Interfacing a unipolar 0-5 V output with a bipolar +/- 0.1 V DAQ system using a differential amplifier and voltage divider

Question

I need to connect a sensor system that outputs 0–5 V_DC to a data-acquisition system (DAQ) that accepts inputs of ±0.1 V_DC (it is designed for a different type of sensor, but I need to repurpose it in this application). (The signals are primarily DC, but there may be AC components. The ranges are the same.) This scenario presents two obvious problems:

1. The 5 V_PP output range of the sensor system needs to be attenuated significantly in order to be compatible with the 0.2 V_PP input range of the DAQ.

   Naively, I figure this could be achieved with a simple voltage divider, performing a 50-times reduction.

2. The sensor system has a unipolar voltage output, whereas the DAQ is designed to work with bipolar voltage inputs.

   This could be simply ignored, if I were willing to sacrifice the loss of half the DAQ's input range (and thus resolution), but given how much the input signal must be attenuated, I think that is not a reasonable approach.

   Therefore, I came up with the idea to use a differential amplifier as a simple way to convert the 0–5 V output of the sensor system to a ±5 V output.

The complete design for the "conversion" circuit is the following:

^{simulate this circuit – Schematic created using CircuitLab}

Note that the sensor system (which generates the input) is modeled here as a waveform generator (far left). It's not actually going to be generating a constant output (in fact, precisely the opposite, or it'd be a terrible sensor), but this is sufficient for simulating the circuit.

According to the CircuitLab simulation, it works, achieving exactly what I want. But I am not an electrical engineer by training, so I'm looking for some expert guidance/review. Is this a reasonable design? Are there any hidden flaws or "gotchas" that I am overlooking?

Is the two-stage design—starting with the differential amplifier to do the unipolar-to-bipolar conversion, followed by the voltage divider to attenuate the bipolar signal to the desired range—correct? Is there a better way of doing this?

Precise and repeatable measurements are important here, to the extent it is reasonably possible. (Obviously, since this is all designed to allow sampling from sensors, the idea is to be able to quantify real-world phenomena as precisely and accurately as possible.) If there's a relatively simple design that would be more precise than this one while achieving equivalent results, then I would be interested in hearing about it. I say "relatively simple" because this is a retrofit/hack. If I have to design a simple PCB that will be inserted as an adapter/converter, then so be it, but both systems (the sensor system and the DAQ) are already existing and can't be easily modified.

Answer 1

Precision resistors required, this circuit uses standard values.

Precision op amp required.

Op amp doesn't need to be rail to rail input because voltage swing at op amp's inputs is quite small.

In an amplifier the offset at the output due to the input offset voltage is equal to the input offset voltage multiplied by the non-inverting gain = Vio( 1+(R5/R2)). In this design R5/R2 is quite small (0.04) and so the offset at the output due to the input offset voltage can only be a maximum of slightly larger than the input offset voltage itself. This is a good thing.

If the op amps input bias currents were equal to each other (which they never are) then the offset at the output resulting from the input bias currents would be almost zero because the impedance seen looking out from each op amp's input is the same = R5//R2 = R3//R4. In reality the impedance seen by the op amp's inputs will not be quite equal to each other because of the output resistance of the sensor adding to the effective resistance of R4, so there will be a small output offset due to that impedance mismatch. In fact the output resistance of the sensor also induces another source of error by forming a potential divider with the combination of R3 and R4, and so there will be a small voltage drop in the output voltage of the sensor when it is connected up to R4, the input of the circuit. With any luck the output resistance of your sensor will be quite small in value compared to R3+R4, minimising these errors.

If the output resistance of the sensor is significant compared to the value of R3+R4 (78k) then you could put a unity gain buffer in between the sensor and R4 to remove the resulting voltage drop. Another option, to remove the error, would be to reduce the value of R4 by an amount equal to the output resistance of the sensor.

I mentioned that the size of the input bias currents is not important (assuming they have the same value) because the impedance seen by the op amp's inputs is almost the same as each other (assuming a low sensor output resistance). What is important in this design is a data sheet parameter called "input offset current" which is a measure of how different the size of the input bias currents can be to each other. When the impedances seen looking out from the op amp's inputs are equal to each other then the source of error is due to the difference in the sizes of the input bias currents.

Answer 2

A lot depends on your requirements for accuracy and stability. But leaving that aside for the moment, your circuit won't work because the OP177 is not an ideal op-amp. It's actually a very good op-amp (much better than most RRIO op-amps) but it has limitations.

With +/-5V supplies it can handle inputs of +/-3V and deliver outputs of maybe +/-3.5V into a load of \$\ge\$ 10k\$\Omega\$. It's not characterized for other than +/-15V supplies.

If you increase the supplies to +/-15V (or +/-12V) then it will function (assuming you come up with a 5V reference voltage). I'm not sure why you would use such high divider resistor values on the output, I would go down by a couple orders of magnitude (eg. more like 10K/200 ohms). Depending on the input impedance and bias current of your +/-100mV input the resistor values could have a significant effect on accuracy. You might even have to add an op-amp buffer.

Using the supply voltage as a reference is generally a bad thing if accuracy matters. If you use +/-15V supplies then you can use a 'canned' reference IC, however note that the reference is called upon to sink current so some series references will be unsuitable without at least some added components (like maybe a resistor to -15V and a Schottky diode for protection). Here is a kind of series 5.00V +/-0.1% or better and low-drift reference that is nominally suitable since it can sink as well as source current (but of course error budget and so on should be considered for your particular case).

Answer 3

You have need high-precision resistors.

Why don't use some high common voltage op-amp as INA149?
Already in and well "paired".