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Let's say I have a power supply 48V/100mA (4.8W). However, I have a push pull solenoid which will draw ~8W (48V/170mA) for about a 50 mS pulse. Can my power supply provide this extra power for that time with a proper capacitor, or do I need to increase the nominal power of the power supply?

  • Well it depends on the power supply. Please give a link to the power supply. – Justme Jun 05 '20 at 16:14
  • Actually, I don't have a commercial power supply. I'm just trying to design one and was confused about that concept. –  Jun 05 '20 at 16:16
  • How frequently do you plan to operate the solenoid? – Dave Tweed Jun 05 '20 at 16:19
  • Once a single pulse is given, the next operation could be after a second. –  Jun 05 '20 at 16:22
  • How much voltage drop it can tolerate? Is the 100mA limit due to thermal limit, overcurrent limit, fuse limit..? – Justme Jun 05 '20 at 16:24
  • I put 100mA because that power supply would supply a very low power microcontroller, which will need at most 35 mA. So I thought that I would't need to overdesign the 48V output just for the solenoid that will not operate continuously. The 48V will be step down to 5V. –  Jun 05 '20 at 16:30
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    So do you have min/max voltage range for the solenoid? But the MCU would not use 48V directly but via a step-down regulator, so it depends on input range limit of your regulator too. Which regulator you use? Which MCU? – Justme Jun 05 '20 at 16:36
  • Here's something related. – jonk Jun 05 '20 at 16:41

1 Answers1

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Short answer - yes. The capacitor size is determined by the minimum holding current required by the solenoid, ignoring inductive effects such as dV/dt and back-EMF. With no supply connected, here is how to calculate the capacitor size. This example does not account for the regulator circuit headroom.

Your solenoid has a DC resistance of (48 / 0.17) = 282 ohms. Let's assume that for the solenoid to hold in whatever force is pulling on the armature, it needs a minimum operating current of 75% of nominal, or 128 mA. At that current the capacitor voltage is (0.128 x 282) = 36 V (duh). So the capacitor is discharging from 48 V to 36 V in 50 ms, into a load of 282 ohms. This is a simple R-C discharge circuit. The exponential equation can be solved for C, the capacitor size in farads.

A quick approximation that you can do in your head is to assume a constant-current discharge rather than exponential. In this case,

EC=IT

The change in capacitor voltage (in volts) times the capacitance value (in farads) = the constant discharge current (in amps) times the discharge period (in seconds).

Rearranging, C = (i x t) / E = (0.170 x 0.050) / 12 = 708 uF

If your supply has a 1000 uF output capacitor, or a 1000 uF bulk filter capacitor driving a regulator that does not go into an overcurrent shutdown, the solenoid will stay held in for at least 50 ms, even if the cap's voltage source is completely removed.

This method always yields a cap value that is larger than the exponential equation value. That "true" value is 616 uF.

Note: Headroom - if the 48 V source is the output of a regulator circuit, and the regulator's input is a bulk filter cap with something like 60 V across it, and the regulator has a minimum 2 V Vin-Vout differential voltage, then the cap's discharge voltage range is 10 V larger and the required cap size is smaller.

AnalogKid
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  • Thanks. The 48V comes from a smps output. It will feed this regulator: https://www.diodes.com/assets/Datasheets/ZXTR2012K.pdf –  Jun 05 '20 at 17:28
  • That part has a 12 V fixed output and can source only 15 mA. – AnalogKid Jun 05 '20 at 19:03