Hello.

Please see the attached image as NPN transistor amplifier circuit found in some tutorial.

Here, C1 and C2 are capacitors to decouple AC input and DC bias. (20 kOhm and 3.6 kOhm compose voltage divider biasing network)

My question is on C_E. What is the role that C_E takes? Is it also to decouple?

What if C_E is removed? Does circuit still work?

 

CE is used to insert a pole in the frequency response at a frequency where the impedance of the capacitor matches the impedance of the resistor. The circuit will work without it, but the frequency will roll off earlier than with it...

Note: Had with and without reversed in the last sentence, corrected now.

 

The gain of the transistor, without the capacitor, in that configuration is roughly Rcollector/Remitter, so say 1.2k/220 = 5.45. With the capacitor the AC signal is shunted around Remitter and therefore the gain skyrockets. So to answer your question, it's there to massively increase gain, if you remove it you are left with approx 5.45 times gain. For audio reproduction it's really not a good idea to have that capacitor there, as the gain will be somewhat non-linear, based on the transistors "resistance"(it's complicated) and the effects of the capacitor.

-
CoolBear

 

Ce bi-passes the emitter resistor so there is no AC emitter feedback and you get maximum gain from the amplifier.
If Ce is removed than the AC gain will be reduced and would be approximately equal to the value of the collector resistor divided by the value of the emitter resistor.
This gain is more stable and less dependent on the value of the transistor Beta or hFE gain with Ce removed.

Sometimes the emitter resistor is split into two resistors with only the bottom one bi-passed by a capacitor. This gives high DC feedback for good stabilization of the bias point, with a lower amount of AC feedback so that the AC gain is still stabilized but at a higher value than the DC (bias) gain.
In that case the AC gain is approximately equal to the collector resistor divided by the top (un-bipassed) emitter resistor.

 

Hi.
The role of C_E capacitor is to act as a coupling capacitor. It does not affect the DC operating point of the bjt. Because capacitors act as open circuit for DC voltages & short circuit for AC voltages( not for all frequencies).
But yes, it will increase the voltage gain of the amplifier circuit. And it will also reduce the input resistance of the amplifier, which is undesirable.

 

The role of C_E capacitor is to act as a coupling capacitor.

Minor correction. CE is a bypass capacitor used to increase AC gain at a frequency where the impedance of CE is 220 ohms. C1 and C2 are coupling caps.

 

I simulated your circuit in LTSpice(highly recommended btw, it's free), without the capacitor, with 100mV/1kHz input, I get 529.2mV output, approx a gain of 5.3, with 0.05% THD.
With a 3uF capacitor(any higher and the gain is so high it distorting horribly) I get 2.21v output, approx a gain of 22.1 with a THD of 2.02%.
If I replace the capacitor with a 100uF and up the supply to 100v(I did not recalculate the bias for this, so results are to be taken with tons of salt), I'm getting peaks of 82v for a gain of 820(!), distortion is somewhat through the roof though, as it's clipping almost half of the bottom away.

 

I simulated your circuit in LTSpice(highly recommended btw, it's free), without the capacitor, with 100mV/1kHz input, I get 529.2mV output, approx a gain of 5.3, with 0.05% THD.
With a 3uF capacitor(any higher and the gain is so high it distorting horribly) I get 2.21v output, approx a gain of 22.1 with a THD of 2.02%.

Simulators are fine as long as you have a grasp of the fundamentals.

The amplifier in question was designed for a gain of about 5.5. The purpose of the bypass cap is to compensate for gain roll off at some frequency.

Check the frequency response without a bypass cap and then try 0.68uF.