I have a general question about watt/heat/current

[Canti]

Led spec:
3v-4v dropout
80mA max current

Power Supply:
12.38v battery

Resistor specs:
40ohm
1/4 watt

Does the heat from the resistor associates with the amount of current that is limiting? The more currents the resistor holds back the hotter the resistor gets? Am I right to think this? I can't seem to wrap my head around this kind of stuff, I have no problem with ohms law and general electronic stuff.

Thanks.

 

[dknguyen]

Sounds about right. The more current that flows through the resistor, the greater the voltage drop across it and the more heat it dissipates.

It doesn't "hold back" current though, so much as it produces a voltage drop across it which causes less voltage to be across the load and therefore less current to flow through the entire circuit. I dunno, probably the same thing, we just think about it differently.

 

[Canti]

Thanks for the help dknguyen.

 

[Canti]

I measure the current between my battery and my Led and it says .212A and the voltage across the Led was like 3.84v

The voltage is within limits for the led, but the current through it is way too high the specs of the led was suppose to be 80mA max and 3v-4v. Is this normal?

 

[jrz126]

Maybe the LED is capable of that high current for a short period of time?

Whats the battery voltage with the led connected? If it's still 12.38V, then that little resistor will be disipating ~1.8 watts of heat. (Power = I^2 * R). Most 1/4W resistors dont appreciate it too well.

Either way, I think you need more resistance.

 

[Canti]

I Like the brightness of Led in the upper 3v range, but my concern is the the life of the resistor. So I would need a resistor that can handle 1.8 watts?

Another idea I have is to use a voltage regulator that outputs a 5v at 1.5amps, so I can hookup 6 of these leds in parrallel. Well this work out well? Keeping in mind that I want to have the leds running in the upper 3v range.

 

[jrz126]

I'd be more concerned about the life of the LED. If the LED is only rated for 80mA, then you should run it at 80mA. It may run at a higher current, but you'll be shortening its life.

If you want some bright LEDs, check out http://www.lumiledsfuture.com/products/family.cfm?familyId=7

**broken link removed** Red LED - 190 lumens
**broken link removed** - 30W blue

Using the 5V regulator will work. but if you are using a linear regulator (like the 7805), you'll have to put a pretty big heatsink on it. (the power disipated by it is = to (12.38-5)*current. )
This isnt a bad idea if you have a bunch of leds that you are trying to drive. I hooked up 62 blue leds in the headliner of my car, ( http://www.cardomain.com/ride/435572/2) and I'm able to turn on each led individually, which means 62 leds in parallel. Running off the car's 12V, I would have needed 1 watt resistors, but using the 5v reg. and a big heatsink, I was able to get away with 1/8watt.

It'd be more efficient to put 3 in series at the 12.38V. then you'll need a 10.75 ohm resistor which will only be disipating 0.0688 watts.

 

[RODALCO]

The LED wouldn't last that long, you will destroy it very soon.
Your ¼ Watt resistor is dissipating 1.8 Watts and will start smoking soon.
If the Led is rated at 80 mA max. which is probably for a pulsed supply, say 1 on 4 off to allowe the LED chip to cool down.
As jrz126 already suggests put 3 LED's in series on the 12.38 Volts supply. and you need a 10.75 ohm resistor ¼ Watts

 

[audioguru]

If the LED is only rated for 80mA, then you should run it at 80mA.

80mA is probably its absolute max current rating which can be used under certain conditions. Conditions such as pulse duration, temperature like no more than 25 degrees C ambient (with no other LEDs nearby) and a certain leads length. Its normal operating current is probably only 20mA. It is probably fried with 210mA though it because the 40 ohm resistor's value was way too low.

 

[Canti]

Well.... I decide to use a voltage regulator (with a good heatsink) that outputs a 5v /1.5A and run my leds parallel. Hopefully lower voltage that is coming through the circuit will reduce the heat at the resistor.

I am still not sure if the heat is cause by the voltage or by the current running through the resistor.

The Luxeon leds look nice, but $4 for one is too much for me.

 

[audioguru]

Well.... I decide to use a voltage regulator (with a good heatsink) that outputs a 5v /1.5A and run my leds parallel. Hopefully lower voltage that is coming through the circuit will reduce the heat at the resistor.

Each LED still needs a current-limiting resistor in series with it. If the LEDs are exactly identical, then they can be in parallel and a single current-limiting resistor can be in series with the bunch of LEDs.

I am still not sure if the heat is cause by the voltage or by the current running through the resistor.

Heat is power wasted. Power is the voltage across the resistor times the current in the resistor. Adding a voltage regulator reduces the voltage across the resistor and therefore reduces its heating, but then the voltage regulator heats instead with the amount of heat that the resistor would have wasted without the regulator.

If the LEDs are in series then far less heat will be produced.

 

[philba]

think of the voltage drop (AKA Vf) as the amount the LED consumes. The resistor has to take care of the remaining voltage. so Vf = 4 (for discussion sake), Vbat = 12 then the resistor "sees" 8V. use ohms law - V = IR or, rearranging, I = V/R = 8/40 = .2A which is pretty close to your measurement. plug more precise numbers in for better accuracy. If you want 20 mA, the resistor should be 400. Power disipation would be 160 mW or within the 1/4 watt rating.

the heat is generated by power which is the product of current times voltage

 

[Canti]

Thanks for the helps guys.

Another questions that bugs me is, What does it me when the voltage regulator manufacturers say "negative" or "positive"?

I assume that it means that the "output" of the voltage regulator is either positive or negative.