I am a final year E&E student and I am trying to built a power meter that must be able to measure quite high DC voltages, up to 1000 V DC. I am measuring with a simple 12-bit ADC that has an input voltage range of 0 - 2.5 V. Would a simple voltage divider and op-amp buffer be sufficient for the application or is there another type of front end analog circuitry needed because the voltage is that high?

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    \$\begingroup\$ It might be educational for you to have a look at how proper multimeters rated for 1000 V are build. These also use a voltage divider for the high voltage ranges. Have a look at some of the videos on the EEVBlog to see examples of safe high voltage circuits. Google for "eevblog multimeter teardown" and you will find many. \$\endgroup\$ – Bimpelrekkie Sep 6 '16 at 13:57
  • \$\begingroup\$ @Eduan Shuda: what is the min. input impedance you can tolerate? That might be an important design requirement. \$\endgroup\$ – Curd Sep 6 '16 at 15:05
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    \$\begingroup\$ Hopefully as part of your studies, you have been given proper training for handling high voltages. To those who are coming across this question from google searches: don't try this one at home! \$\endgroup\$ – Cort Ammon Sep 7 '16 at 4:06
  • \$\begingroup\$ Add a LARGE WATTAGE zener from near the bottom of the divider string to ground. Vzener about 2 x Voltage max at that point. This MAY save your electronics WHEN things go wrong. \$\endgroup\$ – Russell McMahon Sep 7 '16 at 14:15

A resistor divider will do what you want, but at this voltage there are some issues you can normally ignore:

  1. The top resistor has to be able to handle 1 kV. Those are harder to get than "ordinary" resistors, and are often not linear with voltage at the high end.

  2. Power dissipation. Even what would normally be a "large" resistor, like 1 MΩ, dissipates a whole watt when 1 kV is applied to it.

  3. You need physical distance between two points that have a kV between them for safety and to prevent arcing thru the air.

Due to all these reasons, I would implement the top resistor of the voltage divider with multiple more ordinary resistors in series. For example, 0805 resistors are usually rated for 150 V (your job to check the datasheet). Ten 1 MΩ 0805 resistor in series, physically laid out end to end, can be used as a 1 kV 10 MΩ resistor. The voltage across each resistor will be 100 V or less, which keeps them within spec.

All together, the 10 MΩ string of resistors only dissipates 100 mW, so each individual resistor only 10 mW. No problem here.

With a 10 MΩ top resistor, the bottom resistor of the divider would ideally be 25.06 kΩ to get 2.50 V out with 1000 V in. You want to have a little headroom above the maximum input voltage spec of 1000 V, so a 24 kΩ or even a little lower bottom resistor should do it.

The output impedance of a divider with such a high ratio is basically the bottom resistor value. 24 kΩ may be too high for some A/Ds, so you may want to buffer this with a opamp used as voltage follower.

  • \$\begingroup\$ I have done this on the advice of my senior at the time, and it worked well. Multiple high power dissipation resistors for the "top" resistor \$\endgroup\$ – Fuzz Sep 7 '16 at 6:59

Yes, you can use a voltage divider (in fact there are few other practical approaches).

You will need to use a precision resistor for the high value resistor that is rated to operate safely at 1000V. Do not overlook this detail. You will also have to follow recommendations on layout- which might involve milling an isolation slot under the resistor to increase the creepage distance unless the resistor itself is really long, and will definitely involve other PCB considerations at the high-voltage input.

The overall resistance of the divider will be limited by the output impedance you need to achieve, and that will be determined by the ADC if you attempt to go directly into the ADC input. Most likely this will not be desirable because (for full accuracy) the ADC needs to see a few K ohms at its input. Say it is 2.5K. Then you will need to use 1M (or less) for the high value resistor, and it will dissipate 1W (or more) at 1000VDC- not great for accuracy (and it loads the input significantly- 1mA @1kV).

It may be better to use a high performance op-amp buffer at the ADC input, allowing you to use more like 10M and 25K.

If you have higher supply voltages in your system there may be a small advantage in dividing down to a higher voltage, such as 10V with a 15V supply and then buffering and using a second passive divider to get down to 2.5V, but it's probably not necessary with only 12-bit resolution. It would reduce the effect of op-amp offset and offset drift, at the cost of involving two more resistors in the error budget (but the high voltage one should be your main source of concern).


Remember that every resistive divider has a parasitic capacitive divider. Depending on which physical resistor designs are used, the ratio of this divider can be very different from the resistive ratio; this can make surprisingly high voltage spikes appear at your IC inputs, so you should clamp your IC inputs to safe levels with fast diodes and/or compensate the divider (maybe "overcompensate" it with a big capacitor across the lower resistor).

  • \$\begingroup\$ .. best to load your input circuit with a fast square wave and check what your IC input actually gets on an oscilloscope (use a 1:100 or active probe, you don't want probe capacitance to mess with things!)- if there is significant overshoot or ringing, that means your IC inputs could get more than they can take the moment the voltage divider is suddenly connected to something. \$\endgroup\$ – rackandboneman Sep 7 '16 at 11:34

The problem with a divider is going to be V2/R (the power rating). At 1000V, dividing it down to 2.5V, your deltaV is going to be 997.5V. Even if you use a 1 MegaOhm resistor, you're talking about using about a 1W resistor, and in practice you don't want a resistor that large because it's going to be an appreciable fraction of your op-amp input impedance, and throw off your measurement accuracy. At 100kOhms, you're going to be looking more like 10W, and you'll probably need to organize a combination of parallel and series resistors that give you the effective resistance you are going for while distributing the power dissipation requirements.

The other problem is going to be dynamic range. You're going to divide down 1000V to 2.5V, so a factor of 400. That means a natural 1V signal is going to manifest to your ADC as a 0.0025 signal. Your naive voltage resolution with a 2.5V @ 12-bit ADC is 2.5/212 = 0.000610352V/LSB, but your number of effective bits is probably closer to 10, or 0.002441406V/LSB. So you're good as long as you accept that the lower limit of your measurement is going to be around 1V. Averaging techniques can improve your effective voltage resolution, at the cost of reducing your time resolution / distorting your signal in the time domain.

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    \$\begingroup\$ A 1Megohm resistor won't lower the accuracy. Because in reality, what must be compared is the input current leakage of the opamp vs the current flowing through the divider, not the impedances. So at 1000V, the OP should be fine with even larger resistors (10Megs or so). \$\endgroup\$ – dim Sep 6 '16 at 14:18
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    \$\begingroup\$ You can use 10x 100kΩ Instead of 1MΩ to spread the power dissipation by factor of 10. That will make 100mW per resistor. \$\endgroup\$ – Chupacabras Sep 6 '16 at 14:19
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    \$\begingroup\$ @dim: right. IOW, the output impedance of a voltage divider is dominated by the smaller of the two resistors. Which at ≈ 10 kΩ is in an excellent range for jFET inputs. \$\endgroup\$ – leftaroundabout Sep 6 '16 at 14:56
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    \$\begingroup\$ Agreed, input impedance is less of an issue than I originally suggested \$\endgroup\$ – vicatcu Sep 6 '16 at 14:59
  • \$\begingroup\$ @leftaroundabout Actually, I think it's more something like "the parallel combination of both resistors of the divider". Which, if you have a huge one and a much smaller one, is very close to the smallest, indeed. \$\endgroup\$ – dim Sep 6 '16 at 15:15

"Multimeter" way of doing this would be to charge a capacitor with a large resistor and sample it periodically so you can work out the driving voltage.. Obviously you need to clamp the voltage below capacitor max voltage rating and also you need a way to discharge the capacitor. A simple transistor (or mosfet) discharge will not give ideal results as no semiconductor has zero ec or ds voltage. But that's probably getting into too much detail.

The benefit of doing this is that you get a broad workable voltage range, a straight resistor divider suitable for 1kV is not very useful for measuring 1V..

For the megaohm series resistor divider, work out the thevenin resistance and voltage. In essence rth is just the voltage divider top/bottom in parallel and vth is the divider output voltage. This will give you the output impedance and current flowing into the opamp/adc.


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