I have to find the voltage amplification (Av=Vo/Vi) of this circuit (FET amplifier) and output resistance Ro.
So I've simplified it to equivalent pi model like shown below.
In order to find Vo voltage, the current id is needed.

(id-gm1Vgs1)rd1 + idRs1 + (id-gm2Vgs2)rd2 + idRs2=0
Vgs2= Vi - idRs2
Vgs1 ? (I don't know how to express Vgs1)

Vo = (id-gm2Vgs2)rd2 + idR2
Could you please help? Thanks in advance.

 

Your electronic circuit is very bad. Two current generators are connected in series. The currents of these current generators can not be equal. As a result, one transistor (some one) will come out of the amplification mode! The circuit can only work with an upper transistor having a built-in channel! and then, as I said above, there will not be.
See

 

The scheme with the right has the equivalent circuit you need. It is only necessary to close the capacitor in alternating current. The currents of the transistors are equal. For lower frequencies, a larger capacitor should be used. Without resistors at the source, the circuit will amplify the signal.

 

I think that you should split the circuit in half to simplified the calculations.

Let us start with the upper FET's (Q1).

The small-signal equivalent circuit will look like this:



And we can quit easy find the equivalent resistance seen from Q2 drain into Rs1 and Q2.

Rx = Vx/Ix = (Vgs + Vo)/Ix = ( Ix*Rs1 + (Id+Ix)*ro1 )/Ix = ( Ix*Rs1 + (gm1*Ix*Rs1+Ix)*ro1 )/Ix = Rs1 + ro1 + gm1 ro1 Rs1 =
= Rs + (1 + gm1Rs1) *ro1


So now we can relpace Q2 with our eqivalent resistance Rx and your circuit will be simplified to this form:




And finding the voltage gain for this circuit should be easier for you?

Can you do it?

Your electronic circuit is very bad.

But what if both MOSFETs are depletion mode ?

 

I think that you should split the circuit in half to simplified the calculations.

Let us start with the upper FET's (Q1).

The small-signal equivalent circuit will look like this:

View attachment 144842

And we can quit easy find the equivalent resistance seen from Q2 drain into Rs1 and Q2.

Rx = Vx/Ix = (Vgs + Vo)/Ix = ( Ix*Rs1 + (Id+Ix)*ro1 )/Ix = ( Ix*Rs1 + (gm1*Ix*Rs1+Ix)*ro1 )/Ix = Rs1 + ro1 + gm1 ro1 Rs1 =
= Rs + (1 + gm1Rs1) *ro1


So now we can relpace Q2 with our eqivalent resistance Rx and your circuit will be simplified to this form:

View attachment 144843


And finding the voltage gain for this circuit should be easier for you?

Can you do it?



But what if both MOSFETs are depletion mode ?

Thanks a lot. I really appreciate your explanation.

So if I write the Kirchhoff's Voltage Law
idRx + (id - gm2Vgs2)ro2 + idRs2 = 0
idRx + (id - gm2(Vi - idRs2))ro2 + idRs2 = 0
id ( Rx + ro2 + gm2Rs2ro2 + Rs2) - gm2ro2Vi =0
id= gm2ro2Vi / Rx + ro2 + gm2Rs2ro2 + Rs2

Is that the voltage amplification?
Av= Vo/Vi = -idRx / Vi = gm2ro2 Rx / Rx + ro2 + gm2Rs2ro2 + Rs2

I aslo have to find the output resistance (Ro) , is that equal to Rx ?

 

I think that you should split the circuit in half to simplified the calculations.

Let us start with the upper FET's (Q1).

The small-signal equivalent circuit will look like this:

View attachment 144842

And we can quit easy find the equivalent resistance seen from Q2 drain into Rs1 and Q2.

Rx = Vx/Ix = (Vgs + Vo)/Ix = ( Ix*Rs1 + (Id+Ix)*ro1 )/Ix = ( Ix*Rs1 + (gm1*Ix*Rs1+Ix)*ro1 )/Ix = Rs1 + ro1 + gm1 ro1 Rs1 =
= Rs + (1 + gm1Rs1) *ro1

EDIT: Jony130 is correct. I was looking at the wrong end of one of the TS's models. I'll follow up later. Ignore what follows this edit.

Vgs is not as shown in your image. It's like this:



Which gives a different result for the equivalent impedance. You'll appreciate the fact that I noticed this by using the admittance matrix representation for the circuit.

 

For extra credit, consider the case where the gate of the top FET (G1) is connected to small signal ground (AC ground) rather than to D2. Then what is Av and Ro?

 

Look at the modes of transistors (currents and voltages) and you will understand what I was talking about.

 

Hi Bordodynov - that's nice. Have you done a small signal analysis? I know this configuration with BJTs or FETs as a cascode where the upper transistor acts as a high impedance load.

 

Hi Bordodynov - that's nice. Have you done a small signal analysis? I know this configuration with BJTs or FETs as a cascode where the upper transistor acts as a high impedance load.

I have used as the lower bipolar transistor, and instead of the upper transistor I used a combination of field-effect transistor of P-N junction and PNP bipolar trasistor + resistor (multiplication transconductance transistor). I took off the signal from the source. At the source, the output resistance is much smaller. All this was the second cascade. And I stabilized the transistor modes with the help of a common negative feedback.

 

Av= Vo/Vi = -idRx / Vi = gm2ro2 Rx / Rx + ro2 + gm2Rs2ro2 + Rs2

Looks good. But don't forget about the minus sign.

I aslo have to find the output resistance (Ro) , is that equal to Rx ?

No, Ro is not equal to Rx.


We have two paths for a AC current



The first path is through upper FET ( Rx in our case).

The second path is through lower FET ( Ry ).

And this Ry you need to find.



Ry = Vy/Iy

 

Thanks a lot. I really appreciate your explanation.

So if I write the Kirchhoff's Voltage Law
idRx + (id - gm2Vgs2)ro2 + idRs2 = 0
idRx + (id - gm2(Vi - idRs2))ro2 + idRs2 = 0
id ( Rx + ro2 + gm2Rs2ro2 + Rs2) - gm2ro2Vi =0
id= gm2ro2Vi / Rx + ro2 + gm2Rs2ro2 + Rs2

Is that the voltage amplification?
Av= Vo/Vi = -id*Rx / Vi = gm2ro2 Rx / Rx + ro2 + gm2Rs2ro2 + Rs2

Your result for Av is correct except it will not evaluate properly unless you add some parentheses. Also, if you want to indicate multiplication by concatenating variables be careful to leave a space or you will have created a new variable. For example, you have intended for gm2ro2 to indicate multiplication as in gm2*ro2, but what you have done is create a new variable; gm2ro2 is a new variable. So you should have written:

id=( gm2 ro2 Vi )/( Rx + ro2 + gm2 Rs2 ro2 + Rs2)

Is that the voltage amplification?
Av= Vo/Vi = -idRx / Vi =-( gm2 ro2 Rx) /( Rx + ro2 + gm2 Rs2 ro2 + Rs2)

With those changes you have the correct expression for Av.

 

Looks good. But don't forget about the minus sign.

No, Ro is not equal to Rx.


We have two paths for a AC current

View attachment 144912

The first path is through upper FET ( Rx in our case).

The second path is through lower FET ( Ry ).

And this Ry you need to find.

View attachment 144913

Ry = Vy/Iy

Vy = (iy - gm2 Vgs2) ro2 + iy Rs2
Vy = (iy - gm2 (0 - iyRs2)) ro2 + iyRs2
Vy = (iy + gm2Rs2iy) ro2 + iyRs2

So is it that way?
Ro= Ry=Vy/iy= ( 1 + gm2Rs2)ro2 +Rs2

 

So is it that way?
Ro= Ry=Vy/iy= ( 1 + gm2Rs2)ro2 +Rs2

Well, Ry resistance is indeed equal to:

Ry = ( 1 + gm2Rs2 )ro2 + Rs2

But what about Ro? If Our AC current has two paths to AC ground. One through Rx to GND and the second one through Ry to gnd.
So Ro is ?

 

Well, Ry resistance is indeed equal to:

Ry = ( 1 + gm2Rs2 )ro2 + Rs2

But what about Ro? If Our AC current has two paths to AC ground. One through Rx to GND and the second one through Ry to gnd.
So Ro is ?

Ro= Rx*Ry / Rx+Ry ???

 

Ro= Rx*Ry / Rx+Ry ???

Much better now.

 

Much better now.

 

Rx*Ry / Rx+Ry is going to evaluate as:

\(\frac{Rx*Ry}{Rx}+Ry\)

You need to use parentheses like this (Rx*Ry)/(Rx+Ry) if you want it to evaluate as:

\(\frac{Rx*Ry}{Rx+Ry}\)

 

Now that you have solved your problem I can't resist showing you the power of a nodal analysis using a compact linear algebra technique. Perhaps this will pique your interest and motivate you to pursue your study of network analysis to a higher level.

You can number the 4 nodes in your circuit (with ground as the reference node) like this:



Now you can write the 4 nodal equations and format them as an admittance matrix "Y". Node 3 is included to provide a driven node, and a gate resistor Rg is connected from there to ground. The admittance matrix is inverted to become an impedance matrix "Z", and the two things your problem wants you to find are easily derived from appropriate elements of the Z matrix as shown here:



In fact, you can easily find the voltage gain from G2 to all 4 nodes at once. The gain from g2 to itself is just 1, of course. Also, you can find the equivalent resistance from each node to ground all at once. The expressions your problem required are shown in red:

 

Someone will explain to me: what type of transistors are used (with an induced or built-in channel or Depletion ) in the original circuit ? And then I feel like an idiot at someone else's feast !