# What is the predictable difference of the voltage drop between the V+ and V- lines on a shunt resistor?

I am making an electronic speed controller, and the half-bridge driver I am using has a built-in overcurrent protection feature. In order to utilize this feature, I have to supply the pin (a noninverting input of a comparator) with a voltage higher than 540 mV (which is the typical reference voltage supplied to the inverting input of the comparator). I plan to use this shunt with a value of 0.6 mOhm, so that when 65A (the max current I want to flow) flows across the shunt there is a voltage drop 39 mV. I know that the most accurate way to measure current with a shunt is to subtract two sense lines: V+, V-.

I plan to use the l6390's (the driver) built-in op-amp to subtract V+ and V- to get the accurate voltage drop, my issue arises when attempting to select the gain for the op-amp. It's pretty crucial that the amplified voltage is greater than 540 mV when the current reaches 65A. So I was wondering if anyone could tell me a way to estimate the difference between V+, V-. Essentially, will the voltage drop on the V- be significantly lower than the voltage drop on the V+ sense line? If that's the case, I can just do the calculations using an arbitrarily low value, but if that is not the case, I am not sure about how to do the calculations.

Thanks.

• The gain required would appear to be 540mV/39mV=13.8 but the offset needs calibration up to 6mV – Tony Stewart EE75 Jun 12 '19 at 4:40