Go Down

Topic: Why my LED dies when connected directly to a 9v battery? (Read 19245 times) previous topic - next topic

KenF

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.

Grumpy_Mike

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

raschemmel

#17
Jan 18, 2015, 06:47 pm Last Edit: Jan 20, 2015, 05:07 am by raschemmel
@RACEMANIAC,


Quote
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:

Quote
Note: Pulse width ≤0.1 msec, duty ≤1/10.
Quote
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).


5mm RED LED DATASHEET


Quote
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.  
Quote
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.


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

racemaniac

#18
Jan 20, 2015, 11:57 am Last Edit: Jan 20, 2015, 11:57 am by racemaniac
@RACEMANIAC
<lots of interesting stuff>
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 :).

raschemmel

It's good you are trying to learn but I think you still don't see the magic of semiconductors.
Arduino UNOs, Pro-Minis, ATMega328, ATtiny85, LCDs, MCP4162, keypads,<br />DS18B20s,74c922,nRF24L01, RS232, SD card, RC fixed wing, quadcopter

racemaniac

It's good you are trying to learn but I think you still don't see the magic of semiconductors.
Just looked it up, and it's apparantly an exponential function that relates the current to the voltage over a semiconductor (i know it doesn't follow ohm's law ^^). So when the voltage increases a lot, it's indeed pretty close to short circuit. (a bit closer than i thought :) ). I still find it strange to say that it's no load at all, in the end it's still no superconductor ^^.

raschemmel

#21
Jan 20, 2015, 01:47 pm Last Edit: Jan 20, 2015, 06:08 pm by raschemmel
Ok, I'll play your game.
Take an odinary led and connect it with a 220 ohm resistor. Now connect your meter in series in current mode OR connect it across the resistor in VOLTAGE mode. Record the results. Now replace the led with a 170 ohm resistor and repeat the measurement. Now remove the 170 ohm resistor and replace the led. Verify that it still draws thd same current you measured before and ghen put a shorting jumper across the led, shorting the two led pins together and repeat the measurement. Which of the two substitutes draws an amount of current that closest approximates the current of led, the 170 ohm resistior you used in place of the led, or the shorting jumper you put across the led ? How much of a difference in cureent did you see when you shorted the led ?
Arduino UNOs, Pro-Minis, ATMega328, ATtiny85, LCDs, MCP4162, keypads,<br />DS18B20s,74c922,nRF24L01, RS232, SD card, RC fixed wing, quadcopter

racemaniac

#22
Jan 20, 2015, 02:22 pm Last Edit: Jan 20, 2015, 02:25 pm by racemaniac
Ok, I'll play your game.
Take an odinary led and connect it with a 220 ohm resistor.Now connect your meter in series in current mode OR connect it across the resistor in VOLTAGE mode. Record the results. Now replace the led with a 170 ohm resistor and repeat the measurement.Now remove the 170 ohm resistor and replace the led.Verify that it still draws thd same current you measured before and ghen put a shorting jumper across the led, shorting the two led pins together and repeat the measurement.Which of the two substitutes draws an amount of current that closest approximates the current of led, the 170 ohm resistior you used in place of the led, or the shorting jumperyou put across the led ? How much of a difference in cureent did you see when you shorted the led ?
What are you using to type your messages that adds all those annoying new lines, making your text so narrow, it's really distracting.

If i understand your experiment right, it would come down to this (and i'll add my expectations, but if i'm wrong somewhere, i would be very interested :) ):
1. Led + 220 Ohm, would give the expected 20mA
2. 170 Ohm resistor + 220 Ohm: i would expect the same 20mA, and the same voltage drops in the circuit
3. here you go back to sitation 1 if i understand you correctly? so still the same
4. here you short out the led, so only the 220 Ohm resistor is left, i'd expect the current to almost double as the 170 Ohm on the circuit disappears.
and a hypothetical 5: remove the current limiting transistor and connect only the led in the chain: probably a pretty high current (but the led will immediately burn out) since it's current rises exponentially compared to the voltage over it :).

Also, reading up on it a bit more, i'm starting to understand better why they keep using the voltage drop terminology :). It does vary with current, but very little in diodes. So when you put too much voltage on a led, the current will indeed rise very quickly and basically short circuit :).

Is there anything i'm misunderstanding? :)

Paul__B

What are you using to type your messages that adds all those annoying new lines, making your text so narrow, it's really distracting.
I entirely agree.  What is happening, is that a number of people have become so pi$$ed off with the unreliability of the forum software, that they are typing their responses in a separate editor and then using cut-and-past to insert them in the "reply" box, but in the process forgetting - or not comprehending - how to turn off the "line wrap" in that separate editor.

The "Draft" function of the current forum software however makes that unnecessary and I am finding it much more stable than before though I still get caught on occasion and also tend to get the "token" foul-up caused by some truly idiotic code somewhere in the system. :smiley-eek:

Also, reading up on it a bit more, I'm starting to understand better why they keep using the voltage drop terminology :). It does vary with current, but very little in diodes. So when you put too much voltage on a led, the current will indeed rise very quickly and basically short circuit :).
I do believe you are getting the message.  :smiley-lol:

floresta

Quote
What are you using to type your messages that adds all those annoying new lines, making your text so narrow, it's really distracting.
Quote
I entirely agree.  What is happening, is that a number of people have become so pi$$ed off with the unreliability of the forum software, that they are typing their responses in a separate editor and then using cut-and-past to insert them in the "reply" box, but in the process forgetting - or not comprehending - how to turn off the "line wrap" in that separate editor.
He has been doing this for a lot longer than the new forum software has been in use.  I mentioned it to him months ago but since he doesn't have to read his own posts he just doesn't seem to care. 

Don

raschemmel

#25
Jan 20, 2015, 04:17 pm Last Edit: Jan 20, 2015, 06:09 pm by raschemmel
I am using my cell phone and I can't correct the lines on such a small screen. sorry for that.

You need to reread the instructions
Condition-1: led +220 ohm resistor , current approximately 20 mA (+/- difference due to particular led)
condition-2: 170 ohm + 220 ohm = 390 ohm. current I = 5V/390=12.8 mA
condition -3: short + 220 ohm = 5V/220 ohm = 22.7 mA
condition-3 current - condition-1 current = 22.7-20 = 2.7 mA

Point ?
Current for condition-3 is closer to current for condition-1 (2.7 mA verses 7.2 mA)

Quote
170 Ohm resistor + 220 Ohm: i would expect the same 20mA, and the same voltage drops in the circuit
Why ? 170+220=390 ohm. 5V/390 ohm = 12.8 mA (not 20 mA)


Quote
here you short out the led, so only the 220 Ohm resistor is left, i'd expect the current to almost double as the 170 Ohm on the circuit disappears.
The current for 220 ohm is 22.7 mA (not exactly double but considerably more) yet ONLY 2.7 mA MORE than without the short. Why ? Because the led forward voltage keeps the voltage at the turn-on voltage limiting the current. It is not drawing as much current as the short but it is drawing considerably MORE than the 170 resistor by 7.2 mA. The magic of semiconductors is the ability of the led emulate a zener diode by regulating the voltage at the forward voltage, thus dropping the voltage across the 220 ohm resistor from 5V to the forward voltage.  Yet neither the zener diode nor the led can be operated without the current limiting resistor without self destructing.

@Don,
Thanks for sharing...
Arduino UNOs, Pro-Minis, ATMega328, ATtiny85, LCDs, MCP4162, keypads,<br />DS18B20s,74c922,nRF24L01, RS232, SD card, RC fixed wing, quadcopter

racemaniac

#26
Jan 20, 2015, 04:57 pm Last Edit: Jan 20, 2015, 04:58 pm by racemaniac
You need to reread the instructions
I reread them a few times before answering, but they weren't that clear ^^.
I'll give it a try when i got time, seems like some nice experiments :)

What i'm currently thinking: If when a voltage of 3.4V is on a led, it'll allow a current of 20mA trough it, it at that point is equivalent to a 170 Ohm resistor in the circuit, even if it's a magic semiconductor. But if i'm wrong about that, that would be very interesting, and will cause me to reevaluate some things i thought i understood. i'm curious. Atm my line of thought is if it sounds like a duck and quacks like one, it probably is one (although the duck will change into an elephant once i alter the voltage :p).

Grumpy_Mike

Quote
I reread them a few times before answering, but they weren't that clear ^^.
Yes they were clear, you misunderstood them so why not ask about the things you thought were not clear.

Quote
If when a voltage of 3.4V is on a led, it'll allow a current of 20mA trough it, it at that point is equivalent to a 170 Ohm resistor in the circuit, even if it's a magic semiconductor.
Yes in a way, BUT that magic voltage of 3.4V will not apply to every LED you get even from the same manufacturer and same batch. It will change with temperature and age. It only has to change a little bit for there to be a LOT of change in the current. At that point this "effective" resistance starts to shoot down very quickly and the current increases very quickly because you have nothing to limit it.

Quote
Atm my line of thought is if it sounds like a duck and quacks like one, it probably is one
Probably, but in this case it is not.


Paul__B

What i'm currently thinking: If when a voltage of 3.4V is on a led, it'll allow a current of 20mA through it, it at that point is equivalent to a 170 Ohm resistor in the circuit, even if it's a magic semiconductor.
That is still expressed backwards.

What you actually mean is: "If I am feeding a LED with 20 mA and at that current its voltage drop is 3.4V, it at that point is equivalent to a 170 Ohm resistor in the circuit".

racemaniac

#29
Jan 20, 2015, 05:15 pm Last Edit: Jan 20, 2015, 05:15 pm by racemaniac
That is still expressed backwards.

What you actually mean is: "If I am feeding a LED with 20 mA and at that current its voltage drop is 3.4V, it at that point is equivalent to a 170 Ohm resistor in the circuit".

doesn't it go both ways? if i put my lab supply to 3.4V constant voltage mode, and only put the led on it, the led will only allow 20mA trough it. otherwise since most power supplies are constant voltage, and not constant current, getting leds to work would be hell i think.

Go Up