picture A - i think the LED will get burn.
picture B- LED wont get burnt.

question : why picture B wont get burn?
can anyone here help explain?
what will happen if the supply voltage in the picture is 220v is it ok to use 0.1ohm resistor?and what if the resistor is 1Mohm , is it ok to use 0.1v as the supply voltage?

 

An LED will burn out if the current through the LED exceeds the maximum allowable current. A typical operating current for an LED ranges from 1mA to 20mA.

Use Ohm's Law to estimate the current, I = V/R.

An LED needs in excess of 2V for it to turn on.

A supply voltage greater than 3V can destroy an LED if the current is allowed to exceed the maximum allowable current.

 

Looks like a candidate for "Ohm's Law for Noobies".

Look here:http://forum.allaboutcircuits.com/showthread.php?t=69757

 

Looks like a candidate for "Ohm's Law for Noobies".

Look here:http://forum.allaboutcircuits.com/showthread.php?t=69757

i still dont have any explanation or clues from your suggestion.. would you mind eplaining it by answering the question?

 

picture A - i think the LED will get burn.
picture B- LED wont get burnt.

question : why picture B wont get burn?
can anyone here help explain?
what will happen if the supply voltage in the picture is 220v is it ok to use 0.1ohm resistor?and what if the resistor is 1Mohm , is it ok to use 0.1v as the supply voltage?

There is no guarantee that Pic A will "get burn".
There is no guarantee that Pic B "wont get burnt"

Whether a LED 'burns' depends on how much current flows through it as already stated. Please re-read MrChips response several times and think about what he said. He is right on with a simple straight forward explanation. And your pictures do NOT supply enough information to state what will or will not happen.

Do you understand how voltage and current affect circuits? If you have the slightest intention of understanding electricity/electronics it is perhaps the MOST important concept you can learn.

Now sit down for this.
The reason Pic A may not burn is because it requires enough voltage applied to cause enough current to flow to cause something to 'burn'. If the applied voltage is low enough 'NO BURN'!

The reason Pic B may burnt is if the voltage applied is high enough, the resistor may not limit the current to a safe value allowing burnt to extinguish life of LED.

You need to study and understand component properties and how voltage and current affect those components.

 

Looks like a candidate for "Ohm's Law for Noobies".

Look here:http://forum.allaboutcircuits.com/showthread.php?t=69757

There is no guarantee that Pic A will "get burn".
There is no guarantee that Pic B "wont get burnt"

Whether a LED 'burns' depends on how much current flows through it as already stated. Please re-read MrChips response several times and think about what he said. He is right on with a simple straight forward explanation. And your pictures do NOT supply enough information to state what will or will not happen.

Do you understand how voltage and current affect circuits? If you have the slightest intention of understanding electricity/electronics it is perhaps the MOST important concept you can learn.

Now sit down for this.
The reason Pic A may not burn is because it requires enough voltage applied to cause enough current to flow to cause something to 'burn'. If the applied voltage is low enough 'NO BURN'!

The reason Pic B may burnt is if the voltage applied is high enough, the resistor may not limit the current to a safe value allowing burnt to extinguish life of LED.

You need to study and understand component properties and how voltage and current affect those components.

finally an answer ... thank you sir

 

We have no idea whether or not you understand the answer.

To be certain that you understand you should state the answer in your own words.

 

if you add a resister, as in picture B, then you will resist the flow of current, and probably have less current.

Is this some kind of take home quiz? If so then write down "no burn on B because resister is already burned"

 

i still dont have any explanation or clues from your suggestion.. would you mind eplaining it by answering the question?

Copping an attitude is surest way to get ignored or marginalized on this or any other forum. Read the answers carefully and then ask followup questions. Don't just say the answer is insufficient without an explanation. It makes you come off as a rude sponge.

You want to show what efforts you've made to understand the problem and the answers.

 

Copping an attitude is surest way to get ignored or marginalized on this or any other forum. Read the answers carefully and then ask followup questions. Don't just say the answer is insufficient without an explanation. It makes you come off as a rude sponge.

You want to show what efforts you've made to understand the problem and the answers.

One must remember that we all don't have the same grasp on the English language here.. I can just about guarantee that the "tone" of that response was not intended to be "copping and attitude" or "rude".

Just the "would you mind" was enough for me to see the intent..
I've never heard someone say "would you mind kissing my arse"

 

i still dont have any explanation or clues from your suggestion.. would you mind eplaining it by answering the question?

Two rather long pages about how much resistance an LED needs and you don't have a clue? Sorry. I can't do any better than spending a week writing an essay just for beginners. You can only hope some other person here speaks in a way you can understand.

 

picture A - i think the LED will get burn.
picture B- LED wont get burnt.

question : why picture B wont get burn?
can anyone here help explain?

Both circuits in your example can either work or burn the LED; it all depends on the source's voltage and the value of the resistor. As already mentioned, more current than what the LED can handle will burn it, and less voltage or current than what it needs would not light it; so, if the voltage in pic. A is 2V, then the LED will not get burnt, and if the resistor in pic. B is 1M and the voltage source 100000V then it will get burnt.

So, what you really need to consider for the LED to get burnt is the amperage (the current), not the voltage or the resistance. And how do you know the amperage or current that passes through the LED? You calculate it with ohm's law:

A = V / R

Since most common LEDs work best at 20mA (0.02 Amps), then you need to make sure that the combination of voltage and resistor give you that amperage.

what will happen if the supply voltage in the picture is 220v is it ok to use 0.1ohm resistor?

Let's see what would happen using ohm's law...

A = 220V / 0.1Ω = 2200 Amps

You will definitely burn the LED, since what you need is 0.02 Amps.

and what if the resistor is 1Mohm , is it ok to use 0.1v as the supply voltage?

Let's see again...

A = 220v / 1000000Ω = 0.00022 A

So the LED will not even light up, as the current is not enough.

It will also not work with 0.1v, since an LED need a minimum of 2V to light up. This is because the LED is a diode; but that's another story...

 

One must remember that we all don't have the same grasp on the English language here.. I can just about guarantee that the "tone" of that response was not intended to be "copping and attitude" or "rude".

Just the "would you mind" was enough for me to see the intent..
I've never heard someone say "would you mind kissing my arse"

I don't know. Around here (and I don't know how local this might be), the phrase "would you mind" is almost always a precurser to a comment that is meant in a dismissive, rude, hostile, challenging, or condescending manner.

Would you mind driving in your own lane?

Would you mind keeping your dog out of my yard?

Would you mind flushing the toilet when you're done.

Would you mind having not writing me bad checks?

Would you mind answering the question?

Would you mind putting your dirty clothes in the hamper?

But, I could easily see in other cultures that the phrase "would you mind" being meant as a very polite way of making a request. And, to be sure, it can be meant that way here, as well. It's one of those phrases that the tone of voice makes all the difference.

 

A common theme on this forum, and others, is that when posting a question, you need to show that you have made "some" effort to answer the question on your own. Absent that demonstration of effort a response suggesting that some effort be expended seems entirely appropriate.

 

This looked like homework and still does given that the OP bugged out as soon as the answer was sufficiently packaged. If that's the case, he may survive in his class until the next test.

 

Maybe this is not the case, but I remember that when I first began studying electronics the single component that I had more problems understanding was the LED (the diode). My teacher used to explain it to me once and again, but I couldn't get over the fact that an LED would drop a fixed voltage and had no fixed resistance.

What I'm getting at is... with some concepts, when you are a beginner, it doesn't matter if you have access to all the relevant information; you simply won't understand it until someone spells it to you with terms you can relate to.

 

...I could [not] get over the fact that an LED would drop a fixed voltage and had no fixed resistance.

In the water analogy, an LED is a bit like a waterfall. The water drops in height (voltage) with little regard for the current.

What I'm getting at is... with some concepts, when you are a beginner, it doesn't matter if you have access to all the relevant information; you simply won't understand it until someone spells it to you with terms you can relate to.

That's where great teachers come in. There is a big gap between knowing something yourself and being able to teach it to someone else.

It'll be interesting to see if we ever hear from the OP again. Doubt it.

 

In the water analogy, an LED is a bit like a waterfall. The water drops in height (voltage) with little regard for the current.

That's where great teachers come in. There is a big gap between knowing something yourself and being able to teach it to someone else.

It'll be interesting to see if we ever hear from the OP again. Doubt it.

Fortunately I ended up understanding it a long time ago. But I remember that for a while I had a lot of problems analyzing any circuit with an LED because I didn't understand how it would impact the circuit in terms of resistance or where the drop voltage came from.

 

I'm pretty sure that few of us would make the claim that we are great teachers. Some of us are better than others. What does distinguish us is the desire to participate and make an effort to help. We sometimes learn things in the process. The forum is an excellent place to refine your own ideas by explaining them to others. This absolutely could not happen in any reasonable way, with worldwide scope,when I was an undergraduate. How much better is this than the alternative?

My definition of "copping an attitude" is not getting what you expect, and then complaining about it. This is not a matter of "tone" or "culture", but a clear reflection of unmet expectations. We can try to do better, but the OP's need to meet us part of the way.

 

Let's say you have a 9V battery and an LED that operates on 2.4V at 20mA. If you connect the LED across the battery terminals, it will burn out. To prevent this from happening, a current-limiting resistor must be used. Here's how to calculate the value of the resistor using Ohm's Law:

Take the battery voltage and subtract the LED operating voltage. The voltage that remains is the voltage that must be dropped by the resistor.

In this case, 9 - 2.4 = 6.6.

Using Ohm's Law to calculate for resistance, R = E / I where R is resistance in Ohms, E is voltage in volts and I is current in Amperes.

So we have the remaining voltage (6.6) divided by the LED current (20mA or .02A). 6.6 / .02 = 330 Ohms.

Now we have to calculate the wattage of the resistor. That's the amount of heat a resistor has to safely dissipate to prevent the resistor from burning.

Using Joule's Law, the formula is P = I x E where P is power in watts, I is current in Amperes and E is voltage in volts.

So we have .02A x 6.6V = .132 Watts. Since a 1/8 = .125, we cannot use a 1/8 watt resistor. 1/4 = .250 so it would be safe to use a 1/4 watt (or higher) resistor for this purpose.

I hope this clears things up for you.