I'm designing a device which will be off more than 99% of the time. The device is powered from a single AA battery through a DC-DC converter (has enable pin) and comes to life when a button is pressed. The device executed a routine and turns off.

In order to start working I'm thinking of enabling power to the MCU using the button that will latch the DC-DC converter enable pin high and after the MCU is done it will release the latch so the DC-DC enable will go low and the system will turn off. This will enable me to save the maximal battery life.

Anything wrong with this approach?


Here is a latching circuit I've thought about. The user presses SW1, which is momentary, and that pulls the enable pin high. The DC-DC converter powers up and outputs 3.3V which is fed back to the enable pin and keep it high.

An N-channel mosfet is kept open using a resistor from gate to GND. This could be 200k - 500k range. When it is time to shut down the gate which is connected to an MCU pin is pulled high and that drops the voltage at the enable pin.


simulate this circuit – Schematic created using CircuitLab

Update 2 So it seems this DC-DC is a bit strange due to the diode being forward biased all the time. I've picked another device: XT1861. This has the enable switch setup a bit differently so here is an idea how to implement a latching circuit using the new DC-DC (I don't know if this actually works). The idea here is to use the switch to open the pass FET which will start the converter and that voltage will keep Q1 conducting which in turns keeps the FET conducting. Pulling the MCU's pin to ground will close Q1. That's the idea anyway...


simulate this circuit

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    \$\begingroup\$ Nothing wrong with that design. \$\endgroup\$
    – Passerby
    Nov 29 '19 at 7:54
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  • \$\begingroup\$ @Passerby Thanks. The 1st circuit shown in the video at Hackaday is actually pretty close to what I had in mind. I've just noticed another problem... My DC-DC goes down to 0.85V which is good for running the device on a single AA however enable high only above 1.6V :/ \$\endgroup\$
    – user733606
    Nov 29 '19 at 8:12
  • \$\begingroup\$ Can you link to DC-DC you are using? The requirement for EN voltage to be higher than input does not make sense to me. Can you use lithium battery instead of alkaline? \$\endgroup\$
    – Maple
    Nov 29 '19 at 9:27
  • \$\begingroup\$ @Maple datasheet.lcsc.com/szlcsc/… page 4 shows the enable voltage treshold 1.6V although startup is at 0.85V \$\endgroup\$
    – user733606
    Nov 29 '19 at 9:54

Nothing is wrong with your approach except for being somewhat vague.

For example i understand the intent behind "using the button that will latch the DC-DC converter enable pin", but not how a button (I assume you mean momentary switch) can "latch" anything.

In any case, I'd recommend taking a look at TPL5111 chip, which was specifically designed for applications like this.


After looking at the DC-DC specs you've provided all I can say - it is one weird device. To begin with, it's CE pin is usually connected to output (that's how they achieve auto-start, I guess). Then, if it is connected to 0 the output is equal to input?! This is not something that will give you maximum battery life! Quite the opposite, it will drain it in no time. Do not use that pin for power control.

I'd recommend either using lithium battery with TPL5110 and MOSFET, or one of the latching circuits suggested by @Passerby in the comments. In both cases the FET switches power going into DC-DC, not CE pin.

Regarding CE pin voltage - simply connect it to output and try powering DC-DC with 0.9-1.3V. From the datasheet I suspect the way CE works is it disables some parts of the internal control logic. However the DC-DC still begins oscillation when the voltage is above 0.8V, producing some kind of non-regulated output voltage. If this output is connected to CE then when it reaches 1.6V internal regulation is enabled and chip enters normal operation mode. Which means - you should be able to start it even from drained AA.

  • \$\begingroup\$ The latching circuit is not determined yet, however will probably involve a P channel MOS and a transistor that will keep the the gate's state until shut down by the MCU. The problem I find in almost all applications is that the enable input required higher voltage to register ON than the minimal voltage input on the DC-DC which means that I won't be able to max. out the batteries. \$\endgroup\$
    – user733606
    Nov 29 '19 at 9:05
  • \$\begingroup\$ Every single DC-DC I've worked with had EN pin working as expected when pulled up to input voltage. You must have something unique then. You can use TPL5110 which has integrated MOSFET driver if you cannot use EN pin, but I'd check twice before making schematics more complex than necessary \$\endgroup\$
    – Maple
    Nov 29 '19 at 9:13

The boost converter you selected uses a diode, which means the input is always going to be connected to the output. There is no way to turn that diode off. I think that's not what you want.

enter image description here

Your options are:

  • Wire the pushbutton in series with the battery.

So when it is not pushed, there is zero current draw. When pushed, DC-DC receives battery voltage and generates output voltage. Then add a FET to short across the pushbutton to keep the device powered until it wants to turn itself off. You'll need a very low voltage threshold FET though.

  • Use a synchronous boost DC-DC which can disconnect the input from the output.

Like this one or similar. This solves the diode problem. When it is off, the rest of the circuit will not receive any voltage. Check datasheet page 12, it explains how this works.

  • \$\begingroup\$ To be honest I am not sure why they have the diode there to begin with. The specified model conducts when the voltage diff is 0.45V which is way under the minimum start up spec. So this means the diode is always conducting - also when the circuit is off. \$\endgroup\$
    – user733606
    Nov 30 '19 at 13:08
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    \$\begingroup\$ The HT77xx you linked in the comments is an asynchronous boost converter, it only has one switch inside, which means the diode is required for it to work. And yes it will always conduct and there is no way to turn it off. That's why I'd recommend a synchronous boost converter, which has two switches (one FET replaces the diode) and that is both more efficient, and it can be turned off. \$\endgroup\$
    – bobflux
    Nov 30 '19 at 13:12
  • \$\begingroup\$ Agreed. I am looking for an alternative device right now. I am not sure in what applications this circuit is acceptable and can justify the neglectable cost difference. BTW if you don't mind please take a look at my "latching" circuit concept in the original post. Would be happy to have your feedback :) \$\endgroup\$
    – user733606
    Nov 30 '19 at 13:15
  • \$\begingroup\$ Yeah this chip is weird. Your circuit has a mistake: the FET will short the output of the boost converter via the diode. If you replace the diode with a resistor it will work. \$\endgroup\$
    – bobflux
    Nov 30 '19 at 13:20
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    \$\begingroup\$ Sure, but you have to replace D1 in your schematic with a pullup resistor, and also add a resistor in series with the pushbutton. \$\endgroup\$
    – bobflux
    Nov 30 '19 at 13:32

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