Electronic – How to wire 6 LEDs to a 9-V battery

Well, let's analyze the circuit. We know that the power required in a DC circuit is:

P = Vsrc * Iout

We know that

I = (Vout - Vled)/R

and the power delivered to the LEDs is all that matters, so we want to maximize

Pr = (Vout - Vled) * I = (Vout - Vled)^2/R
Pled = Vled * I = Vled * (Vout - Vled)/R

Clearly, we want to minimize Pr and maximize Pled. We can do this without decreasing the current by reducing R and making Vled close to Vsrc.

This is accomplished by putting the LEDs in series.

However, your battery (isn't the 6LR61 a 9V battery?) will go from some nominal voltage (ex 9V) to a lower voltage - 9Vs are spec'd to be dead at 4.8V. This means that a passive solution will go dim while there's still charge left in the battery. For your original schematic, that might mean that you'd end up below the minimum current to turn the LED on, or for the series version, the voltage might go below the diode forward voltage.

A simple way to extract more brightness with the same power is to pulse the LEDs - Human eyes percieve blinking light to be brighter than continuous light, even if the average power is the same. A 555 timer or other oscillator/switch combination will be able to do this, no microcontroller required. Try playing with the duty cycle and frequency of your LEDs to see where it looks the brightest - You may be surprised!

Also, a switching power supply can increase the efficiency of your regulation circuit to 80, 90, or even 95%. However, that will drive up the cost and complexity of the design, and may not be necessary.

If these are 12V LEDs then they most likely have some sort of regulation built in, so you don't have to worry about putting in a resistor. If you have the documentation for them, I would put it in your post.

2) If you wanted to put resistors within each parallel circuit, it would also need to have the same resistance as the one you would put at the beginning or end node, not 1/5. The advantage of doing this is that you could have smaller resistors rated for a lower power than one big resistor rated for a high power.

3) Short answer, no, the resistor is generating heat that would have been made at the LED if the resistor wasn't there.

4) Those don't sound like regular LEDs, but if you mean LEDs like the through hole 3mm or 5mm most of them are indeed designed for 20mA current draw at a characteristic forward voltage.

LEDs are non-linear semiconductor devices, their current draw is non-linear with the voltage across it. Look at this graph to see what I mean. http://en.wikipedia.org/wiki/File:Diode-IV-Curve.svg