0
\$\begingroup\$

I have a microcontroller circuit with a 3.3v linear regulator providing power when run by a 5v supply, the circuit can also be run directly from a 3v battery which connects to the output side of the regulator.

I have found that when powered from the battery the regulator consumes 20mA.

Is there a proper way of designing this type of circuit so that when powered by the battery the linear regulator is isolated and doing nothing?

I am trying to keep the circuit design as low cost as possible so I would like to stick with a cheap 3.3v regulator and change the circuit some other way.

Any advice would be appreciated, Thanks

\$\endgroup\$
  • \$\begingroup\$ Cost versus functionality - you may not have a choice. Target cost for the add-on parts? What current does the microcontroller take (peak)? \$\endgroup\$ – Andy aka Jul 25 '14 at 10:51
  • \$\begingroup\$ The system will draw a few mA when operating with the odd peak when its transmitting data but this should be covered by a capacitor. \$\endgroup\$ – ArthurGuy Jul 25 '14 at 11:52
3
\$\begingroup\$

A few options:

  • find a regulator that tolerates 'back feeding' and consumes less current in that situation
  • isolate the regulator with a (schottky) diode. this will cause some voltage drop that might need compensation. A primitive way to do that is to put the same type of diode in the ground lead of the regulator.
  • automatically switch between the regulator and the battery, for instance with a swiitch (which might double as a on-off switch).
\$\endgroup\$
1
\$\begingroup\$

I have a little thought how to do it: -

schematic

simulate this circuit – Schematic created using CircuitLab

Hope I got this right! When 5V is applied M2 gate is biased on and M1 switches fully on - it wouldn't switch on at all if it were not for the internal diode shown above M1. Choose a low \$V_{GS(threshold)}\$ P channel FET for M1.

When the 5V is removed and 3 applied to the output, there could be enough leakage back thru U1 to start to turn M2 on but I've loaded that side with 100 ohms and quite frankly this could by 1kohm (hard to predict without testing).

If I've made a stupid error please do say but I reckon it should work BUT this sort of circuit is prone to oddities like the input capacitor on U1 (not shown) maybe retaining a bit of charge (hence the 100 ohms for R2). Worth a try possibly.

\$\endgroup\$
  • \$\begingroup\$ Could you put a Schottky diode on the regulator input to prevent back feeding? Then the 100R could be 10K or whatever just to cover off the diode leakage. Still should be enough volts for an LDO. \$\endgroup\$ – Spehro Pefhany Jul 25 '14 at 12:20
  • \$\begingroup\$ @SpehroPefhany - that would work nicely providing the OP is using an LDO regulator. \$\endgroup\$ – Andy aka Jul 25 '14 at 12:22
  • \$\begingroup\$ He did say cheap.. May or may not be LDO. \$\endgroup\$ – Spehro Pefhany Jul 25 '14 at 12:25

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.