I am looking to measure voltage drop across current sense resistor. The waveform which occurs at the resistor is specific and consist of parts which are at voltages of +50V and -50V relative to ground.

The actual current is pretty small, about 50mA. So if I use 0.5ohm resistor, at 50V, there is going to be 25mV voltage drop across it.

Since voltage is changing, the direction of current is changing, too, so for start, I would need bidirectional measurement. I am looking to be as much as possible cost efficient here, too.

There is quite of options which I considered but none I have found to meet all the requirements:

a) using shunt amplifier, for example INA170 - the common mode voltage range is not big enough (best I have found is LMP8601 with -22 to 80V)

b) using shunt amplifier + voltage divider at Vin+ and Vin- pins - I would loose precision, only 25mV difference!

c) using differential bipolar ADC(for example MCP3424) - the max. voltage on pins is too small, too

d) ??? - I have no idea

Thanks on your time, as usual!

EDIT: The +50V and -50V voltage which creates the output is coming from the isolated power supply. If I use shunt amplifiers / ADCs / any other device which is connected to the other ground of the other power supply, the output is not isolated anymore, right?

  • \$\begingroup\$ How often is the voltage changing between +50 and -50? \$\endgroup\$ – user1844 Sep 7 '16 at 16:02
  • \$\begingroup\$ +50V and -50V are only short pulses which last about 400uS. The rest of the time waveform rests at 0V(ground). \$\endgroup\$ – davaradijator Sep 7 '16 at 16:03
  • \$\begingroup\$ I presume you're interested in the current when the voltage is high, correct? \$\endgroup\$ – alex.forencich Sep 7 '16 at 16:09
  • 1
    \$\begingroup\$ In addition to other suggested ideas, you could also possibly use a Hall-effect based current sensing IC. These have an isolated low impedance current path (can be used on high or low side) and provide a voltage output that is linearly related to current. One example is the ACS712, but the same company, Allegro, makes other parts as well. \$\endgroup\$ – mkeith Sep 7 '16 at 16:43
  • 2
    \$\begingroup\$ And why, exactly, are you putting the resistor in the high/low line, rather than the ground return, where the voltage is always 0? \$\endgroup\$ – Asmyldof Sep 7 '16 at 17:34

Well, one possibility is to float the ADC. Call one end of the sense resistor 'ground', power a differential ADC with an isolated DC-DC converter, talk to it through a digital isolator. Analog Devices makes some combination isolated converter/digital isolator parts that could work, but they're a bit pricey.

If that's too expensive, then dividing down is probably the most reasonable method. You'll have to be very careful with the resistor matching and temperature coefficients to make sure the division ratios are very close, if you want to use a differential amplifier. If you convert each 'end' of the resistor separately, then you have some more flexibility in terms of digital calibration after the ADC. This would also require an ADC with a lot of bits.

Perhaps another option would be to divide down both sides of the sense resistor and use a differential amplifier, then feed both the amplified differential voltage as well as the common mode voltage (probably just the absolute divided down voltage on one end of the sense resistor) into two different ADC channels, then calibrate and compensate the common mode to differential gain due to mismatched dividers after the ADC. I'm thinking this may be the best option here.

Alright, let's do the math for that 3rd option. Let's say \$V_a\$ and \$V_b\$ are the voltages on each end of the sense resistor. You want to know \$V_a-V_b\$. So you divide each of them with resistive dividers. The problem is that resistors aren't perfect, so you get a divider with gain \$A\$ for \$V_a\$ and a divider with gain \$B\$ for \$V_b\$. Then you want to get high resolution, so you use a differential amplifier with gain \$G\$ to subtract those two. You get \$V_{out} = G(A V_a-B V_b)\$. Let's rearrange that a bit.

\$V_{out} = G(A V_a - B V_b + A V_b - A V_b)\$

\$V_{out} = G(A V_a - A V_b - B V_b + A V_b)\$

\$V_{out} = G A (V_a - V_b) - G (B - A) V_b\$

Hmm. That's got the term you want, \$V_a-V_b\$, but it also has two extra terms - \$G A\$ and \$G (B-A) V_b\$. \$G A\$ is constant, and so is easy to deal with. But \$G (B-A) V_b\$ involves \$V_b\$. The idea is that by measuring \$V_b\$ (or \$V_a\$) separately, you can calculate \$G (B-A) V_b\$ for each point and compensate for the 'common mode to differential gain' caused by the mismatched dividers. The cost is an additional ADC channel, some math, and some calibration method, but it should be relatively easy to implement.

The problem term is really just \$B-A\$ as \$G\$, \$A\$, and \$B\$ are probably going to be accurate enough by design. You'll have to measure it (directly with a meter or indirectly by applying known voltages and/or currents) on a per-device basis and possibly even do some temperature compensation, depending on the accuracy you're hoping to achieve.

| improve this answer | |
  • \$\begingroup\$ Thanks on quick answer. Third option looks like the best. Could you get into more details with it and give an actual example? \$\endgroup\$ – davaradijator Sep 7 '16 at 16:25
  • \$\begingroup\$ OK, added some math so you can see how it would work. \$\endgroup\$ – alex.forencich Sep 7 '16 at 16:54

You can consider using an isolation amplifier such as the AMC1301.

enter image description here

This kind of device (Broadcom (nee Avago) has some too) is specifically designed to deal with isolating a small signal from a current measuring shunt onto top of a bunch of common mode voltage. You'll probably still get a transient right after switching that you'll have to ignore.

You could also consider a transformer, but it would have to be in the double-digit mH to faithfully transmit a relatively long pulse like that so probably a custom magnetic part, and not cheap in low quantity.

I don't think you're going to get far with a simple differential measurement. Consider 5% of full scale error on a 25mV signal is about 1mV and you will have reject a +/-50V common mode voltage, so about 100dB rejection. To put it another way if a resistor drifts by more than 0.002% (20ppm) your error will exceed 5%. Not completely impossible, but expensive and difficult.

| improve this answer | |
  • \$\begingroup\$ Hmm, those are interesting parts. Cheaper than I expected, but they do have some trade-offs. Namely, they need a floating power supply from somewhere and they basically have a built-in ADC and DAC so they have some delay and probably somewhat limited bandwidth. \$\endgroup\$ – alex.forencich Sep 7 '16 at 17:04

Some options:

  • Here is a difference amplifier with a common mode range of up to +/-270V. That should be able to do what you need it to do. But at $5 in quantity, it isn't cheap.

  • A switched capacitor to remove common mode. Basically a capacitor is allowed to charge over the resistor, reaching the voltage of the resistor, then it is switched down to ground and a second holding capacitor that maintains that value to be read by the ADC or op amp buffer at ground. In some Linear app note there is a circuit diagram that describes this, however I can't for the life of me seem to find it. This also isn't likely to be cheap. You will want optically switched fets to change the capacitor over, and good capacitors with very low leakage. You also run into problems with bandwidth (though with fets, these can get surprisingly fast).

edit: found it, kind of. Here on page 2 is the switched capacitor version. They have redone this circuit in a later document somewhere, where they take the output from a chopper amp (at the frequency of the chopper), and drive the switching of the capacitors at that frequency, reducing noise. Unfortunately, I can't find it.

edit: while looking for the circuit I'm talking about, I ran across a good op amp that might help: http://cds.linear.com/docs/en/datasheet/6091fa.pdf.

more: Here is a one chip instrumentation amplifier with a common mode of -100V to 85V. here uses the op amp mentioned above (ltc6091) to create an instrument amp. And here is another difference amplifier (this one only has +/-250V common mode).

note: The reason that I have been throwing out linear chips is because their datasheets are fantastic and I've been reading a fair number of them lately. They also have an "over the rail" input technology that allows for inputs that are larger than the power rails of the op amp.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.