# understanding an LED's Vf

Discussion in 'General Electronics Chat' started by opeets, Apr 21, 2015.

1. ### opeets Thread Starter Member

Mar 16, 2015
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0
This thread is basically a continuation of a previous discussion about a simple LED project that I am working on with my 5th grade son as part of his science project. I figured that I would start a new thread because we changed around our test equipment today and a clean new slate is always nice to start with.

Some background about me....even though I took EE courses in college I would consider myself a newbie all over again because I've spent the last 20 years as a software engineer. That being said please keep the humorous comments to a minimum.

Tonight we started using a newly purchased regulated DC power supply that provides voltages in the following increments: 3V, 4.5V, 6V, 7.5, 9V and 12V. The max current output of the power supply is 2A.

Once we hooked it up to our breadboard in a simple single-resistor, single-LED configuration we attached a DMM in series after the LED to be able to measure the current in the circuit at any given point during our experimentation.

We worked with three different LEDs, none of which I knew the specs for other than the relative sizes. The first was a 5mm green LED, the 2nd was a 3mm green LED, and the third was a 3mm red LED. My hope was that we would be able to determine the forward voltage Vf of each by taking current measurements at different voltage levels and figuring out Vf based on the formula R = (V-Vf)/I.

We started with a 470Ω resistor for the circuit and measured the current for the 5mm green LED using all 6 voltage settings. After getting all 6 measurements it became clear that Vf was not a constant value. Based on the formula calculation, Vf ranged from ~1.9V to ~2.4 volts (depending on the supplied voltage value) for both green LEDs (even though they were different sizes). The value of Vf for the red LED ranged from ~1.7V to ~2.0V.

Can someone explain why Vf increases as supplied voltage increases? I was under the assumption that it remained constant. How does one know which Vf value should be used when calculating a resistor value for a specific current?

With the red LED still in place, we then replaced the 470Ω resistor with a 100Ω (1 watt) resistor. We measured currents in the range of ~10mA to ~86mA which resulted in calculated values of Vf ranging from ~2.0V to ~3.4V. This was starting to not make sense to me but nevertheless we continued on.

We then replaced the 100Ω resistor with a 10Ω (10 watt) resistor. We measured currents in the range of ~40mA to ~200+mA which resulted in calculated values of Vf ranging from ~2.6V to ~10V. It should be noted that with this resistor the LED was very bright at 3V and 4.5V but at 6V and above it got more and more faint until eventually at 12V it shut down and a few seconds later I saw smoke coming from it. Obviously this was expected (and what we were hoping for).

Can someone explain why the calculated Vf values (based on the current measurements) are so high? Are these calculated values bogus?

See screenshot of a spreadsheet below that I created to record the results of the experiment.

• ###### Vf.JPG
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Last edited: Apr 21, 2015
2. ### MrChips Moderator

Oct 2, 2009
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What gave you the impression that Vf was constant?

An LED is a diode which follows an exponential I-V curve.

With a 12V source into a 10Ω resistor, you are essentially pushing 10V/10Ω = 1A into a device that can handle 20mA.

So there is no wonder that you let the magic smoke out.

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3. ### opeets Thread Starter Member

Mar 16, 2015
103
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First I would kindly refer you to what I said about my flaky background in electrical circuitry. Then I would point of the fact that I've been re-learning basic electrical circuit principles by watching video tutorials and what-not. No tutorial that I have watched has mentioned anything about using a variable Vf value to obtain a certain current value when calculating the resistor value. A magical Vf value is usually just plugged into the formula along with the desired current and the R value is calculated. At that point the closest resistor to the calculated value is chosen. That....is the reason that I thought Vf was a constant value.

But the LEDs I was testing handled currents far greater than 20mA. I keep hearing references to the 20mA magic number and the fact that one should be careful not to exceed it. Why is is that the LEDs I tested allowed current values close to 200mA before fizzling out?

I could have seen that one coming.

4. ### WBahn Moderator

Mar 31, 2012
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Keep in mind that by starting a new thread you also start a new context for the discussion, so everything mentioned in the other thread about what you are trying to do and why you are trying to do it is largely out the window. Many responders to this thread will not have even seen, let along followed, the other thread.

Just like we usually talk about batteries as though they have a single, fixed, constant voltage, we usually talk about LEDs and diodes the same way. That's because, when used like they are intended to be used, the degree to which the voltage varies is a minor, often negligible effect.

For a very long time, most LEDs were used as indicators and the typical current that they were designed to handle while performing well in that roll was 20 mA or something pretty close to that (within a factor of two or so). So unless you know details about a specific LED, it was almost always reasonable to assume that you needed to supply it with about 20mA for it to illuminate well. Today, in the age of battery-powered gadgets, the goal is to reduce that. Many LEDs today work well at 2mA to 10mA. On the other hand, LEDs are seeing an exploding role in primary light devices and the currents in those can be in the many ampere range. But if you get a run-of-the-mill LED indicator lamp you can pretty safely bet that it will work properly with somewhere between 10mA and 30mA.

Just like a resistor, the more voltage you put across an LED the more current you will force through it. That means that there is a relationship between voltage and current. For a resistor the relationship is linear, so if you double the voltage across it, you double the current through it. But for a diode, which includes LEDs (remember, light-emitting diode), the relationship is exponential. For a silicon diode, if you increase the voltage by about 20mV you double the current through it. LEDs aren't very good diodes (they aren't meant to be, they are meant to give off light) and so their characteristics aren't nearly as sharp. You typically see an increase of about 120mV, give or take, for a doubling of the current when you are near the intended operating current. As you get well above that current, resistive effects start coming into play and it takes a larger change in voltage to double the current.

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5. ### crutschow Expert

Mar 14, 2008
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The reason a constant forward voltage is assumed for an LED is that it is sufficiently constant for a given color LED and a given current, to make it a simple equation for determining the required series resistance for that given current, to a sufficient engineering accuracy.
Determining the resistance if you added in the exponential change of voltage with current would make a complex equation to solve with no practical gain in accuracy.

6. ### opeets Thread Starter Member

Mar 16, 2015
103
0
Understood and that was my intent. Thank you for pointing that out.

I understand both of these explanations. However I would still like to understand what happened when I lowered the resistance value to 10Ω. According to my calculations, to obtain the current values that I measured with my ammeter, the Vf for the LED would have had to be in excess of 2.0V which according to the I-V Curve (that Mr. Chirps provided) is not attainable for an LED of this color. For the 9V/10Ω example, the current should have been (9-2)/10 = 700 mA but my ammeter measured 194 mA. Using the 194 mA measurement as the current value in the equation and solving for Vf I get ~7.0V. Should my ammeter be placed between my resistor and LED rather than being after the LED?

Last edited: Apr 22, 2015
7. ### MrChips Moderator

Oct 2, 2009
14,474
4,265
In a series circuit it doesn't matter where you insert the ammeter. The same current has to flow through the LED, resistor and ammeter.
You can insert the ammeter before or after the LED, or before or after the resistor, still same current.

You have to take into account the internal resistance of the ammeter at the specific amp range. Check the meter specs.
What is the make and model of your meter?

The meter inserted in series with your circuit will add resistance and therefore reduce the current flow.

Secondly, Vf = 2V is a guide only at a given current. At higher currents Vf will be higher than 2V.

8. ### dl324 Distinguished Member

Mar 30, 2015
4,944
1,056
Hi opeets,

To better craft responses to your future LED questions, were you an EE major or some other engineering major who had to take some EE courses?

The current in your circuit would be constant regardless of the position of the ammeter. What you're seeing is the "observer effect". By trying to measure the current in the circuit, you disturbed it enough that you can't get an accurate measurement. To prove that to yourself, you can use another voltmeter to measure the voltage drop across each element while the ammeter is in place. You'll find that the ammeter has introduced a resistance, so your assumptions are invalid.

Technically LEDs should be driven from a current source. For simple circuits, a voltage source with a series resistor is sufficient. If you build a 700mA current source, you'll be able to release the magic smoke from that LED quite easily...

9. ### opeets Thread Starter Member

Mar 16, 2015
103
0
That's was my understanding as well....I just wanted to make sure that running the LED at a higher voltage (for a given resistance) didn't cause some type of unpredictable behavior resulting in invalid current readings.

I have a pair of Extech EX330s (one I use as an ammeter, the other to measure voltage drops as needed):
http://www.amazon.com/Extech-EX330-...8&qid=1429711084&sr=8-1&keywords=extech+ex330

According to the I-V curve you posted Vf appears to converge to ~2V very rapidly (relatively speaking) so I don't see how I could get a calculated Vf of 7V (assuming the measured current was correct).

10. ### opeets Thread Starter Member

Mar 16, 2015
103
0
I was a Computer Engineering undergrad (and eventually earned an MS in Computer Science) so my focus was primarily digital design courses, assembly code, OS design, and some higher level programming languages but I did take about 8 or 9 EE courses (most of which is forgotten knowledge). Doing nothing but software development once I graduated college for the next 20 years made me literally forget everything I learned. It feels like a lifetime ago.

I don't quite follow. Is interrupting the circuit and allowing the current to flow through the ammeter so I can get a current measurement considered disturbing it? Please elaborate.

11. ### dl324 Distinguished Member

Mar 30, 2015
4,944
1,056
An ideal ammeter has zero resistance, yours (and all of mine) have resistance. It will vary depending on the current range the meter is set to; just as the input resistance of your voltmeter, and the voltage it applies, will change depending on the range it's set to.

Did you try a 10 ohm series resistor without an ammeter in circuit? That should have allowed you to run 700mA through the LED, at least briefly...

An ideal diode will have infinite resistance until Vth is applied; then it will have zero resistance. You may recall from your EE classes that the resistance of a diode is the slope of the IV curve at the operating point. If you want to run an LED at a constant current, you need to use a current source.

Last edited: Apr 22, 2015
12. ### opeets Thread Starter Member

Mar 16, 2015
103
0
For those that did not read my first (quite lengthy) thread.....what my son and I are trying to accomplish (for his 5th grade science project) is to verify an application of Ohm's Law. The premise of the experiment has evolved over the course of a few weeks but ultimately what we are trying to show is that by using the formula for calculating a resistor value for an LED, R = (V-Vf)/I, we wish to determine a "safe range" of resistor values to light up an LED for different source voltages.

We would first like to prove the sunny day scenario where a specific range of resistor values results in the LED being driven by a safe (optimal, or whatever you wish to call it) current range (perhaps 15-20 mA).

Using a basic 3mm red LED (assuming Vf is ~1.7V for the 15-20mA current range) and a starting out with a 3V power source, we would need a resistor in the 65Ω to 87Ω range. Using the 3V setting on the power supply, we will test a bunch of resistors in this range (perhaps even a combination of smaller-valued ones in series for fun), measure and verify the current values fall within the range, and record the LED brightness.

We then intend to increase the voltage on the power supply (using any resistor in the 65Ω to 87Ω range from the previous step) to 4.5V, 6V, 7.5V, 9V, and then 12V. We would like to show that as the current increases the LED does not behave as designed because the current supplied is too high. Yes, we are intentionally trying to show this misbehavior. If the LED fizzles out, great....that's exactly what we want.

We would then show (using an idential undamaged LED) that it is necessary to re-calculate the resistor value at each voltage level in order to maintain the desired 15-20mA operating value (contrary to what others have told me, what we are effectively showing IMO does prove the importance of Ohm's law). So for the 12V source we would need a resistor in the ballpark range of ~500Ω to ~700Ω. By changing the resistor (from the 65Ω to 87Ω range) to some of those that fit in the new (500Ω to 700Ω) range we will show that the current is indeed within 15-20mA once again and the LED is operating as it was designed to.

Lastly we would show that at really high R values, the inverse observations would be expected. For a given high R value at 12V, the LED would be very dim. As the voltage is decreased down to 3V, we would show it get more and more faint at each voltage increment until (hopefully) it doesn't light up at all.

I hope this clarifies where I am going with this thread and once again thank all that have contributed thus far.

13. ### Papabravo Expert

Feb 24, 2006
11,116
2,168
Perhaps it would be helpful to look at the actual theoretical model of a diode. Now I don't know if they teach things like exponents in the 5th grade, but a standard scientific calculator is a wonderful tool as is the Excel spreadsheet. Using either of these tools you can investigate the amazing behavior of exponential functions.

Here is a link to the actual diode model
http://en.wikipedia.org/wiki/Diode_modelling

Besides leading to fundamental insights, the purpose of an experiment is often to confirm the predictions of a model. Along the way you will discover other interesting things.

If you don't have access to the tools from Mr. Softie, then I highly recommend the LibreOffice suite, which has an Excel clone:
https://www.libreoffice.org/

14. ### WBahn Moderator

Mar 31, 2012
20,205
5,741
You have measurement issues, the biggest of which is using a DMM as an ammeter in circuits having low resistance values. I warned you about that in the other thread, but you keep choosing to use a DMM as an ammeter in circuits with low resistance values. You are also assuming that your supply is staying perfectly at the nominal output voltage as you draw higher currents from it -- maybe it is and maybe it isn't; don't leave it as an blind assumption. So perhaps you will now be willing to improve your measurement methods. Use your two DMMs as voltage meters. Measure the voltage across the LED and measure the voltage across the resistor. Determine the current by dividing the voltage across the resistor by the resistance. Determine the actual voltage of the supply by summing the two voltage readings.

15. ### dl324 Distinguished Member

Mar 30, 2015
4,944
1,056
What is the title of the science project and what is the hypothesis? You mention Ohm's Law, Vf, safe operating current, and brightness. What is it that you're trying to show/prove?

BTW, LEDs don't do well in direct sunlight unless you employ a filter (optical).

16. ### opeets Thread Starter Member

Mar 16, 2015
103
0
No working title yet, but the hypothesis is basically two things:

1) That we studied Ohm's Law and used a variant of it, R = (V-Vf)/I, to operate an LED with correct resistor values. Failure to use a resistor, or randomly choosing (an incorrect one for the supplied voltage) results in improper LED function.

2) That he worked hard and learned something he previously had no exposure to.

Am I dreaming or does everyone seem to disagree with me that this is a reasonable experiment to verify Ohm's law?
The resistor has a voltage drop (based on the current passing through it) and so does the LED (only slightly different based on the current passing through it). Those when added should equal the supplied voltage.

Last edited: Apr 22, 2015
17. ### Papabravo Expert

Feb 24, 2006
11,116
2,168
I don't disagree with you, but we'll see the results eventually.

I did an experiment in the 7th grade to measure the forces on a rotating body in a fluid. My measurement apparatus was flawed and I was unable to measure any forces. I was not able to draw any conclusions, but I got an A on the project due to the application of the method. Science and Engineering are replete with such setbacks. Success is not required, and failure is often a better teacher.

18. ### opeets Thread Starter Member

Mar 16, 2015
103
0
Thank you. I needed that reassurance.

19. ### crutschow Expert

Mar 14, 2008
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If you measure the voltage across the resistor in series with the LED, as Wbahn suggested, and use Ohm's law to calculate the current through the resistor (and LED) then you will perturb the circuit only slightly as compared to putting an ammeter in series with the LED.

20. ### ronv AAC Fanatic!

Nov 12, 2008
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Using a LED to demonstrate ohms law is not a real good idea as the Vf is a complex function and not a constant. The LED has a diode drop that is dependent on temperature and internal resistance. You can approximate the internal resistance from the slope of the curve. If you want to demonstrate ohms law use only resistors to avoid the confusion.

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