Measure a Schottky diode  - Page 1
Forum section: electronics / projects

author: Ronald
author_rank: Newbie
timestamp: March 18, 2018, 06:11:55 pm
content: HiI have a challenge.How do I measure a Schottky diode parallel with a resistor of 0.2ohm .I forward mode the Schottky diode will curse a lower resistance. In backward mode it's only the parallel 0.2,ohm that is operating.It has to be done in circuit!A standard DMM can't do it.I can't post the diagram because it's from my company.I shall also measure the propagation delay over the same circuit.That is done by putting 15v at the input side and measure the time with an oscilloscope.Thanks in advanceBjarne

author: Stray Electron
author_rank: Super Contributor
timestamp: March 18, 2018, 07:29:12 pm
content: The only thing that I can think of is to apply a high enough current load to the output so that the voltage drop across the resistor is significantly higher the Vf of the diode. That will allow you to read the diode Vf and delay time. OTOH I don't know if that will affect other devices in your circuit. In addition, I would pulse the diode circuit at a very low duty cycle ( 0.1 to 1.0%) to minimize heating of the circuit components. I would start with a 10 Amp current so 10A x .2 Ohm = 2 volts at the input and short the diode/resister back to the return side of the pulse generator. 10 amps would create 2 volts across the resister alone and the diode should clamp at more or less .6 volts so it would be easy to see on a scope. Dealing with that low of a circuit resistance would mean that you would need a pulse generator with a very low output impedance and VERY good conductors connecting it to the circuit being tested. Even a 0.1 Ohm lead resistance would throw your results off by 50%. What you really need is a kelvin connection (4 wire) from the pulse generator to the circuit under test to compensate for the lead losses, but I've never seen that on a pulse generator.FWIW I've never dealt with high speed relays but you might be able to use a standard kelvin connected power supply and control the power to the circuit via switching relays. A lot is going to depend on how robust the other devices in the circuit are and how much you need to limit the total energy into the circuit.

author: Marco
author_rank: Super Contributor
timestamp: March 18, 2018, 11:55:26 pm
content: Quote from: Ronald on March 18, 2018, 06:11:55 pmDepends on what you want to measure, but generally speaking with a high amount of current. Or in other words, measure it how it's supposed to be used but with known inputs.QuoteCalling the time it takes to charge/discharge the diode capacitance, propagation delay is probably a misnomer. Unless the signal of interest is a step signal.QuoteThat way you will also be measuring the internal resistance of your voltage source.

author: iMo
author_rank: Super Contributor
timestamp: March 19, 2018, 12:13:31 am
content: QuoteThe measurement method depends on the "circuit"..

author: The Soulman
author_rank: Super Contributor
timestamp: March 19, 2018, 12:36:02 am
content: Quote from: Ronald on March 18, 2018, 06:11:55 pmHow many amps is this diode rated, I'd say get a current source as close to that as possible, high current will swamp the effect of the resistor.As for propagation delay.......................................................................................good luck with that.We could give better advice if you'd describe the current waveform across the diode/resistor combo as in the final product.