We have a bank of LEDs from a flashlight that plug directly into 4.5v. There's 8 LEDs on a circuit. I'm trying to get them to run off a 9v battery by just connecting the wires from the circuit board (it works fine when connected to 4.5v battery pack). So basically I just need to put a resistor in between the 9v battery and the circuit, but I'm not sure which one to use and I don't want to blow the LEDs.
I tried a 270ohm resistor, but they would not light up. I'm thinking that I probably need to add the diode forward current of 20mA together, which would be 160mA, (as there are 8 LEDs in the circuit) this would mean I should use a 33ohm resistor? Is this the right way to think about it?
Can anyone confirm that a 33ohm resistor would make sense in this case, or do I need a different one? It's for a child's project and I don't want to just test it and risk blowing them.
Thank you!
Here's some photos.
Best Answer
You can see from the traces on the bottom of the circuit board that all 8 LEDs are in parallel. There is also a 3 Ω resistor in series.
To find the resistor to drop 9 V to 4.5 V you need to know the current. Since you don't know the exact specs of the LEDs used, the best way to get that is to just measure it. The resistance to add is then the voltage to drop divided by the current.
For example, let's say the current is 160 mA with 4.5 V applied. (9 V - 4.5 V)/(160 mA) = 28 Ω.
However, this is a crappy way to use a 9 V battery to light 4.5 V LEDs. A much better way would be a buck converter to make 4.5 V from the 9 V battery. That has two advantages. First, it will use the energy in the battery more efficiently. Second, the brightness won't go down as the battery drains. It will go out quickly when the battery gets empty.