Help, 12vdc to 9vdc

The voltage drop accross a resistor is given by V=IR.

Rearrange that to give R=V/I

Plug in the numbers..

R= (12-9)/0.01 R= 300 Ohms.

But allways check the power rating...

The power (W) burnt in the resistor is given by: W=IV W=0.01 * 3 =0.03W. which is tiny so a 0.125W (=1/8th Watt) resistor will do fine.

...but perhaps you had best tell us what the 9V will be used for just in case? You might need a capacitor as well!

relsr wrote this in :

Constant load? Yes, then a resistor is the simplest solution. You want

3V drop over the resistor, and have 10mA: 3V/10mA=300ohm should do it. You can also use a 7809, which is a simple IC with 3 legs: In, out and refference (which you connect to - on the source). I think a 7809 will work with 10mA, although not sure. A 7809 costs ~1$, and works even if load differs.

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MVH,
Vidar

www.bitsex.net

300 ohm resistor ?

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Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

In article , snipped-for-privacy@cablespeed-dot-com.no-spam.invalid opined thusly:

About 5 or 6 diodes in series, forward biased, will do the trick and the voltage won't vary with the current draw (much).

dave wrote this in :

Then a 7809 will likely be cheaper, and with better accuracy.

--
MVH,
Vidar

www.bitsex.net