Source: http://ecircuitcenter.com/Circuits/adcin_diff_se/adcin_diff_se1.htm
Timestamp: 2019-04-19 17:05:32+00:00

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Your data acquisition board has an ADC that accepts a single-ended input of 0 - 5 V. However, your signal source happens to be differential. This signal may be generated from a sensor bridge where one leg swings positive while the other swings negative in response to pressure, for example. Or, suppose noise corrupts both a remote signal and its local ground wire equally. Rejecting this common noise would clean up your desired signal.
The Differential to Single-Ended ADC Driver would solve your design challenges. The op amp circuit shown above is remarkably similar to the Single-Ended to Single-Ended ADC Driver circuit covered previously. However, in this case we drive both inputs with the differential input. But, beware! the tolerance of the resistors you choose can make or break its performance!
In a nutshell, this circuit takes the difference of the inputs, applies a scaling factor of R2/R1 and adds an offset VREF.
A single-ended source generates a voltage between one node Vsig and a fixed reference, like ground.
A differential source generates a voltage between two output nodes, Vsig+ and Vsig-, neither one at ground potential!
Vdiff = 2-(-2) = +4 V.
Vdiff = -2-(+2) = -4 V.
The average of the differential voltage is defined by the common-mode voltage represented by VCM.
get R2 = R4 = 6.25 kΩ.
Given the scaling alone, you expect your sensor signal to get attenuated from ± 4 V to ± 2.5 V. Obviously, you're not done yet. Add an offset, VREF = 2.5 V, to force the signal to swing 0 - 5V.
CIRCUIT INSIGHT Simulate the SPICE circuit ADCIN_DIFF_SE1.CIR. Start with these initial circuit values: R1 = R2 =R3 = R4 = 10k and VREF = 0 V. First check the basics. Plot inputs V(11), V(12) and the differential voltage by adding trace V(11)-V(12). Do you see ± 4 V differential? Now remove V(11)-V(12) and add the output and output V(4). Does the circuit find the difference between V(11) and V(12)? You should see the ± 4V at V(4) with no scaling or offset.
Now set R2 = R4 = 6.25 kΩ. Rerun the SPICE file. Did the ±4 V signal get scaled down to ±2.5 V? Finally, set VREF = 2.5 V and rerun the simulation. Does the circuit properly transform the differential input to a 0 - 5 V single-ended signal?
Okay, let's throw in some real world troubles (or challenges if you prefer). There's some 10 kHz noise getting on both the VPOS and VNEG lines. Also, the signal rides on top of a 2 V offset. You can simulate these effects by modifying the VCM statement.
CIRCUIT INSIGHT Rerun the SPICE file. Does the noise and offset at V(11) and V(12) get rejected by the circuit? Awesome, Cool, Wicked!
But, the rejection of noise and offset depends on perfectly matched resistor ratios: R2 / R1 = R4 / R3. Try changing R1 by 5%, say from 10k to 10.5 kΩ. Is the output signal still noise free? To get a better look, plot V(4) in a new plot window. Also, reduce VPOS and VNEG to 0.2 V and rerun the simulation. Do you see any noise slipping past the goalie?
Try reducing the tolerance of R1 to 1 %, or 10.1 kΩ. Did the noise get any smaller? Try replacing R1 with a 0.1% value or 10.01 kΩ. How much of the noise remains? As you can see, noise rejection is a direct function of how well the ratios are matched.
Suppose you've got a new sensor that outputs ±10 V differential.
HANDS-ON DESIGN Start with these initial values: R1 = R2 =R3 = R4 = 10k and VREF = 0 V. What circuit values do you choose to transform the input into to a 0 - 5 V single-ended signal?
Here's around-up of the ADC driver topics to explore.
The last two functions require a couple of op amps. However, take a look at a single device - the Fully Differential Amplifier - that performs the functions.
For a more detailed description of the op amp, see the Basic Op Amp Model.
Browse other circuits available from the Circuit Collection page.

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