Full wave rectification with an offset p-p voltage

Thread Starter

coinmaster

Joined Dec 24, 2015
502
If you have a 100v AC supply and you attach the negative side of a full wave rectifier to ground then you end up with the full AC p-p voltage at the positive side which would be 100v positive DC.
But what If I want 80v positive and 20v negative? Is there a way to offset the waveform so the peaks are shifted partially?
 

Thread Starter

coinmaster

Joined Dec 24, 2015
502
Hold it. 100 Vac from the power line, as in Japan, or 100 Vac from a solid state AC power supply, or what?

Also, 100 Vac RMS or peak-to-peak or what?

ak
100v is just an example number.
I'm referring to that fact that if you have a full wave rectifier that is rectifying 100v of AC then you will have 50v positive and 50v negative on the output of the rectifier because you are getting both halves of the waveform.

If you ground the negative half of the rectifier then the waveform shifts upward so you get 100v DC at the positive terminal (at least this is the way I understand it). What I want to do is only partially shift the wave form so instead of getting either 50% or 100% of the RMS voltage on either side I get something like 80% on the top and 20% on the bottom.
 

crutschow

Joined Mar 14, 2008
24,047

cmartinez

Joined Jan 17, 2007
6,662
That may work for small signal requirements but not for power from a full-wave rectifier.

So the answer is no, there's no way to readily shift the output DC level from a full-wave power rectifier.
The best you can do is use two half-wave rectifiers to give you plus and minus 100V.
Wouldn't a variac be a viable option?
 

cmartinez

Joined Jan 17, 2007
6,662
How will that help you offset the DC output voltage? :confused:
Well... it occurred to me that he could adjust his original AC source using the variac, and then use full bridge rectifiers on each side...

Something like this:

Capture.JPG

VA-B=100 VAC
VA-C=20 VAC
VC-B = 80 VAC​

Is what I'm saying viable? Or did I just put my foot in my mouth?
 

shteii01

Joined Feb 19, 2010
4,647
Nice... but how can that possibly be adapted to shift 100V (say peak-to-peak)? Are there chips out there with that capability? And then what about the amount of current needed for the OP's application? Something tells me that the circuit is a little more complicated than that.
Wait. What current? Nobody said anything about current.
 

ronv

Joined Nov 12, 2008
3,770
100v is just an example number.
I'm referring to that fact that if you have a full wave rectifier that is rectifying 100v of AC then you will have 50v positive and 50v negative on the output of the rectifier because you are getting both halves of the waveform.

If you ground the negative half of the rectifier then the waveform shifts upward so you get 100v DC at the positive terminal (at least this is the way I understand it). What I want to do is only partially shift the wave form so instead of getting either 50% or 100% of the RMS voltage on either side I get something like 80% on the top and 20% on the bottom.
Sorry, didn't read this close enough. Once you rectify it all the peaks are positive with respect to the negative output.
Was this a trick question? :D
upload_2016-2-29_19-59-9.png
 

AnalogKid

Joined Aug 1, 2013
8,251
100v is just an example number.
I'm referring to that fact that if you have a full wave rectifier that is rectifying 100v of AC then you will have 50v positive and 50v negative on the output of the rectifier because you are getting both halves of the waveform.
No, you won't.

First, in a discussion like this you have to be more careful in the way you describe things. 100 Vrms = 141 Vpeak = 282 Vp-p. 100 Vp-p = 50 Vpeak = 35.4 Vrms. It sounds like you mean 100 volts peak to peak, or 100 Vpp.

If the input to a full wave rectifier is 100 Vpp, then the output is 50 Vp (minus diode losses). You have to be careful in the way you describe things because I think when you say 50v positive and 50v negative you mean something very different from what most electronics persons mean. Also, comparing the input to the output is tricky because they cannot have a common reference potential (note I did not say ground) for measurement.

ak
 
Last edited:

WBahn

Joined Mar 31, 2012
25,062
A couple sketches of the waveforms you are talking about would be helpful -- particularly what you have and what you want.
 
Top