Many multimeters have a diode mode and I have always used this mode to measure diodes. Today, I happened upon a Fluke article that explains you can use the ohmmeter mode to measure diodes as well: http://en-us.fluke.com/training/training-library/test-tools/digital-multimeters/how-to-test-diodes-using-a-digital-multimeter.html

In particular, it states that the resistance of a diode (when measured in the forward-bias direction) is on the order of kiloohms and megaohms. I was initially skeptical so I powered up my benchtop Agilent multimeter, and sure enough, a 1N34A diode shows a resistance of 5kΩ.

My first thought was that the ohmmeter's voltage is too low to adequately "turn on" the diode. My diode has a 0.6 V turn-on voltage, which is typical. However, https://electronics.stackexchange.com/a/140531 says that ohmmeters have voltages of 1 or 2 V, which is more than enough for sufficient current to flow.

Could anyone help explain this? I would expect the incremental forward resistance to be on the order of 100 Ω.

  • 1
    \$\begingroup\$ You should measure some different "real" resistors (e.g. 10 Ohms, 100 Ohms, 1 KOhms) and use a second (digital) multimeter to measure the voltage at the inputs of the multimeter while measuring the resistors. Using this method you see what's really going on when measuring resistors. \$\endgroup\$ – Martin Rosenau Oct 23 '16 at 9:28
  • \$\begingroup\$ Okay, I'll try this when I get my hands on two multimeters...but I'm guessing your point is that as I change the range of resistance on the ohmmeter, the test voltage changes drastically, right? I definitely see large differences in the measured resistance as I switch through different resistance modes. \$\endgroup\$ – Tosh Oct 23 '16 at 9:53

The open-circuit voltage from a modern digital multimeter in ohms mode is not very well defined - I measured a few and they varied from less than 0.6V to more than 2V.

Most of them will allow a silicon diode or schottky diode or germanium diode to conduct somewhat so on some range you could get a polarity indication from a common type of diode (though probably not an LED).

The diode function typically puts around 1-2mA through the probes with a compliance of more than 2.5V so it gives you some indication of the forward voltage of diode (and maybe a red or IR LED) at a sensible current.

Your 1N34A is, of course, an ancient germanium diode and has a relatively low forward voltage. Maybe 300mV at 1mA, not 600mV. That drop increases rapidly at higher currents. If you use the diode mode you should be able to see the difference between a 1N34A and, say, a 1N914/1N4148 readily, and a Schottky such as BAT54 or 1N5817/19 will be easily distinguished from, say, a UF4007.

You are correct that the incremental forward resistance of a puny germanium diode is of the order of a hundred ohms at mA-ish currents (order of magnitude) but remember that the ohmmeter is not measuring incremental resistance it is measuring the total voltage drop across the nonlinear device and dividing that by that current I mentioned (measured typically by deriving the ADC reference from a dropping resistor) to give you the equivalent total resistance.

For example, a 'stiff' voltage source of 0.3V (0\$\Omega\$ incremental resistance) would give you a large resistance reading on some range (with the right polarity). (obviously you shouldn't be putting volts into a meter on ohms range generally, but that is what would happen).

As a reminder, the dynamic resistance of an ideal (Shockley model) diode at current I is:

r = \$\frac{\eta V_T}{I}\$

where \$\eta\$ is the ideality factor (between 1 and 2 - ~=1.3 for a 1N34A diode) and I is the diode current, and \$V_T\$ is the thermal voltage - 26mV at room temperature.

So at 1mA r ~= 34\$\Omega\$

To find the 'resistance' a meter would measure- let's assume it's on a 20K range and the meter measuring current is about 50uA (what one of mine uses), and \$\eta\$ = 1.3, Is = 200nA (from a 1N34A SPICE model).

\$R_{EQ} = \frac{\eta V_T \ln(I/I_S)}{I}\$= 3.7k\$\Omega\$

Since your meter is reading a little higher, probably the current is lower than 50uA- maybe 30uA. You can measure approximately by using another meter in current mode.

By comparison, the dynamic resistance at 50uA is less than 700\$\Omega\$, so a huge difference.

The magnitude of the 'resistance' readings you get from a diode in resistance mode are basically meaningless unless you know the internal workings of the multimeter, so all you can divine is that the diode conducts to some reasonable degree in that particular direction. As you have observed they will change drastically from on resistance range to another. In diode mode you usually get a reading that's indicative of a forward voltage drop at a mA or two roughly, which is infinitely more useful. They don't control the current well, they often don't document it well, it's just a free add-on function, but handy none the less.

By the way, almost all digital multimeters (on ohms mode) have the red lead as the positive output in diode/resistance mode but older analog meters often reversed that.

  • \$\begingroup\$ Thanks, Spehro. I wasn't thinking well last night and provided the small signal resistance and not the large signal version. Thanks for taking care of that business. I'm deleting my answer. Thanks! \$\endgroup\$ – jonk Oct 23 '16 at 18:17
  • \$\begingroup\$ @jonk too bad, your small signal derivation was a thing of beauty. \$\endgroup\$ – Spehro Pefhany Oct 23 '16 at 19:56
  • \$\begingroup\$ Perhaps. But still mis-applied. And no point fixing it. I like your answer and upped it. I wouldn't be improving things by repairing my errors and replicating work done already. There's always another day for such derivation. (Speaking of which, I never use the usual \$\frac{\textrm{d} f(x)}{\textrm{d} x}\$ form. Never. I always think in operator form. It's more versatile, forces me to always consider other partials as I go, and I can introduce infinitesimal variables ad-hoc as needed. But I don't see others doing that often.) \$\endgroup\$ – jonk Oct 23 '16 at 20:03

There is no fixed resistance. Remember the I/U graph of the resistor (hint: a straight line through the point of origin.) Resistance is the inclination of that graph. So if you put the I/U graph of the Si diode in, you see the inclination of such a straight line through the operating point is heavily depending on the voltage. A small shift of e.g 0.1V leads to a big increase/decrease of current, which can give you a 1:10 change in resistance.

So, forget about putting a meaning in the resistance measured on diodes. You have to measure the forward voltage at a low, defined current instead.

  • \$\begingroup\$ Janka, okay, but the Fluke article seems to assign some meaning to the resistance measured this way by saying that (last page) the diode is functional if the resistance falls between "1 kΩ and 1 MΩ". So I am assuming Fluke knows something about the typical ohmmeter voltages to get these values, right? \$\endgroup\$ – Tosh Oct 23 '16 at 9:49
  • \$\begingroup\$ And thanks for pointing out the resistance measured this way is the DC resistance (instead of the small-signal AC resistance...) \$\endgroup\$ – Tosh Oct 23 '16 at 9:52
  • \$\begingroup\$ The typical DMM is resistance mode puts a low, defined current on the device under test. The difference to diode test mode is it doesn't show you the forward voltage but the measured forward voltage divided by the measurement current. That's the resistance of the device under test. \$\endgroup\$ – Janka Oct 23 '16 at 12:08

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