Let's say I'm designing a power supply with input 120-600VAC and output 48V/100mA DC. The load for the 48V output is:

  • A 3.3V switching regulator
  • Two LEDS (~2 mA each one)
  • An ultra low power microcontroller
  • Some ultra low power op amps
  • A 48 solenoid with 300 ohm DC resistance. It will be given a 30 ms pulse.

My question is on the solenoid. 48/300 = 160 mA which is greater than my power suppply spec of 100 mA. But the point is that it will be given just a 30 mS pulse. How do I calculate a capacitor that will provide the necessary extra current during the 30 mS pulse? I have to make sure not to have voltage dips during that pulse. Do I need to reconsider my power supply current spec?


2 Answers 2


A drop of 0.5 V can be assumed. with this, with a load of about 160 mA, you can have a capacitor of about 10 mF placed to act as buffer. A factor of two can be considered. so 20 - 33 mF capacitor will definitively hold the voltage supply for that device with in 0.5 V of 48 V. Provided, there will be time for the capacitor to recharge to 48 V back, before solenoid is driven again. Also, be sure about the voltage rating of the capacitor as well while choosing.

If the drop is less, and is okay to drive the system still with the lesser voltage, you may not need the capacitors to be accurate as well. you can monitor once using oscilloscope.

C = (load current)*(load current duration)/(allowed drop in voltage)


Start with the dropout voltage for the relay. This is the minimum voltage that will hold in the relay armature. Let's assume that it is 38 V for a nominal 48 V relay. That is a delta of 10 V, the amount of decrease in the capacitor voltage over 30 ms. You can use the equation for the exponential voltage decrease in an R-C circuit, with 300 ohms for R, to calculate C.

A shortcut is to use the constant-current equation as a first-order approximation:

E C = i t

E = voltage change during time t

C is the capacitor size in farads

i is a constant discharge current

t is the discharge time

For i, use Ohm's Law to calculate the coil current at 48 V and 38 V, and take the average. Use this to calculate C. I get 430 uF; 470 uF gets you a little margin, and 1000 uF is much better if you can handle the size. For long term reliability, use parts overrated by 100%: for a 48 V circuit, use 100 V capacitors, diodes, etc.


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