# Reduce Voltage?

Discussion in 'Electronic Design' started by mopar, Apr 30, 2005.

1. ### moparGuest

I am working on a home surveillance system and I need to do a DC
voltage reduction. I have a supply that is 12v DC @ 80 ma and need to
reduce the voltage to a max of 9v @ 80 & 120 ma (variable load).
As I understand I can do this with a resistor? If so, what size and
should it be in parallel or in series between the supply and device?

2. ### Jonathan WesthuesGuest

Actually you can't do that at all. You have 12V @ 80mA available. That means
that you have P = V*I = 960mW available. If you need 9V @ 120mA, then that
means that you need 1080mW, which is more power than you have available.

If you can live with 9V @ 80 mA output, or if you can come up with 12V @ 120
mA input, then you can use a three-terminal voltage regulator, for example a
7809:

http://www.njr.co.jp/pdf/ae/ae06006.pdf

Hook it up like the datasheet shows in the figure at the bottom left of p.
5. Connect pin 1 to +12V (the positive wire of your +12V supply), connect
pin 2 to ground (the negative wire of your +12V supply), and you will get
+9V between pin 3 and pin 2.

The datasheet shows some capacitors at the input and output; it might work
without them but it would be better if you put them in. The values are not
critical; anything over 0.1uF will do for either cap.

You can buy the 7809 from Digikey (http://www.digikey.com/), but it is very
common so you can probably find it locally too. Get it in the TO-220F
package (see the picture on the first page of the datasheet).

A resistor would drop the voltage, but the voltage drop would depend on the
current, so the voltage at the load would depend on how much current the
load was drawing. That probably isn't what you want, though it might still
work.

Jonathan
http://cq.cx/

3. ### Homer.SimpsonGuest

mopar said
You're asking the resistor to drop the excess voltage at the current
of interest. That makes the resistor between (12V-9V)/80mA =~38 OHMS
to (12V-9V)/120mA = ~25 OHMS. The power rating should be (12V-9V)*
120mA=360mW. It's placed in series with the load.

That being said, that is a poor way to drop the voltage. It ignores
things such as surge current which may yield an excessive voltage
drop under transient loads. A better idea might be to series several
diodes (12V-9V)/0.6V= ~5 diodes.

Note you're going to violate the current rating of your 12V PS. I'd
pick up a cheap 9V PS from Rat-Shack or something. They're dirt
cheap.

hth