# Perplexed newbie (strange DMM results?)

Discussion in 'General Electronics Discussion' started by NuLED, Jan 7, 2012.

1. ### NuLED

294
0
Jan 7, 2012
Hello everyone!

I disassembled a few pieces of spare equipment at home. A battery operated fan, an LED flashlight, etc.

For my experiment, I got several white LEDs from the flashlight. The LEDs seemed to have been connected in parallel and the flashlight used 3 "watch" batteries in series, so that looks to be 4.5V total.

Using a breadboard, I put in one of the LEDs and connected a 1.5V AA-battery. Nothing. Not surprising. I then used the battery holder from the electric fan, which provides 3 batteries in series, giving 4.5V. This lit up the LED.

I then put two more LEDs in parallel on the breadboard and they all lit up. Nothing special.

Now, I wanted to see what is the current in the circuit (indeed, what is the current going through one LED) and what are the LED resistances (each) and the total circuit resistance.

Using my \$50 (or so) DMM, I measured the voltage of the battery pack (around 4.5V as expected), and as far as I can determine, the current in the LED circuit was 3.45 mA.

Is this right? From websites, I seem to read that most LEDs need 20 mA of current, but the ones I have here measure only 3.45 mA?

Also, using Ohm's law, the resistance comes to around 1.3 KOhms. 1,300 Ohms. Does that seem right? (with the 3.45 mA + 4.5V figures).

I tried measuring resistance across the circuit wiring only - the breadboard and the jumper wires only, not with the LEDs plugged in, and without the battery connected and the DMM gave strange numbers, going from 0 Ohms counting up to 0.9 Ohms and then back down again.

Also, I was not able to measure any resistance directly across an LED. Does that make sense? I switched polarities of the LED and nothing on the DMM. I am assuming there is insufficient voltage/current from the DMM to cross the LED "gap"?

The circuit continuity test on the DMM also does not show anything for an LED.

Any help to clarify things for me would be much appreciated!

Thanks!

Last edited: Jan 7, 2012
2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010

LEDs don't need a particular current to light up, any forward current is enough. The magnitude of the current determines how bright they are (and at high currents, how fast you destroy them).

Don't bother trying to read the resistance across a LED, it's meaningless for determining the current through the LED. (There can be other reasons to try to do this, but determining a value of resistance s not one). It is possible that the voltage on the probes is less than Vf of the diode (or that the polarity was reversed).

The resistance that limited current to your LED was probably the internal resistance of the battery (maybe the cells are flat) or the LED has an inbuilt resistor (probably less likely).

294
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Jan 7, 2012
Many thanks!

4. ### NuLED

294
0
Jan 7, 2012
It seems thermal runaway is crucial to the whole LED discussion but I got confused.

Everything prior to the below I understood, but:

Why does forward voltage fall? As the LED heats up, does that mean resistance decreases? I read somewhere else that Vf means the minimum voltage needed to power up the LED. (BTW Does resistance decrease for all metals as temperature rises?)

I don't understand why there is suddenly a conceptual jump from the Vf falling, to "the same Vf" with a larger current? I think this is the mental gap that I stumbled over, losing the train of thought for the rest of the discussion. In other words, I don't understand those sequence of words.

I don't understand "This leads to a further increase in I to maintain the Vf." What does "maintaining" the Vf mean? I thought Vf is just a threshold below which the LED doesn't light up. This phrase makes it seem like there is an active process going to stabilize the Vf, whereas previously I thought Vf just decreases as the LED heats up (from current, with ostensibly lowered resistance at higher temperature)?

Thanks!

Last edited: Jan 7, 2012
5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
No, it's not resistance. It's the forward voltage drop.

The forward voltage varies with current and temperature.

So as the current rises, the Vf rises (but rather slowly, and not linearly like a resistor)

The particular section you're reading talks about driving a LED from a constant voltage. As it heats up, the Vf at a particular current falls. If the voltage remains constant, the current increases (often dramatically) to the point at which the Vf equals the supply voltage.

Vf is often thought of as a fixed voltage, but if you look very carefully you will always find that it is specified at a particular current and (often) temperature.

6. ### NuLED

294
0
Jan 7, 2012
Hi Steve - ok thanks for your reply but apologies if I am sounding a bit dull on this point. Perhaps there are some fundamentals I don't quite understand or know yet. For the forward voltage, this is the first time I've heard the term after studying basic electronics theories (from at least 3 books so far). In your most recent reply, are you saying that Vf is a phenomenon of the LED, and that it is a voltage phenomenon that occurs only because of the nature of an LED that creates a difference in the voltage across it? If so, how does that happen? That is, in my mind, any component has a voltage drop, but the total circuit voltage has to remain constant, as per Kirchhoff's voltage law. Thus, I am confused where is the voltage going, if you say that it is varying due to the LED changing it (since the voltage source would, for the sake of argument, put out a constant voltage).

Said another way, if the voltage source puts out, say, 5 volts, then with the LED being the only component in the entire circuit (say), whatever the LED does, that total circuit still can only have 5 V, no more no less. With any other component not varying much (e.g., a resistor), whatever that LED has as a voltage drop should remain constant.

If the LED changes voltage drop, then the other component (e.g., a resistor) would have to experience higher voltage (?). But it is also a bit confusing if you are saying the LED effect is not resistance.

It is also confusing to me when you say "driving a LED from a constant voltage. As it heats up, the Vf at a particular current falls. If the voltage remains constant, the current increases (often dramatically) to the point at which the Vf equals the supply voltage."

As the LED heats up, I thought the current RISES due to a reduced "resistance" (even though you say it is not called resistance; whatever it is that impedes voltage) and it is the current rising that causes additional heating. But your statement says the current falls (???). The other part, that if V remains constant, current rises as the temperature rises also in a vicious circle, that I understand, but not the immediately preceding statement.

Sorry if I seem to harp on this but I am just not understanding something here and hopefully I was able to further pinpoint the concept that has me confused. Your advice continues to be most appreciated.

Thanks!

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
A LED is just a sort of diode.

Ask yourself the question of what would happen if you placed a normal silicon diode (Vf approx 0.7V -- but if you look at the spec sheets you'll see it varies with temperature and forward current) across that 5V supply.

Have you asked yourself that question?

OK, did you realise that something (possibly something bad) is going to happen?

In a perfect world, the current would rise to the point at which Vf was 5V. There are calculations you can make to determine when this would happen. It may require hundreds or thousands of amps.

But will that happen? Probably not.

More likely, one of the following (or a combination of them) will happen:

1) The diode will be destroyed, going first short circuit, then going open circuit, possibly with a bang.

2) The power supply will not be able to supply the required current, the voltage will fall (and hence the required current) and things will stabilise at some voltage less than 5V, but equal to the Vf of the diode at the voltage it stabilises at.

3) The wires will exhibit sufficient resistance to drop the difference between Vf and the power supply voltage at some current, so the excess voltage is lost across the wires.

4) The power supply blows a fuse or fails in some (possibly nasty) manner

A resistance gives you a simple relationship between voltage and current as per Ohms law.

Double the voltage and the current also doubles. Halve the voltage and the current halves.

Semiconductor junctions are not resistors. They exhibit a relationship that is not linear -- in some cases they exhibit regions which appear to be *negative* resistance.

Look at the graph of voltage vs current in the LED sticky. If a LED were a resistor that would be a straight line.

Let's say we start with a voltage that gives a particular current. If we halve the voltage, the current will fall almost to zero. If we double the voltage, the theoretical current will rise astronomically (perhaps tens or hundreds of times)

Because of this relationship (and the variation with temperature and even between parts, the voltage across a LED is best seen only as an approximate guide. Attempting to use a constant voltage across a LED will only result in tears.

If I said the current falls, then either you're misunderstanding, I've not written clearly, or I've typed the wrong word. I'll try to find where that is, but you may have to provide the direct quote.

is it this part? "As it heats up, the Vf at a particular current falls"? That does not say the current falls. It says that when it gets hotter, for a particular current (remember that we should aim for a constant current through LEDs) the Vf falls. The consequence of this is that if you maintain the Vf, the current will rise, and due to the non-linear relationship, by a disproportionate amount.[/quote]

This will only lead to a better understanding for you *and* improvements so the next person will have an easier time.

8. ### NuLED

294
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Jan 7, 2012
Thanks again Steve, I think I've got it now!