0
\$\begingroup\$

While all your great answers to my previous question are simmering in my brain (I printed out all your answers and placed a copy on my work bench to have another go at it), I have one more:

As I mentioned in a previous post, I have a new model UNI-T UT89XD meter. A unique feature is that it generates up to over 11 volts in "diode test" setting. The question I asked an expert was this:

How does the 11 volts of LED testing not burn out a 3 volt LED?

He initially answered: "The output is controlled so it only sends out the required voltage."

And being the curious noob I am, I asked: "Can you supply a brief, laymans explanation of how it senses voltage capacity of a low voltage LED?"

He responded: "The forward voltage works by trying to force a constant current of (lets say) 1 mA through the device under test (diode). Forcing the current through the DUT will drop voltage (forward voltage for a forward-biased diode). The maximum voltage is then limited to whatever the multimeter can output in diode mode. Rather similar to how resistance is measured, but instead of calculating the resistance from the voltage and the current using Ohms law, it displays the forward voltage directly. So if the 1 mA results in a drop of 0.5 V, the resistance range would display 500 Ohm (0.5 V / 1 mA), while the diode range would display 0.5 V."

Now I suspect his answer makes sense to most of you guys but not so much to me. But I didn't want to press my luck asking him to clarify that for a noob thinking that he might be thinking his attempt would be something akin to explaining the theory of relativity to a fly.

So, can anyone out there in electronics land give a more "layman" oriented answer to how that meter can deliver up to 11 volts to an LED without blowing out a 3-volt LED?

\$\endgroup\$
  • 1
    \$\begingroup\$ it does not deliver 11 V to a 3 V LED ... the answer above says ... it slowly turns up the voltage until 1 mA flows through the diode \$\endgroup\$ – jsotola May 14 at 17:39
1
\$\begingroup\$

The 11V is what is called the compliance voltage. Pretty much if no current is flowing, it will rise to 11V.

He kind of butchered the explanation.

The diode test mode uses a constant current source. For example, 1mA. For say a 3V led it will drop 3V at 1mA so will measure the forward voltage at 3V.

Same for a normal diode. At 1mA it drops 0.7V so the meter reads 0.7V. If you measure a resistor e.g. 1K, it drops 1V when 1mA flows through it. So would measure 1V.

If you measured say a 20K resistor, the current source cannot supply more than 11V to drive that 1mA so it would show OL or 11V depending on how they have it set up in software. In this case the current would be less than 1mA because it can only supply up to 11V.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

It can be more complex actually but you can think it's really the same thing than powering a 3V LED from a 9V battery. You put a resistance in series, to limit current. With 1k resistor, there's 6mA flowing.

So in reality, the open circuit voltage from the meter is up to 12V when nothing is connected, but connecting the probes to a 3V LED will have 3V over the LED, and the rest of the circuit, which may simply be a 2400 ohm resitor inside the multimeter, will have 9V over it, and if it is a 2400 ohm resistor, 3.75 mA would flow.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ This meter apparently automatically places the correct resistance to make up the difference between the voltage of the LED and the 11 volts the meter produces. How does it discern the amount of resistance to impose in the circuit? \$\endgroup\$ – Gfmucci May 14 at 16:19
  • \$\begingroup\$ We don't know how it works internally. It might have a constant current source with up to 12V compliance. Or it might just have a 12V supply with 2400 ohm resistor. All that the manual says is that 12V is the maximum open circuit output voltage and that the maximum short circuit current is max 5mA. \$\endgroup\$ – Justme May 14 at 16:45
1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The test current source with voltmeter and the DUT (device under test).

Let's analyse his response.

He responded: "The forward voltage works by trying to force a constant current of (lets say) 1 mA through the device under test (diode).

As shown in Figure 1. The constant current source can be made with a voltage source and a couple of transistors and resistors or with a more elaborate circuit.

Forcing the current through the DUT will drop voltage (forward voltage for a forward-biased diode).

Any diode requires a bit of push to get current through it. This is its forward voltage.

enter image description here

Figure 2. The diode check-valve analogy. Source: What is an LED?.

If you look at the check-valve in the figure above, it should be clear that the spring normally keeps the ball in position and prevents back-flow. When “forward-biased” the ball shut-off can be moved against the spring but it will take some initial pressure to move the ball. This results in a pressure drop across the valve: the pressure downstream will be less than the inlet pressure.

In a similar manner the PN junction of a diode causes a voltage drop. For silicon it is about 0.7 V. For an LED it will vary from about 1.4 V for infrared to 3.5 V or so for blue.

The maximum voltage is then limited to whatever the multimeter can output in diode mode.

It should be fairly obvious that if the test circuit is powered by a 12 V supply then the constant current source can only maintain its current as long as the voltage required to drive that current through the DUT doesn't exceed, say, 11 V (because the constanct current circuit needs a volt or two to operate).

Rather similar to how resistance is measured, but instead of calculating the resistance from the voltage and the current using Ohms law, it displays the forward voltage directly. So if the 1 mA results in a drop of 0.5 V, the resistance range would display 500 Ohm (0.5 V / 1 mA), while the diode range would display 0.5 V."

And the reading is simply a voltmeter across the DUT.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

It could (conceivably) damage an LED when you test it in reverse, because it will exceed the absolute maximum reverse voltage of most LEDs (typically 5V). In practice a brief test is unlikely to cause noticeable damage, but in some cases (a space application, perhaps) you'd want to discard the LED after such a test.

In the forward direction, the meter will try to force a constant current (1.00mA is a standard) through the LED and the voltage you read will be determined by the I-V response of the LED. For example, for the Panasonic LNJ437W84RA this is the datasheet curve:

enter image description here

At 1.00mA the typical forward voltage will be about 1.86V (at 25°C, it will decrease with increasing temperature). The voltage will vary with the LED "chemistry" and color, typically it increases with decreasing wavelength so an IR LED might have a 1.1V Vf and a 405nm UV LED might be several volts. Some devices (such as large LED digits) use several LED dice in series so this DMM would be able to read their forward voltage.

Many inexpensive DMMs use only a few volts, and many use a simple resistor so the current is not constant and some won't even allow you to measure a common white/blue LED. In this answer, I measured the actual voltage and voltage/current characteristic of a number of meters. The highest open-circuit voltage was the Agilent at 7.062V, which is a lot closer to the 5V typical limit than the Uni-T, but still exceeds it. One was as low as 1.54V which is okay for testing silicon diodes or Schottky diodes but pretty much useless for LEDs.

One caution is that (as you can see from the above graph) the LED voltage increases significantly with increasing current. Fortunately, if you calculate the LED series resistor based on the measured Vf from your meter (at ~1mA +/-50% for most) you'll get a conservative value. The error will increase the closer your supply voltage is to the LED Vf, and of course you would prefer to have a significant voltage across the resistor to control the current well (and you'd prefer to have a low voltage to maximize efficiency so it's an engineering trade-off).

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

I'm not an EE, but one of my goals for coming here was to be able to translate engineering-ese into layman-ese, based on the premise that if I've learned it recently, I should be able to put it into the words needed for the learner. So it may seem like I'm rambling on, but it is actually purposeful, as I am trying to paint the picture over and over again, from slightly different angles. So, here it goes...

It is my hypothesis that a simple resistor is used in most cases by most multimeters, and in one of the answers posted by Spehro Pefhany, he referenced another of his answers that gives some evidence to support my hypothesis. Just one of the multimeters he tested took the higher road and used a true constant current source.

So, going with the resistor-in-series theory...

As long as the 11 volts is held back by a suitably large resistor, the diode or LED (or LEDs) will remove their forward voltage first -- not first in terms of time, but first in terms of priority (or calculation, or mental model).

So, let's say your mystery LED is from a 60-watt equivalent LED bulb. You open it up, and your multimeter dutifully sends current through one of the LED's -- and you find that the forward voltage is 6.0 volts. This means that internally, inside the LED itself, there are actually two chips put in series, and this helps the manufacturer to accomplish their design without having to put too many 3-volt LED's on the circuit board (9 vs. 18).

So, let's assume that the resistor in series with the 6.0-Volt-LED is a 10K resistor. Then, if we do the calculations, V=IR, so I=V/R = (11V-6.0V) / 10K = 5/10,000 = 0.5mA which should be visible.

Notice how small that is. The resistor keeps the current small. Hence no blown-up LED.

Now let's test a 2N3904 NPN, the PN Base-Emitter diode junction, which we'll assume is 0.6 volts forward voltage. The calculation is (11.0V - 0.6V) / 10K = 10.4 / 10,000 or 1.04 mA. Even though the forward voltage being tested is really small, it doesn't really matter much because the resistor is comparatively much bigger.

If we just touch the test leads together and run the diode test, we get (11.0V - 0.0V) / 10K = 11.0 / 10,000 which results in 1.1mA -- the maximum current that can flow in a diode test, because there is no forward voltage taking a bite out of the total voltage range. This shows how, even though there is a total potential of 11 volts, it will never kill any LED or diode -- because 1.1mA won't kill anything.

Make your own LED tester with two 9-volt batteries in series, using a 10K resistor like we've been assuming. That's 18 volts total that we have to drop. And attach a voltmeter to the parts that go across the "diode" under test. So, the numbers testing a 3-volt white LED would be: (18V-3V)/10K = 15/10,000 = 1.5mA. And when you do this, put your volt-meter across the LED and measure the voltage, then go look up where on the V-I curve (from the datasheet for the LED) it corresponds to. Play around with different combinations of LED's and other diodes in series, and see how it works, but also do the math for a few cases, and match it to what you're measuring and seeing. If you look up LEDs on Digi-Key for instance, you will see that some LEDs have a 12-volt forward voltage. That means that, internally, such an LED has 4 chips in series (which your UNI-T may not be able to test very will). So it helps to have a decent voltage range for testing diode voltage drop -- it may be more than one diode in series (you should also be able to test some zener diodes -- 0.6 volts forward and testing backwards, something like 4.7 volts for a 1N750).

If you use a 1-ohm-resistor in series with your UNI-T meter while it is testing a 3-Volt white LED, you should be able to measure the voltage drop across the resistor to figure out how much current the UNI-T meter is actually using. This will be confirmation that there is another resistance in series with the LED, limiting the current.

Keep the 1-ohm-resistor in the chain, and track how much current is used at each step, and start adding 3-Volt white LED's in series and retesting until no current flows, then remove the last one that you just added so that current once again flows. Keep adding to the total forward voltage, now by adding RED LED's (you should need only 1, but maybe 2, in series). Then, in the same way, add garden variety 0.6 volt diodes (like 1N4148) until you've added as much forward voltage as you can, and still have current flow. In this way, you can find out if that 11 Volt maximum is real. The more LED's and other diode voltage drop you place in the test loop, the less current will flow if it is really a simple resistor in the loop. If you keep getting just about the same current flowing no matter how much total forward-voltage you insert, then you have a constant current source delivering a set amount of current, which is really the best way to implement the multimeter circuit, as it gives the widest voltage range, and also the best apples-to-apples comparison (more comparable brightness for LEDs).

Even though the diode test could be implemented using a microcontroller, with a more active, involved process, why make it so complicated, when what I've demonstrated will do the job. So that's my bet. Let us know what you find out, if you can.

Here is a constant-current-source (sink, actually) that you can try out with the two nine-volt-batteries:

Higher voltage LED and diode tester schematic and scenarios

Notice that in each case (short-circuited, one 3-volt-led, four 3-volt-LED's) the current flowing is about 1 mA.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.