Diode test principle - Page 1
Forum section: electronics / beginners

author: marekpasek
author_rank: Contributor
timestamp: September 06, 2017, 09:07:17 pm
content: Hi, I have a question. I was wondering how the diode tester works. I came with my solution, how would I do that but I would like to compare that with the solution used by multimeter manufacturers and, you know, proffesionals. So there is my solution.I would use some constant current source which should output voltage equal to the diode's voltage drop in some particular current (if I would do enough measurements I would be able to draw the L shape curve of current through the diode, would I not?)So I would like to know if my solution would work and If so, what current should I use because as long as voltage drop is proportional to the current through and all the multimeters shows only on value when testing diodes, there must be some widely used normal current which makes the voltage drops of all the diodes comparible (in the case that they are not using some wiser way to measure it without some standardized values).Thanks!

author: KL27x
author_rank: Super Contributor
timestamp: September 06, 2017, 09:23:18 pm
content: I dunno how the average multimeter does it. But it seems to me you would need to use two different current levels to remove resistive voltage drop from the measurement.I would suspect a typical DMM powered by 2xAA or a li coin cell would put 3V through the probe with a series resistor to limit current and to act as shunt for measuring the voltage drop across it. Then it should need to change to a different value series resistor/shunt in order to verify that the Vcomponent is actually due to diode, vs resistive drop. A drop due to diode will be relatively static, hence the voltage drop across the two different series resistors would be approximately the same. If you tested with just one level of current without this check, you would return an "Fvd" value for any resistor you measured.

author: macboy
author_rank: Super Contributor
timestamp: September 08, 2017, 02:36:08 pm
content: You're over-thinking it. It's really very simple. The meter uses a constant current source of 1 to 2 mA (usually the same current source used for one of the lower/lowest resistance ranges), and it measures the voltage at the terminals. This current is enough to get past the "knee" of the V-I curve for Si diodes, so the forward voltage will be around 0.6 V. It may be a little lower for large and/or low reverse voltage rectifiers, and a little higher for higher reverse voltage ones. If the diode is connected reversed, then the terminal voltage is over full-scale, so usually "OL" or "1 " or similar is displayed.Obviously, the voltage displayed is only 'accurate' at the exact test current used, so it's just a nominal value. It can be useful, like when testing a bridge rectifier for example, you should get all four diodes within a few percent of each other (since they are identical) on a good rectifier. Otherwise you start to suspect a bad diode.You can directly measure the current used by your meter by setting it to diode mode, and connecting the leads to a second meter set to measure mA. Incidentally, the indicated reading on thediode test meteris the burden voltage of the current testing meter, at that current level.

author: MrAl
author_rank: Super Contributor
timestamp: September 08, 2017, 03:55:23 pm
content: Hi,It all depends how well you want to test the diode. For a simple static test a constant current can work ok, but for example if you want to test the speed of the diode then you need a scope and wave generator.Assuming you just want a simple test, then a constant current source through the diode and then measure the voltage. The voltage should be typical for that kind of diode at that current level.You should also check the reverse conduction too though to make sure it is not overly leaky. This involves a voltage near the rating of the diode while measuring current. This tells you if the diode works good in reverse.So one test applies a current, measures the voltage, while the other test applies a voltage and measures the current. These are simple minimum tests but they should both be done not just one or the other.If you want a little more comprehensive test, run a higher current though the diode and check the voltage. For a 1 amp diode for example that might mean pumping nearly 1 amp through the diode, then take a voltage reading.

author: David Hess
author_rank: Super Contributor
timestamp: September 08, 2017, 06:53:39 pm
content: The ohms converter in a multimeter will already provide a constant current so that is used and the display is calibrated in millivolts or volts instead of ohms. There is no standard test current but it is always low enough that lead resistance contributes insignificant error; 1 to 10 milliamps is typical. Modern multimeters limit the test compliance voltage so that it is high enough to test LEDs but low enough to avoid breaking down a base-emitter junction.If you want a more thorough test, then use a curve tracer or source meter or make your own test setup. A power supply, resistor, and multimeter can be used to test reverse leakage for instance.

author: KL27x
author_rank: Super Contributor
timestamp: September 09, 2017, 04:17:30 am
content: QuoteIf this is all it does, then open circuit would show as an FVD equal to Vsupply. A 0 ohm connection between the probes would show as 0V. And with the right resistance, you could produce "FVD" of anything between. There would be no distinction between a resistor and a diode. And there would be resistor values that would return an "FVD" of between 0.3 and 0.6V, just like a diode.It is very simple to verify this. Put the DMM on DC volt and put a 12V battery with 2K resistor in series between the probes. Then put a 100K pot between the probes to simulate the component under test; adjusting the pot, you can reproduce any voltage approaching the rails and everything between. This setup, BTW, is a simple way to test FVD of LEDs and reverse breakdown of zeners, BTW, so long as you know what you're testing. But a DMM needs to do better than that. It differentiates between diode and resistor and it won't show a measurement for a resistor. And it doesn't show Vsupply while the probes are open circuit.This isn't a perfect analogy, but the result is the same. If you have a lab supply, you can limit current to a few mA and do it that way. Or you can build a constant current circuit to take place of the 2K series resistor and 12V battery.

author: David Hess
author_rank: Super Contributor
timestamp: September 09, 2017, 05:34:01 am
content: Quote from: KL27x on September 09, 2017, 04:17:30 amOhms mode works by connecting an "ohms converter" which is just a crafted current source to the volt-ohms lead and measuring the voltage produced. If the voltage rises above a defined level, then overload is indicated.The crafted current source has a maximum compliance voltage which is lower than the supply voltage and almost always deliberately limited to prevent excessive output voltage which could damage some parts like bipolar transistor base-emitter junctions. (1) It also has output protection so that applying a high voltage to the ohms terminals will not damage the current source. It is interesting to contemplate how to arrange a 1 to 10 milliamp test current while being immune to high voltages now and in the past.(1) This really is needed because diode mode is very useful for testing transistors and a reverse current of just a couple milliamps through the base-emitter junction will immediately damage a bipolar transistor. Base-emitter breakdown commonly occurs just above 5 volts but for some transistors it may be as low as 3 volts. In the past tunnel diodes were often destroyed this way as well.

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: September 09, 2017, 07:14:55 am
content: "..a reverse current of just a couple milliamps through the base-emitter junction will immediately damage a bipolar transistor. "Can't say I've ever encountered this. The reverse biased e-b junction behaves like a zener diode, albeit a relatively inaccurate one whose voltage depends rather more on current than for the proper component. I've seen cases where transistors were used as crude voltage stabilisers this way.The main situation where reverse Veb destroys transistors is where a relatively large coupling capacitor is used between stages, with no series resistance. The classic multivibrator is one example. If the Vcc is well over 5v, then the collector of the other stage going to 0v will 'crowbar' the fully charged coupling capacitor across the b-e junction, dumping its charge into it with a huge current.Result is that the timing varies wildly due to the caps being partly s/c discharged by the e-b zener on each cycle. With large timing caps, one of the transistors will eventually throw the towel in under this pummeling though.

author: Shock
author_rank: Super Contributor
timestamp: September 09, 2017, 10:49:32 am
content: Quote from: KL27x on September 09, 2017, 04:17:30 amIf the multimeter is showing 2V all the time people would think they have a faulty diode. The analog measurement is sampled so thresholds and ranging can easily be applied.QuoteIt does on my multimeter shows 0.000 since there is not much drop in the leads.QuoteIf I measure a 3k resistor I see 1.1V. Checking with another multimeter it appears to just be measuring the voltage drop across the resistor. With a constant current supply on a diode, maximum current flow would cause the voltage to lower to the FV so that makes sense as well.QuoteWell according to my multimeter I have a 1.1V diode with that 3k resistor. Diodes like most conductors are resistive when conducting.QuoteI think it's more likely you haven't found the right resistance, but your mutlimeter may be different. Again showing the supply voltage would be like showing the upper threshold on the ohms range with an open circuit, it would just confuse the user.

author: David Hess
author_rank: Super Contributor
timestamp: September 09, 2017, 05:33:58 pm
content: Quote from: IanMacdonald on September 09, 2017, 07:14:55 amThe damage mechanism is via hot carriers causing dislocations. Annealing fixes it but that is hardly practical unless you want to attach a heater to a ceramic package.I have run across it a couple of times where an operational amplifier or differential pair had a mysterious offset which is an order of magnitude or more worse than expected or where a transistor operating as an amplifier simply had low hfe. Testing will reveal either a high input bias or offset current.QuoteWhat you are describing is damage through overheating and very similar to secondary breakdown. They make special PN junctions for handling this but of course that does not apply to transistors as far as I know. Although some are avalanche rated, I do not think that applies to reverse breakdown of the base-emitter junction.This is different and happens at low currents with the result that hfe, especially hfe at low current, is degraded significantly because carrier lifetime is shortened lowering collector efficiency. Transistors operating at high current and gold doped devices (1) which have low hfe anyway are not affected *as much* but it is fatal in any application which relies on low base current like the differential input stage of an operational amplifier. This is why bipolar operational amplifiers include shunt protection across their input differential pair; even momentary breakdown permanently ruins their precision although they still function.You can observe the effect in a standard 2N3904 or 2N3906. Run a safe reverse current through the base-emitter junction, a milliamp will do, and measure the hfe before and after. The damage levels off over time but eventually the hfe at low collector currents may be only 1/3rd of the original and even lower at lower collector currents.We had a big discussion here on the forum not long ago with various references but I found searching for it futile. A search for "hot carrier reverse base emitter breakdown" will find things.(1) RF transistors and fast saturated switches fall into this catagory. They already have a low carrier lifetime so are in essence "pre-damaged" as far as hfe at low currents.