Measuring high current (180A) indirectly with shunt resistor method

Thread Starter

jim0000

Joined Oct 28, 2020
32
Hello,

I am trying to measure high current and I do not have a measurement device rated high enough to directly measure 180A, is this a good method? ----> I was reading about the use of shunt resistors connected in parallel with multimeters and although the multimeter may not be rated for 180A, it could still measure the current indirectly by putting it on the voltage setting and measuring the voltage drop across the shunt resistor. Since the internal resistance is so high compared to the shunt resistor the multimeter wouldn't see anywhere near the 10A fuse rating. The shunt resistor I have says 500A and 50mV on it so I assume it has a resistance of 0.1 milliohm. The multimeter I have is a fluke 117 true rms multimeter and after looking online it looked like the internal resistance was roughly 10Mohms. My reasoning is below:

Side note: Also I am curious, would I need special test leads or can I use what came with the fluke?
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Last edited:

Hymie

Joined Mar 30, 2018
1,002
Your reasoning seems correct – one observation, it is normal practice for the current shunt to be in the 0V/return line, minimising any stray voltage effects that may be present with the measurement in the high side of the circuit.
 

Thread Starter

jim0000

Joined Oct 28, 2020
32
Your reasoning seems correct – one observation, it is normal practice for the current shunt to be in the 0V/return line, minimising any stray voltage effects that may be present with the measurement in the high side of the circuit.
Oh okay so instead of putting it on the hot wire put it on the ground side?
 

Ian0

Joined Aug 7, 2020
3,754
A few more observations.
1) If you use the 10A current range on the meter, then the resistance is not 10MΩ - that would be silly because the voltage drop across the meter would be 100,000,000 Volts. Use the meter on the 200mV Voltage range, and measure the voltage across the shunt resistor. Multiply the mV by 10 to get the current in Amps.
2) The tolerance on a shunt will be no better than 1%, possibly 0.1% if you buy a really expensive one.
The current through the meter is 10^10 less than the current through the shunt. To all intents and purposes the meter resistance is infinite.
 

dcbingaman

Joined Jun 30, 2021
495
If you want to keep with the sense resistor, in order to have any accuracy be sure to use a '4 wire current sense resistor'. These devices contain two terminals that are designed to handle the main current and two smaller terminals that are used for the sensing. The manufacturer of such devices can keep the accuracy across the sensor element at much greater values.
 

dcbingaman

Joined Jun 30, 2021
495
The LEM sensors are great, provided you never need to measure a small current. They are 42dB noisier than a shunt and its associated amplifier, and have a bit of an offset problem.
Yea, the LEM sensors look like a good choice for many current measurement applications.
 

dcbingaman

Joined Jun 30, 2021
495
I know it's not my thread, but thanks for that - that looks really useful, but not cheap!
Honestly neither will the shunt sense resistor be cheap. Because it must hold to a very small resistance and be a 4 wire device, it tends to be rather expensive as well.
 

dcbingaman

Joined Jun 30, 2021
495
Here is a 4 terminal current sense resistor that is in stock and available from digikey. It happens to be 750uOhms so your scaling factor will be a little different:

https://www.digikey.com/en/products/detail/ohmite/SHD1-100C075DE/2776198

At around 20 bucks not to bad.

Oops, never mind. This one will not meet your current requirement. If is only capable of up to 7.5W that is I^2*R = 200^2*0.00075 = 24Watts.
 
Last edited:

Ian0

Joined Aug 7, 2020
3,754
Here is a 4 terminal current sense resistor that is in stock and available from digikey. It happens to be 750uOhms so your scaling factor will be a little different:

https://www.digikey.com/en/products/detail/ohmite/SHD1-100C075DE/2776198

At around 20 bucks not to bad.

Oops, never mind. This one will not meet your current requirement. If is only capable of up to 7.5W that is I^2*R = 200^2*0.00075 = 24Watts.
This one's rated for 200A
https://uk.farnell.com/hobut/shb200a75-class-1/shunt-75mv-200a/dp/1015894
 

dendad

Joined Feb 20, 2016
3,939
In a couple of my products, I used INA196 chips. They are high side current sensor amps.
https://www.ti.com/product/INA196
INA196AppCct.png
Here is a part of my circuit. R1 is sized to suit the current. R2 and R3 scale U1's output to suit the 3.3V PIC32MX795F512L processor I used.
Along with R2 and R3, C4 is a filter, and D2 is an over voltage clamp for the PIC input.

INA196_Example.png
 
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