? For Electrical Geru

[EddieS'04]

I need to reduce the voltage on two DC power supplies.

what resistor do I need for;



1) 13.3 vdc to 6or7 vdc

2) 20.4 vdc to 6or7 vdc



amperage draw is no more than 300ma



Thanks, Eddie

 

[Gavin Allan]

V=I/R?

 

[Thomas Rogers]

Last time I went to school, Gavin, that was:



V = I * r



P.S. Changed it to make r small case, which is the convention I remember

 

[Gavin Allan]

Thanks. It has been a long while.



Still remember:



Bad boys rape our young girls, but violet gives willingly...

 

[Fred]

black brown red orange yellow green blue violet gray white

 

[Bill Ellis]

Resistors are not the way to go unless you know the actual current draw and know that it will not fluctuate. Your best bet is a voltage regulator. Most have three leads, input, output and common [ground]. An L78M06CV will handle 500 ma without a heat sink and cost about $1.00. Here's a link to a data sheet. These are available from Digikey.com

 

[Andy G]

Bill-E has the best answer; that way you aren't killing tons of power running your resistor divider.



If you need a more tunable voltage, the LM317 is also very easy to hook up, and will give you an adjustable output from 1.2V to V+rail - 1.5V. So on a 13.5 V system, your range will be between 1.2 and 12V.



You will need one other resistor (220ohm typ) and a 5K pot. These are all radioshack parts.



Also available at Digikey, or Mouser.

 

[Adam Smith 2]

Voltage regulator is indeed the correct answer if you have a variable load.



In order to use a current-limiting resistor to get the right voltage and current to your load, the load has to have a fixed and known resistance. Consider if your load is 20 ohms, such that 300mA through it gives 6 volts across it. Then you need a 47-ohm resistor to burn off the extra 14.4 volts, which is also going to be expending about 4.2 watts. The little resistors at the rat shack with the colored lines on them (color code quoted above) are 1/4 or 1/2 watt typically. Not gonna cut it! The type you would need would most likely look like a little ceramic block the shape of a french fry.

 

[Ryan Schaecher]

Definately go with the regulator. A voltage divider circuit is very expensive on the power side.



Also, what kind of load are you powering? If it's LED's, then there is no problem. But if you are powering IC's then you may need to add a few small capacitors to the regulator for stability. I usually use a pair of 22uf ceramics. One gets tied to the high side of the regulator, then the other is tied to the output. Both are referenced to ground.



Rat Shack has a mini-perf board that is pretty hand for mounting everything on if needed. It can be cut down quite a bit as well. Just remember to insulate the assembly from touching anything.



A heat sink may be required on the regulator as well, depending on your load.

 

[EddieS'04]

R Shek...Iam trying to use one of two existing AC/DC converters to power a caller ID unit, instead of batteries.

 

[Richard L]

Yep. Voltage regulators are the way to go to insure your voltage is right on the money.



...Rich

 

[Adam Smith 2]

I am trying to use one of two existing AC/DC converters to power a caller ID unit, instead of batteries.

I was assuming this was a DC-to-DC automotive application. Now I'm wondering why you aren't just getting a 6 VDC wall wart. I don't think you're going to get good advice here without detailing your exact situation.

 

[Bill Ellis]

The batteries in a caller ID unit should last a very long time. Not really worth the trouble to make a power convert for it....unless, you just want to experiment and learn something new. Then it's OK. You probably only need a 6V zener diode and resistor since the current draw on the ID unit is going to be very small. Look up zener diode on google and you will learn how to make a low current, fixed voltage power converter.

 

[EddieS'04]

This caller ID eats 4AAA batteries every month. I was going to set it up on the garage phone.

 

[Bill Ellis]

OK, now we know more. AAA batteries are good for about 1000 mah. There are about 720 hours in a month so it would seem that your display is only drawing 2 ma or so which would right it if is a liquid crystal display. If you use a 6v zener diode and a 13v supply, you would be sinking 7 volts thru the zener. If you limit the zener current to 5 ma, you would need a 1.4 kohm resistor [1400 or so ohms].

 

[EddieS'04]

Thanks Bill, I will give that a try....