I have designed a power delay switch which is intended to cause a short delay while a reservoir capacitor fills up before giving power to my microcontroller due it needing a short burst of high current before settling down to it's operating current of around 5uA.
The power delay circuit (see below) itself works fine and gives me the delay I need and everything powers up as it should, however my issue arises when the circuit loses power.


simulate this circuit – Schematic created using CircuitLab

The circuit gets 2V from a regulator which regulates batteries (3V) down to the 2V I want to power the MCU as this will ensure that the supply voltage is constant which makes ADC readings and calculations easier to calculate.
Now the problem that I am having is that when the batteries are disconnected (circuit loses power) the two capacitors (C1 and C2) take a long long time to discharge (about 3 minutes) - this isn't a problem is power is reestablished quickly (under 2-3 seconds) which is how long it takes the voltage to drop below the MCU's minimum supply voltage, after this point, however, the MCU completely shuts off and won't power back on unless the capacitors fully discharge (<100mV) and then the power can be applied again and it will work as it should. So I feel I need a way of completely draining the capacitors in the event of power being lost. It might be worth noting that the capacitors initially discharge fairly quickly down to about 0.6/0.7V before beginning the painfully slow discharge process - if anyone has any bright ideas as to how to resolve this, that would be great.
Another alternative is to possibly design a slightly better delay switch that will cut the power line from the MCU when power is lost - as you might notice this one will not do that, but again I can't think of any ideas for that as current consumption is a big thing and this current design draw virtually no current once running properly.

  • \$\begingroup\$ What about bleeder resistors? \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 6 '15 at 10:41
  • \$\begingroup\$ What is the required drop-out voltage of the 2 volt regulator? \$\endgroup\$ – Andy aka Jan 6 '15 at 10:49
  • \$\begingroup\$ The regulator doesn't drop out until the input starts dropping below 2V - the requirement is just as low as possible but unlikely it would be required to actually work below 2.5V. A bleeder resistor would waste current which I do not want \$\endgroup\$ – MrPhooky Jan 6 '15 at 11:18

Once the reservoir capacitor voltage goes below 0.6V there's not much you can do, because any transistor you try to use to discharge it further will also be turned off. However you could discharge the capacitor in the time delay circuit before the reservoir capacitor goes below 0.6V. This will completely remove power from the MCU so it should reset properly, and when the battery is reconnected the power on delay will be reapplied.

A supervisor IC could be used to discharge the timing capacitor, but are any available which have a low enough operating current and trip voltage? Alternatively you could make a discrete circuit using two FETs and a resistor, like this:-

enter image description here

Note: Q8 must have a low Gate threshold voltage to ensure that it stays turned on below +2V (BSS138LT is 0.5~1.5V).

  • \$\begingroup\$ Thanks for a good answer - my only issue is, when I have been testing the circuit as it is, to avoid having to wait 3 minutes before I can try again I just shorted out C2's pins - effectively what this would be doing - but it doesn't seem to shut off the output / turn on Q1 so I have to short it out more than once before it actually gets to zero volts... \$\endgroup\$ – MrPhooky Jan 7 '15 at 9:19
  • \$\begingroup\$ Until C1 is completely discharged it can recharge C2 through R1 and R2. There should only be about about 0.6V left after the reset circuit gives up, which hopefully won't be enough for the MCU restart. If it is a problem then insert a small Schottky diode between +2V and R6, and a capacitor from R6 to ground. This will hold up the voltage on R6 to keep the reset circuit activated, even after the voltage on C1 drops below 0.6V. \$\endgroup\$ – Bruce Abbott Jan 7 '15 at 9:49

You could try with a relay which is powered by the batteries and is placed in series with a bleeding resistor. Of course, powered = open switch. The relay will also draw some current, but there's not much you can do without additional power consumption.

Also, you could try with a PNP transistor that has the base connected directly to batteries and has a resistor between base and ground (a pull-down resistor): whenever powered, base is high and PNP is open; whenever power is removed, base is at ground, PNP is closed and the capacitor discharges through an additional bleeding resistor. Power consumption is basically as low as it gets.

  • \$\begingroup\$ The PNP solution would only allow the caps to discharge to ~0.6V as the base of the PNP needs to be 0.6V less than the emitter and the caps take longest to discharge from 0.6 - 0. \$\endgroup\$ – MrPhooky Jan 6 '15 at 13:46
  • \$\begingroup\$ You are right and also the relay is not an option, if your current for the MCU is 5uA: the relay would consume WAY more than that and it would be inefficient. \$\endgroup\$ – FarO Jan 9 '15 at 10:36
  • \$\begingroup\$ What about MOSFETs? your batteries are at 3V and MOSFET may conduct above 2V: removed the batteries, they act as open switch and the power to the MCU is cut. You would need to place one after the MCU, of course, to get a >2V potential on the gate. \$\endgroup\$ – FarO Jan 9 '15 at 10:38

Perhaps you are looking for a power management supervisor IC (they have funky names and flavours)? Essentially something that holds the IC's reset/enable until the voltage rail is sensed to reach a certain level. Similarly they can be made to begin a shutdown process if they sense voltage falling below a safe level. Something like this?



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