I read once that a 'load' gets only the current it needs from the battery / voltage supply.. So, why the LED gets more than it needs and dies when connected directly to the battery?
Another question, when I put a 470 ohm , how much milli Amps does it get from the 9v battery?
V= IR
I=V/R
9/470 = ?
(This assumes zero internal resistance for the battery)
Oh damn, I know this equation..
So, 9/470 = 0.019 ... that means 19 milli amps?
Also, the first question needs an answer as well , every body!
Also, the first question needs an answer as well , every body!
Seriously ? Have you heard of a datasheet ? Look at "Maximum ratings " (for the led in question)
The answer to your question is :
I = V/R
Let V= 9V
Let R = 0
9/0 = ? (unlimited current (well, not exactly since the battery has internal resistance so it will try to draw the short circuit current of a 9V battery but the led will vaporize long before it even gets close)
; POOF !! There goes the led junction , vaporized in a uS)
I also don't understand the "load gets only the current it needs". If you keep increasing voltage the load will eventually be damaged because it'll take current it doesn't need???
cybefox:
I read once that a 'load' gets only the current it needs from the battery / voltage supply.. So, why the LED gets more than it needs and dies when connected directly to the battery?
This is in fact incorrect.
Very very incorrect.
In fact... it is very very very very incorrect.
If you connect a battery to an LED with no current limiting resistor you will burn out the LED.
I read once that a 'load' gets only the current it needs from the battery / voltage supply.. So, why the LED gets more than it needs and dies when connected directly to the battery?
This is a very vague statement. If the load is a resistor it draws the current defined by Ohm's Law.
If it is a led , without a current limiting resistor, there is nothing to limit the current so it will draw as much current as the source can supply until it melts (or vaporizes) the junction.
I read once that a 'load' gets only the current it needs from the battery / voltage supply...
You are applying this statement out of the context in which it was written.
If you have a device that is rated at 9v @ 100mA you can connect it to a 9v supply rated at 100mA and it will work correctly.
If you connect the same device to a 9v supply rated at 200mA it will still draw only 100ma as it is drawing only the current it needs from the supply.
Don
I note the OP has not shown their face (or whatever that avatar depicts) again, but ...
A LED is not a "load". It is a component. With the exception of some which have an integrated resistor for this specific purpose, it is not ever intended to be connected directly to a power supply, and I do mean ever.
Therein is the simple misunderstanding.
LEDs are not like normally electronic devices in that you can’t just apply a voltage to them and they work, they have to be fed the correct voltage and current to keep them happy.
A better statement would be:
LEDs are not like household appliances in that you can't just plug them in and they work,
Paul__B:
I note the OP has not shown their face (or whatever that avatar depicts) again, but ...A LED is not a "load". It is a component. With the exception of some which have an integrated resistor for this specific purpose, it is not ever intended to be connected directly to a power supply, and I do mean ever.
Therein is the simple misunderstanding.
Why on earth would a led not be a load?
I've seen many responses here, but half of them read as if people assume the resistance of a led is 0 Ohm?????
If a led's spec says it draws 20mA @ 3.4V, then you can just as well apply Ohm's law:
U=I*R, R = U/I, R = 3.4 / 0.02 = 3.4 * 50 = 170 Ohm
Now a led isn't a real resistor, @ 3.4V it acts as a resistor of 170 Ohm, i don't think the relation between voltage and current is as linear in a led as it's with a resistor. But it'll certainly never have 0Ohm resistance at any point (or you'd have invented the room temperature super conductor, better patent it immediatly).
@OP, the statement you said there is true for a certain voltage. If you read ohm's law, if you put a load on a supply, depending on the resistance of your load, your load will only allow a certain current to go trough it as a result of the voltage difference. Even if your supply is capable of giving 100 A at 3.4V, if you put a white led on it, it'll only allow 20mA trough itself due to its resistance. If you put a higher voltage on your circuit, that voltage will cause an increase in the current (according to Ohm's law), and if that is more current than your circuit can handle, it will break(/burn) things :).
Tamulmol:
I also don't understand the "load gets only the current it needs". If you keep increasing voltage the load will eventually be damaged because it'll take current it doesn't need???
You fail to understand the context of that quotation.
A 5V part supplied from a 5V supply "only gets the current it needs" however
much current the supply can produce.
A 2V part connected to a 9V supply burns out, always.
racemaniac:
If a led's spec says it draws 20mA @ 3.4V, then you can just as well apply Ohm's law:
LEDs are simply not specified in that fashion. You have it backwards. 
The bottom line is that you just can't apply Ohm's law to an LED in spite of the fact that more than one of the 'answers' tried to do so.
Ohm's law defines the linear relationship (the constant of proportionality) between voltage and current that certain types of devices (conductors) have. LEDs (semi-conductors) are not one of those devices.
Don
cybefox:
I read once that a 'load' gets only the current it needs from the battery / voltage supply
You were probably reading about a specific circuit. It's not one of the laws of physics.
cybefox:
I read once that a 'load' gets only the current it needs from the battery / voltage supply.
Yes you read that but you totally misunderstood it.
That is only true if the voltage is what the load is designed for. So you power a 5V logic chip from 5V and it will only take the power that the logic chip was designed for.
An LED is not designed for a specific voltage, this is because it is a non linear device. in theory if you gave it exactly the right voltage then it would only take enough current to keep it safe. HOWEVER, that "right" voltage changes with the age of the device and the temperature. If you look at the data sheet this is not a single value but a range. This range is where the sweet spot is not the sweet spot is anywhere in the range. Trying to provide that exact voltage is just stupid, you have to find some way to limit the current. A resistor is the simplest way but not the only way.
See:-
http://www.thebox.myzen.co.uk/Tutorial/LEDs.html
If a led's spec says it draws 20mA @ 3.4V, then you can just as well apply Ohm's law:
U=I*R, R = U/I, R = 3.4 / 0.02 = 3.4 * 50 = 170 Ohm
Clearly you need to do a little more electronics homework.
Fact-1: A led is a diode, which means, like any other diode it is essentially a direct short when forward
biased. (even zener diodes burn up when used without a current limiting resistor)
Fact-2: The datasheet shows a peak forward current of 100mA (0.1A)
See Note:
Note: Pulse width ≤0.1 msec, duty ≤1/10.
Note 1. Pulse Width = 100μs, 10% duty cycle (1/10th).
This tells the reader that the led can tolerate 100mA (0.1A) for a maximum of 100uS (0.1mS) with an ON time duty cycle of 10 % (1/10th) !)
Fact-3: To obtain the current of 20mA @ 5V, a 90 ohm resistor must be used.
VResistor=Vcc-VF
Let Vcc = 5V
Let VF= 3.2V
Let IF= 20mA
then,
VResistor=5V-3.2V = 1.8V
To drop 1.8V across a resistor at 20mA, the resistor value must be:
R=1.8V/0.020A= 90 ohms.
Without the resistor, the led, like any other diode acts like a direct short across the power supply.
If you wouldn't put a diode across your 5V power supply with the cathode to ground then you shouldn't
connect a led the same way (without a resistor).
Now a led isn't a real resistor, @ 3.4V it acts as a resistor of 170 Ohm, i don't think the relation between voltage and current is as linear in a led as it's with a resistor.
i don't think the relation between voltage and current is as linear in a led as it's with a resistor.
No it's not because the relationship for a led is infact NON-LINEAR. (as Mike mentioned)
Equally untrue. (except for the part mentioned above) A led is nothing like a resistor. How can it be ? It's a DIODE ! You need to understand the difference between a diode (a semiconductor) and a resistor (non-semiconductor). Your understanding is all wrong and your math is all wrong.
I've seen many responses here, but half of them read as if people assume the resistance of a led is 0 Ohm?
Not 0 ohms but , like other semiconductors very low since it is designed to conduct and being a diode , the only thing to prevent it from self-destructing when connected directly across a power supply is a current limiting resistor.
@Nick,
Great link ! (Care & feeding of Leds)
Still no response from the OP. She should at least have the courtesy to acknowledge the time spent by
members answering her post.
raschemmel:
@RACEMANIAC
I see you're thinking basically the same things as me. Probably due to my limited experience the ways i word it isn't how i should word it :p.
I probably don't have that good of a view yet of how a led acts when putting 9V on it, but saying it's no load at all and a pure short circuit doesn't seem that correct to me (although for the led it might as well be that, it'll burn up nearly instantly XD) .
And the point i tried to make with assigning it a resistance, is that at the correct voltage for that led, there is no difference between the led, and a resistor of 170 Ohm (so at the right voltage, it's a load of 170 Ohm in your circuit, and not some magic semiconductor device that does strange things).
It's nice however to see the replies here go deeper into that and give me some more terminology and vocabulary to talk about things like this :).
It's good you are trying to learn but I think you still don't see the magic of semiconductors.