On Nov 26, 5:09*pm, "P E Schoen" wrote:
"Jeff Liebermann" *wrote in message

...

That's because of the 12uA typical quiescent current, where the
chip draws about the same current as the caliper load. *For equal
currents, that's 50% maximum efficiency. *The TPS62054 shows
50% efficiency at 2.7V in and 1.8V out (See Pg 8 Fig 4).
The chips do have a shutdown pin that cuts the quiescent
current to "less than 2uA". *Still high, but much better.

A step-down regulator/converter could be made from a Microchip PIC18LF14K22
(http://ww1.microchip.com/downloads/e...Doc/41365c.pdf) which has a
quiescent current of 34nA and an operating current of about 10 uA at 1.8
VDC.

And it may be even more efficient to use a low power linear regulator such
as the TPS71501 (http://www.ti.com/product/tps71501) which has 3.3 uA
quiescent current. If the input voltage is, say, 2 VDC and the output is 1.6
VDC at 12 uA, the overall efficiency is (1.6*12)/(2*15.3) or almost 63%.
Even at 3 VDC input it is 42%.

Paul

The LED shunt regulator saves the 2uA, so it's simpler, cheaper, and
even more efficient.

--
Cheers,
James Arthur