# MOSFET differential amplifier gain

1. Jun 16, 2014

### etf

Hi!
For differential amplifier on scheme calculate voltage gains Av1=vout1/(vin1-vin2) and Av2=vout2/(vin1-vin2). MOSFET's M1 and M2 have same characteristics. MOFSTE's M3 and M4 have same characteristics. All MOSFET's are in saturation. Assume that gm*rds>>1.

How to use fact that gm*rds>>1?
I did AC Analysis of this amplifier and found vout1 in function of vin1 and vin2 and vout2 in function of vin1 and vin2. For the sake of simplicity, I used variables a,b,... . I got these results:

$vout1=\frac{bgn-cfn+dfm-dgl}{afm-agl-bem+bgk+cel-cfk}\, \, \,\, vin1+\frac{chl-bhm+bgo-cfo}{afm-agl-bem+bgk+cel-cfk}\, \, \, \, vin2$
$vout2=\frac{cen-agn-dem+dgk}{afm-agl-bem+bgk+cel-cfk}\, \, \, \, vin1+\frac{ahm-chk-ago+ceo}{afm-agl-bem+bgk+cel-cfk}\, \, \, \, vin2$

My question is, how to get rid of vin1 and vin2 in Av1 and Av2? V1 and v2 shouldn't figure in gains Av1 and Av2.

Last edited: Jun 16, 2014
2. Jun 16, 2014

### rude man

I don't think there is enough info to determine the gains.

But if you feel you know those four fractions which I will call F1, F2, F3 and F4, then
Av1 = Vout1/(Vin1 - Vin2)
1/Av1 = Vin1/Vout1 - Vin2/Vout2
and Vin1/Vout1 = 1/F1 etc.

It should be apparent that F1, F2, F3 and F4 all have the same magnitude.

3. Jun 17, 2014

### The Electrician

You could replace those complicated expressions you have with single variables x and y, and make use of the extreme symmetry of the circuit. You get this:

$vout1=x*vin1 - y*vin2$
$vout2=-y*vin1+x*vin2$

Also make use of the fact that for purely differential input, Vin1 = -Vin2

I think that then you can find a solution.

4. Jun 18, 2014

### The Electrician

Why do you have about 14 single letter variables for this problem? Have you assigned a variable to each branch current? If that's what you've done, I would recommend using some other approach. Your circuit only has 6 nodes; wouldn't it be simpler to use nodal analysis rather than branch current analysis?

5. Jun 18, 2014

### etf

I used nodal analysis. I will post complete work very soon.

6. Jun 18, 2014

### etf

7. Jun 18, 2014

### The Electrician

So is your final answer what you have shown in the last page, expressions involving a11, a12, a21 and a22?

I would have thought you would need expressions in terms of the MOSFET parameters, such as gm1, gm2, gm3, rds1, rds2, rds3, etc.

Your problem stated that gm*rds>>1, and you asked how to use that fact. Wouldn't you expect that expressions involving gm*rds would appear in your algebra, and that's where you could make use of the strong inequality? But I don't see any expressions like in your work. I wonder why?

I find it hard to follow your work. For example, on page 2 you have a subexpression you've labeled q. It looks like the subexpression is:

(||rds1 + ||(2 R) + gm3/2 + ||rds3) "phi"d1

What are those 2 vertical bars in front of rds1 (why are they there?), and what is the character I've labeled "phi"? Are the vertical bars || really 1/ (1 divided by)?

Also on page 2 is what looks like a signal flow graph. Is it supposed to be a signal flow graph, or is it just a representation of the circuit topology to be used as an aid in setting up the nodal equations?

Between nodes d1 and s1 I see a branch with an arrow in series with rds1; beside the arrow I see what looks like qds1. What is qds1?

8. Jun 18, 2014

### etf

I have bad handwriting :) I used "phi" to represent node potentials...There are no vertical bars, "(||rds1 + ||(2 R) + gm3/2 + ||rds3) "phi"d1" is (1/rds1 + 1/(2 R) + gm3/2 + 1/rds3) "phi"d1 :) I will rewrite solution and upload it again :)

9. Jun 18, 2014

### The Electrician

What about my question regarding the form of your final solution? Shouldn't your answer involve things like gm1, rds1 and not just a11, a12, etc.

10. Jun 18, 2014

### etf

Firstly, we draw original circuit with shorted Voltage sources (VB, VSS, VDD):

Then we draw 5 small-signal MOSFET models and connect them according to first circuit:

Here is result of "simplification" of second circuit:

Now we form system of 3 equations with 3 uknowns:

Here is solution (I didn't write potential for node S1 since it isn't required):

Now we can find gains:

11. Jun 18, 2014

### etf

Now I should replace variables a11, a12, a21 and a22 with firstly introduced a,b,c,... and then replace them with gm1, rds1,... and use fact that gmrds>>1 ? :)

Last edited: Jun 18, 2014
12. Jun 18, 2014

### The Electrician

I'm off to lunch with some of my former EE professors at the local University.

I'll have a look at your work when I get back.

In the meantime, answer me this. Do you have access to Matlab, Mathcad, or some similar modern program to do the tedious algebra involved in this problem?

If you don't have any of those, consider downloading the free package Maxima and learning how to use it:

http://maxima.sourceforge.net/

13. Jun 18, 2014

### etf

Ok :)
I use Matlab. It would be very exhausting to do all these calculations by hand :)

14. Jun 18, 2014

### etf

Ok :)
I use Matlab. It would be very exhausting to do all these calculations by hand :)

15. Jun 18, 2014

### The Electrician

Don't bother replacing your subexpressions with a, b, c, ...

Just plug the complicated expressions into the Matlab solver and let it do the work. Look at your results and see if you see any thing like this: (1+gm*rds). If you do, then you can replace that with (gm*rds). That may help simplify your results.

16. Jun 19, 2014

### The Electrician

By the way, do you have the correct answer for this problem, from the text or from your instructor?

17. Jun 19, 2014

### etf

I don't have solution for this problem in my book...
Below is code and final result from Matlab. I did DC analysis and got gm1=gm2, gm3=gm4, rds1=rds2, rds3=rds4.
>> syms gm1 gm2 gm3 gm4 gm5 rds1 rds2 rds3 rds4 rds5 R
>> rds1=rds2;
>> rds3=rds4;
>> gm1=gm2;
>> gm3=gm4;
>> a=1/rds1+1/(2*R)+gm3/2+1/rds3;
>> b=-(1/(2*R)-gm3/2);
>> c=-(gm1+1/rds1);
>> d=-gm1;
>> e=-(1/(2*R)-gm4/2);
>> f=1/rds2+1/(2*R)+gm4/2+1/rds4;
>> g=-(1/rds2+gm2);
>> h=-gm2;
>> k=-1/rds1;
>> l=-1/rds2;
>> m=gm1+1/rds1+gm2+1/rds2+1/rds5;
>> n=gm1;
>> o=gm2;
>> a11=(b*g*n-c*f*n+d*f*m-d*g*l)/(a*f*m-a*g*l-b*e*m+b*g*k+c*e*l-c*f*k);
>> a12=(c*h*l-b*h*m+b*g*o-c*f*o)/(a*f*m-a*g*l-b*e*m+b*g*k+c*e*l-c*f*k);
>> a21=(c*e*n-a*g*n-d*e*m+d*g*k)/(a*f*m-a*g*l-b*e*m+b*g*k+c*e*l-c*f*k);
>> a22=(a*h*m-c*h*k-a*g*o+c*e*o)/(a*f*m-a*g*l-b*e*m+b*g*k+c*e*l-c*f*k);
>> Av1=simplify((1/2)*(a11-a12))

Av1 =

-(R*gm2*rds2*rds4)/(2*(R*rds2 + R*rds4 + rds2*rds4))

>> Av2=simplify((1/2)*(a21-a22))

Av2 =

(R*gm2*rds2*rds4)/(2*(R*rds2 + R*rds4 + rds2*rds4))

Last edited: Jun 19, 2014
18. Jun 19, 2014

### The Electrician

That's the result I get:

You have made your solution more cumbersome than it needs to be by introducing all those single variables a, b, c, ... m, n, o.

As you can see, the most compact way to do it is with the matrix form of your equations. I have more equations than you do because I set up the general problem. I provided equations for the g1, g2, and g3g4 nodes, and I allowed for the possibility of a gate resistor Rg, which turns out not matter in the end. Also we see that rds5 doesn't matter because of the symmetry.

You can see 2 matrices; that's because I kept the matrix for M1 and M2 separate from the matrix for M3 and M4. They could be combined into one matrix; in fact as you can see, I added them, but didn't bother displaying the single matrix which is the sum of the two.

You can use your same technique to solve the CMRR problem thread. You will need to label the MOSFETs corresponding to M1, M2 and M3, M4 of this problem. You will need more equations (at least one more), but the equations you used in this problem will apply in the CMRR problem, possibly with different designators.

The reason I asked you in the CMRR problem if you need to keep the various parameters for each MOSFET separate, is because if you do, the solutions become very complicated. For example, in this problem, if you keep gm1 and gm2 separate, and rds1 and rds2 separate, the answers become (now we see that rds5 appears, because the symmetry is lost):

If you let all of the MOSFETs have only a single gm and rds, the result becomes very simple.

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19. Jun 19, 2014

### The Electrician

The results shown in red above were obtained in another Matlab session. If you would just let Matlab do the work of solving with your original equations, you should get the same answer.

What I mean is this. You set up some equations for Matlab to solve, using variables a, b, c...m, n, o. Just leave the original subexpressions in place. Everywhere you have the variable a, just put 1/rds1+1/(2*R)+gm3/2+1/rds3 instead. Do the same for all the other single variables. Put the subexpressions for each in the equations Matlab is solving.

If you were doing this by hand, it would be very tedious, but Matlab can do it without mistakes.

20. Jun 19, 2014

### etf

Thanks a lot! One more question for thread about CMRR: Can I analyse diff. amp. with Iss source from b), calculate CMRR for that circuit and use that CMRR to calculate CMRR for diff. amp. with Iss source from a) (i.e. seting for M2 gm2=0, rds2=∞) ? More precisely, for CMRR of diff. amp. with ISS from b) I will got CMRR in function of gm1, rds1, gm2, rds2 and gm's and rds's of other MOSFETS so I will just put there gm2=0 and rds2=∞ and that would be my CMRR for diff. amp. with source Iss from a)?

Last edited: Jun 19, 2014