Go Down

### Topic: Why my LED dies when connected directly to a 9v battery? (Read 19255 times)previous topic - next topic

#### cybefox

##### Jan 18, 2015, 12:12 am
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?

#### AWOL

#1
##### Jan 18, 2015, 12:15 am
V= IR
I=V/R

9/470 = ?

(This assumes zero internal resistance for the battery)

#### cybefox

#2
##### Jan 18, 2015, 12:19 amLast Edit: Jan 18, 2015, 12:20 am by cybefox
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!

#### raschemmel

#3
##### Jan 18, 2015, 01:11 amLast Edit: Jan 18, 2015, 01:14 am by raschemmel
Quote
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)

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)
Arduino UNOs, Pro-Minis, ATMega328, ATtiny85, LCDs, MCP4162, keypads,<br />DS18B20s,74c922,nRF24L01, RS232, SD card, RC fixed wing, quadcopter

#### Tamulmol

#4
##### Jan 18, 2015, 01:39 am
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???

#### ChilliTronix

#5
##### Jan 18, 2015, 01:47 am
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.

#### raschemmel

#6
##### Jan 18, 2015, 03:27 am
Quote
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.
Arduino UNOs, Pro-Minis, ATMega328, ATtiny85, LCDs, MCP4162, keypads,<br />DS18B20s,74c922,nRF24L01, RS232, SD card, RC fixed wing, quadcopter

#### floresta

#7
##### Jan 18, 2015, 03:38 am
Quote
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

#### Paul__B

#8
##### Jan 18, 2015, 08:08 am
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.

#### nickgammon

#9
##### Jan 18, 2015, 08:17 am
The care and feeding of LEDs

Quote
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.
Please post technical questions on the forum, not by personal message. Thanks!

#### Paul__B

#10
##### Jan 18, 2015, 08:24 am
A better statement would be:
Quote
LEDs are not like household appliances in that you can't just plug them in and they work,

#### racemaniac

#11
##### Jan 18, 2015, 09:03 am
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 .

#### MarkT

#12
##### Jan 18, 2015, 12:17 pm
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.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### Paul__B

#13
##### Jan 18, 2015, 01:04 pm
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.

#### floresta

#14
##### Jan 18, 2015, 05:44 pm
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

Go Up