Need to design an op-amp to subtract AC voltages....

[MetalMat]

This is what I have, two AC voltages coming out of two op amps. I need some way to subtract the voltages. I have tried designing an op-amp to do this with what I found, but it keeps adding the two voltages. Also, do the voltages have to be "exactly" in phase? Thanks.

 

[PottedMeat]

can you invert one and add it to the other?
or if they have to be in phase, invert one, buffer the other and add?

 

[TheLonelyPhoenix]

I know there's a subtractor circuit for DC circuits but no clue if it works for AC the same way. Op amps confuse me. 🙂 Can go find it if you like.

 

[MetalMat]

Originally posted by: PottedMeat
can you invert one and add it to the other?
or if they have to be in phase, invert one, buffer the other and add?

Inverting did not help, already tried that 🙁

 

[MetalMat]

Check out page 6 on this pdf file, I tried designing one just like that, but it keeps adding the voltages together. I dont understand what the V^(-1) = V^(+) means at the top and the V^(+) means at the bottom (Is this just showing what the chip needs to be powered with?).

PDF file

 

[busterTI]

If I am not mistaken a comparitor will do this.

 

[TheLonelyPhoenix]

Originally posted by: MetalMat
I dont understand what the V^(-1) = V^(+) means at the top and the V^(+) means at the bottom (Is this just showing what the chip needs to be powered with?).

Yeah, those are just the input high/low voltages that don't usually get penciled in explicitly.

That circuit is the one I was taking about before, btw. No idea why it would add (unless you're a perfect 180 deg. out of phase at the same frequency?)

 

[PottedMeat]

Originally posted by: MetalMat
Check out page 6 on this pdf file, I tried designing one just like that, but it keeps adding the voltages together. I dont understand what the V^(-1) = V^(+) means at the top and the V^(+) means at the bottom (Is this just showing what the chip needs to be powered with?).

PDF file

------------------


it is saying that the opamp needs equal bipolar supplies ( i.e. V+=+12V, V-=-12V relative to a ground).


EDIT:

what the hell is going on? why are the supplies attached to the inv/noninv inputs?, i guess those should be on the opamp itself, but the circuit should work assuming you have bipolar supplies, and get rid of the dc offset with capacitive coupling on both inputs

 

[MetalMat]

Originally posted by: TheLonelyPhoenix

Originally posted by: MetalMat
I dont understand what the V^(-1) = V^(+) means at the top and the V^(+) means at the bottom (Is this just showing what the chip needs to be powered with?).

Yeah, those are just the input high/low voltages that don't usually get penciled in explicitly.

That circuit is the one I was taking about before, btw. No idea why it would add (unless you're a perfect 180 deg. out of phase at the same frequency?)

Nah, it seems to be around 30 degrees out of phase.

 

[TuxDave]

Originally posted by: PottedMeat

Originally posted by: MetalMat
Check out page 6 on this pdf file, I tried designing one just like that, but it keeps adding the voltages together. I dont understand what the V^(-1) = V^(+) means at the top and the V^(+) means at the bottom (Is this just showing what the chip needs to be powered with?).

PDF file

------------------


it is saying that the opamp needs equal bipolar supplies ( i.e. V+=+12V, V-=-12V relative to a ground).

Since the op-amp is connected with negative feedback, the gain of the opamp will force the positive and negative input terminals to be roughly the same. This does not imply that terminals v1 and v2 necessarily are the same. You do NOT attach the power supplies to those terminals. I repeat, DO NOT. The place to attach the supplies is not drawn in that schematic.

When you say the two AC signals are adding, is this through a real circuit or is this via spice?

 

[MetalMat]

Originally posted by: TuxDave

Originally posted by: PottedMeat

Originally posted by: MetalMat
Check out page 6 on this pdf file, I tried designing one just like that, but it keeps adding the voltages together. I dont understand what the V^(-1) = V^(+) means at the top and the V^(+) means at the bottom (Is this just showing what the chip needs to be powered with?).

PDF file

------------------


it is saying that the opamp needs equal bipolar supplies ( i.e. V+=+12V, V-=-12V relative to a ground).

Since the op-amp is connected with negative feedback, the gain of the opamp will force the positive and negative input terminals to be roughly the same. This does not imply that terminals v1 and v2 necessarily are the same. You do NOT attach the power supplies to those terminals. I repeat, DO NOT. The place to attach the supplies is not drawn in that schematic.

When you say the two AC signals are adding, is this through a real circuit or is this via spice?

Yeah, I fried enough LM741's to know not to do that 🙂 This is a real circuit though, part of my senior design that being a pain in the ass.

 

[TuxDave]

Originally posted by: MetalMat
Yeah, I fried enough LM741's to know not to do that 🙂 This is a real circuit though, part of my senior design that being a pain in the ass.

I believe this is a case of "SPICE is being a bitch" I'm pretty sure that it's working properly and spice is just having issues with subtracting AC signals. If you change the AC signals into some transient signals, are you getting what you want?