Hello Electrical Engineers and Fellow Hobbyists. Im trying to figure out how to power an array of 12 leds off 8 volts by buckling the 9 volt battery down to 8 volts and keeping it at a constant voltage. I have a buck boost converter but its the only one i have left so I was going to use some Zener diodes but im not sure if that will work. I've uploaded the material i have and the array.

 

Using a regulator will waste energy unnecessarily and the battery you've chosen doesn't have much to begin with. Just choose a suitable resistor value to give you the desired LED current.

 

Using a regulator will waste energy unnecessarily and the battery you've chosen doesn't have much to begin with. Just choose a suitable resistor value to give you the desired LED current.

Wont that make it last for a shorter amount of time?

 

Wont that make it last for a shorter amount of time?

You're going to have to dissipate the power somewhere. Even a buck regulator is going to consume power. Zener diode and/or resistors will still consume power.

How much runtime are you expecting from a 9V alkaline battery? Eveready datasheets are lacking in detail. At a 75mA drain, you'll be lucky if you get a few hours.

 

What do you want to use the array for?

How bright does it need to be?

Your schematic shows the use of 'regular' 20 ma leds.

If this amount of brightness is acceptable to your design goal, consider using 'super bright ' 3.2 v leds.

They are meant to be used at 20ma @ 3.2v and provide for a very intense release of photons , but they are quite visible at 1ma. At this level their forward voltage drop is about 2,5v

You can obtain the same amount of brightness from these parts using only 1ma of current as you can from the 'old' regular ones at 10 - 20ma.

A group of 12 leds arranged in a 3 x 4 series / parallel configuration could be used with a minimum current of 3 ma.