dc bias voltage from the output signal. But how can this be? Once an a.c. signal superposes with a d.c., it still remains an a.c., right (like the one above)?

2) Bypass Capacitor:

" Bypass capacitors provide low impedance path to unwanted AC signals." Below is a circuit diagram.

How are these "unwanted" a.c. signals filtered from the output? In the first place, do these signals even come from the output?

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• "Bypass capacitors provide low impedance path to unwanted AC signals" That is a very confusing statement, and so is the labelling on the diagram, because in your CE circuit the AC component IS "the signal", not "noise". Perhaps you should find a better website or book to read. – alephzero Dec 28 '18 at 16:47
• Given the linearity of many circuits, we use the superposition theorem, and we separate the constant (operating point?) DC conditions from the signal (AC) conditions. The DC conditions require an initial movement of charge (often called displacement currents, by Maxwell himself), and then there is no more charge movement, except from the signal (AC) input. – analogsystemsrf Dec 29 '18 at 3:17

First of all, understand that an ideal capacitor has infinite impedance to DC signals, so the two waveforms you drew will generate identical current across a capacitor (while they would generate very different currents across a resistor).

1) A concrete example: given a sine wave that is 1V peak-to-peak, 0 DC bias would be it varying from -0.5 to +0.5; with 1V of DC bias, it would range from +0.5 to +1.5. An ideal capacitor has infinite impedance to DC, so either of those waves would look identical to the input of this circuit.

Notice that R1 and R2 create a voltage divider. C1 not only prevents any DC bias from entering the circuit, it also prevents this DC bias voltage from appearing to the signal source.

This bias voltage is needed in order for the transistor to work at all (for there to be collector-emitter current, there must be a smaller current from the base to the emitter). What will happen is that the base-emitter current will be proportional to the AC component of the input signal, plus the DC current established by the path from R1, through the base-emitter junction, to Re.

1a) (describing how Ce works) Remember that this amplifier works by having the base-emitter current (small) control the collector-emitter current (large). Without Ce, both of those currents would always have to pass through Re, which limits the gain of the amplifier. Having Ce there means that any high-frequency components of the signal do not pass through Re (since a capacitor has zero impedance to a high-frequency signal).

1b) You didn't ask this, but Re is there in the first place to stabilize the biasing of the transistor. If no Re was needed, then Ce wouldn't be needed either (because the whole point of Ce is to allow AC currents to bypass Re).

Lastly, you might want to read the wikipedia page on BJT biasing. It's not perfect, but it walks you through all of the circuits that are simpler than the one you posted, until it finally gets to the one you posted, and it provides some (minimal) explanation for each step.

2) I'm not sure why they show that as a bypass capacitor. Bypass capacitors are used because real world circuits have resistance and inductance, so the large spikes in currents caused (in particular by digital signals, and especially with MOSFET digital logic) will cause the voltage to "sag" which can mess things up. Capacitors provide a very low impedance path for these currents, which keeps the voltage more consistent

• Why would the emitter-collector current pass through Ce? Shouldn't it contribute to the output (V out)? – Gokulakrishnan Shankar Dec 29 '18 at 6:07
• @GokulakrishnanShankar All current passing through the emitter has to pass through Re, Ce, or both. There is no other path for the current. – Jason Dec 29 '18 at 6:19
• So, for an NPN transistor in common emitter mode- thinking about it in terms of motion of electrons- the electrons that come out from the collector pass through Rc and are returned back to the emitter through Re, Ce, or both, right? Please correct me if I am wrong. – Gokulakrishnan Shankar Dec 29 '18 at 7:05
• @GokulakrishnanShankar The diagrams here: allaboutcircuits.com/textbook/semiconductors/chpt-4/… show electron motion. Note that all of the electrons enter the transistor through the emitter, but exit through the base and the collector. This means that the sum of the base and collector currents equal the emitter current (This is true for PNP transistors as well, but the direction of current is reversed). – Jason Dec 31 '18 at 18:50
• @GokulakrishnanShankar I realized my previous comment didn't answer your specific question, and it's too late to edit. Regardless of which direction you draw the arrows, the total of the base current and collector current is equal (but opposite) to the emitter current. The path from the emitter to ground must go through one or both of Re and Ce. However you are neglecting the input signal as another path for current that will also go through some combination of Re and Ce. – Jason Dec 31 '18 at 19:01

"unwanted ac signal" is a bit confusing - but not wrong. The ac voltage drop across the emitter resistor RE is - indeed - in some cases "unwanted".

Why? Because the ac voltage drop across RE (caused by the ac current through the transistor which is identical to the output current caused and controlled by the input ac voltage at the base) acts as a negative feedback signal. As a consequence, the ac voltage gain would be reduced to approximately A=-RC/RE.

If we want a larger gain value the feedback effect for ac signals must be cancelled - but the DC feedback effect (caused also by RE) should be retained (for better stability of the DC operating point).

Hence, the resistor RE must be ac-wise shortened using a sufficiently large capacitor CE.

But note: In same cases we want to have "some" negative feedback also for ac signals (better linearity, better stability of the gain value). In this case, only a part of RE is bypassed using a capacitor CE (in practice: RE consists of two separate resistors).

1) The point is that any DC component will drive a net current across the stages. This is unwanted, as only the part that carries information (which is the AC), should be passed along.

2) The resistance of a capacitor becomes lower with increasing signal frequency. Thus, the very high frequency components have very little resistance at the capacitor and will take the path through $$C$$ to ground, rather than through $$R_E$$.