Voltage reduction

kungfugeek · Apr 27, 2005

Hello,

I am trying to understand the math behind voltage reduction. I have read up on Ohms law, but it not quite clicking yet. Basicly I want to calculate what size resistors I need to add in series to a particular voltage to reduce it to a predetermined voltage. e.x. reduce my 48v phantom to say 9v. I read some stuff about voltage diving as well, but not sure that is what I am after. Any feedback would be gladly welcome.

thanks!,
kfg

CJ · Apr 27, 2005

Ohm's law has three variables. So you need two variables to get the third. You know your voltage , you want to find ohms, so guess which parameter you are missing. :?:

kungfugeek · Apr 27, 2005

Watts, ok, so how do I figure out the wattage on any given voltage line I have?

alk509 · Apr 27, 2005

[quote author="kungfugeek"]Watts[/quote]

Try Amperes... You really want to do some reading on Ohm's law again. Is CJ's Electronics 101 thread that far along yet?

Peace,
Al.

CJ · Apr 27, 2005

Have not covered watts yet. Promise to do 3 or 4 more chapters this weekend.

Anyway, dropping 48 to 9 is with resistors is probably not the best way to go.

You might be better off with a voltage regulator.

You can use a high voltage regulator, or you can use the zener in the ground leg of a standard part like a 7808.

Anyway, the math:

Lets say you need 100 milliamps at 9 volts from a 48 volt supply.

OK, where to start? Well, we need to knock the voltage down with resistance. How much voltage do we need to lose? 48 minus 9 equals 39 volts.

Ohm's law says E=IR so fill in the knowns to get the unknown:

39=0.1 amps times R, the unkown value we wish to find. A little algebra:

39/0.1=390.

So 390 ohms is the resistor value needed.

But what pwr rating?

There are two ways to find out watts.

Two formulas for watts:

P=VxI
or
P=I^2xR (I squared R)

The easiset is just a plain old Volts time Amps calculation.

How much voltage is going to be across the resistor?

39.

How much current is going thru the resistor?

0.1

39 times 0.1 is 3.9 watts.

Thats a lot of juice. I would use a ten watt power resistor for long life.

The second method:

I^2R=0.1x0.1x390=.01x390=3.9 watts. Luckily, we came out with the same answer.

mwkeene · Apr 27, 2005

What you are doing, by adding resistors in series is creating a voltage divider (I assume diving was a spelling error?)

What is the load resistance?
Adding resistors in series will reduce the voltage, but you need to make sure that you know what the load resistance is, and if that load ever changes, the voltage you get across it (your "output") will be different.

It would be a good idea to learn voltage dividers and have an intimate knowledge of how they work. They are one of the most basic, yet important ideas in electronics. Pound it into your head until you can just "see" it.
-Mike