Discussion in 'General Electronics Chat' started by brenande, Jul 3, 2007.

1. ### brenande Thread Starter New Member

Jul 3, 2007
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If I have a device that is, say, rated at 200mA at 5v and I have a power supply that provides 1.9A at 5v, am I going to burn out the device? Or does the device only draw what it needs? (I'm specifically talking about an Arduino board, but the question is more general than that). How about an LED that is rated at 20mA at 3.3v and I feed it 2.1A at 3.3v?

Thanks for the help.

Apr 20, 2004
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3. ### Joe24 Active Member

May 18, 2007
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If you provide 2.1A to that LED, you will have to search your tool box for another one. Anything significantly above 20ma will burn it out, so make sure you use an appropriate resistor.

4. ### recca02 Senior Member

Apr 2, 2007
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ohm' law.
the device draws what it needs, else we wud always have had a hard time limiting the current (if the manufacturer didnt do it himself).
so if u apply rated voltage rated current will be drawn.
the current specification for the power supply is the limiting value of the current/load the power supply can handle without damage to its components.

5. ### mrmeval Distinguished Member

Jun 30, 2006
833
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VS = supply voltage
VL = LED voltage Check the Datasheet!
I = LED current (e.g. 20mA), this must be less than the maximum permitted Check the Datasheet!

R=(VS-VL)/I

Convert MA to A by dividing the 20 by 1000
So if VS is 5 and diode is 2.1 and current is .020 then the resistor is 145 ohms or 150 would be good and it's less current , you can use higher for dimmer , lower than the max is bad and the poor bairns will flash and die.

For better life in a harsh environment where voltage, noise, spikes etc can be a problem you'd want a more sophisticated current limiter with some filtering.
These are usually specially designed switching regulator circuits so that you can use less than the LED's rated voltage (it's boosted) as well. Especially important for the very expensive high intensity white LED chips when used in flashlight or mobile lighting applications.

6. ### thingmaker3 Retired Moderator

May 16, 2005
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I = E/R

The odd thing about LEDs & other diodes is that R is not constant.

7. ### brenande Thread Starter New Member

Jul 3, 2007
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Thanks everyone.. I think my question has been answered - and more!