Voltage dropping resister?

[Old Iron]

I have a 3 volt Dc power supply and I need to drop the voltage to 1 1/2 volts. Need a little help figuring out what resister to use and at what wattage?

Thanks
Paul

 

[John Hasler]

I have a 3 volt Dc power supply and I need to drop the voltage to 1 1/2 volts. Need a little help figuring out what resister to use and at what wattage?

Thanks
Paul

The resistance needed to produce a particular voltage drop depends on the current. Please tell us exactly what you are trying to do.

 

[AlanR]

Use 2 diodes in series. The voltage drop across a diode is .6 to .7 volts. Very common technique.

 

[Codered741]

Use 2 diodes in series. The voltage drop across a diode is .6 to .7 volts. Very common technique.

If you use this technique, make sure that you take into account the ampacity of the diodes. Of course, it would still help to know exactly what your application is.

-Cody


Sent from my iPad using Tapatalk HD

 

[Wireaddict]

Resistance=Voltage across it [1.5V in this case]/ the Current that'll flow through it, &

Resistor Power rating in watts=1.25 X the Voltage across it [1.5V] X the current through it. Hope this helps!

 

[tigtorch]

As us aging EE's have long opined:

"twinkle twinkle little star, E is equal to I R"

 

[Old Iron]

It is a liquid bath and I would guess the more parts I add the more currant it will draw. I really don't want to say what it is till I see if it works or not. The best voltage would be about 1 1/2 volts at 300 milliamp.

I have a variable voltage and currant power supply coming, If and when FedEx delivers it its been on there truck 5 days. They say it is because of the winter weather but the ice has been gone since Wednesday.

So that problem will be fixed, I still want to lower the voltage on the other one for a friend to use if it works then I'll do a write up on it.

Thanks
Paul

 

[John Hasler]

It is a liquid bath and I would guess the more parts I add the more currant it will draw. I really don't want to say what it is till I see if it works or not. The best voltage would be about 1 1/2 volts at 300 milliamp.

I have a variable voltage and currant power supply coming, If and when FedEx delivers it its been on there truck 5 days. They say it is because of the winter weather but the ice has been gone since Wednesday.

So that problem will be fixed, I still want to lower the voltage on the other one for a friend to use if it works then I'll do a write up on it.

Thanks
Paul

At 300 ma a couple of 1 amp diodes will give you approximately the right voltage drop.

 

[Pmedic828]

From a previous post, I understand that you want to drop the voltage by half. If half is what you need, use 2 of the same sized resistors in series(from 1 power terminal, to 1 resistor, then to the other resistor, and then to the other terminal) and power your device across 1 leg and the center junction of the 2 resistors in series. To determine the amperage or current that will be available, note the available amperage on your power supply. If it is 1 amp, then this equals 1000 ma. R=P/I squared so if your power supply is 12 volts and can output 1 amp, it would be 12/1 squared = 12 watts Now determine R=E squared/p so R=12 squared is 144/12 = 12 ohms so you would need 2 6 ohm resistors in series for 12 volts and 12 watts.
See this site for formulas: http://www.elec-toolbox.com/Formulas/Useful/formulas.htm
Without exact knowledge about power available, voltage and current needs, you can refer to this site. If this is a LED project, you can use an online calculator found at: http://ledcalc.com/#calc.

 

[Old Iron]

Ok thanks for all the help, I got my new power supply in today the only thing is it doesn't work.

I caller the company and there sending me another one, And they don't want the old one back.

Thanks again
Paul