# Symmetrical DC Offset - See Dual Op Amp Ckt.

• posted

The amplifier circuit linked below produces two identical outputs, one each above and below the DC line.

Hover over the right bottom corner and click the icon to enlarge.

What I would like to do is create a greater symmetrical DC offset between the two. Something like 3 volts. IOW the low side of channel A would rest at 1.5V above DC, and the high side of channel B would rest at 1.5V below.

Can anyone tell me how to modify the above circuit to do this, without reducing the input signal level?

Also, the spec for this amp is 10ma output. What would be a good substitute part with a rating of 50ma?

Richard Clarke

• posted

The 10k strings R5/7, R1/6 generate the offset. Reduce R1/R5 to decrease the offset.

It's been a while....I'll await the definitive answer.....

• posted

Actually, I wanted to increase the offset not decrease it. So I took your cue and changed R6, R7 to 8.2K and raised the input signal to 2V. That works well.

But I was fishing for something less convnetional. To be somewhat independent of changing parts values and input signal levels, I was wondering about applying an offset voltage to one of the signals via a variable DC supply. That could shift its DC offset up or down as desired.

Does anyone know how this can be done?

Richard Clarke

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there are not many real reasons for doing it this way, what is the REAL reason ?

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I mean there are too many easier ways to do this.

• posted

You can't use a dual op amp running off two different supplies like you have shown there. There's no reason why both amps cannot be connected between the +/-6V supply, then you can use a dual op amp.

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There are some composite amplifier schemes that use this kind of voltage windowing. Two immediately come to mind. One is a scheme where you try to operate a low voltage circuit over a wide voltage range. For instance, if you need a high voltage buffer, you could create a crude window with high voltage components around where the output voltage will be with enough headroom to run a precision buffer circuit comprised of low voltage components. The other is a combination of class D and linear amplification, where you have a class D generated window around which you operate your linear circuit (A or AB).

• posted

Sure, If you can add another opamp then take the 'offset voltage' and invert it, and then send it and the inverse to your two non-inverting inputs.

Without another opamp you could make two differential amps. one does (offset - Vin) the other (Vin - offset)... Oops that flips the phase of Vin... I guess you've got to add another opamp. (they're cheap enough.)

George H.

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Thanks Jim. I will build this circuit up over the weekend. Definitely looks better than mine.

Roger Clarke

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I will try these variations on my original circuit as a learnng experience.

Roger Clarke

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It might just be me, but I don't like using power supply rails as voltage references. However, the circuit I'm thinking of would be more complicated. Probably buffer the signal source then pull current through resistors to get the offset, then buffer those signals.

You are in a world full of phase shift with those 10uF caps. This really should be a DC circuit. I don't like caps in the AC path if I can avoid them.

• posted

Do you want the input to be a variable resistor, or a programming voltage, or a fast-varying signal? The basic plan, as I see it, is to take an input signal X, and produce amplified signals aX +b and aX - b. You can replace the R5/R7 with an input terminal driven with 'b', and use an inverting op amp to produce '-b' to replace R1/R6.

If the DC input level is known, and the op amps have suitable compliance range, the coupling capacitors are unnecessary and failure-prone.

• posted

The way the original circuit is done, you need the caps. Think of the op amps being in the inverting gain configuration. The DC value on the plus pin is setting the output voltage. But if the signal source is at ground, there will be DC current flow in the input resistor.

Just look at the top op amp. He splits the rail, so there is 3V at the positive input. Now ground the free side of R2. You have current of

3V/10k=300ua. But you are pulling that through r3, causing it to rise to 22K*300ua=6.6V With the op amp being a virtual short, that puts the output at 9.6V.

Granted I haven't run spice on this, nor have a written anything on the back of an envelope. Just eyeballing the circuit.

I still think the caps are a bad idea, but I would use a substantially more complicated circuit since I don't like using the power supply rails as a reference. While supplies should be regulated, circuitry should also have power supply rejection.

• posted

Oh, I was thinking of the four-resistor difference amplifier configuration for each of the two op amps. The addition of a third op amp (to produce symmetrical positive and negative 'b' signals) was intended to keep the offset voltages at a different junction from the input signals, for power sequencing reasons. aX +b and aX-b amplification is just NOT that hard to arrange.

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