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Emitter Bypass Capacitor in a CE Amplifier

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How exactly is the emitter capacitor C3 helping to increase the voltage gain in the common emitter amplifier below? Is it because for AC signals the emitter looks like a short to ground and that essentially increases the voltage drop across the collector resistor? Thanks.

Common Emitter

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Is it because for AC signals the emitter looks like a short to ground and that essentially increases the voltage drop across the collector resistor?

Yes.

The gain of a common emitter amplifier is approximated by the ratio of R3 to R4 - this is when there is no emitter decoupling capacitor. When you add the emitter capacitor, the impedance to ground seen by the emitter is very much smaller for AC signals than DC hence the gain increases. But it comes with a price - distortion is the main issue and the next issue is gain changing with temperature.

Inside the emitter (for simplification) there is an "internal" resistance called $r_E$ and this is the only limitation on voltage gain when using a large emitter capacitor but, $r_E$ is temperature dependent and emitter current dependent: -

$$r_E = \dfrac{26 \text{ mV}}{I_E}\text{ at room temperature}$$

Note - The 26 mV changes with temperature

So, if $I_E$ is 1 mA DC, $r_E$ will be about 26 ohms at room temperature but, as the emitter/collector current rises up and down from the 1 mA quiescent level (due to a signal being present), $r_E$ falls and rises from 26 ohm and this produces a significant non-linear gain with signal amplitude.

Recommendation: Don't use an emitter decoupling capacitor - if you need more gain, use a 2nd stage.

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Thanks for the response. Always good to validate my ideas with people more experienced. ‭jm567‭ 16 days ago

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With the capacitor C3 disconnected, the stage has a minimum gain and with C3 connected it has a maximum gain. There is a compromise solution with a moderate gain - to connect a resistor in series to C3.

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