# Using Capacitor Formula To Help With Power Supply Voltage Drop

#### odm4286

Joined Sep 20, 2009
246
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!

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#### odm4286

Joined Sep 20, 2009
246
Bump for morning viewers. Hopefully that's ok mods

#### oz93666

Joined Sep 7, 2010
737
The way I would approach this is by energy calculations ...

energy stored in a capacitor is 1/2 CVV .. what level is acceptable for the voltage to drop to ?? I will have to drop ...11V ???

calculate energy to cover 40nsec ... and energy stored in capacitor between 11 to 12 v ...

What capacitors do you have at hand?? 40nsec is not long , I would think 470uF would cover it , bigger the better of course

#### shortbus

Joined Sep 30, 2009
8,206
40 Nano Seconds? You're kidding right? How can you even see that, unless you're watching a oscilloscope? For this to be a problem it seems your leaving a lot of things out in your question.

#### Hymie

Joined Mar 30, 2018
928
Fundamentally your calculations/formula are correct – but you are in error in how you are applying these to your motor supply circuit.

The calculation you have made indicated the capacitive hold-up value that is currently in the circuit (giving a 1V drop over 40ns due to a 4.75A current draw).
You can use your formula to determine the circuit capacitance to say reduce the voltage drop to 0.1V over the 40ns;

So we get C = 4.75 * 40*10^-9 / 0.1 = 1.9uF

Since the relationship between voltage drop and capacitance value is linear, we can see that 0.01V drop would be achieved with a capacitance value of 19uF.

Capacitors charge/discharge linearly if the current is constant; if the load is resistive then the discharge follows an exponential decay curve due to the current reducing as the capacitor voltage decays.