Film-Tech Forum ARCHIVE: N3 IREM Rectifier

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Posted by Paul G. Thompson (Member # 655) on 07-22-2003, 02:08 AM:

It looks like the diodes are in backwards. The cathodes of the diodes are grounded and the center taps of the star winding are connected to the anode of the xenon bulb. Are my eyes failing me or my mind? Or, am I missing something? Or, is the bulb polarity a missprint?

Posted by Dave Macaulay (Member # 813) on 07-22-2003, 11:51 PM:

If this is the rectifier I'm thinking of...
all diodes are reverse polarity.
nothing is grounded, the heatsink is isolated.
The circuit is unusual to our non-Italian minds but it works quite well.
I have much more trouble trying to understand the boost circuit!

Posted by Ken Lackner (Member # 1002) on 07-23-2003, 12:35 AM:

Uh....please forgive the ignorance, but how can a diode be reverse polarity? I've been trying to figure this circuit out for days, and I just can't understand how it can work with the diodes drawn the way they are. (I have a Big Sky manual in my possesion, and the Irem manual is on this site.)

Posted by Paul G. Thompson (Member # 655) on 07-23-2003, 01:49 AM:

Dave, on that boost circuit....Join the crowd. I never did figure that thing out.

As far as reverse polarity on the diodes, that may be true. But the cathodes are all grounded in the IREM roadmap I was looking at, including the bulb's anode. The center tap of the star was feeding the positive side of the bulb.

I think either the doides are drawn incorrectly or the polarity of the bulb is reversed for some reason. I wonder if the Italians used the theory that current flows from positive to negative...as some colleges teach.

[ 07-23-2003, 08:53 AM: Message edited by: Paul G. Thompson ]

Posted by Phil Hill (Member # 371) on 07-23-2003, 02:53 AM:

I'm confused as to why yous are confused. By looking at the N3 schematic on this site, it is a straight-forward "negative" power supply. The center taps are the "+" terminal and the diode anodes are the "-" terminal. The diodes are shown correctly.

BTW: The boost ckt is also correct.

I agree that most PS are a "positive" configuration, but there is nothing wrong with the schematic.

>>> Phil

Posted by Paul G. Thompson (Member # 655) on 07-23-2003, 08:43 AM:

Many apologies....My tri-focals have deceived me. When I orginally glanced at the schematic of the N3G3 IREM, somehow I saw the cathodes of the diodes tied to the cathode of the xenon lamp. I am still wondering how I saw that incorrectly. Bad eyeballs strike again. [Embarrassed]

The diodes and the bulb are correct in that schematic.

Posted by Ken Lackner (Member # 1002) on 07-23-2003, 02:45 PM:

Paul, I am glad you understand how it works, but I still do not! Please help!

Posted by Sam Hunter (Member # 1119) on 07-23-2003, 03:01 PM:

Ken, just think of it as a SERIES circuit and that current flows in one direction meaning that it doesn't matter which side the diode is in as long as it gives you the correct flow of current. Personally I like my rectifiers and pass transisters on the plus or supply side myself as I have trouble understanding some of those backwards circuits myself.

Posted by Paul G. Thompson (Member # 655) on 07-23-2003, 05:56 PM:

Ken, maybe this will help.

Remember there is phase reversal from one side of the secondary with reference to the other side. The center tap is common to them both. Above is basically what IREM is doing, now all you have to do is add the other 4 doides and secondary windings of the star transformer. The capacitors across the diodes are RF suppression capacitors. As you can see, the power supply is not referenced to ground. The capacitor on the negative end of the bulb is a .022 mfd at 1500 volts. It is another RF suppression capacitor, and not a filter capacitor. The filter capacitor is not shown on my diagram. It is connected across the RA unit as the schematic shows in the manual. The circuit designation of the filter capacitor is C7.

Remember the diodes do not conduct when the cathodes are driven positive with respect to the anode.

Posted by Phil Hill (Member # 371) on 07-23-2003, 06:36 PM:

Below is what I earlier emailed Ken, While not as elegant as Paul's diagram…yous get the gist. It's a simple 1/2 - wave explanation.

The arrows represent "conventional current flow."

Also, Sam's suggestion is excellent.

>>> Phil

Posted by Paul G. Thompson (Member # 655) on 07-23-2003, 08:33 PM:

Ken, I think Phil's explanation will also help you. If you still have problems understanding it, please let us know. [Wink]

Posted by John Pytlak (Member # 331) on 07-24-2003, 07:37 AM:

Links to some on-line tutorials on power supply design:

http://www.tpub.com/neets/book7/index.htm

http://www.tpub.com/neets/book7/27.htm

http://www.tpub.com/neets/book7/27b.htm

http://members.aol.com/_ht_a/RAdelkopf/rect.htm

http://www.eleinmec.com/article.asp?18

Posted by Ken Lackner (Member # 1002) on 07-24-2003, 01:27 PM:

Thanks for the links, John. I think I am beginning to understand it. I learned about half-wave, full-wave, and bridge rectifiers back in school, but we did not study negative supplies. Below is a diagram of a full wave rectifier similar to that shown in my text book. So I guess the only difference between that and what Irem is doing is that the Irem supply is negative? Am I correct?

So now I still don't understand something: why are the center taps connected to the bulb? In the circuit above, the center tap is grounded. That is the only use of a center tap that I am familiar with.

Posted by Gordon McLeod (Member # 33) on 07-24-2003, 01:39 PM:

The centre tap does not have to be grounded. But it is the low reference to the two outside windings But it doesn't have to be actually grounded

Posted by Phil Hill (Member # 371) on 07-24-2003, 04:35 PM:

Ken, in your diagram:

1) Draw a line between the xmfr center tap and the bottom of the load resistor.

2) Erase both "ground" references.

3) Replace the load resistor with a xenon lamp…ANODE at the top.

That's a basic " positive ", full-wave, xenon supply.

Remember the "grounds" indicate a common connection, not necessarily an earth or chassis ground.

NOW:

1) Reverse the diodes.

2) Invert the xenon lamp... CATHODE at the top.

That's a basic " negative ", full-wave, xenon supply.

>>> Phil

Posted by Paul G. Thompson (Member # 655) on 07-24-2003, 06:39 PM:

Very good explanation, Phil. I'll bet that will make sense to Ken without any problem.

Now, a question I have: Can you explain the theory of operation of that damned RA unit? I never understood how that worked. Apparently, some others are in the same boat. I know it has to be simple, but I am just missing a key.

Thanks

Posted by Phil Hill (Member # 371) on 07-24-2003, 07:01 PM:

Paul, it's a bit of a joke on my part...sorry. [evil]

There is no separate "boost" circuit. The power supply relies on a high no-load open-voltage for the "boost". Thus the boost circuit is also correct since it's the same as the main ckt. (It's all in their excellent transformer design & saturation)

That RA crap is nothing more than an automatic surge protector to limit the charging current to the filter cap.

>>> Phil

Posted by Michael Schaffer (Member # 1204) on 07-24-2003, 11:18 PM:

We had actual "boost circuits" installed on some of our FP30Es. I am not familiar with the details of the circuit, unfortunately. A friend of mine came up with the idea because he wanted to try using welding rectifiers. The experiment wasn`t too successful though.

Posted by Ken Lackner (Member # 1002) on 07-24-2003, 11:41 PM:

Paul: You are correct; Phill's explanation helped greatly.

Phill, Like this:

?

Pretend, of course, that the resistor is a xenon bulb. When I run this simulation and connect a scope, I only see a half-wave rectified signal. Why?

Posted by Phil Hill (Member # 371) on 07-25-2003, 12:08 AM:

Ken, it sounds like you are referring to circuit simulation in software rather than a "breadboard"?

Either way, the scope connection "dots" are located in the wrong place.... that would indeed only give you a 1/2-wave ckt and waveform since the bottom diode is not included in the scope-monitoring point of the load. You are looking at the voltage across the bottom diode....

The scope should be connected across the load resistor.

Mike, yes as a result of low line voltages in the field, many had a "real" boost ckt to ummm... boost the initial voltage. [Wink]

>>> Phil

Posted by Sam Hunter (Member # 1119) on 07-25-2003, 12:09 AM:

Because the diode acts as a one way valve and that there is no filtering of the rectified voltage. The simplest filter is just a large value capacitor connected between the rectifier out and return . Nominal values are anywhere between 50-100000uF depending on the design and use of the supply. The filter cap smoothes out the half wave waveform.

Due to my late night memory fog I have deleted the incorrect portion of this post and hope that FT will forgive my oversight [Confused]

[ 07-25-2003, 11:50 AM: Message edited by: Sam Hunter ]

Posted by Ken Lackner (Member # 1002) on 07-25-2003, 12:19 AM:

Yes, I am using Electronics Workbench. The dot to the lower left was not a scope connection, it was a connection for a grouding point. I removed all the grounds as you said, but when I went to run the simulation, a box popped up that said the circuit requires a grouding point for simulation. So I added one, without giving much thought as to its location. I moved the ground to the bottom of the resistor, and now I get a full-wave rectified signal.

Posted by Phil Hill (Member # 371) on 07-25-2003, 12:39 AM:

But Ken, your scope is ground referenced so the left dot was indeed the scope monitoring point... [Wink]

So, in that case you were looking at a 1/2-wave ckt...and got the correct waveform for it

>>> Phil

Posted by Ken Lackner (Member # 1002) on 07-25-2003, 12:49 AM:

Phill, now to make sure I fully understand how the circuit works, I followed your second suggestion and reversed the diodes. Indeed, I got a negative full-wave rectified signal. So I can see that it does work. But I still don't understand how it works. [Confused]

I return to my original question: As the current leaves the transformer, how can it get past the dioes if they are facing the way they are?

Posted by Phil Hill (Member # 371) on 07-25-2003, 01:14 AM:

Remember, the transformer "takes in" the current as well as put's it out. It has to in order to form a complete, closed circuit. (God forgive me for saying this... Kirchoff's Law) [Big Grin]

So, in the case of a positive supply, the diodes only allow current to flow "from" the xmfr. The center tap is the "into".

In the negative supply, the diode only allow the current to flow "into" the xmfr. The center tap is the "from".

The circuit is closed and complete only while a diode is conducting.

>>> Phil

Posted by Sam Hunter (Member # 1119) on 07-25-2003, 12:10 PM:

Please refer to my last but anyway on a center tapped full wave only one diode cunducts at a time as the ac between the center tap and the two other ends are 180 Deg out of phase. Refer to my somewhat better drawings for a visualization of current flow and phase relationships.

P.S. Phil, its been a while but I remember this shit and corrected myself [Wink]

Posted by Daryl C. W. O'Shea (Member # 1303) on 07-25-2003, 11:19 PM:

Jerkoff's Law! You shall never be forgiven, Phil.

You know this is probably one of the best uses of the upload feature so far... I guess you're forgiven then.

Posted by Ken Lackner (Member # 1002) on 07-26-2003, 12:43 AM:

Wachu got against Kirchoff's Law?

Posted by Paul G. Thompson (Member # 655) on 07-26-2003, 12:49 AM:

How you define Kirchoff's Law depends on your frame of mind and whether or not there is a full moon. [Razz]

Posted by Phil Hill (Member # 371) on 07-26-2003, 02:22 AM:

Daryl: That sound's like a character from "Flesh Gordon". [beer]

Ken, if your referring to me, I was merrily interjecting some humor cuz of my reference to "text-book" stuff.

I hope all the time that many of us have spent in trying to help you out, did.

>>> Phil

Posted by Michael Schaffer (Member # 1204) on 07-26-2003, 07:18 AM:

quote:
You know this is probably one of the best uses of the upload feature so far...

Indeed, and somewhow Sam`s diagrams seem to have a Picasso feel to me. I can`t explain why though. Maybe some day they are worth tons of money.

Posted by Sam Hunter (Member # 1119) on 07-26-2003, 01:34 PM:

[fu]

Posted by Ken Lackner (Member # 1002) on 07-27-2003, 01:36 AM:

Indeed, the time you have spend has helped me a great deal. Thanks!