# Current measurement op-amp calculation

I am trying to measure the current using sense resistor method.

Low Side Current Sensing method:

• I am using shunt resistor in the low side, there will be a voltage drop across the resistor.
• Since the voltage drop is very small, using op-amp we can amplify to the required voltage level for microcontroller's ADC input.

The ratings are I = 30 & R = 0.001 (2 watts)

Voltage drop across the resistor,

Vrsense = IR
= 30 * 0.001
= 0.03 volt


Power dissipation in resistor,

P = I * I * R
= 30 * 30 * 0.001
= 0.9 Watts


Op-amp gain,

Vout = Vrsense * (1+ Rf / R1)
= 0.03 * (48)
= 1.44


My doubt is LM358 has a maximum power supply of 36 volts, but I am using 50 to 60 volt. I thought that the input channel will allow zero current, so by giving such a supply will not be a problem. Since I am new to op-amp correct me, friends, if I am wrong. Or I have to use some other op-amp.

Figure 1. Low-side current measurement voltages.

• Since (2) is connected to ground the voltage at that point is 0 V, not 49.97.
• With 30 A flowing through the load the voltage at (1) will be + 0.030 V, not 50 to 60 V.
• If you power the chip at (3) with 3 to 36 V you will be OK. I would be generous and use 5 to 30 V.
• Your circuit has TWO ground nodes. How to keep those at the same ("zero") voltage? As the 30 amps exits the left GROUND region, the current will cause voltage drops everywhere. Does that cause an error? Feb 21, 2019 at 9:20
• This will work fine (@analogsystemsrf the differential amplifier configuration should cancel out small ground differences provided the resistors are well matched) however OP should note that the LM358 is not very appropriate to this application. With a Vos of 3mV max at 25°C there will be +/-3A error due to that alone, and it may not start reading anything until the current exceeds 3A since the output cannot swing below ground. Plus temperature drift. A "zero drift" op-amp would be much better. Feb 22, 2019 at 11:49

[Note: This answer was written before the question was revised to include a schematic]

My doubt is LM358 has a maximum power supply of 36 volt, but I am using 50 to 60 volt.

Very likely, you will damage the LM358. The maximum supply voltage limitation is most likely to prevent damage to the IC due to breakdown (too high an electric field somewhere that current is not supposed to flow). None of the other parameters you mentioned has anything to do with this.

If you want this circuit to be reliable, you'll need to either find an op-amp that can run on 60 V (of which there really aren't very many), or reduce the voltage you're supplying to the LM358.

• Using voltage divider in lm358 input channel (inverting and non inverting) could be solvable for this problem?
– Bud
Feb 21, 2019 at 6:46
• @Nihal, Just connecting 60 V to the power pins will probably damage the part. Feb 21, 2019 at 15:25