Cascode Amplifier Small Signal Model

[CoolDude420]

Homework Statement

Hi,

I'm just trying to derive the small-signal model of the cascode MOSFET amplifier circuit. In the lectures, we were just given the small-signal model directly. I'm trying to derive it.

I have almost the exact same thing except where I put the output voltage. In the lectures, it is put between D2 and S1,G2. However when I was deriving it, I put it between Ro2.

My derivation attempt:

Homework Equations

The Attempt at a Solution

 

[scottdave]

Isn't Vo supposed to be in reference to ground, not the other transistor?

 

[CoolDude420]

Isn't Vo supposed to be in reference to ground, not the other transistor?

Ah yes! Thank you.

One more question, how would I go about finding the voltage gain Vo/Vi? I have no idea where to start with it because Vo is at such an awkward position.(between D2 and gnd)

I'm assuming its a KVL loop involving ro1 ro2 and vo?

 

[scottdave]

2 loops should do it.

 

[CoolDude420]

2 loops should do it.

In the lecture notes we have,
it says that

$$V_{ro2} + V_o + V_{gs2} = 0$$
$$ V_{ro2} = (g_m)(v_{gs2})(r_{o2}) $$

How can you assume that all of the $$(g_m)(v_{gs2})$$ current flows through ro2??

 

[donpacino]

How can you assume that all of the

(gm)(vgs2)​

(g_m)(v_{gs2}) current flows through ro2??

Where else is it going to go?

 

[CoolDude420]

Where else is it going to go?

through r01?

 

[donpacino]

through r01?

how would it do that, current has to flow in a loop

 

[CoolDude420]

how would it do that, current has to flow in a loop

Ah yes. Just another question. If the gmvgs2 current only goes through ro2, and since there is no ground in that loop. Where does the current leave? I'm used to most circuits with a ground. I'm assuming it just disspitates through the resistor as heat?

 

[donpacino]

If the gmvgs2 current only goes through ro2, and since there is no ground in that loop.

You don't need to have ground in a current loop.

your kvl, kcl, and ohms laws still apply here. The current doesn't need to "leave" it just needs to 'complete its loop'

 

[scottdave]

I'm not where I can draw a picture, right now. In the loop which has r02: look at Node D2. The current entering and leaving that node must be equal, so the current through r02 must equal the source current.

 

[CoolDude420]

You don't need to have ground in a current loop.

your kvl, kcl, and ohms laws still apply here. The current doesn't need to "leave" it just needs to 'complete its loop'

Ah perfect. I get it now. Thank you!

 

[CoolDude420]

I'm not where I can draw a picture, right now. In the loop which has r02: look at Node D2. The current entering and leaving that node must be equal, so the current through r02 must equal the source current.

That makes perfect sense. Thank you.

 

[CoolDude420]

I'm not where I can draw a picture, right now. In the loop which has r02: look at Node D2. The current entering and leaving that node must be equal, so the current through r02 must equal the source current.

I just had a second thought. Doesn't that sort of disagree if we apply KCL at the other node(S2, D1)?

I apply KCL at the the middle node on the right side(S2, D1).
KCL at S2, D1 $$ g_mv_{gs,2} + i_{r_{o1}}= i_{r_{o2}} + g_mv_{gs,1}$$
Rearranging,
$$ g_mv_{gs,2} = i_{r_{o2}} + g_mv_{gs,1} - i_{r_{o1}}$$

That is telling me that in fact, not all of the current is going through r_o2.

What am I missing?