# Common Emitter Amplifier Help

Discussion in 'Homework Help' started by Denny1234, Apr 23, 2008.

1. ### Denny1234 Thread Starter Member

Feb 17, 2008
27
0
Hi, could you help?

In the following circuit

http://www.ecircuitcenter.com/Circuits/trce/trce.htm

Which capacitor would correspond to my high frequency cut off and which would correspond to my low frequency cut off? And also in each case what resistor values would I use to calculate the cut offs in each case.

Thanks for any help

2. ### Caveman Active Member

Apr 15, 2008
471
0
C1 and CL are both high frequency cutoffs. The low frequency cutoff in this circuit is actually defined by the transistor properties. CE extends the bandwidth of the circuit a bit.

3. ### Denny1234 Thread Starter Member

Feb 17, 2008
27
0
Oh didnt know that, thanks a lot, I thought that CI and CL would correspond to low cut offs.

SO does that mean when i try to work out the capacitor vlaues I need.. via 1/2piRF

For CI, I just input into the equation the desired upper cut off frequnecy (1Mhz) and the parallel combination of R1 AND R2?

For CL, I just input into the equation my desired upper cut off frequnecy (1Mhz) again and the sum of the resistor values RC and RL?

4. ### Caveman Active Member

Apr 15, 2008
471
0
Yeah standard single pole high pass would do it. At least for a general frequency.
For C1, you also have the reflected emitter resistance. If I remember correctly, it is the emitter resistance multiplied by Beta. So if it is really big, you can probably ignore it. Otherwise, just put R1 and R2 in parallel.

For CL, I had to work it out. You have the frequency correct, but don't forget that there is also frequency independent loading from RL as well.

Vout = Vcollector*Rc/(Rc + RL)

Does this make sense?

5. ### Denny1234 Thread Starter Member

Feb 17, 2008
27
0
I havent covered Beta in my amplifier so will check that out, it would be great if I sussed it and was able to include that. If not then i will mention that as its in parallel to my R1 and R2 that its very large value meant i could ignore it.

Im unsure on the second statement, does that mean that RL on its own is acting as another source to my cutoff?

Sorry to hassle ya, thanks so much for your help on this!

6. ### Caveman Active Member

Apr 15, 2008
471
0
Remember that Beta is not very well controlled so use a minimum value for your calculation.

What I meant by the second statement was that you use RL parallel with RC and 1/2*PI*RC to calculate it, with R=RL||RC and C=L.