I am driving an OLED display via two AAA batteries plus a charge pump to step up the voltage to 9V. I was wondering if there is likely to be any significant difference in efficiency between (a) using the two batteries in parallel and using a charge pump to step up from 1.5V to 9V, or (b) connecting the batteries in series and stepping up from 3V to 9V.
Electronic – 9V from two AAA batteries plus charge pump: batteries in series or parallel
batteriescharge-pumpoled
Related Solutions
I'd consider using a boost regulator to take you to a voltage suitable for your micro. For instance: -
All the devices above can produce 5V AFAIK and once you have your micro running you can drive the charge pump circuit.
Finding a booster that works from about 1V and takes you to 7V is a little tricky, that's why I opted for this method.
simulate this circuit – Schematic created using CircuitLab
Here's one circuit you could use to switch from series to parallel. Check that both switches are break-before-make or there could be sparks. The diode protects the charger if the batteries are switched into series. This will waste a little power, but it should still leave about 4.3V which is plenty for charging the cells in parallel. You could get a bit fancier with a 5V relay if you want, which would mean no external switch is required at all. I wouldn't bother with mosfets because they would increase the component count unnecessarily and require a higher gate voltage than you have at hand for high-side switching.
Best Answer
You can calculate the efficiency by using \$n = 1 - \dfrac{p_{in}}{p_{out}}\$ and get the power \$p_{in}\$ by measuring the current from the batteries times the batteries voltage and the power \$p_{out}\$ by multiplying the current to the LED times the voltage across the LED(s) (at the output of the converter).
This assumes the converter has a filter capacitor. Use a scope and a shunt resistor of about 0.1 Ohms if not. Although trickier, you can estimate the current consumption by averaging the pulses height times the duty cycle.
$$I_{avg} = I_{peak} \times \frac{t_{on}}{t_{off}}$$
From my experience doing a lot of this sort of thing, using a higher voltage into a boost converter will improve efficiency.