1
\$\begingroup\$

This site says: "To increase the gain for AC signals the emitter resistor bypass capacitor is added. This should be calculated to have a reactance equal to the emitter resistor at the lowest frequency of operation".

I assume that I have 1K ohm emitter resistor. If I added a parallel capacitor of reactance 1k ohm. The total impedance would be 0.5 K ohm.

In this case,

Should I increase the emitter resistor to be 2K and increase the reactance to be 2K so that I get 1K ohm (Which is the original value of emitter resistor before adding the bypass cap)?

Or it is ok to have an impedance of 0.5k ohm although the original value of emitter resistor is 1K?

Will a bypass capacitor affect the bias voltage of the biasing resistors?

Thank you very much,

\$\endgroup\$
2
  • 1
    \$\begingroup\$ You don't need to change the emitter resistor at all when you add a bypass capacitor. The reason for choosing an equal reactance is to define the zero of the bandwidth. However 1k is probably too low for other reasons. \$\endgroup\$
    – user207421
    Feb 1, 2016 at 21:03
  • 1
    \$\begingroup\$ It depends on the bandwidth you want the amplifier to achieve. A simple thumb rule is that a capacitor is seen as a "short-circuit" at higher frequencies and an "open-circuit" at DC. Therefore, the gain of a common-emitter is about the load connected to the collector divided by the load connected to the emitter. Therefore, you can change the AC gain of the amplifier using a capacitor in parallel without changing the bias point of the transistor. \$\endgroup\$
    – lucas92
    Feb 1, 2016 at 21:29

2 Answers 2

1
\$\begingroup\$

It depends on the bandwidth you want the amplifier to achieve. A simple thumb rule is that a capacitor is seen as a "short-circuit" at higher frequencies and an "open-circuit" at DC.

Therefore, the gain of a common-emitter is about the load connected to the collector divided by the load connected to the emitter. So, you can change the AC gain of the amplifier using a capacitor at the emitter in parallel without changing the bias point of the transistor. A .1 uF in parallel with a 1k resistor will put a low cut frequency of about 1.6 kHz. Therefore, higher values should be used for lower frequencies.

\$\endgroup\$
-1
\$\begingroup\$

This site gives you wrong information. The bypass capacitor (\$C_E \$) do not influence the DC bias condition. This capacitor "short" the emitter to ground for AC signals (removes the negative feedback from the circuit). And thanks to this the AC gain increases from

\$ A_v = \frac{Rc}{Re}\$ to

\$ A_v = g_m*Rc \approx 40*Icq*Rc\$.

where \$I_{cq} \$ is the collector current at quiescent point.

The capacitor value can be found like this

\$Xc <\frac{1}{gm}\$ or

\$ C_E = \frac{1}{2 \pi f\frac{26mV}{I_{cq}}}\$

where f is "the lowest frequency of operation" and \$g_m \$ is the transistor transconductance equal to \$ \frac{i_{cq}}{26 mV} \$.

\$\endgroup\$
1
  • \$\begingroup\$ The site doesn't say anything about the bypass capacitor affecting the DC bias. \$\endgroup\$
    – user207421
    Feb 1, 2016 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.