Op Amp circuit analysis

voltage

Ok, sounds like fun...

Rin= (r3+r4)(r1*(A+1)+r2) ---------------------- (r3+r4)(A+1) - m*A*r4

Haven't checked it, but seems right. Like your use of 'm', should cover both common-mode and differential-mode input impedances nicely, as well as more general case!

-f

DC gain of Ao, single pole roll-off crossing axis at GBW... so behavior will depend on closed loop gain selection ;-)

"Floating" between the two inputs.

...Jim Thompson

You didn't answer the immediately previous question. Is there more to it than just a standalone opamp? If so, what is the rest of the circuitry? Just above you say "behavior will depend on closed loop gain selection", so what is the rest of the circuitry?

voltage

I'm impressed!

How would you check it?

I'm not going to say whether I think it's right just yet in case anybody else plans to give it a try.

You've done so well, you, and anyone else who wants to participate, deserve some more practice exercises. :-)

Go back to the circuit without the extra buffer:

R1 R2 1 ___ 3 ___ Vin- o----o-|___|-o---o------o-|___|---. | | | | | |\\ | '---|-\\ | |A >---------o--- Vout .---|+/ | |/ | | ___ | ___ Vin+ o----o-|___|-o---o------o-|___|---o 2 4 | R3 R4 o

Assume A has a finite value for all that follows.

1. Derive a general expression for the differential impedance that a floating source, such as the floating transformer secondary, would see between Vin- and Vin+.

1. Connect Vin- and Vin+ together and derive a general expression for the driving point impedance seen at the point of common connection. I think this would be called the common mode impedance.

2. With Vin- and Vin+ still connected, derive a general expression for the differential impedance between nodes 3 and 4 in the schematic above (the plus and minus inputs, in other words).

R1, R2, R3, R4 as in the _almost_ original version.

...Jim Thompson

It seems to me that there is more than one _almost_ original version. But, I'll choose one. Assuming this circuit:

R1 R2 1 ___ 3 ___ Vin- o----o-|___|-o---o------o-|___|---. | | | | | |\\ | '---|-\\ | |A >---------o--- Vout .---|+/ | |/ | | ___ | ___ Vin+ o----o-|___|-o---o------o-|___|---o 2 4 | R3 R4 o

Let R1, R2, R3 and R4 be equal (the gain of one circuit Eero-Pekka referred to as his second case). I'll let them all be 100000 ohms.

The opamp has the open loop characteristics you gave, "GBW=20MHz and Ao=100K".

Then the differential impedance between Vin- and Vin+ is shown in plots posted over on ABSE with the subject "Opamp differential impedance".

If this isn't what you wanted, then describe in particular detail what you do want.

above

Single source (-) Terminal to Node 1, (+) Terminal to Node 2, otherwise floating like your (IIRC) comment about removing the center-tap of a transformer drive.

NO simulations, hand-derived expressions only ;-)

...Jim Thompson

I'll

The plots didn't come from a simulation. They're plots of this expression:

A*(R1+R3)+R1+R2+R3+R4 Zdiff = --------------------- A+1

wa where A = A0 * ------ , Ao=100000 and wa = GBW/Ao s + wa

I'll

Almost...

GBW was expressed in _Hz_, so wa = 2*pi*GBW/Ao ;-)

I'd be interested in knowing if anyone, other than us old farts, can do that calculation... by hand??

...Jim Thompson

I noticed this boo-boo last night when I was reviewing my post. The Zdiff expression is correct, I just slipped up on the 2*PI.

Didn't you notice the post by Frank Miles? He thinks it's fun. :-)

He must be an old fart, too!

I missed his post. Is he posting from Googlegroups? Post the message ID and I'll whitelist him.

...Jim Thompson

It looks like:

snipped-for-privacy@homer23.u.washington.edu

voltage

It IS fun ;-)

...Jim Thompson

Repeating myself: I'd be interested in knowing if anyone, other than us old farts, can calculate the general input impedance... by hand?

I don't think so. I believe that's why there are so few electronic discussions here... truly incompetent college grads :-(

...Jim Thompson

.... guilty.

-f

Glad to meet you!

...Jim Thompson