# Super low power DC/DC conversion

I was inspired by heliosoph's "Arduino powered by a capacitor" blogs and wanted to replicate it to see how far I can get. To have some function I made a darkness detector - it blinks about once a second while it is "dark" and once 8 seconds when it is "light" (internal pull-up connected to ground via LDR decides). Using external wake up I manage to get about 2uA power consumption at ~2V but the power grows quickly with growing voltage.

To keep current consumption low even with charged capacitor (~5V) I made a "LDO" from 3 transistors and a few LEDs (voltage reference ;)). It has quiescent current less than 1uA so it helps my detector to work longer. But then I was calculating:

Capacitor with capacity C charged to 5V has 12.5C of energy. Dropping to 2V and using what is possible I burn about 6C of energy over load, 4.5C over LDO and 2C stays in the capacitor unused. It is less than 50% eficiency.
If I charged the capacitor to higher voltage the efficiency would be much worse. So I decided I NEED switched DC/DC converter. But even super-extra low power ICs I have found have quiescent current in order of uAs (because they are able to supply mAs of current) so I need to make one from discrete parts. At first I wanted to make charge pump but it would need 4 switches and I am not sure if it were possible to use in this application anyway. So I decided a buck-boost converter is the best way, possibly inverting topology - in theory I would be able to drain all energy from the capacitor. But I don't know where to start. Do you know (if it is even possible) some simple schematic of buck-boost inverter from discrete parts I can try and possibly modify? Or is there another simpler approach I can use?

Be sure to use the word micropower in all your searches - this is the standard term for ultra low
power consumption electronics. At these power levels a simple CMOS inverter is a good half-H-bridge,
note.

Hi,

Are you saying that you use most of the cap energy but some stays in the cap, and that you are using ALL that energy as the basis for an efficiency estimate?
If i understand you right, then that may be a mistake because energy that ‘stays’ in the cap is not energy consumed, it is energy stored, and stored energy is not part of the efficiency calculation.
Another view on this is that if you have X amount of energy still in the cap, when it comes to the charge time that means less energy has to be put back into the cap in order to charge it back up to Y amount of energy, and you only use energy between X and Y, not from between Y and zero, so you actually may be using less energy than you think.

MarkT: tried to google "micropower". I found some ICs but they can supply hundreds of mA and have quiescent current around 20uA. 5uA of output current is more than enough for me and more than 1uA of quiescent current will make it useless. I dobt there will be an IC doing this (but some clock applications may find it useful).
MrAl: It depends how you define efficiency. My definition is (energy used for application)/(all available energy). Your definition is (energy used for application)/(all used energy). Since energy to refill the cap is "free" any unused energy left in it is as good as wasted: if I manage to get the energy stored in the last 2V of charge with only 10% efficiency it is still energy gain. Surely if it were autonomous solar powered system where refilling the cap would cost much energy from solar panel it would be dubious if it is worth it. But I am getting the energy from my computer's USB port - the computer eats much more energy while I write this than the "darkness detector" for its whole life.

Still I wonder if something like this is possible. To reduce current I am using ~1M resistors for my "logic". But then I have slow rising and falling signals. This will lead to inefficiency of MOSFET and low frequency. I added estimated values to a calculator and it suggested something like 100kH inductor. So I will use my 33uH inductor (made a tiny mistake while ordering) and see what will happen.

Use CMOS logic, no pullups or pulldowns in micropower designs.

MarkT:
Use CMOS logic, no pullups or pulldowns in micropower designs.

While playing with discrete RTL I followed your advice and bought 74AC14 Schmitt inverter. I got it yesterday and had very little time so I made quick power consumption estimation. With all pins (except power ofc) unconnected the consumption was negligible. When I connected pin 1 (input of one inverter) to "signal" from pot I got a strange result: while signal was slowly falling (for a few seconds) the consumption rised to a few uA just before switching. But on rising signal however slow no rise in consumption happened. I looked at some schematics of CMOS Schmitt triggers and they seem symmetric. What is going on?

1. My measuring is wrong, rise in consumption is on both edges?
2. It is feature of AC family - "standard" HC has rise on both/no edge?
3. Measuring is invalid because other pins (including measured pin's output) were open circuit?
4)?
Thanks for help...

I Googled nano power buck converters.
Several hits with idle currents as low as 350nA.
Leo..

Apart from having fun with the project, have you find out any special reason to use a capacitor over a small coin cell for real life applications?

Smajdalf:
While playing with discrete RTL I followed your advice and bought 74AC14 Schmitt inverter. I got it yesterday and had very little time so I made quick power consumption estimation. With all pins (except power ofc) unconnected the consumption was negligible. When I connected pin 1 (input of one inverter) to "signal" from pot I got a strange result: while signal was slowly falling (for a few seconds) the consumption rised to a few uA just before switching. But on rising signal however slow no rise in consumption happened. I looked at some schematics of CMOS Schmitt triggers and they seem symmetric. What is going on?

1. My measuring is wrong, rise in consumption is on both edges?
2. It is feature of AC family - "standard" HC has rise on both/no edge?
3. Measuring is invalid because other pins (including measured pin's output) were open circuit?
4)?
Thanks for help...

Probably asymmetric noise on the signal. In the direction with spikier peaks the transition happens
while the circuit is only spending a small time near the threshold, so less current consumption.
The current consumption probably increases close to the threshold from either direction, due to
noise current flowing into/out of the gates/interconnect of the output stages of the trigger.