# How to find Rsense value?

I want to use a current sense resistor (R54 in circuit) for overcurrent sensing. The rated drain current of the MOSFET is 17 A. Bus voltage is to be 100 V. Then would my sense resistor be R = V/I = 100/17 ohm? Or would the sense resistor be decided by the OC pin voltage?

U15 is the gate driver : UCC21732 from TI

For the current R54 value of 350 mOhm, I am getting a huge power rating of $$I^{2}R$$ = $$17^{2}*0.35$$ = $$101.3 W !!$$

• Rsense = V(OC pin) / I. See page 10 of the datasheet : Vocth = 0.7V. So the shunt resistor is set to trigger OC at 0.7V/0.35R = 2A.
– user16324
Jan 30, 2021 at 17:59
• L5 = 1 pH really?.... .5nH/mm Jan 30, 2021 at 18:06

You choose the resistor value R54 not to protect the MOSFET but to prevent current saturation in your 50 μH coil called L1. Your MOSFET maximum current rating should be at least 50% higher in rating than what the maximum current is that you need to flow in L1.

So, working backwards, if the MOSFET is 17 amps, then I'd expect the coil current to be no greater than about 11 amps. This means that R54 would be: -

$$R_{54} = \dfrac{0.7}{11} = 0.064\Omega$$

But, the current waveform is sawtooth in nature and you can't just assume that the peak current of 11 amps dictates power dissipation. So work out the RMS current value for a sawtooth waveform.

If the positive peak is 11 amps falling to zero, the RMS of the equivalent AC biased sawtooth is: -

$$\dfrac{11/2}{\sqrt3} = 3.1754 \text{ amps}$$

This is then added (as square terms) to the average value (5.5 amps) then square rooted to get an RMS current of: -

$$\sqrt{3.1754^2 + 5.5^2} = 6.351 \text{ amps}$$

So now, the power in R54 is $$\6.351^2 \times 0.064\$$ = 2.581 watts. Use a resistor of wattage greater than 3.872 watts is my advice (50% margin on power).

Set your R54 to $$\ R = \frac V I = \frac {0.7}{17} = 40 \ \text m \Omega\$$ if 17 A is what you want.