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I need to measure the voltage difference between two high-voltage points (~500 Vdc). I'm thinking of using a differential amplifier with a gain of 1/100. While it behaves as I want it to in the simulation, I want to be sure about picking the correct op-amp for this task.

Would a general purpose op-amp like 741, LM321, OP07, etc. work? (of course, the op-amp will receive the low voltage supply: 12 Vdc).

enter image description here

$$V_{out}=\frac{500V-431V}{100}=690mV$$

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  • \$\begingroup\$ There are purpose-built high common-mode differential amplifiers designed for this application, like the AD8479 (±600V). \$\endgroup\$
    – Theodore
    Mar 17 at 20:05
  • \$\begingroup\$ @Theodore I'm aware of the AD8479, but its gain is fixed to 1. \$\endgroup\$
    – David
    Mar 17 at 20:13
  • \$\begingroup\$ "HV differential probes" for oscilloscopes already do this specific thing, and are certified etc. They are very expensive, so people make their own: youtube.com/watch?v=_OZ5Xer84eo Divider network discussed at 6:54 \$\endgroup\$
    – Miron
    Mar 17 at 21:13
  • \$\begingroup\$ Another video: youtu.be/GOlgaEK2Hsk?t=945 Both use a battery as the power source, since it has to be floating, I don't know, but maybe an isolated DC-DC converter could be used too \$\endgroup\$
    – Miron
    Mar 17 at 21:24
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    \$\begingroup\$ Your problem is under-defined. Please state: (1) the frequency range of the voltages, (2) the absolute range of each of the two voltages, i.e. minimum and maximum voltages you expect, (3) the current sourcing capability of those voltage sources, and (4) the output voltage range and how you can allow it to be mapped to the differential voltage. This has large impact on feasibility of various solutions. The example you show does not warrant 1/100 gain, since you only measure 70V differential: a 1/10 gain would produce 7V on the output, and you said nothing that would make this invalid! \$\endgroup\$ Mar 17 at 21:46

2 Answers 2

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The maximum specified input voltage of a 741 op amp is only around 15 volts (depends on supply voltage). This is going to be true for ordinary op amps which were never intended to handle such high voltages. Your best bet is to use a voltage divider before the differential amplifier to bring the levels to less than the supply voltages.

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    \$\begingroup\$ If the differential output will not be more than 5 V (\$\frac{500V-0V}{100}\$), why should I worry about the input voltages at the Inverting and Non-Inverting? [Edited by a moderator to fix the Latex delimiters.] \$\endgroup\$
    – David
    Mar 17 at 20:11
  • \$\begingroup\$ @David What a voltage divider/divider network linked in my earlier comment does that the opamp circuit above doesn't, is maintain a safe input voltage for the opamp, even when power is off. If the battery runs out during measurement, your circuit is toast \$\endgroup\$
    – Miron
    Mar 17 at 21:31
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Concerns of note:

  1. Common mode voltage range - set by the 500V common mode voltage divided by 1/100. Should be OK for pretty much any op-amp that runs from a single 12V supply.

  2. Output voltage range - most "jellybean" general purpose bipolar op-amps do not swing close to either supply voltage, so you'll need to move that closer to mid-supply. Instead of connecting the topmost resistor to 0V, connect it to 6V, for example.

  3. The matching of the resistors is critical, and if there's any AC component to the measured common or differential mode, the AC CMRR adjustments will prove necessary - typically trimmer capacitors in various places.

A different approach would convert the voltages to currents, and perform subsequent operations on the currents. A high-level scheme of the approach would be:

schematic

simulate this circuit – Schematic created using CircuitLab

VCCS1 and VCCS2 use a resistor to convert voltage to current I10 and I20 respectively, and then mirror that current. CCCS3 mirrors I10 and injects its into the current summing node. Out of that node flows current I20 and the offset current I1. The net current into the current summing node is mirrored by CCCS4 and subsequently converted to voltage on R1.

This could be implemented using op-amp based current sources/mirrors.


Alternatively, a somewhat hilarious implementation of such a circuit, with DC CMRR mostly dependent on thermal tracking of R1 and R2, is shown below.

I1 can be a current reference or a current source driven with a resistor from 12V. The common mode extends to 0V, although full accuracy is only available above input voltages of about 20V.

The bandwidth of the circuit can be adjusted by scaling the currents. The current consumption from the 12V supply is maximum 0.9mA with the currents shown. The output voltage would be typically buffered, scaled and fed to an A/D converter.

The current mirror design comes from “Current mirror circuit with accurate mirror gain for low β transistors” by Chen, Whiteside and Geiger, and performs admirably with discrete transistors.

With some care in trimming and transistor pair matching, the accuracy at room temperature should be better than 0.1%, but you absolutely should verify that for yourself. Some AC trimming may be necessary as well, depending on the desired bandwidth of operation and the common mode frequency content. A common mode rejection choke could be used to improve AC CMRR.

The trimmers should be initially set to the center point position, and then adjusted as needed. The 10kOhm bias resistors are not critical and can be 5%. Other resistors should be 0.5% or 1% and 50ppm/K tempco or better.

Note that this circuit does not claim to be the best in any way nor even practical for a product. A well implemented discrete in-amp would use potentially fewer transistors and have better AC performance anyway. It's meant to be funky and illustrate how discrete current sources can be harnessed for some degree of DC accuracy.

schematic

simulate this circuit

D1 and D2 bring the load voltage on the output of the current mirrors closer to the reference voltage. This minimizes the offset of the current mirror and stabilizes its current gain. The follower Q45+Q46 does the same job. It couldn't be a Zener diode, since the voltage on the other end of it varies - that's the voltage proportional to the current.

In a physical circuit, the performance may be potentially improved by connecting the collectors of Q21, Q44 to ground, and of Q7, Q14 to 12V, instead of connecting them to their respective bases as shown. This will be especially the case if the beta of the transistors is low (<50). With high beta transistors - the simulated ones have beta=100, the connection as shown seems to provide most accuracy.

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