I was looking through a popular circuit on Sparkfun and trying to lower the power consumption of the device. I found that the outputs of their regulators are shunted to ground with a 300-ohm resistor.
Any idea why they shunted the output to ground with a resistor? Does it have to be such a low resistance?
I'm trying to reduce the current consumption of this small device as much as possible. Thanks!
You should find some of the better LDO IC's for example those offered by Micrel (like the MIC5205) which do not need a load resistor.
For as low power as possible, make sure the input voltage is also as low as possible to what you eventually need it to be. Like, if you want 2.7V as your system power rail, then use a 3.2V Li-ion coin cell or something, until a 2.7V fixed LDO regulator like the one I suggested earlier.
But yes, for the load resistor you see there on Sparkfun, that is because the linear regulator IC cannot maintain the voltage level on the output unless there is >= 10mA or so. This is terrible for power efficiency, and you should avoid that device. The internal pass element (probably a MOSFET) and it's feedback loop (Opamp to the Gate of MOSFET) obviously cannot handle very low duty cycle (no-load condition) properly. This could be due to cheap components or perhaps bad bandwidth of the feedback circuit. If it has no load, it may oscillate/overshoot or just cut off entirely until there is enough load/demand to continue normally.
It's a little hard to say without complete information but I would guess that this is a shunt voltage regulator used to provide a precise reference voltage for A/D or D/A converters. The regulator may require a certain amount of current to be drawn in order to maintain precise regulation.
Without a part number, we're just guessing, but my guess is that's a cheapie LDO regulator (perhaps an AP7115) to generate an analog supply voltage of 2.8V from the 3.3V. Probably not very accurate.
The ~9.3mA shunt may be to prevent the 2.8V from rising if they allow current from a higher voltage output into the inputs of a 2.8V chip, but that's just a guess.
