Efficiency of a voltage reference to drive a load?

[Lumberjack]

I'm looking at using a series voltage reference to drive a resistive load.
I have no experience with these other than "powerless" uses such as providing Vref to an opamp. But it seems there are some parts out there that can handle some mA's of current. The example part I'm looking at is TI's REF3012 which provides 1.25V out at 25mA max load.  http://www.ti.com/lit/ds/symlink/ref3012.pdf

As an example, if I have an output load of 125 ohm (for easy math), that will draw 10mA current at 1.25V.
My input supply to the REF3012 will be, say, 5V. What's the current draw from the 5V input?
Is it simply power in = power out, plus the quiescent current?
That would mean, 1.25V * 10mA output, divided by 5V input; 1.25V*10mA/5V = 2.5mA draw from 5V? Plus the 42 uA quiescent current for a total draw of ~2.542mA @ 5V

This seems to give an efficiency of 42uA/2.542mA = 98% ? Seems too good to be true.


FWIW, I'm thinking of using this to provide the filament supply for a 6418 vacuum tube, stepping down the voltage from rechargeable AA's. If I have the math correct, it looks like it's a low cost, low component count, and very efficient solution. What am I missing?

Tube datasheet: https://frank.pocnet.net/sheets/127/6/6418.pdf

[SeanB]

A lot cheaper to simply use a LM317 voltage regulator to drive the tube, as the output voltage will be right for it, and the LM317 will be perfectly fine driving the 10mA load as well. It will be a lot cheaper as well.

[Paul Moir]

Is it simply power in = power out, plus the quiescent current?
That would mean, 1.25V * 10mA output, divided by 5V input; 1.25V*10mA/5V = 2.5mA draw from 5V?

No, only switching regulators can do that.  This is effectively a linear regulator, so efficiency will be 1.25V * 10mA out / 5V * 10mA in.  So 25% - quiescent power.  If you needed a bunch of rails then you might want to design your own switching power supply.  If not you can buy one for $5 from Digikey that does 5V->1.2V at 70%.

[Lumberjack]

A lot cheaper to simply use a LM317 voltage regulator to drive the tube, as the output voltage will be right for it, and the LM317 will be perfectly fine driving the 10mA load as well. It will be a lot cheaper as well.

True the LM317 will work fine, I've used it before, though I'll say it's not really much cheaper. Some of the voltage references I've looked at cost <50 cents in single unit quantity and pennies each for larger quantities. An LM317 also require some external passives as well, and though those are cheap, more board space = higher overall cost.

Is it simply power in = power out, plus the quiescent current?
That would mean, 1.25V * 10mA output, divided by 5V input; 1.25V*10mA/5V = 2.5mA draw from 5V?

No, only switching regulators can do that.  This is effectively a linear regulator, so efficiency will be 1.25V * 10mA out / 5V * 10mA in.  So 25% - quiescent power.  If you needed a bunch of rails then you might want to design your own switching power supply.  If not you can buy one for $5 from Digikey that does 5V->1.2V at 70%.

Right, of course. That's what I was missing in my over-thinking things - it's a linear regulator and efficiency is simply vout/vin, less the (rather trivial) quiescent loss. It would be rather efficient with a lower supply voltage, though. A lithium AA which is around 1.4V would yield 80% or better. Of course, with a non-rechargeable battery I should simply drive the tube directly from the cell and not lose any power to the regulator, since the tube can handle quite a bit of voltage swing (+/-20%) without troubles.

SMPS is of course efficient, but can be tricky with tubes due to RF or AF noise from the switching.

Appreciate the clarification!