can I burn out an LED without burning out the resistor

blocco a spirale

Joined Jun 18, 2008
1,546
How are you getting the numbers in your last two statements?

How do you figure that at 12V with a 1Ω/10W resistor that you would see 1.2A of current?

Hint: Isn't the whole point to see the importance of applying Ohm's Law correctly?
I had hoped that by post #79 Ohm's law would be well understood.:rolleyes:
 

Thread Starter

opeets

Joined Mar 16, 2015
103
How are you getting the numbers in your last two statements?

How do you figure that at 12V with a 1Ω/10W resistor that you would see 1.2A of current?

Hint: Isn't the whole point to see the importance of applying Ohm's Law correctly?
For the next to last one.....
I = V/R
I = 3V/1 Ohm = 3A (but since the power supply only is capable of 2A, that is all it will draw). Or is this a wrong assumption? That's exactly why I'm asking.

For the last one I miscalculated. It should have been 12A.
I = V/R
I = 12V/1 Ohm = 12A. But as asked above, what happens in the case where it exceeds the power supply's capabilities?
 

Thread Starter

opeets

Joined Mar 16, 2015
103
To all contributors (even the wise guys who give us all a chuckle) thanks for your continued support through 82 posts.
 

WBahn

Joined Mar 31, 2012
29,979
For the next to last one.....
I = V/R
I = 3V/1 Ohm = 3A (but since the power supply only is capable of 2A, that is all it will draw). Or is this a wrong assumption? That's exactly why I'm asking.
Your basic thinking is correct. The details of how it plays out are anyone's guess. Using a 1Ω resistor will result in you reaching the 2A rating of the power supply. But that does NOT mean that the supply will simply put out 2A into your resistor. It might. Good supplies will transition to a current-limiting mode and drop the output voltage so as to hold the set current limit. But lesser supplies only "guarantee" that they can output 2A while maintain the full voltage. They can generally put out something more than that before the voltage starts dropping and they may be able to put out significantly more than 2A by the time the output is getting dragged down by 25% or more. Many of these supplies will have thermal protection so that if you are pulling too much power for too long they will "trip" somehow. That might take the form of the current getting limited or it might take the form of the supply shutting down completely until it cools. Other supplies take a simpler route -- we'll deliver as much current as we can until the fuse blows.

Unless you KNOW how a supply will react when its specs are exceeded, it is best not to try to exceed the specs (unless the intent is to find out how it will react, allowing for the possibility that it might react by becoming a flaming brick).

For the last one I miscalculated. It should have been 12A.
I thought you had made the oft-seen mistake of going from the power rating and trying to calculate the current based on that (though that would have given you a different result, but we've seen countless times when someone would go 12/10 = 1.2, oh, so it must be 1.2A. Glad that wasn't the case here.
I = V/R
I = 12V/1 Ohm = 12A. But as asked above, what happens in the case where it exceeds the power supply's capabilities?
Thanks for tracking your units. PLEASE teach your son to ALWAYS do that!
 

WBahn

Joined Mar 31, 2012
29,979
To all contributors (even the wise guys who give us all a chuckle) thanks for your continued support through 82 posts.
I, and I think many of us, applaud what you are trying to do. Even the ones that are shaking their heads at how you are going about it. Some might think that there are better ways, but I think almost everyone acknowledges that there is merit in the attempt -- and a whole lot of learning to he had along the way.
 

Thread Starter

opeets

Joined Mar 16, 2015
103
Your basic thinking is correct. The details of how it plays out are anyone's guess. Using a 1Ω resistor will result in you reaching the 2A rating of the power supply. But that does NOT mean that the supply will simply put out 2A into your resistor. It might. Good supplies will transition to a current-limiting mode and drop the output voltage so as to hold the set current limit. But lesser supplies only "guarantee" that they can output 2A while maintain the full voltage. They can generally put out something more than that before the voltage starts dropping and they may be able to put out significantly more than 2A by the time the output is getting dragged down by 25% or more. Many of these supplies will have thermal protection so that if you are pulling too much power for too long they will "trip" somehow. That might take the form of the current getting limited or it might take the form of the supply shutting down completely until it cools. Other supplies take a simpler route -- we'll deliver as much current as we can until the fuse blows.

Unless you KNOW how a supply will react when its specs are exceeded, it is best not to try to exceed the specs (unless the intent is to find out how it will react, allowing for the possibility that it might react by becoming a flaming brick).
Interesting and certainly good to know. I don't intend on creating a hazardous condition but at least I know not to attempt drawing more than 2A.
Have you had a chance to see my general questions about resistors in post #80?
 

Thread Starter

opeets

Joined Mar 16, 2015
103
I, and I think many of us, applaud what you are trying to do. Even the ones that are shaking their heads at how you are going about it. Some might think that there are better ways, but I think almost everyone acknowledges that there is merit in the attempt -- and a whole lot of learning to he had along the way.
Both my son and I area learning a lot from this experience. It is certainly taking a lot more time and effort (and not necessarily progressing too far in the process to date) than other kids in his class are doing on their own projects so keeping him on track has been an effort, a lot of it being trial and error due to my own shortcomings in the area of electrical circuitry. At least now with a regulated power supply in lieu of household batteries, the experimentation phase should become more streamlined.
 

WBahn

Joined Mar 31, 2012
29,979
Interesting and certainly good to know. I don't intend on creating a hazardous condition but at least I know not to attempt drawing more than 2A.
Have you had a chance to see my general questions about resistors in post #80?
Actually, I didn't see that post at all. Thanks for bringing it to my attention.

Some basic questions about resistors while I'm on the subject....

1) How is it that a resistor draws current from a power supply when in fact it actually resists it?
Some of this gets into word semantics. When we think of the word "draws" we envision the resistor pulling electrons out of a supply. A better description is that, when connected across a potential difference, a resistor "allows" electrons to be pushed through it. The higher the resistance the more voltage difference it will take to push electrons through at the same rate.

Is it more correct to say that the resistor accepts a certain amount of current (which is fixed based on the resistor value but limited by the resistor's power rating)?
This is NOT a good way to look at it. The amount of current that a resistor will allow is not dependent on the power rating, only on the applied voltage and the resistance. If the current exceeds the amount that the resistor is rated for, then bad things happen.

2) Does it then follow to say that based on the maximum current a resistor can accept (e.g. 50mA for a 1/4 watt 100 Ohm resistor) that if the maximum voltage (5V in this case) is exceeded the resistor will (at least start to) burn up?
Yes and no. It depends on why the rating was set where it was -- and the damage mode is usually not due to the current, but rather the heat that results. If you put that 100Ω resistor in a circulating coolant bath, you could put a lot more than 50mA through it without damaging it because you are carrying the heat away as it is created instead of letting it build up and cause an excessive rise in temperature. Some ratings are based on keeping the resistance value within specs. Others are based on keeping temperature affects reversible. For instance, high current precision resistors used for current sensing should not be allowed to exceed a certain temperature, such as 145°C, otherwise their resistance value, even after they have cooled off, may be permanently changed. So exceeding the rating does cause the unit to change very noticeably (as far as appearance and basic behavior is concerned), but bad things are still happening.

3) When a resistor burns up does the circuit short?
It depends -- mostly on the type of resistor and on just how the excess current is applied. Usually run of the mill resistors will fail as an open as some part of them physically burns up. If you really pump excessive power into a resistor it can literally vaporize.
 

Thread Starter

opeets

Joined Mar 16, 2015
103
The amount of current that a resistor will allow is not dependent on the power rating, only on the applied voltage and the resistance. If the current exceeds the amount that the resistor is rated for, then bad things happen.
Don't quite understand how it is dependent on the applied voltage.

Here is my thinking, please tell me where I go wrong....

P = VI
P = (IR)*I
I = √(P/R)

Since Pmax is 0.25 watts for the 100Ω resistor,
I <= √(0.25/100) = 50mA

This tells me that this resistor's capabilities are limited to 50mA flowing through it, which for a 100Ω resistor means that the max applied voltage is:

V = IR = (.05)(100) = 5V

Any voltage less than 5V is certainly acceptable.

So the above being said....it would appear to me that the amount of current that a resistor will allow is indeed dependent on the power rating.

Unless..... the power rating is only a guide to let you know that even though it is possible to exceed it, there is no guarantee that the resistor will still function properly or even safely at that point.
 

tjohnson

Joined Dec 23, 2014
611
Don't quite understand how it is dependent on the applied voltage.

Here is my thinking, please tell me where I go wrong....

P = VI
P = (IR)*I
I = √(P/R)

Since Pmax is 0.25 watts for the 100Ω resistor,
I <= √(0.25/100) = 50mA

This tells me that this resistor's capabilities are limited to 50mA flowing through it, which for a 100Ω resistor means that the max applied voltage is:

V = IR = (.05)(100) = 5V

Any voltage less than 5V is certainly acceptable.

So the above being said....it would appear to me that the amount of current that a resistor will allow is indeed dependent on the power rating.

Unless..... the power rating is only a guide to let you know that even though it is possible to exceed it, there is no guarantee that the resistor will still function properly or even safely at that point.
Resistors do not have a current rating, but a power rating. Since P=VI, a resistor with a power rating of 1 watt can handle 1e6 amps of current if the applied voltage is 1e-6 volts. This is obviously an extreme example, but the point is that the amount of power is directly affected by the applied voltage.

Hope this helps.
 

WBahn

Joined Mar 31, 2012
29,979
Don't quite understand how it is dependent on the applied voltage.
Think about a straw. Take two straws of about the same length but considerably different diameters (say a soda straw and a coffee/tea straw).

Pick up the large straw and blow through it.

You'll get a certain amount of airflow, right?

Now blow harder. You get more airflow, right?

Now blow softer. You get less airflow, right?

So the amount of air you get through the straw is dependent on how hard you try to push air through the straw, right?

Well, if you apply more voltage across a resistor, you are trying harder to push current through it and, not surprisingly, you will get more current being pushed through it.

Now pick up the small straw and blow through it about as hard as you originally did with the larger straw. You get quite a bit less airflow through it than you did the larger straw, right? The small straw has considerably more resistance to air flow than the larger straw did, so the same pressure results in less current. The same with resistors.

Here is my thinking, please tell me where I go wrong....

P = VI
P = (IR)*I
I = √(P/R)

Since Pmax is 0.25 watts for the 100Ω resistor,
I <= √(0.25/100) = 50mA

This tells me that this resistor's capabilities are limited to 50mA flowing through it, which for a 100Ω resistor means that the max applied voltage is:

V = IR = (.05)(100) = 5V

Any voltage less than 5V is certainly acceptable.

So the above being said....it would appear to me that the amount of current that a resistor will allow is indeed dependent on the power rating.

Unless..... the power rating is only a guide to let you know that even though it is possible to exceed it, there is no guarantee that the resistor will still function properly or even safely at that point.
Exactly. If you buy a truck with a load rating of 1000 lb, there is nothing to prevent you from exceeding it, but bad things will happen at some point. All the manufacturer is doing with the rating is saying that as long as you don't exceed 1000 lb, bad things won't happen.
 

WBahn

Joined Mar 31, 2012
29,979
Resistors do not have a current limit, but a power limit. Since P=VI, a resistor with a power rating of 1 watt can handle 1e6 amps of current if the voltage is 1e-6 volts. This is obviously an extreme example, but the point is that the amount of power is directly affected by the applied voltage.

Hope this helps.
But the point here is that a power limit is more of a "rating" that you are cautioned not to exceed, as opposed to some kind of physical limit that the resistor somehow enforces and makes it so that you can't exceed it.
 

Sensacell

Joined Jun 19, 2012
3,432
I think you will find the the LED will light with VERY low currents, so the point where the "LED does not light at all" will be hard to reach.
 
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