0

enter image description hereI have a 1400ma constant current power supply (plenty of power to drive all loads at 100% duty), and I want to PWM 4 separate loads that are all driven optimally at either 350ma, 700ma, or 1400ma. The voltage drop across the loads vary. If this is possible, is there a most-efficient way to do this?

I am no EE, but I was considering some sort of current mirroring circuit with resistive current dividers (since loads are common factors of 1400ma) but I can't figure out an efficient way to do it when any of the individual loads is turned all the way off (0% duty). Any help would be greatly appreciated, and I can provide more details if this gains interest.

Wesley Casper
  • 1
  • 1
  • Please add a schematic or diagram of your idea. – Aaron Apr 14 '22 at 14:20
  • Ok, thanks for the tip. I added one – Wesley Casper Apr 14 '22 at 14:33
  • 2
    You have a 1400mA CC supply and one of the loads by itself is 1400mA. Where is the 1350mA for the other three loads going to come from? Is your PWM going to be synchronized so that the loads are not all getting power at the same time? But then you say 'plenty of power to drive all loads at 100%', so apparently not? – GodJihyo Apr 14 '22 at 14:37
  • I would like to be able to pwm all the loads separately while being driven from the same power supply. I am not sure how to do it, and this diagram is what I came up with, but I don't think it will work. I just attached it to show what I had been thinking of. I am stuck. But the 1350ma could be rounded up to 1400ma and I could run both 'legs' on the right of the current mirror at 1400ma. The PS has some headroom even at 100% duty of all loads. – Wesley Casper Apr 14 '22 at 14:44
  • 1
    Constant current means that as the load varies the current stays the same while the voltage changes. Currents in parallel add, so if you have two 1400mA loads in parallel they would need 2800mA. In series they would needs 1400mA, but the voltage would be higher. I'm not sure what you mean by the PS having headroom, do you mean voltage or current? It would help us to have specs for the supply and loads. – GodJihyo Apr 14 '22 at 15:25
  • @WesleyCasper Just skimming, I find \$2\cdot 350+2\cdot 350+2\cdot 650 + 1400=4100\$ or about \$4.1\:\text{A}\$ total. And each of these appear to have different compliance voltage requirements. (For speaking purposes, assume worst case and 100% duty cycle for all of them unless you can promise that will never happen under any conceivable circumstances.) Do you get the same result as I show here? – jonk Apr 14 '22 at 17:52

0 Answers0