voltage drop of bridge rectifiers - Page 1
Forum section: electronics / beginners

author: Simon
author_rank: Unknown
timestamp: September 08, 2011, 11:59:14 am
content: the voltage drop on bridges on the Farnell website is specified, presumably this is the total voltage drop ? as the brindge rectifier is essentially to paralleled sets of 2 diodes in series. The datasheet say's 1.1V drop per "bridge element" is this a crafty way of calling a 2.2 V drop a 1.1 drop ?While I'm at it, if I'm rectifying 7.2amps is a 10A rectifier (with 200A surge) man enough ?

author: ejeffrey
author_rank: Super Contributor
timestamp: September 08, 2011, 12:59:16 pm
content: 1.1 volt is the voltage drop of one of the diodes at the rated current.

author: Simon
author_rank: Unknown
timestamp: September 08, 2011, 01:07:11 pm
content: well that is a con, looks like I'm going for individual Schottky diodes

author: ElektroQuark
author_rank: Unknown
timestamp: September 08, 2011, 01:08:21 pm
content: I think it is the combined voltage drop of the two diodes.

author: FreeThinker
author_rank: Frequent Contributor
timestamp: September 08, 2011, 01:24:15 pm
content: AFAIK it will be the voltage drop across the bridge. As you will have a diode in the forward and return legs of your current flow you will drop approx 0.6v across each one ~ 1.1v.

author: deephaven
author_rank: Frequent Contributor
timestamp: September 08, 2011, 01:53:17 pm
content: It's per diode. If you look at the spec of a typical 10A rectifier, each diode has a drop of between 0.7V to 1.1V over the current range of 200mA to 10A which is what you would expect from one silcon diode.

author: Simon
author_rank: Unknown
timestamp: September 08, 2011, 02:49:38 pm
content: so presumably the vast majority of rectifiers are silicone, there is nothing on the datasheet about the diode type, Schottky it is

author: david77
author_rank: Frequent Contributor
timestamp: September 08, 2011, 03:35:01 pm
content: Why not use two double schottky diodes like the MBR20H200CT the datasheet claims a Vf of 0.88V @ 10A load.Also they're in a handy TO220 case so easily mountable for cooling purposes.Sorry bullshit.

author: Simon
author_rank: Unknown
timestamp: September 08, 2011, 04:43:44 pm
content: I can get some 10A 100V diodes with 640mV drop for £0.65+vat, I think that's my ticket, I'm not looking to be extravagant but as I'm building my first decent power supply based on a 500VA transformer with dual 35V outputs I want to do make the most of it, whittling away power on unnecessary heat dissipation is not my game

author: Unknown
author_rank: Unknown
timestamp: September 08, 2011, 05:48:21 pm
content: Quote from: Simon on September 08, 2011, 04:43:44 pmYou could also consider active rectifiers, a concept based on discret components can be found here:http://projectcircuit4u.blogspot.com/2010/04/power-mosfet-active-bridge-rectifier.html

author: Simon
author_rank: Unknown
timestamp: September 08, 2011, 06:09:49 pm
content: Quote from: AcHmed99 on September 08, 2011, 05:46:41 pmSo your saying that Iwill only get 250 W from a 500VA transformer ? I might need two transformers then, I was thinking of a modular design whereby I plug various sections together to get different voltages etc.What does active power factor correction involve ? I can get a transformer for £35

author: Simon
author_rank: Unknown
timestamp: September 08, 2011, 06:16:25 pm
content: Quote from: Alex on September 08, 2011, 05:48:21 pmI had considered something like that, except not many opamps handle 50V, still maybe some level shifting can be done so that I can run the opamps at say24V

author: gregariz
author_rank: Frequent Contributor
timestamp: September 08, 2011, 06:53:33 pm
content: If you are using a full wave bridge on a AC transformer with a decent smoothing cap you will probably have noticed your DC being higher than your AC RMS transformer voltage(DC~1.4xVunregRMS). You can easily measure it. You should be losing about 1.4v (two diode drops) through the bridge but you will make up due to smoothing on the peaks.Anyways thats just a tip assuming my memory is working properly

author: Simon
author_rank: Unknown
timestamp: September 08, 2011, 09:00:30 pm
content: Quote from: AcHmed99 on September 08, 2011, 07:18:32 pmmy thoughts on the active bridge are that the opamp driving it all needs to work at the supply voltage - I've got tons of 100V mosfets

author: Psi
author_rank: Super Contributor
timestamp: September 09, 2011, 01:49:07 am
content: I noticed that you made that other thread about paralleling up transformer outputs.Meaning the transformer you want to use can easily be connected as a center tap output.If you're worried about the rectifier voltage drop you could ground the tap and put a diode to each side.That way you'd only have the Vdrop of one diode instead of two and the waveform would still be fullwave.eg.

author: Simon
author_rank: Unknown
timestamp: September 09, 2011, 05:25:54 am
content: Won't that mean that I don't get full [power from the transformer ?, I'm fine with a 4 diode rectifier, I just don't want to waste power

author: gregariz
author_rank: Frequent Contributor
timestamp: September 09, 2011, 06:47:52 am
content: Assuming you had a 12 - 0 - 12 transformer @ 1A winding, Psi's arrangement would give you a 12VDC @2A output or thereabouts whereas the non- centertapped 4 diode bridge would give you 24VDC @ 1A. both would have the same power capacity ie 24W. I wouldnt worry about diode drops for the reason mentioned in my previous post.

author: Psi
author_rank: Super Contributor
timestamp: September 09, 2011, 09:57:13 am
content: Quote from: Simon on September 09, 2011, 05:25:54 amAs gregariz said, you will get the same power either way.When one diode and one side of the tap is working the other side isn't, then it switches around.So one side is carrying double the current but at a 50% duty cycle, so it averages out and all is fine.It's an easy way to increase efficiency, especially for very low voltages were the diode Vdrop is a significant factor.

author: ejeffrey
author_rank: Super Contributor
timestamp: September 09, 2011, 10:20:47 am
content: Transformer efficiency is considerably worse in a full-wave center tapped rectifier. Since each winding only conducts a half cycle, the copper losses in the secondary will be twice as high (I^2*R: if you conduct twice as hard for half as long, the losses are twice as high). Generally, copper losses are more significant than diode losses, so double the former to halve the latter is a bad deal except at low voltage.Schottky diodes are a good solution, but watch out. Some (mostly lower current) rated schottky diodes have high forward drops at maximum current. Their reverse leakage can be also problem especially at higher voltages -- make sure to check the leakage at the expected operating temperature (including the temperature rise from the forward conduction). They also create unnecessarily fast switching transients in low frequency rectification applications that can increase noise. None of these problems is particularly serious, just make sure to check the ratings before you drop them in your circuit.Personally, since you are talking about 50 V, I would just eat the second diode drop using standard high current silicon diodes. It is a drop in the bucket as far as efficiency of a line transformer linear power supply.

author: Psi
author_rank: Super Contributor
timestamp: September 09, 2011, 10:31:12 am
content: I don't think the copper losses are quite that bad.I remember going from 75% -> 80% efficiency on a 12v 10A power supply i was building when i moved to a center tapped rectifier.But yeah, the move voltage you have the less of an effect you get from removing one diode.

author: Zero999
author_rank: Super Contributor
timestamp: September 09, 2011, 03:50:01 pm
content: Quote from: ejeffrey on September 09, 2011, 10:20:47 amYes that's true.Assuming the load is perfectly resistive and the diodes are ideal, the transformer should be de-rated to 70.7% of the winding's VA rating. It's Ohm's law.

author: ejeffrey
author_rank: Super Contributor
timestamp: September 10, 2011, 02:24:14 pm
content: Quote from: Hero999 on September 09, 2011, 03:50:01 pmIt isn't as bad as that. There are more factors that go into a transformer power rating. The total power dissipation including losses in both primary and secondary windings as well as iron losses is often more important than the power dissipation of a single winding, as is core saturation.The 1/sqrt(2) factor would be correct if you had a multi-voltage transformer with a low-power winding that was rated well below the total VA rating of the transformer.