Can anyone help me find the input resistance of the circuit? I have been stuck for many hours and can't solve it.

I find the equivalent resistance of the circuit by applying a voltage source at the input and solve for the current I.

The input resistance:

Rin = V/I

However, I don't know why this method doesn't work. Rin is always a function of V and that is not right.

 

Show your work.

 

How can we possibly help you find what you did wrong when you don't show us what you did at all? We are not mind readers.

Show us the work for the best effort that your several hours produced.

 

Let's call the voltage across R3 is Vx.
Apply a voltage source V to the input and let's call the potential at the bottom is V1 => The voltage in the left side of R1 is V1 + V.

Apply KCL:

\(\frac{ V_{1} + V - V_{x} }{ R_{1} } = A*V + \frac{ V_{x} - V_{1} }{ R_{2} } + \frac{ V_{x} }{ R_{3} }\)

\(V_{x} ( \frac{1}{ R_{1} } + \frac{1}{ R_{2} } + \frac{1}{ R_{3} } ) = \frac{ V_{1} + V }{ R_{1} } - A*V + \frac{ V_{1}}{ R_{2} } \)

\(V_{x} = \frac{\frac{ V_{1} + V }{ R_{1} } - A*V + \frac{ V_{1}}{ R_{2} } }{ \frac{1}{ R_{1} } + \frac{1}{ R_{2} } + \frac{1}{ R_{3} } }
\)

\(I = \frac{ V_{1}+ V - V_{x} }{ R_{1} } = \frac{ V_{1}+ V - \frac{\frac{ V_{1} + V }{ R_{1} } - A*V + \frac{ V_{1}}{ R_{2} } }{\frac{1}{ R_{1} } + \frac{1}{ R_{2} } + \frac{1}{ R_{3} }} }{ R_{1} } \)

\(R_{in} = \frac{V}{I }= \frac{V*R_{1}}{ V_{1}+ V - \frac{\frac{ V_{1} + V }{ R_{1} } - A*V + \frac{ V_{1}}{ R_{2} } }{\frac{1}{ R_{1} } + \frac{1}{ R_{2} } + \frac{1}{ R_{3} }} } \)


Rin is a function of V and V1. What I am wrong here?

 

I don't see any close loop for R3 resistor current? So how can any current flow through R3?

 

I solve for Rin and I get this result Rin = (R1 + R2)/(1 + gm R2)

 

Rin is a function of V and V1. What I am wrong here?

No, it's not. It is a function only of the circuit parameters.

EDIT: If you read the original version, I thought the OP was someone other than anhnha. Sorry.

 

I don't see any close loop for R3 resistor current? So how can any current flow through R3?

Jony sorry for the confusion. I am solving for a cascode amplifier. The first stage is common source and the second stage is common gate and this stage has a resistance between gate and drain.

I want to know the voltage at the output of the first stage. If I know the input impedance, Zin of common gate stage then the voltage at drain of the common source is -gm1*Zin*Vin.

I think this maybe should consider as there is a voltage source V1 + V connects the left side of R1 to ground and another voltage source V1 connects the bottom of R2 to ground.

PS: the second I think it is not a common gate but a configutation in which drain and gate is connected by a resistance.

 

Let's call the voltage across R3 is Vx.
Apply a voltage source V to the input and let's call the potential at the bottom is V1 => The voltage in the left side of R1 is V1 + V.

Apply KCL:

\(\frac{ V_{1} + V - V_{x} }{ R_{1} } = A*V + \frac{ V_{x} - V_{1} }{ R_{2} } + \frac{ V_{x} }{ R_{3} }\)

\(V_{x} ( \frac{1}{ R_{1} } + \frac{1}{ R_{2} } + \frac{1}{ R_{3} } ) = \frac{ V_{1} + V }{ R_{1} } - A*V + \frac{ V_{1}}{ R_{2} } \)

\(V_{x} = \frac{\frac{ V_{1} + V }{ R_{1} } - A*V + \frac{ V_{1}}{ R_{2} } }{ \frac{1}{ R_{1} } + \frac{1}{ R_{2} } + \frac{1}{ R_{3} } }
\)

\(I = \frac{ V_{1}+ V - V_{x} }{ R_{1} } = \frac{ V_{1}+ V - \frac{\frac{ V_{1} + V }{ R_{1} } - A*V + \frac{ V_{1}}{ R_{2} } }{\frac{1}{ R_{1} } + \frac{1}{ R_{2} } + \frac{1}{ R_{3} }} }{ R_{1} } \)

\(R_{in} = \frac{V}{I }= \frac{V*R_{1}}{ V_{1}+ V - \frac{\frac{ V_{1} + V }{ R_{1} } - A*V + \frac{ V_{1}}{ R_{2} } }{\frac{1}{ R_{1} } + \frac{1}{ R_{2} } + \frac{1}{ R_{3} }} } \)


Rin is a function of V and V1. What I am wrong here?

What you have is correct, as far as it goes. You need to apply KCL at either the ground node or at V1, in which case you will discover that the current in R3 is zero. Then it is just a matter of slogging through some algebra. But you are definitely doing it the hard way.

 

Jony sorry for the confusion. I am solving for a cascode amplifier. The first stage is common source and the second stage is common gate and this stage has a resistance between gate and drain.

I want to know the voltage at the output of the first stage. If I know the input impedance, Zin of common gate stage then the voltage at drain of the common source is -gm1*Zin*Vin.

I think this maybe should consider as there is a voltage source V1 + V connects the left side of R1 to ground and another voltage source V1 connects the bottom of R2 to ground.

PS: the second I think it is not a common gate but a configutation in which drain and gate is connected by a resistance.

Well, you definitely need to decide if current can flow into or out of the left hand terminals other than just from one to the other and, if so, represent that in the schematic.

 

Here is the small signal model of the circuit. I need to know Zin.
Then the voltage at the output of common source will be equal to -Gm1*(ro||Zin)

PS. The first stage is common source degradation.

 

Can you show "normal diagram" of this cascade amplifier.
And if the second stage is common gate what is Zin for CG stage zin = 1/gm2 ??
If so the voltage at Q1 drain is equal to:
Vin * (1/gm2)/(1/gm1 + RS)

 

Can you show "normal diagram" of this cascade amplifier.
And if the second stage is common gate what is Zin for CG stage zin = 1/gm2 ??
If so the voltage at Q1 drain is equal to:
Vin * (1/gm2)/(1/gm1 + RS)

Jony, here is the circuit. I need it for a project. I have modified and simulated it a lot but it doesn't work. Now I decide to learn the theory carefully before make any simulation.

 

Simulation don't work because you made a error in the diagram.
Look where you connect M2 source. Also I don't see any capacitor between M2 gate and GND?

 

Simulation don't work because you made a error in the diagram.
Look where you connect M2 source.

Thanks, yes, D and S of M2 should be swapped. I made the mistake here but in the simulation it is correct.

Also I don't see any capacitor between M2 gate and GND?

yes, you are right. I used capacitor in place of R2 in the topology. However,
last week, I go to a university and asked a friend of mine for help. He said that
I can use resistance instead of a capacitor provided that they have the same impedance. Therefore, I draw a resistor here.

 

I think that your small signal model should look like this

 

I think that your small signal model should look like this

Yes, can you tell me how to calculate Rin of the second stage?

 

I don't know how you can find Rin for the second stage. I think that you need to analyze the whole circuit. Because R2 is connect between M2 base and M1 source. So M2 gate is not grounded, so you cannot split the circuit.

 

Jony, can you tell me steps to analyse this? I don't know where to start. It is too complicated to me.

 

It is pain in the a... to solve this circuit manually.
So I write four nodal equations and with help of math software solve all of them.

For Vs node
Vs/Rs = (Vin - Vs )*gm1 + (Vg - Vs)/R2

VG node
(Vg - Vs)/R2 = (Vout - Vg)/R1
For Vout node
(0 - Vout)/RD1 = (Vout - Vg)/R1 + ( Vg - Vd)*gm2

For Vd node
( Vg - Vd)*gm2 = (Vin - Vs )*gm1

And finally solution for Vout and Vd