Calculating gain of a CE with a bypass cap

Your approach is basically correct. However, this will give the AC gain at a specific frequency only, as Xc varies so does the gain.

Pick Re and Ce so that the gain is essentially constant in the frequency range you are interested in (think of how the parallel combination varies with Re and Xc.)

The DC gain independant of Ce.

SOA

Well I have one constant frequency going in. I'm simulating this on Multisim.

I'm using a 1kHz @ 1.414 p-p on the input. Rc is 5k, Re is 470ohms, Ce is

400nf with an r'e of about 20.37 so my gain should be about 21.2 and I'm getting something like 15 on multisim.

I've run through the calculation and agree with your numbers. Do you have any information on the model that Multisim uses?

Have you tried it without the C? The gain should then be 5000/470 +/-. If not, maybe Multisim is using a different approach .

SOA

I'm not sure I can post the file to the group. so I'll give you all the values.

input 1000Hz at 1.414 p-p

1.25uF input cap top resistor on base is 170k bottom resistor on base is 17k Rc 5k Re 470 bypass cap 400nF input voltage was orginally 20 volts, but I changed it to 30 volts just so I can get some more gain and figure out where I'm going wrong. once I tried the bypass cap, I started getting clipping, so I just upped the input voltage and set the scope to AC coupling.

the transistor is a virtual BJT NPN with 100 for a foward beta

Steven, did you give up on this????

Sorry Peter,

I've not been back to the group.

I've looked at your values and (other that a small correction for the Z of the input C) all are as I would expect

If you remove the Bypass capacitor, does the gain DROP to what you expect (Rc/Re)? If not, suspect the model that the package uses.

Removing this C should not cause clipping, as the amplitude of the o/p should drop!

SOA

By removing the bypass cap, the gain is exactly what it should be. I even put a meter on the output to measure the Ie current so I can correctly add in my r'e factor as well.

But once I put the capacitor in there, the gain is not what I expect. I'll email my instructor and see if he knows. I thought this would be an easy one.

Thanks

i saw your post a few days ago and thought it should be easy too.

doesn't bypassing with the cap change from CE to someother configuaration ? Then your formulas need to change somewhat }:o

since you already know your gain is on the nose for the plain CE configuration, you must take into account what the cap is doing to the transistor and calculate the value you are looking for forthwith from a different point in the circuit., not within the junction. it's been a while:-)

Emailing the instructor };-) that is so cool., i wish this was around when i started :-)

maybe I'll go back with a new sight for the finish line.....

Good luck there Peter

Roy