1
\$\begingroup\$

I have a diode meter on my DMM Powerfix Profi and I want to measure the forward voltage. (actually it is not so profi at all).

So I tried the steps needed, and found the Vforward of a red led to be 1.7 V which is as expected.

However, when measuring a 3mm and 5mm white LED it did not show anything. Probably because the LED has a higher forward voltage and the batteries inside the DMM (I think 2x1.5V) are too less to power the LED.

Is there a way to measure a white LED anyway with the diode function of a DMM?

\$\endgroup\$
2
  • \$\begingroup\$ I think without any modification of the DMM or external circuitry it won't work. CMIIW. \$\endgroup\$
    – Unknown123
    Apr 5 '19 at 22:21
  • \$\begingroup\$ This is normal for a DMM Diode test to apply a constant current such as 1mA but full scale is normally 1.99V. \$\endgroup\$ Apr 5 '19 at 23:39
3
\$\begingroup\$

... Probably because the LED has a higher forward voltage and the batteries inside the DMM (I think 2 x 1.5 V) are too less to power the LED.

You are on the right track but 9 V meters will show the same problem because they have split-rail internally. (You can test this by switching to 20 V DC range and touching the red probe to the multimeter's battery + and -. The difference between the readings will add to the battery voltage. e.g. +3.2 - (-5.4) = 8.6 V.)

The diode test function will use some sort of constant current generator to drive the device under test (DUT).

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Imagineered simplified schematic for a 0 - 199.9 mV LCD meter.

If there are only a few volts headroom due to the split-rail power supply and a volt or two is lost in the constant current source then there may be only two volts or so available on the open-circuit test leads. R1 and R2 will be sized to divide the voltage down to the 200 mV full scale (but with decimal point adjusted to scale by 10) so a 2 V diode test would seem a rational limit.

enter image description here

Figure 2. LED IV curves. Source: LED IV curves.

According to my IV curves a diode tester with a low current - 1 mA or so - and a 2 V voltage limit would be able to test IR, R, O and G but might fail with anything else.

schematic

simulate this circuit

Figure 2. Use another meter, VM2, to measure the open-circuit voltage on the multimeter's diode test function.

You can check your diode test voltage if you have another meter. Switch the meter to be tested to diode test and measure the voltage on the open circuit probes using another meter on DC V. You might well find that the diode test does actually go higher than 2 V but due to the fixed voltage divider that it goes OL (overload) at 2.0 V.

Is there a way to measure a white LED anyway with the diode function of a DMM?

Maybe. Figure out which of the battery's terminals has a voltage of greatest magnitude relative to COM. Connect a current limiting resistor to that and connect the LED between it and COM. Switch to DC V and touch the red lead to the junction and measure your VF. Is it worth the trouble? No, unless it's a life or death situation that having a way of measuring the VF of a blue or white LED when you have only one meter and no other PSU.

enter image description here

Figure 4. The meter in question is the 200 mV display type. All readings are scaled down to 200 mV full-scale.

\$\endgroup\$
4
  • \$\begingroup\$ What do you mean by "200mV full scale"? I've never seen a digital meter that didn't just give you the unscaled reading with the units determined by the range setting (whether automatic or manual), but is that what you mean here? \$\endgroup\$
    – Hearth
    Apr 5 '19 at 23:38
  • 1
    \$\begingroup\$ Yes, the basic LCD module is reads 1999 when a 200 mV (199.9 mV) signal is applied to its input. On the 200 mV range the signal can be fed straight in (through the protection circuit) but all higher voltage ranges will require a voltage divider of 10:1, 100:1, 1000:1, etc. The current ranges should have the shunts organised to give 200 mV on full scale but I suspect that the cheap ones use shunts for more than one range so the voltage drop can be quite high. The three decimal points are switched by rotating the knob. \$\endgroup\$
    – Transistor
    Apr 6 '19 at 7:51
  • \$\begingroup\$ Thanks for your answer; especially the graphic is really useful (I was not aware about the big dependency of forward voltage versus current. \$\endgroup\$ Apr 6 '19 at 10:55
  • \$\begingroup\$ @Transistor Ah, so you meant the inputs are scaled down, not that the displayed value is. \$\endgroup\$
    – Hearth
    Apr 6 '19 at 10:59
2
\$\begingroup\$

The current source in a typical inexpensive DMM seems to be just a resistor, chosen to give about 1mA.

If you use a 5.0V wall plug adapter and a series resistor of about 2-3K you can do something similar.

schematic

simulate this circuit – Schematic created using CircuitLab

I checked a few multimeters I happen to have within reach, and these are the open-circuit voltages on diode range:

UNI-T M890G 2.809V

UNI-T UT56 2.96V

Fluke 8026B 3.03V

Agilent bench top 7.062V

ANENG AN8008 3.23V (super cheap 11/11 singles day purchase)

Chinese AC/DC clamp-on meter 1.54V

Short circuit (into mA meter) currents were 1.20/1.51/1.09/1.000/1.60mA/0.51mA

Only the Agilent appears to have a real constant current source. That's also the only one with enough voltage output to reliably bias a white or blue LED to anywhere near 1mA.

Note that the 7V will exceed the typical 5V abs. max. reverse voltage of most LEDs when the LED is reversed. Doubtful any measurable harm will result from a short test, but it is a violation.

\$\endgroup\$
1
  • \$\begingroup\$ Thanks for your answer; I used the circuit you showed too (but hoped there was an easy solution for a DMM). \$\endgroup\$ Apr 6 '19 at 10:49

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .