So here are the results of tonight's experiments (using a pair of identical Extech DMMs)....

1. Using a 9V battery (which measured 9.24V on Extech DMM1) we used a pair of 180 ohm resistors in series to get us a calculated current value of 20.67mA. Using Extech DMM2 we measured the current as 19.46mA when placed in series with the anode of the LED and the battery lead. The red LED lit up nicely. The voltage measured across R1 was 3.54V, across R2 was 3.55V, and across the LED was 1.815V (very close to spec). However if the measured battery voltage was 9.32V, what happened to the other 0.4V? The difference in the measured and calculated current certainly accounts for it but then why was my current calculation off by over 1mA?

2. We then replaced both R1 and R2 with a pair of 100 ohm resistors in parallel so that Rt would be 50 ohms. The DMM measured 103.5mA but we were expecting ~150mA. Any idea what happened here with our calculation? The LED did not burn out.

3. We then added R3 (another 100 ohm resistor) in parallel with R1 and R2, so that Rt would be ~33 ohms. The DMM measured 134mA instead of the expected ~225mA. The voltage measured across each of the three resistors was 4.35V which also didn't make sense to us. The LED did not burn out.

4. We then replaced R1, R2, and R3 with a pair of 50 ohms resistors R1 and R2 producing an Rt of 25 ohms. The measured current was 151MA. Again, this did not add up. LED did not burn out.

5. We then added a third 50 ohm resistor producing an Rt of ~16 ohms. The LED didn't light up.

6. After disconnecting the battery, we then replaced the burned out LED with another identical LED using the setup described in Step 5. The LED did not burn out this time and the measured current was 183mA.

7. We then added a 4th 50 ohm resistor producing an Rt of ~12.5 ohms. The LED briefly lit up and then went out. The current on the DMM at that point kept fluctuating between 198mA and 210mA.

Conclusions:
- Several measured current values did not match our expected values.
- Burning out an LED does not appear to be a good demonstration of Ohm's law in action since they can withstand high amounts of current.

So now what do I do?

 

So here are the results of tonight's experiments (using a pair of identical Extech DMMs)....

1. Using a 9V battery (which measured 9.24V on Extech DMM1) we used a pair of 180 ohm resistors in series to get us a calculated current value of 20.67mA. Using Extech DMM2 we measured the current as 19.46mA when placed in series with the anode of the LED and the battery lead. The red LED lit up nicely. The voltage measured across R1 was 3.54V, across R2 was 3.55V, and across the LED was 1.815V (very close to spec). However if the measured battery voltage was 9.32V, what happened to the other 0.4V? The difference in the measured and calculated current certainly accounts for it but then why was my current calculation off by over 1mA?

You said that the battery measured 9.24V. Where is the 9.32V coming from?

In either case, your measured the open circuit terminal voltage of the battery. As soon as you start drawing current from a battery, the terminal voltage will drop. This is due to internal processes that are usually modeled as the "internal resistance" of the battery. Think of the battery as being an ideal voltage source in series with a small resistance. As you draw more current, more voltage gets dropped across this internal resistance.

You also have a voltage drop that you didn't measure, namely the voltage drop across the DMM that you were using as a current meter.

Your resistor is nominally 180Ω, but it has a tolerance, probably 5% is my guess. Notice that an error of 1mA in 20mA is a 5% error. What does your meter say the actual resistance of that resistor is?

2. We then replaced both R1 and R2 with a pair of 100 ohm resistors in parallel so that Rt would be 50 ohms. The DMM measured 103.5mA but we were expecting ~150mA. Any idea what happened here with our calculation? The LED did not burn out.

You need to measure the battery voltage under load. 100 mA is a pretty heavy load for a 9V alkaline battery. It's life at that draw is only three hours or so. So unless you are working with a very fresh battery, your terminal voltage is probably dropping pretty significantly.

3. We then added R3 (another 100 ohm resistor) in parallel with R1 and R2, so that Rt would be ~33 ohms. The DMM measured 134mA instead of the expected ~225mA. The voltage measured across each of the three resistors was 4.35V which also didn't make sense to us. The LED did not burn out.

You are excessively loading the battery and dragging it down. I thought you were going to use AA batteries.

 

You said that the battery measured 9.24V. Where is the 9.32V coming from?

My mistake. I think I initially measured it as 9.32V and then later on when I checked again it was 9.24V. Not a significant factor though in my expected vs measured results.

In either case, your measured the open circuit terminal voltage of the battery. As soon as you start drawing current from a battery, the terminal voltage will drop. This is due to internal processes that are usually modeled as the "internal resistance" of the battery. Think of the battery as being an ideal voltage source in series with a small resistance. As you draw more current, more voltage gets dropped across this internal resistance.

Understood. So how does one measure the actual voltage supply before introducing it to the circuit.

You also have a voltage drop that you didn't measure, namely the voltage drop across the DMM that you were using as a current meter.

Okay, is that measureable (using DMM1)?

Your resistor is nominally 180Ω, but it has a tolerance, probably 5% is my guess. Notice that an error of 1mA in 20mA is a 5% error. What does your meter say the actual resistance of that resistor is?

The measured resistances were actually between 180 and 182 ohms, so pretty close to spec.

You need to measure the battery voltage under load. 100 mA is a pretty heavy load for a 9V alkaline battery. It's life at that draw is only three hours or so. So unless you are working with a very fresh battery, your terminal voltage is probably dropping pretty significantly.

Our plan is not to let it sit for three hours, just long enough to get our expected results.

You are excessively loading the battery and dragging it down. I thought you were going to use AA batteries.

I could have used AA batteries but I decided to start with a 9V battery. How does this factor in anyway? I would have just used different resistors if I used AA batteries.



None of this changes the fact that our goal is to show that Ohm's law plays a critical role in determining the right current value to use. If we can't burn the LED out without the use of excessive current we may need to rethink our approach to this. What else can we do to demonstrate that using Ohm's law the "correct" resistor value to use is in a much smaller range.

 

The "three hours" was mentioned to point out how heavy a draw that current is for that battery type. Using AA batteries (so four of them in series to get your 9V) allows you to have a considerably smaller internal resistance.

You can get an estimate for the internal resistance of the battery (and it's resistance goes up as the state of charge goes down) by measuring the open circuit voltage and then putting a known load on it comparable to what you want to use and measuring the loaded circuit voltage. Use that information to determine the effective internal resistance.

Why don't you think that Ohm's Law plays a critical role in using a proper current? Yes, you have a wide range of current that you can use and not destroy an LED immediately, but you used Ohm's Law to determine the range of resistor values that will put your current in that range. Also, keep in mind what was said about the effect of excessive current on LED life. The range of current that you can use and have the LED last long enough to be useful is much narrower than the range it can tolerate for a few seconds.

Most things are designed, when possible, to be tolerant of a wide range of operating conditions. That's a good thing.

If you really want to show a simple circuit in which small changes in resistance produce large changes in operating point, then something like a Widlar current source would do it. But that would involve using a transistor (and three resistors and your LED) and the meaning would be muddied without a basic understanding of how a transistor works.

 

The range of current that you can use and have the LED last long enough to be useful is much narrower than the range it can tolerate for a few seconds.

Which is more or less what we were hoping to prove. It's not a necessity that the LED has to blow out instantaneously. We just want it to show that (within a reasonable amount of time) an LED will stop functioning due to its tolerance being exceeded. Perhaps we should run two experiments simultaneously. One with the standard operating current and one with a much higher current and see how long the 2nd LED lasts. We just need some evidence to prove that because we applied Ohm's law in one case the LED is still working and because we neglected to apply Ohm's law in the other case the LED did not last very long.

Using AA batteries (so four of them in series to get your 9V) allows you to have a considerably smaller internal resistance.

If I'm understanding you correctly. by using AA batteries we will have less internal resistance and be able to overcome the LED's tolerance with less effort (i.e. fewer multiple resistors in parallel). If so, what causes the difference in the internal resistance in a 9V single battery configuration and one with four AAs?

Thanks again for your help.

 

If I'm understanding you correctly. by using AA batteries we will have less internal resistance and be able to overcome the LED's tolerance with less effort (i.e. fewer multiple resistors in parallel). If so, what causes the difference in the internal resistance in a 9V single battery configuration and one with four AAs?

Thanks again for your help.

Mostly the size. For a given chemistry, the internal resistance is largely a function of the plate areas available for participating in the chemical reactions.

 

Mostly the size. For a given chemistry, the internal resistance is largely a function of the plate areas available for participating in the chemical reactions.

Okay, we will try to repeat the experiment from last week using AA batteries tonight.

 

Look forward to seeing your measurements.

 

Burning out an LED does not appear to be a good demonstration of Ohm's law in action since they can withstand high amounts of current.

You've said this a few times, but just so you're clear...

Ohm's Law always applies. It's a physical law describing the relationship between voltage, current, and resistance. Failing to burn out the LED doesn't make it a bad demonstration. And succeeding in burning out the LED doesn't make it a good demonstration.

Burning out the LED simply shows you that you need to adhere to the "max current" device specification. Some LEDs can tolerate more current. And, as you've seen, some batteries can't provide a lot of current. It's all Ohm's Law.

The best demonstration of Ohm's Law is measuring the change in current as resistance is varied and seeing the manifestation of this in the change in LED brightness. You can even plot the results on a graph to see the relationship between R and I.

Even with the 9V battery it's still a good demonstration, because you can show that the battery can't provide a constant 9V. And its reduction in voltage is why you weren't getting the current output that you calculated.

I hope you and your son have fun.

 

The best demonstration of Ohm's Law is measuring the change in current as resistance is varied and seeing the manifestation of this in the change in LED brightness. You can even plot the results on a graph to see the relationship between R and I.

I like this idea but how can you quantify brightness of an LED? By using multi-colored LEDs? I don't know much about them because I've never used them.

 

For your purposes, it is probably sufficient to only qualify the brightness -- i.e., have two LEDs and be able to say that A is brighter than B. To do that, it is best to have both LEDs on at the same time side by side. That's a fairly easy thing to do. First, however, you need to account for the fact that not all LEDs are the same, meaning that even with the same current they might not be the same brightness. But if you use two LEDs from the same pack, they will probably be pretty close. You can test this by putting the LEDs in series and then putting three or four currents through them at the low, middle, and high (but not too high) ends of the spectrum and seeing if they appear to be equally bright. Then put them in parallel, each with its own current limiting resistor, and you are all set to go.

 

We finally got back to some more experimenting last night. Not an easy thing to get a kid to do on a Friday night. This time we decided to use a 5mm red blinking LED. The local Radio Shack was going out of business so a few days ago I stocked up on a bunch of components all at 50% off.

The front of the blinking LED package said the LED's operating values were 35mA and 3V. On the back of the package it specified the "absolute max rating" as 35mA. Though we didn't try to exceed that just yet, we decided to go in the opposite direction first by calculating a resistor for a 30mA current.

Using 4.5V as our source (3 AAA batteries in a holder with leads) we first calculated that we would need a 46 ohm resistor to get 30mA (the battery voltage was 4.38 volts measured). We ended up using a 47 ohm resistor and the LED blinked as expected. We then kept progressively increasing the resistor value (100 ohms, 330 ohms, 680 ohms, and then we actually connected a bunch of 680 ohm resistors in series (4 of them actually so 2.7K total) and still saw the LED blinking but more faint each time we added another resistor. We ended the experiment with a single 10K resistor which barely turned the LED on but it definitely did not blink. The was the equivalent of ~0.1 mA. How could the LED light up at such a low current?

Tonight we plan to decrease resistor values now that I actually have some 1 watt resistors and even a 1 ohm 10 watt wire-wound resistor.

 

We finally got back to some more experimenting last night. Not an easy thing to get a kid to do on a Friday night. This time we decided to use a 5mm red blinking LED. The local Radio Shack was going out of business so a few days ago I stocked up on a bunch of components all at 50% off.

The front of the blinking LED package said the LED's operating values were 35mA and 3V. On the back of the package it specified the "absolute max rating" as 35mA. Though we didn't try to exceed that just yet, we decided to go in the opposite direction first by calculating a resistor for a 30mA current.

Using 4.5V as our source (3 AAA batteries in a holder with leads) we first calculated that we would need a 46 ohm resistor to get 30mA (the battery voltage was 4.38 volts measured). We ended up using a 47 ohm resistor and the LED blinked as expected. We then kept progressively increasing the resistor value (100 ohms, 330 ohms, 680 ohms, and then we actually connected a bunch of 680 ohm resistors in series (4 of them actually so 2.7K total) and still saw the LED blinking but more faint each time we added another resistor. We ended the experiment with a single 10K resistor which barely turned the LED on but it definitely did not blink. The was the equivalent of ~0.1 mA. How could the LED light up at such a low current?

Tonight we plan to decrease resistor values now that I actually have some 1 watt resistors and even a 1 ohm 10 watt wire-wound resistor.

The wavelength of light tends to get broader as you increase the current on an LED. The center of that wavelength will also move to a lower frequency (shift to the near infrared) where you cannot see. Therefore at some high current, above the manufacturers maximum and the time it burns out, you will likely see the LED get less bright as you increase current. I just wanted to point out that so you have an explanation when your son asks. It won't be easy to explain (just say they are less good at making red light when they get hot).

If you would have started with a Green LED, you would have seen this wavelength shift as a color shift (green to greenish-yellow to yellow to orange).

 

The wavelength of light tends to get broader as you increase the current on an LED. The center of that wavelength will also move to a lower frequency (shift to the near infrared) where you cannot see. Therefore at some high current, above the manufacturers maximum and the time it burns out, you will likely see the LED get less bright as you increase current. I just wanted to point out that so you have an explanation when your son asks. It won't be easy to explain (just say they are less good at making red light when they get hot).

If you would have started with a Green LED, you would have seen this wavelength shift as a color shift (green to greenish-yellow to yellow to orange).

modern superbright leds dont shift so much in color, and you can use them with 2.2k to 4.7k for indication.

 

Tonight we plan to decrease resistor values now that I actually have some 1 watt resistors and even a 1 ohm 10 watt wire-wound resistor.

So tonight we got back around to more experimentation. This time we increased the current across our 3V/35mA blinking LED. We calculated R for ~40mA, ~65mA, and ~140mA. The resistors used were 36, 22, then 10 ohms. The power supply was still three AAA batteries that measured 4.42V across the leads of the battery housing before connecting it to our breadboard.

For all three of the current values mentioned above, the LED did not burn out. It continued to blink.

We then tried a 1 ohm resistor expecting the LED to shatter with 1.42 amps going across it but it did not to our surprise.

We then replaced the blinking red LED with a standard 20mA red LED. It also did not blow out. Hmmpphhh.....frustration setting in.

I then told my son that if we removed the resistor and connect the positive end of the battery directly to the LED it would certainly blow out. WRONG AGAIN. With no resistor in the circuit neither the blinking LED nor the non-blinking LED blew out.

Now our plan for this science project is in serious jeopardy. What else could we possible do to verify Ohm's law that would produce visual results that would catch the eye of a casual observer??

Should we try increasing the voltage to 9V (using 6 AA batteries instead of a single 9V battery) and repeat what we've done to date? I am running out of ideas here.

 

You are STILL using a single 9V battery? The internal resistance of a 9V battery is pretty high, particularly at higher current draws and particularly if it is not fresh.

Take a 10Ω resistor (or thereabouts) and connect it directly across your 9V battery and measure the voltage across it. What results do you get?

 

You are STILL using a single 9V battery?

No. We were using three AAA batteries. I believe it was you that suggested not using a 9V battery due to its internal resistance. Unless I misunderstood you.

 

That's correct. I would recommend at least AA batteries or even D cell batteries. If you can get your hands on them, NiCad batteries can deliver pretty high currents (amps) even from AA batteries (in fact many D-cell NiCads just have an AA-cell NiCad inside a D-cell sized carrier).

 

@opeets: Sorry if I missed something, since I'm jumping into this thread, but I'm wondering why you're trying to burn out an LED using high current? When I first started experimenting with electronics, I burned out several LEDs due to high voltage before I understood that I needed to use a resistor in my circuit. To do a project like you're attempting, I would simply use a 9V battery with and without a 470Ω resistor to power an LED (any color should work).

 

@opeets: Sorry if I missed something, since I'm jumping into this thread, but I'm wondering why you're trying to burn out an LED using high current? When I first started experimenting with electronics, I burned out several LEDs due to high voltage before I understood that I needed to use a resistor in my circuit. To do a project like you're attempting, I would simply use a 9V battery with and without a 470Ω resistor to power an LED (any color should work).

By applying a voltage without a current limiting resistor you are allowing high currents to flow through the LED -- provided the voltage source's internal resistance doesn't prevent it -- and it is the high current that does the damage. The resistor is there not to limit the voltage, but to limit the current. Of course, like in a resistor, the voltage and current in an LED is an invertible function (at a given temperature). If you tell me the voltage across it I can tell you the current through it and vice-versa (assuming I have the voltage-current characteristic for that LED).