Variable Gain Amplifier Analysis help

  • #1

Homework Statement:

Find the gain for this VGA

Relevant Equations:

I(sat) = k(V_gs - V_t)^2
M1 = M2 = M3
1592076104940.png


I need to find A_v, I want to relate Vin to Vout. To do this I did KVL on the major loop:
$$V_{DD} = I_D*R_D + V_{SD1} - V_{SD2} - I_D*R_D$$
Since M1 = M2 then everything would cancel out and I would just get V_dd = 0 which is probably wrong. This is assuming V_sd1 = V_sd2, is that correct?
 

Answers and Replies

  • #2
berkeman
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Sorry, your post is very disjointed to me. Vdd is a power supply voltage that has little directly to do with the voltage gain ##\frac{V_O}{V_I}## of this circuit.

And the jumble of equations that you have posted to right of the circuit make little sense, especially since you have not defined terms like ##I_3## etc.

Try again?
 
  • #3
Sorry, your post is very disjointed to me. Vdd is a power supply voltage that has little directly to do with the voltage gain ##\frac{V_O}{V_I}## of this circuit.

And the jumble of equations that you have posted to right of the circuit make little sense, especially since you have not defined terms like ##I_3## etc.

Try again?
Sorry, those equations to the right are what the author already solved. I am trying to find out how he got the gain of ##g_{m1}R_D##. Vdd is in my equation because I thought I could use KVL to solve for Vout/Vin but from your feedback I can tell that I am wrong. How would you find the gain?
 
  • #4
berkeman
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What is the traditional voltage gain for a common-emitter or common source transistor amplifier? Why?
 
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  • #5
DaveE
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It looks like you are assuming that the drain currents are the same in M1 & M2. In that case you are correct, everything cancels and the output is 0 (not the gain!). Maybe consider KCL too? Also, notice that the gate voltages may be different.

Start over but next time be more careful about naming each current and voltage. Just because they may (or may not) have the same value, that doesn't mean that the currents, or voltages, are the same. Cancel terms after you have made the equations, not while you are constructing them.

edit: If I was working this problem, there wouldn't be Rd or Id, there would be Rd1, Rd2, Id1, Id2, Id3, Vg1, Vg2....
 
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  • #6
Joshy
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I've grown stale on this topic and had to pull out my book a few times after this. Great question.

I haven't read a copy of Razavi's book on this topic although I do hope to purchase it one day as I've thoroughly enjoyed a few chapters of RF Microelectronics. I have Microelectronic Circuits by Sedra and Smith. Are you familiar with a small-signal model? It looks very similar to the op-amp models where there is a dependent source inside of it; I also feel like when they talk about gain, that they are talking about the small-signal and so when you try using KVL things get a bit confusing because your sources look different in the model.

It's not to say that KVL is wrong, but they superimpose that later.

Anyhow: To approach Berkeman's question I would draw a common source amplifier and replace the MOSFET with its small signal model. What happens to the DC supply (##V_{DD}##)? Solve for its ##v_o/v_i## and you'll mysteriously stumble upon solutions Razavi is sharing on the right side. I think once you try that out for yourself that'll really help clear things up.

DaveE is totally right about the gates not being the same voltage: The two gates aren't the same voltage. You can see right in your picture one of them is ##V_{in}^+## and the other one is ##V_{in}^-##. I wouldn't get too twisted over noticing that right now, but just to clear up the air further about things possibly "canceling out" when it's not.
 
  • #7
I've grown stale on this topic and had to pull out my book a few times after this. Great question.

I haven't read a copy of Razavi's book on this topic although I do hope to purchase it one day as I've thoroughly enjoyed a few chapters of RF Microelectronics. I have Microelectronic Circuits by Sedra and Smith. Are you familiar with a small-signal model? It looks very similar to the op-amp models where there is a dependent source inside of it; I also feel like when they talk about gain, that they are talking about the small-signal and so when you try using KVL things get a bit confusing because your sources look different in the model.

It's not to say that KVL is wrong, but they superimpose that later.

Anyhow: To approach Berkeman's question I would draw a common source amplifier and replace the MOSFET with its small signal model. What happens to the DC supply (##V_{DD}##)? Solve for its ##v_o/v_i## and you'll mysteriously stumble upon solutions Razavi is sharing on the right side. I think once you try that out for yourself that'll really help clear things up.

DaveE is totally right about the gates not being the same voltage: The two gates aren't the same voltage. You can see right in your picture one of them is ##V_{in}^+## and the other one is ##V_{in}^-##. I wouldn't get too twisted over noticing that right now, but just to clear up the air further about things possibly "canceling out" when it's not.
I am familiar with the pi model for the mosfet but thought that would be overdoing it since the mosfets are equivalent and thus I would be able to cancel out terms using KVL. I see now that the gate voltages are different. Ill try it again.
 
  • #8
DaveE
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I am familiar with the pi model for the mosfet but thought that would be overdoing it since the mosfets are equivalent and thus I would be able to cancel out terms using KVL. I see now that the gate voltages are different. Ill try it again.
Yes, use the simplest models first. Then, IF NECESSARY, make your models more complex. I would start with a voltage controlled current source model first, since it's the simplest thing I can think of.
 
  • #9
I've grown stale on this topic and had to pull out my book a few times after this. Great question.

I haven't read a copy of Razavi's book on this topic although I do hope to purchase it one day as I've thoroughly enjoyed a few chapters of RF Microelectronics. I have Microelectronic Circuits by Sedra and Smith. Are you familiar with a small-signal model? It looks very similar to the op-amp models where there is a dependent source inside of it; I also feel like when they talk about gain, that they are talking about the small-signal and so when you try using KVL things get a bit confusing because your sources look different in the model.

It's not to say that KVL is wrong, but they superimpose that later.

Anyhow: To approach Berkeman's question I would draw a common source amplifier and replace the MOSFET with its small signal model. What happens to the DC supply (##V_{DD}##)? Solve for its ##v_o/v_i## and you'll mysteriously stumble upon solutions Razavi is sharing on the right side. I think once you try that out for yourself that'll really help clear things up.

DaveE is totally right about the gates not being the same voltage: The two gates aren't the same voltage. You can see right in your picture one of them is ##V_{in}^+## and the other one is ##V_{in}^-##. I wouldn't get too twisted over noticing that right now, but just to clear up the air further about things possibly "canceling out" when it's not.

Following your advice and using the small signal model I come up with the following:
$$V_{in} = V_{GS1} - V_{GS2}$$
$$V_{GS1}gm_1 = I_1$$
$$V_{GS2}gm_2 = I_2$$
$$I_1 = -I_2$$
$$gm_1V_{GS1} = -gm_2V{GS2}$$

Rearranging the second and third equation and subbing to one gives me:

$$V_{in} = \frac{I_1}{gm_1}- \frac{I_2}{gm_2}$$

and using equation 4:

$$V_{in} = \frac{I_1}{gm_1}- \frac{-I_1}{gm_2}$$

Can I assume gm1 = gm2 since they are the same op amp?
If so then I would get:

$$V_{in} = 2 \frac{I_1}{gm_1}$$

and using the fact that ##I_1 = I_{D1}## and ##I_{D1} = \frac{V_{DD} - V_{out}}{R_D1}## then:

$$V_{in} = \frac{2*(V_{DD} - V_{out})}{gm*R_{D1}}$$

And here I do not know where to go.
 
  • #10
Joshy
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I meant just one common source amplifier. Try that one out :)
 
  • #11
I meant just one common source amplifier. Try that one out :)
Oh ok I had forgotten that in small signal analysis I had to shut off all DC sources. Now I get the equations I need. If I was not doing small signal analysis then I could take the derivative of the Vout/Vin to get the gain?
 
  • #12
Joshy
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I can't say I know it or am comfortable with that information, but here's what I see in my book if this is what you meant by the derivative. I'm just copying it straight from Chapter 5 of Microelectronic Circuits, 6th Ed. by Sedra and Smith.

$$\left. A_v \equiv \frac{dv_{DS}}{dv_{GS}} \right|_{v_{GS}=V_{GS}}$$

Their notation was ##v_{GS}## is total signal and it's equal to ##V_{GS}## DC plus the small-signal ##v_{gs}##. This means that they set the results of that derivative to the biasing or operating point. Similar notation approach was used for ##v_{DS}##. That little extra bar on the right is saying that not only do they perform a derivative, but they set the input to equal the biasing or operating point.

I can hardly remember much from my class, but I feel confident we did not use this approach often. Small-signal model was definitely the way to go. I took an upper division version at my university and later took a non-degree masters module on it as well (admittedly it concentrated on layout for full-custom IC design) and we still didn't use this approach.

I don't feel comfortable with it so I can hardly question it, but my guess is that it's not the preferred approach for solving the problems. I probably could see someone who has a graph, simulation results or measurement they might throw it into a computational program like MATLAB to get the gain.

This topic isn't one of my strengths, but I will be taking another class on it this fall so thanks for the great questions
 
  • #13
LvW
819
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I think, the circuit cannot work because you have applied Vin BETWEEN both gates - without any ground difference. Both gates should get its own input signal (Vin1 resp. Vin2). Both can be equal or opposite or assume arbritrary values.
 
  • #14
LvW
819
220
Sorry...instead of "difference" please read "reference".
 

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