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Check my math: resistor to obtain specific current

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RogerTango

New Member
I have a 12 volt power source, I want to provide 700ma of current to a load.

R=V/I

R=12/.7

R=17ohms


Is this correct? I wasnt sure if I am supposed to use the source voltage, or try
to calculate the voltage after the resistor drops it to determine the current.

If I sound confused, please forgive me...

Thanks,
Andrew
 

carbonzit

Active Member
Not quite.

If your total load was 17 ohms, it would draw 700 mA at 12 volts.

Presumably your load requires 12 volts, correct? If you put a resistor in series with a 12-volt power supply, the load will get less than that.

I'm not sure what you have, but I think I'd figure it this way:

Your load requires 12V @ 700 mA. If you need a series resistor to limit current, obviously you'll need > 12V supply. Get one, or assume the voltage. Let's say 15 volts.

You want the resistor to drop 3 volts (15-12 = 3). So size the resistor to draw 700 mA at that voltage:

R = 3 / 0.7 = 4.3 ohms.

By the way, are you sure you need to limit the current to your load? Some devices (notable LEDs) require current limiting, while others, like most motors, do not: you can just let them draw what they want to draw. What is your load?
 

RogerTango

New Member
Thank you kindly for viewing and replying to my post.

The source is 12v battery or power supply.

The load is a Star 3W LED, datasheet says deliver .7a - .8a.

I assume, the voltage will be... whatever the voltage is, past the resistor and is not important,
mostly that the LED is getting 700ma.

Corrections to my thinking are welcomed.

Andrew
 

BrownOut

Banned
I assume, the voltage will be... whatever the voltage is, past the resistor and is not important,

Never assume. The voltage is important. There should be a value given on the data sheet. Use that value, call it VD. The your equation is

R(series) = (12V-VD)/.7
 

RogerTango

New Member
Ahhhh... that makes a *LOT* more sense to me!

Thanks a lot, your explanation was of great help!

Andrew
 

RogerTango

New Member
This is from one source:

Forward Voltage : 2.3V~2.5V
Forward Current: 700mA

You exampled:
R(series) = (12V-VD)/.7

Thus:
R=(12-2.3)/.7
R=9.7/.7
R=13.9ohm

Thanks much!
Andrew
 

RogerTango

New Member
Sorry for another "stupid" question, if I have say 2X of these LEDs wired together, should I wire them in series, and double the current delivery, 1.4a instead of 700ma?

A few applications will need X1, X2 or X3 of these LEDs to be powered at one time.

Thanks,
Andrew
 

RogerTango

New Member

audioguru

Well-Known Member
Most Helpful Member
700mA is the absolute maximum allowed current when it is "perfectly cooled" (with an infinite size heatsink plus liquid nitrogen?).
All the spec's are listed when its chip is at 25 degrees C somehow.
But you probably will not use an enormous heatsink plus liquid nitrogen so you must reduce the current to a reasonable amount that will not allow the LED to melt when it has a reasonable heatsink.

LEDs in series all use the same amount of current, not doubled nor tripled.
 

RogerTango

New Member
700mA is the absolute maximum allowed current when it is "perfectly cooled" (with an infinite size heatsink plus liquid nitrogen?).
All the spec's are listed when its chip is at 25 degrees C somehow.
But you probably will not use an enormous heatsink plus liquid nitrogen so you must reduce the current to a reasonable amount that will not allow the LED to melt when it has a reasonable heatsink.

LEDs in series all use the same amount of current, not doubled nor tripled.

Very well then Guro, thank you for your contribution to the thread. Every time I learn
something new, its exciting!

Andrew
 
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