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# ADA4700 - DAC-controlled dual polarity voltage source

I've selected the ADA4700-1 in an application to drive a static charge upon a stainless steel plate (conductivity measuring instrument).  For this purpose I have two supply rails, one at +38V and the other at -38V.   The ADA4700-1 will be configured as an inverting gain stage.  There is virtually no current load on the output.

I'd like to drive the ADA4700-1 from a microprocessor's DAC output (0 to 3.3v) such that the ADA4700-1 presents +38V when the DAC drive is 0 (or +3.3) and -38V when the DAC drive goes the other way.

To make this work, I think we need to swing the DAC drive above and below 0V.  Is there any way to avoid this conversion through biasing the ADA4700-1?

Suggestions on the best way to achieve our desired result most gratefully appreciated.

## Top Replies

• Hi.

Apologies for late response.

I have done something very similar to your application in the past. So here it is.

You will be needing a gain of 23.0303 (76/3.3) to get the output you need. R and C are optional for low pass RC filter. The output transfer function will be:

OUT = IN ( 1 + RF/RG ) - 1.725V ( RF/RG).

For RG you can have a parallel combination of 29.8K//30.5K or 15.073K. For RF use 332K.

So,

@ IN = 0 V (ZS)

OUT = 0 - 1.725 (332/15.073) = - 37.9951V

@ IN = 1.65 V (MS)

OUT = 1.65 ( 1 + 332/15.073) - 37.9951 = -0.00196 V

@ IN = 3.3 V (FS)

OUT = 3.3 (1 + 332/15.073) - 38.9951 = 37.9917 V

The accuracy of conversion (gain up) will depend on how tight/close the values of the resistors with respect to the computation.

Please be careful with the supplies. ADA4700-1 is not true rail-to-rail, it has headroom in each rail. For safety please power it with 3V allowance for each output.

Regards.

• Pretty nifty!

My design is a bit more complex.  Unfortunately we don't have much headroom... +-38V is it.  However, with this circuit, we achieve very close to +-37V with a buffered DAC drive.

• Hi.

I see. If that's your intended schematic, I would like to suggest this one.

This will be your single stage solution. In which the use of Buffer and a Level Shifter are unnecessary, so you will save two op-amps.

This circuit will follow a circuit transfer function of:

OUT = IN (1 + RF/RTH) - VTH (RF/RTH)

To compute for the values, see below.

>>For  RF/RTH

Gain(non-inverting) = (36 - -36)/3.3 = 72/3.3 = 21.81818182

Gain(inverting) = Gain (non-inverting) - 1 = 20.81818182 = RF/RTH

>>For VTH ;

@ IN = 0 , OUT = -36 V

-36 V = 0 - VTH (RF/RTH)

VTH = 1.729257642 V

>>For R1, R2 and RF;

VTH = 3.3 V (R2 / (R1 + R2))

1.729257642 V = 3.3 V (R2 / (R1 + R2))

R2 / (R1 + R2) = 120/229

So: R2 = 12.0 kΩ and R1 = 10.9 kΩ.

RTH = R1//R2 = 5.711790393 kΩ.

R= 20.81818182 * RTH = 108.909 kΩ

Now:

OUT = IN (21.81818182) - 36 V

@ IN = 0V ; OUT = -36 V

@ IN = 1.65 V ; OUT = 0 V

@ IN = 3.3 V ; OUT = 36 V