# Is there any solution to high-precision high-side sampling current measurement?

I want to do some high-precision current measurement. (Voltage up to 20V and requires current to be shown in 0.1mA resolution, current can be up to 4A.) The simplest is low-side sampling by MCP3421:

simulate this circuit – Schematic created using CircuitLab

This one is easy and accurate, but it breaks the ground connection.

Using an MCP3421 for high-side sampling is not possible, because it will break its input voltage limit(VCC + 0.3V.) Our upstream voltage can be up to 20V.

Another possible solution is by using op-amps:

Link to pdf

Page 18 describes how to use op-amps to do the conversions. I wonder if the resistors can be inaccurate. Ageing is a killer of accuracy.

There's one more solution: using a high side current sense amplifier like TSC101 or LTC6101. This sort of ICs often has a high input offset voltage (Vos) up to approximately 1mV. 1mA on 15mohm sampling resistor is 15uV, so the offset seems to be unacceptable.

Are the above solutions possible? Are there any other solutions?

• What are your rails? please draw the entire circuit, it makes a difference. Sep 13, 2016 at 15:45
• You have not explained why interrupting the ground connection kills your otherwise acceptable 1st solution. High-side sampling interrupts the current loop too, and it seems to be acceptable to you. Sep 13, 2016 at 15:53
• How about an instrument amp monitoring the high side resistor? You'll need a power supply for the Int. amp. that is above the highest supplied voltage. Sep 13, 2016 at 15:56
• You need to clarify to yourself the difference between accuracy and precision, and which one you need in your application. 1 mV input offset voltage doesn't imply that you won't have 0.1 mA resolution in your measurement. Offset voltage is an accuracy error, not a precision error. Sep 13, 2016 at 16:44
• Lots of questions, no answers yet from the OP. And why a $15m\Omega$ sense resistor? Does he really want to require resolution to $\Delta 1mA\cdot 15m\Omega=\Delta 15\mu V$? That seems extreme. That's just $60mV$ drop, full scale. Doable. But can't the high side supply provide a larger difference voltage for sensing? And still no info on the available rails for opamps, for example.
– jonk
Sep 13, 2016 at 18:01

## 7 Answers

The simplest is low-side sampling by MCP3421. This one is easy and accurate. But this breaks the ground connection.

If you are content with this method then use one more chip that can provide isolated power and isolated buffers for SCL and SDA. I'm thinking here of the ADuM54xx series from ADI: -

(source: bdtic.com)

You can get versions with different IO configs to suit clock in and data out from the ADC. Possibly the ADuM5402 will be of most interest to your application.

Basically use the isolation chip and float your ADC up to 20V to make the measurement.

• Can anyone explain the down vote? Sep 14, 2016 at 11:00

For PRECISION measurement, I see two options:

1. Since your voltage is not very high, use a good instrumentation amplifier. With high CMRR it will do the job. The AD8222 is great. A drawback: you may need to create a voltage rail higher than the input voltage.
2. For higher voltages you can use an isolated ADC. Meaning you put isolated power and ADC right near the measured resistor, so the common mode component is 0 related to them. A good example is the AD7400.

There are tons of high-side current-sense amplifiers which already include the precision resistors needed for good CMRR and offset.

I suggest you spend a moment using DigiKey's search engine, look here:

Then select "Current sense" in the "amplifier type" box.

For ultra low offset, you can search for a chopper (or auto-zero) instrumentation amplifier like LTC1100, however you are unlikely to find one which can sense a voltage beyond its rails.

If the load can be switched on/off, then a simple way to null offset is to use double sampling: measure the current with load ON, then with load OFF, and substract both. This works extremely well.

If you don't want to be using a dedicated chip, then a VCCS is the usual way to go.

A precision op-amp with a PMOS transistor on the output is configured as a voltage-controlled-current-source (VCCS). This converts the voltage on the high-side sense resistor into current, with a useful gain. Then, convert the current back into voltage using a resistor, connected to any common level you need, as long as it's at least a couple volts below the supply voltage. The C->V converting resistor usually can feed the ADC directly, without further buffering.

To make the circuit practical and to protect the ADC from startup transients, add a ballast resistor between the drain of the PMOS and the ADC input. The ballast can drop much of the voltage between the supply voltage and the maximum input level of the ADC. That way, the PMOS will stay cool, and the ADC input gets inherent impedance protection. No Zener diodes are usually needed then - and they are leaky anyway when hot, so that's not great.

Look at the Allegro Microsystems current sensor ICs. These use Hall-effect sensors and can be used in high-side applications.

• Is there a 3.3V power supply version for ASC712? Also, what's the positive minimum measurable current of it? Sep 13, 2016 at 22:55

Go ahead and use a high-side monitor, and if the offset voltage bothers you, preload the current, then apply an offset at the output to null the result with zero load. The TSC101, for example, with a 0.050 ohm sense resistor, can handle the 4A.

Up to 2.5 mV of offset in the chip can be treated by shunting 0.0025V/.050ohm = 50 mA from the output (so that the offset voltage will never saturate the sense amplifier).

A negative power supply will simplify the level-translation (and allow you to continue to monitor current below the 2.8V 'input common mode' limit). I'd draw the 50 mA with a grounded-base NPN transistor.

simulate this circuit – Schematic created using CircuitLab

• This is unacceptable because of the extra power consumption. Sep 14, 2016 at 10:43

Maybe INA226 is a good choice. You can choose to ignore its alarm pin signals, though I don't know if ignoring is a better choice.