i have attached an assignment below with the answers included. i have to do a test.

Since posting this q, i have found some errors. so here are the corrections.

Q2 - answer is - reverse biased so that quiescent Drain Current . Idss/2=Id

Question 3 (a) Should Read

Idss = Drain to Source Saturation Current.

3(d) should read gm = forward transconductance = ^ Id/^Vgs

^ stands for delta. the actual delta sign is a equilateral triangle, but you knew that.

delta stands for change. so change in drain current over change in gate source voltage equals gm.

Question 7 (a) should end up with answer of 7.4 or there abouts




When the assignment was checked by the teacher, i wasn't there.

i would certainly appreciate any comments you may wish to add.

the only question i really had trouble with was Question 7. This deals with a circuit, which i think is known as "Common Source Amplifier" circuit. the question is asking about the AC parameters of the circuit, from the look of the drawing.

part (a) asks for the Av Gain of the circuit. i have shown the formula for a load connected.

Then at part (d) they ask for Av again, but with bypass capacitor removed.

I can't see how you can remove bypass, and still be connected to load?



Anyhow, I've appreciated the help I've got from this site so far very much. So anymore is a bonus. I look forward to hearing from you.

watt"s what?

 

Originally posted by rodn.m@May 5 2006, 11:29 AM
i have attached an assignment below with the answers included. i have to do a test.

Since posting this q, i have found some errors. so here are the corrections.

Q2 - answer is - reverse biased so that quiescent Drain Current . Idss/2=Id

Question 3 (a) Should Read

Idss = Drain to Source Saturation Current.

3(d) should read gm = forward transconductance = ^ Id/^Vgs

^ stands for delta. the actual delta sign is a equilateral triangle, but you knew that.

delta stands for change. so change in drain current over change in gate source voltage equals gm.

Question 7 (a) should end up with answer of 7.4 or there abouts
When the assignment was checked by the teacher, i wasn't there.

i would certainly appreciate any comments you may wish to add.

the only question i really had trouble with was Question 7. This deals with a circuit, which i think is known as "Common Source Amplifier" circuit. the question is asking about the AC parameters of the circuit, from the look of the drawing.

part (a) asks for the Av Gain of the circuit. i have shown the formula for a load connected.

Then at part (d) they ask for Av again, but with bypass capacitor removed.

I can't see how you can remove bypass, and still be connected to load?
Anyhow, I've appreciated the help I've got from this site so far very much. So anymore is a bonus. I look forward to hearing from you.

watt"s what?

[post=16828]Quoted post[/post]​

The capacitor C2 would be refered to as a coupling capacitor, CS is the bypass capacitor. When you remove CS the load is still connected The thing to keep in mind is that with a large enough bypass capacitor, source feedback is small making the gain much large. When you remove CS the gain will be reduced to about RL||RD/RS. I think you can fill in the details. Good luck.