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I have a microcontroller with a program that remains in "sleep" mode 99.9% of its life. During this sleep mode, it drains less than 1 microamp with an operating voltage of 2.0V to 3.5V. When it finally is awake, it has a short burst of 2.2 mA consumption that lasts about 10 ms.

Elsewhere in the circuit, I have a 12V battery being used. I'd like to save physical space by eliminating my 3V battery and just tapping off the 12V battery with a 12V to 3V converter.

It seems like most linear regulators and switching converters I've found have very very awful efficiency at low current consumption.

Does anyone have a novel solution to regulate a low power device, relatively efficiently?

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    \$\begingroup\$ Have a look at the answers in this question \$\endgroup\$ – Doodle Sep 13 '16 at 13:27
  • \$\begingroup\$ Search for switching regulators with Light Load Efficiency feature. These are design to give higher efficiency in these low current regions. \$\endgroup\$ – Bence Kaulics Sep 13 '16 at 13:43
  • \$\begingroup\$ LTC1574, LTC1174, ... and similar devices. \$\endgroup\$ – Marko Buršič Sep 13 '16 at 13:51
  • \$\begingroup\$ @user2913869 - does an external event pull the microcontroller out of sleep? Or does the microcontroller only need to do its task "occasionally" ? \$\endgroup\$ – glen_geek Sep 13 '16 at 14:21
  • \$\begingroup\$ Forget about efficiency - you are going to get poor efficiency in this application; more importantly is what the normal load is on the 12V battery and how much of a problem is caused by a few extra uA taken from it. \$\endgroup\$ – Andy aka Sep 13 '16 at 17:05
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If cost is not too much of a concern, a switching regulator such as the LTC3388 may suffice. It draws well under 1uA (Iq, no-load) with a 12V input.

Your average consumption (at 3V) will be about 3.2uA, assuming the 0.1% duty cycle number is accurate, so the average from your 12V battery should be less than 2uA (but read the datasheet carefully).

A simpler and potentially cheaper solution would be to use a linear CMOS regulator with (say) 1uA Iq. The average current draw would then be ~4.2uA from the 12V battery. That may in fact be much less than the self-discharge current of the 12V battery so it might not make much difference (to the battery life) which you choose, despite the more than 2:1 decrease in efficiency.

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  • \$\begingroup\$ It seems that, according to the datasheet, the LTC3388 has an efficiency of 20% at 10Vin and 1uA Iload. Which means 25uW of power consumption, or more than 2uA form the 12V, am I wrong? \$\endgroup\$ – CasaMich Sep 13 '16 at 14:22
  • \$\begingroup\$ A larger inductor will benefit the higher voltage inputs, but I would expect the efficiency to be higher with 3V than with 1.8V output. Typical Iq with 12V in in regulation is 800nA so I think 2uA is too high an estimate for supplying 1uA out. \$\endgroup\$ – Spehro Pefhany Sep 13 '16 at 14:35
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If your microcontroller is tolerant on the 3V level, you could use a 9V zener diode in series to derive the 3V form the 12V. This solution is not efficient in terms of percent of power dissipated, but when the microcontroller absorbs 1uA, you will use only 12uW. I don't think a buck converter can work with so low power.

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    \$\begingroup\$ The only thing left is to find a zener working with 1 uA. I heard those grow somewhere near the unicorn meadows ;) \$\endgroup\$ – Dmitry Grigoryev Sep 13 '16 at 14:18
  • \$\begingroup\$ @Dmitry Grigoryev Actually the avalanche starts at 2.2mA. Obviously there must be some capacitance on the 3V side. \$\endgroup\$ – CasaMich Sep 13 '16 at 14:36
  • \$\begingroup\$ I have seen zeners which work with as little as 1 mA, but a device going 3 orders of magnitude below that would genuinely impress me. \$\endgroup\$ – Dmitry Grigoryev Sep 13 '16 at 14:49
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    \$\begingroup\$ @DmitryGrigoryev It might indeed work, since 9V zeners are pretty good at lower currents (especially some of the Japanese ones) but I suspect this is not really a practical solution since a small drop in battery voltage leaves nothing for the MCU and a small increase will leave too much. \$\endgroup\$ – Spehro Pefhany Sep 13 '16 at 14:55
  • \$\begingroup\$ THe zener will work correctly at startup, when it charges the capacitor, and during the burst at 2.2mA; during the low current period, the voltage on the capacitor will certainly drop. Also the battery voltage variation can be a concern, this is why I started my answer with the condition of tolerance on the 3V supply. \$\endgroup\$ – CasaMich Sep 13 '16 at 15:00

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