||question: what is the simplest attenuverter/offset
| br>i have a brief question to the experts (I know I should try on a breadboard and find out myself, but I thought I ask here first ... and after all not all problem/drawbacks of a design are apparent even when trying it).
I was wondering what the simplest way to do an attenuverter with offset. I started with a differential opamp, using the same voltage on both inputs. In this way, I get an attenuverter with a single OpAmp. In fact, this is exactly the circuit fonitronic uses on the Attenuverter chicklet. It works and is a very useful design.
If I also need an offset, however, is it possible to use the negative input of the OpAmp as in a conventional (inverting) mixer? Does the negative input act here as a virtual ground, such that there is no bleed from the offset to the input? Are there any (obvious and not-so-obvious) flaws in the schematic?
Here is what it looks like:
without the offset, it is a differential amplifier (with unit gain) that acts as an attenuverter (identical to the fonitronic chicklet). The offset is supposed to add/substract some voltage.
And, yes, I can/will use a dual OpAmp but i need the other half for something else. Comments are greatly appreciated! br> br>
| br>The negative input is only a true virtual ground when the positive input is at ground potential. Otherwise the negative input will track a fraction of the input signal set by the VR1 potentiometer so the current in R4 will change with the input signal and will effectively change the gain. You can still trim the gain to 0 but it will be at 33% of the potentiometer setting instead of 50%. By adding a 50kohm resistor between the input and the top of VR1 you can move 0 gain back to the center of the VR1 potentiometer.
This still assumes the total resistance of the offset circuit is 100kohm, but actually is R4 in series with the two portions of the offset potentiometer in parallel so it can be from 100kohm at the ends to as high as 125kohm at the center so the position of the VR1 potentiometer to get 0 gain will vary with the offset setting. To reduce the problem the offset potentiometer value can be reduced at the expense of higher current consumption or eliminated by buffering the signal from the potentiometer. Also, if you mostly use offset voltages near 0V, you can assume the resistance of the potentiometer is around 25kohm so you can reduce R4 to 75kohm so the total resistance of the offset circuit is the correct 100kohm most of the time. br> br>
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