Welloilbeefhooked

iTrader: (3)

I have a switch with an LED that runs at 2.2v, 25ma. Maplins site says that I need a 390 ohm resistor, in series I would imagine, to drop from 12v to 2.2v.

I cant remember the calculation that confirms this.

Help please!

Cheers,

Wayne.

PS The volt meter still shows 15v from a car battery when the resistor is in series or not.

alcazar

iTrader: (2)

Go with maplin's.

To measure the voltage, you need to do it in parallel with whatever component.

Put the resistor in circuit, and the voltmeter across the ends of the diode.

Alcazar

Dill_typeR

V=voltage (volts)
I=current (amps)
R=resistance(ohms)

Basic equation is V= IR

V= I x R

I= V / R

R= V / I

Hope this helps, maplin shud know what they're on about tho

Jay m A

iTrader: (2)

I'd have thought 25mA is its maximum current, it should run happily with less.

if you're using a car battery then aim for 13.8V and 20mA when doing the calc, this works out at 690 ohms, so grab a 680R IMO