12 volt DC source and 9 volt DC 15 Watt appliance What do you need to do to reduce the 12Volt to 9 volt?

Answer 1

I'm assuming you don't want to rip apart and modify your power supply. I'm also assuming that this question isn't just a hypothetical, and that you actually want to hook up an appliance to a supply. All you need to do is put a resistor in series with your appliance and the 12V source. It would only cost a few bucks at most. "In series" means... Well here:

_______________Resistor___

+ +

12V Source Appliance

-_______________________-

The resistor lowers the voltage by dissipating power when the circuit is on. You'll need to figure out what resistance you need. I'll take you through the math, it's simple. There's always more than one way to do the math, you'll always get the same answer. Or you can just skim near the bottom for the answer ;)

Basically, you'll need to figure out what current should flow in the above circuit to give you 15 watts of power at 9 volts. From that info, you can find out what resistance you should add to get the right amount of current flowing.

Some background...

Watts (W) are the standard unit of Power (P). Amps (A) are current, Volts (V) are... electical pressure, sort of.

Resistance (R) is measured in Ohms. I'll use ^2 to mean "squared"

In general...

V = I*R; I = V / R

P = I*V = I^2 * R

For your appliance:

P = 15Watts

V = 9 Volts

So to get the current your appliance must use,

I = P / V = 15 / 9 = 1.667 A

When you put resistors in series, there is a voltage drop across each one. You want your appliance to have a 9 V drop from one terminal to the other. Since your source is 12 V, you need a resistor to make a 3 V drop (12-9 = 3) so your appliance gets the 9V instead of a whole 12 V.

V = I*R;

R = V / I = 3 / 1.667 = 1.8 ohm *******Resistor value

Keep in mind that this resistor will be absorbing and dissipating that extra power from the circuit as heat. If the power is too much, the resistor will BURN.

Since the current in the circuit will be 1.667 Amps, the power dissipated by the resistor will be:

P = I^2 * R = 1.667*1.667*1.8 = 5 Watts *****SMALLEST power rating

Resistors typically have a wattage-rating, so make sure the rating is AT LEAST 5 Watts. Unless you'd like to watch it literally go up in smoke...

Note about resistors:

One problem you might have is that resistors don't always come in the values or the power rating you need when you're at Radio Shack or some other hobby store.

What you can do is use a bunch of larger resistance, lower power rating resistors in parallel with one another. This isn't ideal, but it will work IF you solder them all together well.

You can look up "resistors" in wikipedia to learn more about how the resistance changes when in series or parallel. I strongly recommend using only the same value resistors for this (it keeps the math simple, no room for mistakes). Here's my quick solution.

Basically, take the value of the resistors you will use and divide by the 1.8 ohms you need.

For 15 ohm resistors you'd need 15/1.8 = 8.33 resistors in series. (Just use 8, it gives 1.875 ohms) Since you need at least 5 watt power rating, you'd look for 1 watt rating resistors (giving you an 8 watt rating)

Lets say you only have 33-ohm resistors. 33/1.8 = 18.33... resistors

So if you connect 18 of them in parallel, you should get roughly 1.8 ohms (1.8333)

This would work ASSUMING they were 1/2 Watt resistors (18 * .5 = 9 watt rating)

If you only have 1/4 watt resistors, you'd have to use more, so maybe 47 ohm resistors... 47/1.8 = 26.11...

Granted, you probably don't want to solder together 26 resistors... but it would be good for 6.5 watts.