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We need to measure +-50Amps of current (PWM) on a PCB. We need the measurement to be isolated, so I thought of using an ACS758, and feeding its output to an ADC. However, this hall effect sensor is "limited" to 120Khz bandwidth and in our application, we need around 200Khz of bandwidth.

I suppose another option is to use a shunt resistor and an op-amp feeding an ADC. Aside from an isolation problem, at 50Amps, assuming a shunt resistor value of even 1mohms, that's 2.5Watts dissipated on the poor resistor. Just another thing to cool on the board.

What are other possible solutions, if any?

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  • \$\begingroup\$ How heavy is your PCB copper? My first thought was to make a resistor out of a trace, but 50A is a lot of current... \$\endgroup\$ – Kevin Vermeer Jul 18 '11 at 20:40
  • \$\begingroup\$ It will probably be 105 microns, through at least 2 layers. \$\endgroup\$ – SomethingBetter Jul 18 '11 at 20:43
  • \$\begingroup\$ You write about "continuous current", then about the bandwidth. What do you mean exactly? How is shaped the current? \$\endgroup\$ – Mario Vernari Jul 19 '11 at 5:45
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    \$\begingroup\$ 100 micro-ohms sense circuit. 5 mV drop at 50 A !. Data: allegromicro.com/en/Products/Part_Numbers/0758/0758.pdf \$\endgroup\$ – Russell McMahon Jul 19 '11 at 7:21
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    \$\begingroup\$ People "would be well advised" to look at the Allegro datasheet before commenting. Allegro are very competent and this is an extremely real; solution. In practice it is liable to be a very serious solution to have to compete against with things like PCB track drop. It's calibrated [tm] 100 micro-ohm current path and isolated sensing, factory trimmed parameters and formally specified rise times are not going to be trivially matched by 'bits of copper track' and an op amp. Better solutions may exist, but they are not one liners - unless the one line is a part number. \$\endgroup\$ – Russell McMahon Jul 19 '11 at 19:21
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This started as a comment but grew into an answer.

Summary: I'd try to characterise that IC in the current application, possibly with advice from Allegro. It's a beautiful solution and you'd be hard put to better it if you can work out how to live with the bandwidth issue. [I have no relationship with Allegro apart from having been an occasional very small scale satisfied customer].

People "would be well advised" to look at the Allegro ACS758 datasheet before commenting.

Allegro are very competent and this is an extremely real; solution. In practice it is liable to be a very serious solution to have to compete against with things like PCB track drop. It's calibrated [tm] \$ 100 \mu \Omega \$ current path and isolated sensing, factory trimmed parameters and formally specified rise times are not going to be trivially matched by 'bits of copper track' and an op amp. Better solutions may exist, but they are not one liners - unless the one line is a part number.

Here is Allegros range of High current sensors

Note that the ACS758 is at the top of the range both for current and for bandwidth.

The datasheet specifies bandwidth as being \$ \frac {1}{3 \times T_{rise}} \$ and \$T_{rise}\$ is typical. Performance is in the order of right to rather marginal. Given the otherwise superb nature of the part I'd take a very close look at how the device behaves at target frequency. There will certainly be "roll off" but how much. Is something like a single pole, which can be happily used an octave or even two above notional cutoff, or is it an 8-pole-boxcar-go-away-use-something-else response? I'd suspect more the former than the latter.


If I was doing this and wanted unlimited freedom of manouver I would indeed start with a resistive voltage drop solution. But I'd not be surprise if the chase was long and hard. For any sort of accuracy across temperature I'd probably want to use an add in resistive shunt, and something of the magnitude of Allegros \$ 100 \mu \Omega\$ shunt would seem about right. (\$50A \times 100 \mu \Omega = 5 mV\$ drop. \$(50A)^2 \times 100 \mu \Omega = 250 mW \$ loss. Note that a \$1 m \Omega \$ shunt takes 2.5W and a \$ 1 \Omega \$ shunt takes \$25W\$. Even \$2.5W\$ may be considered "intrusive" depending on the system Voltage.

\$5 mV\$ full scale drop = \$20 \mu V\$ per bit at 8 bits. Not "hard", but offset voltages become important. But with devices PWMing 50A nearby, using an off theshelf solution that had dealt with such issues looks more attractive than sometimes. At $US7 in 1's and half that in 1000's the ACS758 looks like a good start.

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This application sounds perfect for a LEM current transducer, I searched for a good example on their website.

Bit more pricey however, but they are good. At this current I wouldn't use a shunt resistor, also I would think twice about running this much current through PCB copper for long traces.

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I think the simplest way is to measure the PCB resistance between two precise points, then playing around the voltage drop across. Maybe it's not a heaven, but at least it's not intrusive.

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I have had success with using short copper land patterns. I amplified the land shunt to a usable level into an ADC. Was useful @ 80A for around 30ms and the highest recorded was 1KA @ 50µs. If your driving the PWM on a pcb anyway, just tap a section of your land pattern then amp it accordingly.

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