Abnormal Waveform of Schottky Diode - Page 1
Forum section: electronics / beginners

author: Sainteus
author_rank: Contributor
timestamp: March 15, 2017, 08:04:39 pm
content: We are doing a project on Linear Power Supply and we are trying to choose the diodes with better performance. Thus, we use a gate driver (Power MOSFET) to test the forward characteristic of Schottky Diode (Part Number: SBRT15U50SP5) but we notice that there are these few abnormal symptoms:Abnormal waveform for Schottky Diode.Forward voltage differs significantly compared to the datasheet.Below is the legend of the graph we get from the circuit.Vgs which is Yellow colored signal (Channel 1)Forward Voltage of Schottky Diode which is Blue Colour signal (Channel 2)Load Resistor (2 Ohm) which is Pink colored signal (Channel 3)Hoping anyone can clear our doubts. Thanks in advance.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 15, 2017, 08:41:22 pm
content: The current signal still looks good. So I don't see any trouble with the diode. However the horizontal speed is set rather low: even with slow Si diodes the recovery is seldom more than 10-50 µs.The signal for the diode more looks like a probe problem, or maybe due to AC coupling. How is it measured ?There is some tolerance in the forward voltage. So a 60 mV difference is not that unusual.

author: Seekonk
author_rank: Super Contributor
timestamp: March 15, 2017, 09:01:00 pm
content: Why don't you explain what those three signals are and your actual circuit. I think it is something you are introducing.

author: Benta
author_rank: Super Contributor
timestamp: March 15, 2017, 09:07:16 pm
content: If your test circuit is just the battery, the resistor and the diode, trying to measure recovery time makes absolutely no sense.For this, you'll need to go from forward conducting to reverse blocking. Reverse blocking means an actual reverse voltage over the diode. You don't have this.Perhaps you have a more detailed schematic of your test setup?

author: Ammar
author_rank: Regular Contributor
timestamp: March 15, 2017, 09:16:52 pm
content: Quote from: Benta on March 15, 2017, 09:07:16 pmAgreed. The diode has a capacitance associated with it turning on and off. If you just open the switch in your schematic, that capacitance will discharge slowly.Variation in forward voltage is to be expected. However, you may be wondering why it is different from the datasheet, but the same for two diodes. The two diodes will likely have come from the same section of the same silicon wafer and therefore be closely matched. The manufacture of semiconductors uses probabilistic processes and requires lots of testing. Typical yields can be around 30% to 40%.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 15, 2017, 10:40:25 pm
content: Your text says "gate driver" but your circuit shows "switch".The circuit cannot produce the waveforms shown.Tim

author: Benta
author_rank: Super Contributor
timestamp: March 15, 2017, 10:52:47 pm
content: Quote from: T3sl4co1l on March 15, 2017, 10:40:25 pmWhich is why I asked for a better schematic. But after one post, OP seems to have lost interest. Let's leave it there and move on.

author: David Hess
author_rank: Super Contributor
timestamp: March 16, 2017, 02:50:05 am
content: You could use a fast recovery rectifier for comparison but my guess is that you used a wirewound 2 ohm resistor and are seeing the inductive kickback.Reverse recovery tests require bipolar drive to reverse the current through the diode and careful high frequency construction including decoupling of the power sources. The use of a gate driver for the source is a good idea.

author: Sainteus
author_rank: Contributor
timestamp: March 16, 2017, 11:11:03 am
content: A very sorry for late reply because I am still new in this forum and there are many features that I am not familiar with. We are still doing the experiment and we will keep updating for this subject.

author: danadak
author_rank: Super Contributor
timestamp: March 16, 2017, 12:22:53 pm
content: Measuring approaches -http://www.edn.com/design/components-and-packaging/4441348/Easily-measure-diode-capacitance-and-reverse-recoveryhttp://www.jedec.org/sites/default/files/docs/jesd282b01.pdfpages 90 - 92https://www.onsemi.com/pub/Collateral/NSR0340P2-D.PDFtest circuit for a specific diodeRegards, Dana.

author: w2aew
author_rank: Super Contributor
timestamp: March 16, 2017, 02:16:14 pm
content: For what it's worth, I did a video on measuring reverse recovery time:

author: orolo
author_rank: Frequent Contributor
timestamp: March 17, 2017, 01:24:10 am
content: If you are using a mosfet as the switch, and turn it off letting Vgs fall from 12V to 0V, then the source terminal is pulled to negative -12V by the gate-source capacitance. Since your oscilloscope has a range of 2V in the blue trace, and the capacitance pulls the voltage below that range, it appears flat for so long: in reality, it's below -2V. Then leakage through the mosfet/schottky recharges the source terminal to 0 volts. The diode has nothing to do with the whole proccess, except working as a open switch.

author: Sainteus
author_rank: Contributor
timestamp: March 20, 2017, 07:16:31 am
content: Quote from: Kleinstein on March 15, 2017, 08:41:22 pmHi, Kleinstein. Thank you very much for your reply. We also had the same concern as you. The recovery time of diodes surprise us and according to our research, Schottky Diode shouldn't have a recovery period. By the way, we are actually using DC power supply.

author: Sainteus
author_rank: Contributor
timestamp: March 20, 2017, 07:44:55 am
content: Quote from: David Hess on March 16, 2017, 02:50:05 amHi, David Hess. Thank you very much for your reply. You are correct I am using a wirewound 2 ohm resistor in the circuit. After considering you advise, I replace it with a film resistor but the waveform I get is still the same.

author: Sainteus
author_rank: Contributor
timestamp: March 20, 2017, 07:46:23 am
content: Quote from: T3sl4co1l on March 15, 2017, 10:40:25 pmHi, Tim. Thank you very much for your reply. I am very sorry that I had confused you with our schematic diagram. We actually use a Power Mosfet in gate driver to conduct the experiment.

author: w2aew
author_rank: Super Contributor
timestamp: March 20, 2017, 12:44:02 pm
content: Quote from: Sainteus on March 20, 2017, 07:46:23 amIf you're connecting your probe as shown in your schematic, then you're shorting the 2 ohm resistor to ground - unless your scope is floating or you're using a differential probe.

author: Sainteus
author_rank: Contributor
timestamp: March 21, 2017, 02:45:17 pm
content: Hi, w2aew. Thank you very much for your reply. I'm not getting what you means actually. But I agreed that it might due to our scope or probe. Hope you can teach me more. I am using Tektronix TPS2014 Digital Oscilloscope, 4 Channels, 100MHz.http://www2.tek.com/cmswpt/psdetails.lotr?cs=psu&ci=13304&lc=ES-MX

author: StillTrying
author_rank: Super Contributor
timestamp: March 21, 2017, 04:48:41 pm
content: Quote from: w2aew on March 20, 2017, 12:44:02 pmIf the current then limits at 10A, that'll explain the 360mV Vf across the diode.