2

I'm currently working on a project where I need the MCU to be able to keep running for at least a few seconds after the power have been interrupted (in case of power outage for example).

So I choosed a 470uF cap for the output of the regulator (AP7333-33), but unfortunatly when the power is interrupted, the cap discharge quickly in the regulator (and through the regulator, in the wall adapter).

I'd like to prevent this phenomenon, but I can't find a satisfying solution...

I tried to put a schottky diode between the regulator and the cap, but this is far from an ideal solution, as voltage drop and reverse leakage current lower the time my MCU have before shuting down.

I also tried various mosfet and BJT configuration, but could not find anything good.

I feel a bit stupid, ok I'm a beginner, but I'm sure such a simple problem have a simple solution !

schematic

Chenard
  • 31
  • 5
  • Schematic? And why are you worried about the wall adapter receiving this charge? Is it poorly designed? Details about it? If you really want to discharge locally use a bleeder resistor, but it will draw power all the time. The typical regulator protection schemes do assume that you can discharge back into the upsream supply. – Fizz Oct 19 '15 at 19:02
  • By typical schemes I mean http://electronics.stackexchange.com/questions/148340/how-to-protect-lm317-from-output-short – Fizz Oct 19 '15 at 19:09
  • You should perhaps firstly look into lowering the power consumption of your MCU by deactivating peripherals (c.f. datasheet), if it is possible. – Mister Mystère Oct 19 '15 at 19:13
  • 4
    I'd move the diode and capacitor to the input side of the regulator. This will allow the regulator to drain down the capacitor until it drops out, allowing the circuit to power your MCU with a stable voltage for longer. – Jon Oct 19 '15 at 19:16
  • @Respawned I'm not worried about the wall adapter. I'm worried about any element eating up the charge in my output cap, should they be before the regulator, or the regulator itself. Actually if I unplug the wall adapter from the wall, the cap discharge in 500ms, but if I disconnect the wall adapter from the circuit, the cap discharge more slowly. This is why I use a Schottky just before the cap, so that it will not discharge trough or in the regulator. – Chenard Oct 19 '15 at 19:48
  • @Mister Mystere The probleme is not comming from the MCU, I already applied all the required techniques. And it works fine, as long as the regulator is no longer connected to the cap. – Chenard Oct 19 '15 at 19:49
  • @Jon Thanks for your suggestion, it adress the voltage drop issue. – Chenard Oct 19 '15 at 19:49
  • "any element eating up the charge in my output cap, should they be before the regulator" So what do you have before the regulator? The usual scheme (see other question linked) is a diode in parallel with the regulator so that it's reverse blocked in normal operation but opens when output voltage exceeds the input. With that you see no drop at all from the diode in normal operation. – Fizz Oct 19 '15 at 19:55
  • You could increase the value of R4 and R8, if you do not mind the increased noise. That's not on the output, but that wee bit of energy saved on the first cap could make its way through the regulator to the second cap. – Mister Mystère Oct 19 '15 at 19:58
  • But I see you want use the cap as "UPS"... I think the usual solution for this is a bigger (than output) "tank" cap at the board's input, before the reg. – Fizz Oct 19 '15 at 19:58
  • You could try a diode in the common leg of the regulator and a diode back on the output. The regulator output will be lifted by a diode drop which will become subtracted by the output diode. Might need a resistor to Vin to put a little current thru the diode. – Andy aka Oct 19 '15 at 20:03

1 Answers1

1

Ok thanks for your comments. The tank cap before the LDO seems to be the right way. I'm gonna test this solution.

By the way I also found this question, and the answers interesting : Help with power-loss protection using capacitor

Community
  • 1
Chenard
  • 31
  • 5
  • What about the idle current of the reg .Ye olde 78 series are hungry.Deal to thev reg as well as what you are doing already and your cap value and hence size and cost should be OK.Remember 1 millifarad =1 milliampsecond per volt. – Autistic Oct 20 '15 at 08:54