Center-tapped rectifier without diodes?

[pantaz]

I can't decide what the 'throbbing colours' are supposed to represent on the first sim.

The description of the applet says:
"The green color indicates positive voltage. The gray color indicates ground. A red color indicates negative voltage. The moving yellow dots indicate current."

They seem to imply that, even with equal values of 'series' resistors, current will flow in the load. This is clearly not the case because the 'bridge' is balanced and that resistor shouldn't be throbbing.

IMHO, the simulator is not for serious design and analysis, and since that particular circuit really doesn't follow normal design rules, the extraneous "throbbing" is probably just an artifact of the underlying Java code.

If you want an accurate answer, you would have to ask the applet's creator.

Also, why not just call it Current, rather than electron flow? You are just adding another level of (irellevant) complexity; all the electrical calculations I have ever come across use V and I (defined in a way that there is no misunderstanding).

I apologize for any confusion. It was almost 1:00 am, I was very tired, and I chose my words poorly.

 

[sophiecentaur]

The reason that I took exception to the applet, looking again, was that there was an implication that the centre tap voltage changes. I now realize that the secondary circuit actually has no ground, so the voltages are a bit meaningless as they are not referenced to anything. It would have helped if there had been a ground connection on the CT and then the colour of the middle resistor could have been grey all the time (which is surely what the whole thing is about). Any ground connection would have helped. The non-moving charges in the middle leg make a good point.

Sorry for the grumpy bit about electron flow but I have a 'thing' about it. Bringing it into every electronics explanation is guaranteed to confuse at least a few of the readers. It ranks with 'the photon explanation' as something that can do as much harm as good in most circs.

 

[Studiot]

Violent Corpse,
AlephZero made a comment you have not taken on board.

Armed with your newfound knowledge I suggest you revist his post.

He said you do not have to use diodes to achieve rectification.

However resistors, (or capacitors or inductors) will not do.

You could replace the diodes with switches. If you could switch them on and off at appropriate intervals you could achieve rectification.

This is not as silly as it sounds as in the past this was actually done when suitable diodes were unavailable for power reasons.

 

[ViolentCorpse]

Violent Corpse,
AlephZero made a comment you have not taken on board.

Armed with your newfound knowledge I suggest you revist his post.

He said you do not have to use diodes to achieve rectification.

However resistors, (or capacitors or inductors) will not do.

You could replace the diodes with switches. If you could switch them on and off at appropriate intervals you could achieve rectification.

This is not as silly as it sounds as in the past this was actually done when suitable diodes were unavailable for power reasons.

Yes sir, I have taken that comment on board. I understand that if we can make any other non-linear component work as a switch, it should work as a rectifier. However, my main concern was, with the original circuit diagram still in mind, why couldn't we use ANY component for the purpose?

Thankfully, my question has been answered for equal resistors. All the posts have been very informative, and I've been religiously following this thread for any new comments. :D

 

[dlgoff]

You could replace the diodes with switches. If you could switch them on and off at appropriate intervals you could achieve rectification.

And to put a picture with it:

http://upload.wikimedia.org/wikipedia/commons/b/b6/Schaltskizze_Elektromechanischer_Zerhacker.svg

 

[sophiecentaur]

Yes sir, I have taken that comment on board. I understand that if we can make any other non-linear component work as a switch, it should work as a rectifier. However, my main concern was, with the original circuit diagram still in mind, why couldn't we use ANY component for the purpose?

Thankfully, my question has been answered for equal resistors. All the posts have been very informative, and I've been religiously following this thread for any new comments. :D

1.) these components have to be non-linear. If you use R' L' and C's, you will get nothing (if the bridge is balanced) or you will get AC (in some phase and amplitude or another) if it's not.

2.) you presumably want the rectification to be reasonably efficient, so just any old non-linear component wouldn't be a lot of use for the purpose. You need something that, ideally, has zero resistance for one half of the cycle and infinite resistance for the other half. A good diode does the job best.

 

[ViolentCorpse]

^ I understand these. My point was still related to the current path. I didn't understand why alternately working switches are necessary for producing a unidirectional current through the load, since my mistakes were leading me to believe that any component should be able to do the job based on the configuration and assembly of the circuit alone (which, I admit, is stupid). It was the third terminal, a.k.a the center-tap that caused all my confusion. It's the first time I've encountered a circuit involving three terminals.

 

[Studiot]

You should look at two port networks next.

http://web.eecs.utk.edu/~green/notes/Two port networks.ppt