Hello everyone, I'm looking to solve a problem I'm having with a DC motor circuit. The motor is driven by a 12V source and has a stall/inrush current of 4.75A. My supply sags by about 1V during inrush for about 40nS. I want to help smooth this out by using a properly sized capacitor, a bit of a hack but this is more of a learning exercise than anything.

My gut told me to start out with a 470µF capacitor but I want to use C = (dt/dv)i to find something sized a bit more appropriately. Can someone doublecheck my understanding of the formula here?

C = ((40*10^-9)/1)4.75

= 1.9*10^-7 = 190nF the closest standard value being 220nF

Also, if I use T=RC to calculate discharge time I get the following time this value would supply 12V while my power supply catches up.

Solving for R in a worst case/in rush situation 12 = 4.75/R. R ≈ 2.53Ω. T = 2.53(220*10^-9). T = 556nS. So according to this formula, the capacitor would take about 556nS to fully discharge. If my approximation is correct, there are other loads involved, could I do a rough calculation of what the voltage across the capacitor would be after 40nS? If I remember correctly capacitors do charge/discharge linearly. Thank you!

My gut told me to start out with a 470µF capacitor but I want to use C = (dt/dv)i to find something sized a bit more appropriately. Can someone doublecheck my understanding of the formula here?

C = ((40*10^-9)/1)4.75

= 1.9*10^-7 = 190nF the closest standard value being 220nF

Also, if I use T=RC to calculate discharge time I get the following time this value would supply 12V while my power supply catches up.

Solving for R in a worst case/in rush situation 12 = 4.75/R. R ≈ 2.53Ω. T = 2.53(220*10^-9). T = 556nS. So according to this formula, the capacitor would take about 556nS to fully discharge. If my approximation is correct, there are other loads involved, could I do a rough calculation of what the voltage across the capacitor would be after 40nS? If I remember correctly capacitors do charge/discharge linearly. Thank you!

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