Does a diode really have a small current flow even when reverse biased - Page 1
Forum section: electronics / beginners

author: Adhith
author_rank: Frequent Contributor
timestamp: July 20, 2018, 08:15:42 am
content: Hello everyone..By diodes i meant the general purpose diode 1n4007 and the schottky diode 1n5819. I just reverse biased the 1n4007 with a 12v dc supply(from a wall adapter) and found that even though its reverse biased it has a voltage of 0.7V at the other end. where as the 1n5819 has a voltage of 2.5v at the other end. What does this indicate..are my diodes are bad or does it mean even though it has a voltage flow in reverse bias the current will be very low so that it almost is like an open circuit??

author: IanB
author_rank: Super Contributor
timestamp: July 20, 2018, 08:22:38 am
content: Quote from: Adhith on July 20, 2018, 08:15:42 amI think you need to elaborate on this statement, because as it stands it does not have any meaning. A voltage is measured between two points. Between which two points are you measuring 0.7 V ? And what does your complete test circuit look like?In particular, if you apply 12 V DC across a diode in reverse bias, the voltage across the diode will be 12 V. Are you putting a resistor in series with the diode and measuring across the resistor perhaps?As for your subject question, yes, all diodes have a leakage current. For some diodes, like Schottky diodes, it might be significant.

author: borghese
author_rank: Regular Contributor
timestamp: July 20, 2018, 08:27:51 am
content: Is not clear; can you post the diagram of your setup?

author: Raj
author_rank: Frequent Contributor
timestamp: July 20, 2018, 08:31:41 am
content: yes,diodes leak some current but I need to see a diagram to know what exactly what you are doing.Normally,if there is only a diode attached to the source of power,there should be negligable current flow through it and power accross it should be a little less than source (just a few microvolts)Unless you don't have stable voltage source.In that case,the diode will act somewhat like a capacitor.

author: leeatljs
author_rank: Contributor
timestamp: July 20, 2018, 08:34:10 am
content: Yes, all diodes have some reverse leakage.The 1N4007 can 'leak' up to 5uA when reverse biased. Your 1N4007 is leaking about 700nA into the resistance of your meter.Schottky diodes leak more when reverse biased.

author: Adhith
author_rank: Frequent Contributor
timestamp: July 20, 2018, 04:36:28 pm
content: Thank you everyone for all your replies. I'm sorry that I didn't posted an appropriate schematics and I have attached one in this post.

author: Raj
author_rank: Frequent Contributor
timestamp: July 20, 2018, 04:45:35 pm
content: v=ir,typical multimeter is 1mega ohms impedience.0.7v would mean 0.0000007 amps.Perfectly fine...nothing wrong here.peobably bad power supply with jitters or just diode leakage.Nothing to worry.

author: Adhith
author_rank: Frequent Contributor
timestamp: July 20, 2018, 04:55:06 pm
content: Quote from: Raj on July 20, 2018, 04:45:35 pmThank you very much for the help

author: Adhith
author_rank: Frequent Contributor
timestamp: July 20, 2018, 04:56:09 pm
content: The reason that I'm so much concerned with this reverse current is that I'm actually using some diodes to restrict the current flow to some LED strips but LEDs are actually lighting up with very low brightness and not turing off completely.I have attached a diagram showing what I'm doing. I have made a VU meter based on LM3915 IC, which has 10 outputs and each output pin is connected to a small led strip section through a transistor switch and thus making a row of 10 LED strips. but I also want to include another option where the VU meter circuit is turned off and the LEDs are driven separately by a wifi LED driver. In order to do that I should isolate the 10 output pins between each other but also want the current to flow through all the output terminals so that the driver could lightup the LED strips. So I made my mind to place 10 diodes at the 10 output terminals which solves the problem i guess. But now all the LED strips are lighting up with a very low brightness while turning on the VU meter and I doubt that it would be the reverse current flow through the diodes that cause this problem.In the circuit diagram, wifi output case 1 indicated that the vu meter is turned off and the LEDs are lit up through the 10 diodes to the 10 LED strip rows. case 2 means the power supply through the the diodes mentioned in case 1 is turned off and instead the power supply to the transistor is turned on and thus the vu meter could light up the led strips according to the music

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: July 20, 2018, 05:19:07 pm
content: Quote from: Raj on July 20, 2018, 04:45:35 pmKinda oftopic, but most handheld meters are 10 Mohm, and most bench meters have much higher (Gohm) impedance.

author: David Hess
author_rank: Super Contributor
timestamp: July 20, 2018, 09:01:02 pm
content: A typical meter with a 10 megohm input resistance will read 100 picoamps/millivolt while typical diodes have a reverse leakage of nanoamps to microamps so reverse leakage can be very apparent even with basic instrumentation.

author: Raj
author_rank: Frequent Contributor
timestamp: July 21, 2018, 04:12:41 am
content: still,his diode is normal

author: Nusa
author_rank: Super Contributor
timestamp: July 22, 2018, 08:24:12 pm
content: A pulldown resistor to ground on the power 1 line would ensure the leakage current from power 2 won't cause the line to float when power 1 is off. I bet that'll solve your issue.