Wanting to power ~20 LED's; which small, common, & commercially available battery will give the longest life?

Question

20 White Leds, each rated at a nominal voltage of 3.4V and a nominal current of 20mA.

I want to use battery power exclusively. I'd like to use commercially available batteries. So AA, AAA, 9V, etc.

Not really looking to make my circuit any more complicated than a magnetic reed switch, resistors, and the leds themselves.

My two questions are interdependent and so would like to hear your thoughts. 1) best way to wire them series/parallel and combination thereof 2) best battery source for the longest life using common commercially available batteries (includes common commercially available Li-ion batteries (AAA, AA, etc.))

I used a calculator which produced the following results:

ignore V & R values in schematic, the left one was a suggested repeating pattern for the 4.5V source for all 20 leds, the right was suggested for 9V source which repeated for the remaining leds

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Selecting the longest-lasting battery is easy. For a given chemistry, the energy content is proportional to size, so get the biggest one you can. D cells will last longer than AAA cells. A 9V is just six very small cells in series.

There are many battery chemistries out there, but it's hard to beat alkaline batteries for long life at a reasonable cost.

It doesn't really matter how you arrange the LEDs, series or parallel, from an efficiency standpoint. Consider these two circuits:

^{simulate this circuit – Schematic created using CircuitLab}

Each has two LEDs, with the resistors and battery voltage selected such that 20% of the voltage is dropped over the resistors, and 20mA in the LEDs.

The power in a resistor is given by \$P=I^2 R\$.

Power in R1: \$ (20mA)^2 85\Omega = 34mW \$

Power in R2 and R3: \$ (20mA)^2 \cdot 42.5\Omega \cdot 2 = 34mW \$

This is the rate at which the resistors are converting electrical energy (and indirectly, your battery's chemical energy) to heat. As you can see, it's the same for either circuit.

There is also some power lost in the batteries, by their internal resistance. But it can be seen that the current through any of these batteries is the same. BAT1 and BAT2 must supply 20mA. BAT3 and BAT4 must supply 40mA to power the two parallel LEDs, but since the batteries are also in parallel, each supplies only 20mA. Since the current is the same in all the batteries, the losses are also the same.

However, if you really want to stretch your battery life, a resistor is not the way to ballast your LEDs. See How can I efficiently drive an LED?