Absolute Value Circuit

[corey415]

Hello folks,

I am new to the forum and I have a question. I was wondering how one could build a circuit that behaves like a bridge rectifier with ideal diodes. From what I have initially researched, one could use op amps along with diodes; but that's about all I know.

In short, I need to construct a absolute value circuit, but the 4 diode bridge rectifier setup does not fulfill my design requirement because in my case, with non-ideal diodes, small voltages below .7V yields a 0V output.

P.S. - The circuit would have a DC input in the range of 200V and 50A.


I appreciate any help or insight you could give me.
thanks!

 

[corey415]

accidental double post

 

[evandude]

perhaps you could attach an op-amp wired as a voltage follower (ie- a gain of exactly 1) except with two diodes in series in the feedback... then it should make the output always 2 diode drops greater than the input... thus making up for the diode drops you would have through the bridge rectifier.

just a quick thought I had, I haven't tried it out so I don't know if it would work but it seems like it should.

but now that I read your post more carefully, there's no way this is going to function at 200V or 50A... are you really planning on using a 10KW power supply???

 

[corey415]

Yeah, that's the problem that I have, its such a large input. I personally am not going to test it, I just need a design of how one could do it. It is possible right?

 

[evandude]

I don't think it's something you could do reasonably...

no way are you going to get a 10KW op-amp... We have some power supplies here at work that behave like an op-amp (voltage output corresponds to voltage input) and they are pretty beefy, they go up to 200W (20V at 10A) but that's nowhere NEAR to 10KW. I don't even think they would MAKE something like that that could handle what you want and it would be completely cost-prohibitive.

why do you need such low-voltage precision on such an incredibly beefy power supply anyway? are you planning on using a 10KW power supply to drive just a couple of volts or something?

 

[corey415]

I don't think it's something you could do reasonably...

no way are you going to get a 10KW op-amp... We have some power supplies here at work that behave like an op-amp (voltage output corresponds to voltage input) and they are pretty beefy, they go up to 200W (20V at 10A) but that's nowhere NEAR to 10KW. I don't even think they would MAKE something like that that could handle what you want and it would be completely cost-prohibitive.

why do you need such low-voltage precision on such an incredibly beefy power supply anyway? are you planning on using a 10KW power supply to drive just a couple of volts or something?

Ok, I was confused about what I was trying to achieve. The circuit is needed for the GFCI. What I meant to say is that the load will be 200VDC and 50A. The AC input to the circuit should be a much smaller voltage.

 

[Oznog]

I think "synchronous rectifier" is the proper term, someone correct me if I'm wrong.

You could use a MOSFET and a comparator (or op amp) on Vds to switch on Vgs. Diodes are not a part of it.

It's a bit harder than that, since when the MOSFET's on, Vds is ideally quite small but it is the only way you can determine when to shut off the gate to prevent the current from reversing. The offset inside the comparator is a big part of the issue.

Also MOSFETs have an inherent body diode inside them, thus they will conduct current as a diode when Vds < -0.7V. So you will need to use them in reverse so that the diode will only conduct in the normal direction of forward current, when the gate should be turned on anyways leaving it conducting fully.

This can be a very effective way to make an exceptional rectifier, but it's not a simple or cheap task. Are you SURE you need such a thing that badly? 1.4V of rectifier drop doesn't seem significant in a 200V system. Usually you'd use this if you were extremely serious about conserving power and/or deal with low system voltages where 1.4V of rectifier drop is a problem.