Just wondered if any of you have ideas on
designing a very low voltage low power SMPS. It's for a 1 off home build
fun project for my shed. I've inherited a lovelly old Sifam analogue
meter reading 0 to 10V at a current of 1 mA for 10 V reading. If I put
this meter straight across the battery powering the shed lights (with a
bit of circuitry so 10 to 20 V suitable for a 12 V battery shows as 0 to
10 V on meter) it's going to take maybe 0.5 to 1 mA. This is quite a
permanent drain on the battery and speed its discharge.
Inside the meter there's a 10k resistor in series with the actual meter movement. At 1 mA for a 10 V reading the movement itself only drops 71 mV. So I thought about using an SMPS to convert the 12 V battery supply to the 71 mV for the meter itself. So let's say I want 10 to 20 V from the battery to show as 0 to 10 V on the meter. At 100% conversion efficiency this means the 71 mV @ 1 mA for the meter would require 71 uW, from the battery this would be 3.5 uA at 20V. Now 3.5 uA from the battery is likely to be less than its self discharge rate so I'm not going to mind that.
But how to design an SMPS to get the 20 V to 71 mV at good efficiency?
A simple SMPS output with a Si diode rectifier would give an efficiency of only 10% if I assume a diode drop of 0.7V. A Schottky diode would drop maybe .5ish V and have lower switching losses so a step in the right direction. But still not good.
This is a challenge! Any ideas on a higher efficiency design would be welcome.
Couldn't you just put another 10K resistor in series with the meter? That would basically cut the current in half and make it a 0 to 20V meter.
That's not how it works. The meter needs 1mA to go full scale, so if you just do the job with series resistors it needs 1mA whether you're measuring 0-1V or 0-1kV.
Ah yes, it seems I completely misread the OP--his goal is to drive the movement itself directly. Sorry about that.
There's certainly not a dedicated part!
Here's some possibilities:
Go shopping for the lowest output current lowest standby current buck converter that I could find that works with a 24V input, then use an op-amp circuit to translate the output voltage at the meter into an appropriate feedback voltage to the converter. The op-amp circuit would still have to run off of your 20V supply unless you go for a two-stage circuit, so you'd also need to shop for a low-power op-amp.
Go shopping for the most efficient 24V to 1.8V buck converter, then use an op-amp voltage follower running off of 1.8V to supply that 1mA to the meter. This probably isn't going to make you happy (you're still at 1.8mW, after all), but it's better than 1mA from 20V, and a super-low-power op-amp should be easier to find.
Interesting ideas. Had wondered about 2 stage system and 1.8V is not too far from the geometric mean from 20 V to 71 mV, so 1.8 V might be a reasonable intermediate voltage. I'm clearly out of touch, had not realised low power 1.8 V buck converters were available off the shelf. Useful info, thanks.