PSU Transformer and rectification related question. - Page 1
Forum section: electronics / beginners

author: Mint.
author_rank: Frequent Contributor
timestamp: February 07, 2012, 11:37:37 pm
content: Hello,So I recently bought "Tab electronics guide to understanding electricity and electronics". In there was a project on building a power supply and all the steps and processes were explained. However I came across a rather interesting aspect that perhaps at other times I have missed. As seen in the attached photo when the diode is reverse biased it only passes half of a cycle of the sine wave. However what I don't understand why is there a 0.7v positive when it is reversed biased. Where does it get the 0.7 volts from?

author: amspire
author_rank: Super Contributor
timestamp: February 07, 2012, 11:40:59 pm
content: The 0.7V is when the diode is forward biased. If you look at the resistor voltage above, it corresponds to the half cycle when the resistor has a voltage across it.When the diode has the half sinewave across it, that is when the diode is not conducting (reversed biased), and no current is getting to the resistor.Richard.

author: electrode
author_rank: Regular Contributor
timestamp: February 08, 2012, 12:15:38 am
content: Yep, notice that the top waveform has a 16.3 V peak, and the 2nd has a 17 V peak (= 16.3 + 0.7).Incidentally, I also have that book. Not terribly keen on it though - maybe just because I was young when I read it a few times and didn't understand everything too well.

author: IanB
author_rank: Super Contributor
timestamp: February 08, 2012, 12:23:59 am
content: Yes, the 0.7 V is the voltage drop across the diode when it is conducting. Therefore you have to bear in mind that if the diode is passing 1 A then it will be dissipating 1 A x 0.7 V = 0.7 W during that period. It doesn't take much for diodes or rectifiers to start getting warm if you pass large currents through them. They may even need heat sinking.Also remember that the 0.7 V is a convenient fiction. With larger currents the voltage drop could be 1 V or higher.

author: Mint.
author_rank: Frequent Contributor
timestamp: February 08, 2012, 05:29:58 am
content: I understand now! I misunderstood the text slightly. Thanks for your help!

author: FenderBender
author_rank: Super Contributor
timestamp: February 09, 2012, 03:38:57 am
content: Quote from: IanB on February 08, 2012, 12:23:59 amYeah I'm not quite sure why textbooks try convincing me that all silicon diodes magically have this "set" forward voltage of 0.7V. Even some of the jellybean diodes you see around can have vDrops of 0.4-1.4V.

author: vk6zgo
author_rank: Super Contributor
timestamp: February 09, 2012, 04:08:53 am
content: Most of the time,small signal silicon diodes & transistor junctions do have approx 0.7 v drop.Some books quote 0.6v.,others 0.5v.It's an approximation that works most of the time,so that,if you are troubleshooting a circuit,you can usually read something closeto that value across silicon semiconductor junctions.For germanium junctions ,it is 0.2v!

author: Mechatrommer
author_rank: Super Contributor
timestamp: February 09, 2012, 04:18:46 am
content: Quote from: FenderBender on February 09, 2012, 03:38:57 amto make easier for you to pass the exam

author: SeanB
author_rank: Super Contributor
timestamp: February 10, 2012, 09:05:17 pm
content: Silicon diodes generally are 0.7V, at low current. If you are using a a few A then it will be higher, and if you ever look at high voltage diodes ( 1kV plus) they can have forward drops of over 40V, and pretty high leakage currents as well.