I am looking for a method to drive/pulse (ocasionally) an inductive load from a battery. The problem is, that the battery does not deliver enough power the load needs. But, energy is not a problem. I thought about boosting the voltage and charging a capacitor, then almost instantaneously releasing that energy to the load. These are the batteries' specs:

  • I_max = 200 mA
  • U_bat = 3 V

Regarding the load:

  • L = 500 mH
  • I = 0,5 A
  • P = 6 W
  • t_drive = 30 ms

Of course, if something is missing, please point out and I will be glad to supply that information.

What I'd like to ask is if you see any pitfalls or would propose another approach?

@Neil_UK: The load is a selenoid. Which specs are missing you think?

@Transistor: 3 V is the battery voltage. The capacitor's voltage I want to boost up to maybe 16 V

@Joribama: hence I'd like to boost up the voltage since: $$E=\frac{1}{2}C U^2$$

@pipe: I do not have the datasheet yet :/, just those numbers posted above.

@pjc50: Yes, boosting up the voltage is what I was thinkgin of, too.

  • \$\begingroup\$ charging an intermediate capacitor is an excellent method of energy storage to drive a load that needs high power pulses. Your load is not yet well enoiugh specified to tell what the pitfalls might be. \$\endgroup\$ – Neil_UK May 10 at 5:25
  • \$\begingroup\$ 3 V x 0.5 A = 1.5 A. Where is 6 W coming from? \$\endgroup\$ – Transistor May 10 at 5:41
  • \$\begingroup\$ Please also link the datasheet to your load, to avoid even further questions \$\endgroup\$ – pipe May 10 at 8:12
  • \$\begingroup\$ Your numbers don't add up very well, as @Transistor pointed out. Assuming you need 6W for 30ms only, this represents 0.18J of energy. A capacitor to store that amount of energy @ 3V would have to be at least 40,000uF. Maybe a super-cap can do the job as long as its ESR is not too high. \$\endgroup\$ – joribama May 10 at 8:20
  • \$\begingroup\$ This strategy - boost converter charging a capacitor - works very well for cash drawer solenoids @ 12/24V. \$\endgroup\$ – pjc50 May 10 at 9:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.