The attached file is common-source stage. The circuit has a gain:
\(A_{v}=-g_{m}R_{D}= -\sqrt {2\mu _{n}c_{0x}\dfrac {w} {L}I_{D}}\dfrac {V_{RD}} {I_{D}}\)
Then in order to maximize the voltage gain of common-source stage, my book say that:
1. Increase \(\dfrac {W} {L}\) and other parameters are constant
2. Increase \(V_{RD}\) and other parameters are constant
3. Decrease \(I_{D}\) and other parameters are constant
I understand #1 completely but both #2 and #3 are really confusing me.
I want to know how can you do that, for example, increase \(V_{RD}\)
but keep \(I_{D}\) constant or decrease \(I_{D}\) and keep
\(V_{RD}\) constant.

 

Change the number of ohms in the resistor, Rd.

 

Change the number of ohms in the resistor, Rd.

Thanks,
\(V_{RD}=I_{D}R_{D}\), then if \(V_{RD}\) increase and keep \(I_{D}\), \(R_{D}\) has to increase.
In order to keep \(I_{D}= const\) the voltage between G and S \(V_{GS}\) must be kept constant. But \(V_{GS}\) include both DC and AC (small signal) and the ac signal varies in time. Therefore, \(V_{GS}\) also varies in time. Could you tell me what I am wrong here?

 

You are mixing AC with DC. Your reasoning is perfectly good as long as you think only in terms of DC. As soon as you apply an AC signal to the gate, the Vgs must change and therefore violate the terms of the DC analysis. Pick one. This problem only works in DC thinking.

 

We can increase Rd and kept Id constant by simply using active load (current source) instead of a Rd resistor.

 

You are mixing AC with DC. Your reasoning is perfectly good as long as you think only in terms of DC. As soon as you apply an AC signal to the gate, the Vgs must change and therefore violate the terms of the DC analysis. Pick one. This problem only works in DC thinking.

Sorry, I still haven't yet get it. Do you mean that the circuit is only used to amplify DC signal?
I am confused because the purpose of amplifier is amplifying signal input (I see most of them are ac signals). If this circuit only used to amplify DC signal, then how ac signal will be amplified?
And in my book it said that the formula above is the gain of the small signal circuit.

 

What I'm saying is that you stated the problem in two different ways. From post #1, how can you increase the voltage across the drain resistor while keeping the drain current constant? By changing the resistance of the drain resistor, or, from Jony130, make the drain resistor an active circuit called, "a constant current circuit" which automatically changes its apparent resistance as the voltage changes.

Now (from post #3), in order to keep the Id constant, the Vgs must be constant, but the input to the gate contains an ac signal. How do you keep the Vgs constant while applying an ac signal between the gate and the source? Either you don't (because resistors don't work that way), or you add a constant current circuit between the source and ground to force the drain current to be the same at all times. If the source of your transistor is connected directly to ground and you apply an AC signal to the gate, there MUST me a change in the volts, gate to source. If the source of your transistor is connected not to ground, but to a constant current circuit, the Vgs will remain the same when AC signal is applied to the gate. The voltage fluctuation will be seen as a change of the voltage at the top of the constant current circuit, and not as a change in the volts, gate to source.

Trying to answer both questions at once is very difficult. I can do one at a time, but I can't do both at the same time because the configuration of the circuit must be different for each of the two questions.

 

What I'm saying is that you stated the problem in two different ways. From post #1, how can you increase the voltage across the drain resistor while keeping the drain current constant? By changing the resistance of the drain resistor, or, from Jony130, make the drain resistor an active circuit called, "a constant current circuit" which automatically changes its apparent resistance as the voltage changes.

Now (from post #3), in order to keep the Id constant, the Vgs must be constant, but the input to the gate contains an ac signal. How do you keep the Vgs constant while applying an ac signal between the gate and the source? Either you don't (because resistors don't work that way), or you add a constant current circuit between the source and ground to force the drain current to be the same at all times. If the source of your transistor is connected directly to ground and you apply an AC signal to the gate, there MUST me a change in the volts, gate to source. If the source of your transistor is connected not to ground, but to a constant current circuit, the Vgs will remain the same when AC signal is applied to the gate. The voltage fluctuation will be seen as a change of the voltage at the top of the constant current circuit, and not as a change in the volts, gate to source.

Trying to answer both questions at once is very difficult. I can do one at a time, but I can't do both at the same time because the configuration of the circuit must be different for each of the two questions.

Thanks a lot! Yes, your reply is complete and I understand now.