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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.

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  • \$\begingroup\$ The gain required would appear to be 540mV/39mV=13.8 but the offset needs calibration up to 6mV \$\endgroup\$ – Sunnyskyguy EE75 Jun 12 at 4:40
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The only voltage drop you will see on V+ and V- sense lines will be due to the 2 front-end resistors you select for the differential amplifier configuration, multiplied by the input bias current of the op amp.

Checkout Table 7 of the datasheet:

enter image description here

Keep in mind that these will change part-to-part, and will depend on voltage, temperature, the resistor you select, etc..

Also keep in mind Note 2, which says this current is actually flowing OUT of the IC. Bias current can either be due to a current sink or source, and in this case it is a source inside the op-amp above the inputs.

The best way to reduce any error by this is selecting resistors in the 1k-10k range, no higher.

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  • \$\begingroup\$ Sorry, Im a bit confused, does that mean that I cant amplify the voltage using the op amp, as the resistors will throw off my current induced voltage drop? \$\endgroup\$ – Kobe Young Jun 12 at 3:55

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