(V1 + V2) - (V3 + V4) from one op amp ?

Schoolboy problem ?:-)

It's trivial. Think on it! Hint: Superposition ;-)

...Jim Thompson

You don't need a second op-amp.

Use R1 between V1 and non-inverting input R1 between V2 and non-inverting input R1 between V3 and inverting input R1 between V4 and inverting input R2 between non-inverting input and ground R2 between inverting input and output

and Vo=(R2/R1)*[(V1+V2)-(V3+V4)]

Make R1=R2, if you need gain 1.

Best,

Aw-w-w-wh! You told! The student didn't have to think. Another "engineer" on his way to graduation and a job where he copies other people's work without knowing that it's about to fail :-(

...Jim Thompson

It's important that we have students that aren't capable of thinking. Otherwise, we'd run out of engineering managers.

Bob

This is first-chapter, first year opamp stuff and probably should have been posted to sci.electronics.basics. Is this homework?

The final function you want is (-). Opamps do that, inherently. So place an opamp on the paper to start. You have your (-) behavior there already. The opamp also sets the two inputs to be at the same voltage.

Assume for now that you've got both input pins at the same voltage (opamp 'wants' that.) This makes summing easy. Sum currents from the V1 and V2 into one of the nodes and currents from V3 and V4 into the other node. Since both nodes are assumed to be at the same voltage for now, and held there by some (assumed) opamp behavior, this is easy. Just use the same value of R between each V-source and either the (+) or (-) node. So four resistors, all the same value, for now. Two from V1 and V2 to one node. Two from V3 and V4 to the other node. Their contribution current to the nodes will depend on their voltages. But the currents will be summed, as they must be.

Which node for which pair of voltages, though? Well, V1 and V2 are on the positive (+) part of your equation so their resistors should connect to the (+) terminal of the opamp. V3 and V4 are negated, so connect their resistors to the (-) terminal.

Now you still have a problem. We've assumed the opamp somehow keeps the input nodes at the same voltage. But the output has no way to feed backwards and do that. We've another problem, whose solution solves both problems. This other problem is the accumulation of current at the input nodes. There needs to be some way to flush out that accumulation so that charge doesn't build up. Feedback needs to counter input tendencies here, so we'd guess that the output needs to somehow be connected to the (-) input via a resistor. So put one there. This allows a pathway for the current to flow out of the (-) input node, because the output can either sink or source current (which ever is needed.) So that seems to be a good idea. This still leaves the (+) node, though. How to get current to leave it? Well, just sink it to ground with another resistor.

The summed current in the (+) node now must all go through the resistor to ground and that creates a 'voltage' at that input node. To compensate and set both inputs to the same voltage, the opamp with adjust its output terminal to cause the same voltage to appear at the (-) node, drawing current out through the feedback R connecting the opamp output to the (-) node. All currents have a place to go, now, and the opamp has a way of ensuring that the voltages are the same at both input nodes. All problems solved.

Assume all R values are the same. What's the output going to do? Both nodes are at the same V, call it Vx. V1 through its R will create a current equal to (V1-Vx)/R. V2's current will be (V2-Vx)/R. V3's current will be (V3-Vx)/R, and V4's current will be (V4-Vx)/R. Currents from V1 and V2 will sum to the (+) node, so that incoming current is (V1+V2-2*Vx)/R. All of that must go through the R to ground. So the voltage at the (+) node will be R times that current or just V1+V2-2*Vx. Therefore Vx=(V1+V2)/3. That value is set independently and only depends upon V1 and V2 and the resistors used on the (+) node. The other node is going to have to match it.

So meanwhile, current entering the (-) node is (V3+V4-2*Vx)/R and we already know the voltage there must be the same as for the other node, or (V1+V2)/3. The current must leave by way of the feedback resistor going from the (-) input to the opamp output, so the voltage across that resistor is [-(V3+V4-2*Vx)/R]*R or -V3-V4+2*Vx. That adds to the (+) node's voltage, Vx, to get the final Vout of the opamp:

Vout = (-V3-V4+2*Vx) + Vx = -V3 - V4 + 3*Vx

But Vx = (V1+V2)/3, so,

Vout = -V3 - V4 + 3*(V1+V2)/3 = V1 + V2 - V3 - V4 The desired output with one opamp and the idea is scalable.

Jon

Bad! Bad! You almost ruined a wonderful Italian meal/wine ;-)

...Jim Thompson

be

I've never liked circuits that summed on the non-inverting input. (I'm not sure why?)

George H.

There are more than enough failed politicians to take up the slack.

V3

s be

The signal from one input can effect the other input.

That's why you use the dual- or quad-pack op-amps -- buffer whatever signals don't already look like a reasonably low impedance source (e.g., a voltage source). :-)

Offsets, input bias currents, GBP, etc, etc. That's why you usualy see at least three in an instrumentation amp.

Cheers! Rich

My bad : I had tried the one op amp design (of course) but had a bad powr supply that made it seem to not be working. Thanks to those who provided a suggestion that agreed with my attempt and thus prompted me to check it.

als

If you doing a "mad man Muntz" design, you just reduce the resistor values to compensate for the nonzero impedances. If you can, the highest impedance input will be scaled to make that resistor come out to zero so you can have one less part in the design.

=20

(V3

be

Your cynicism does not become you.