Reviving an old full wave rectifying valve

[ppppenguin]

Alan, perhaps I'm being a bit dim but I still don't understand the reasoning for loading an unused winding. Whether it's elevated to HT or not is irrelevant.

Perhaps our resident transformer expert, Ed DInning, could cast some light on the subject.

[Brianc]

I tend to agree with Jeffrey, Alan. I see no reason why a mains transformer low voltage winding should not be left with no load. We are not dealing with impedance issues here and any small load taken off the primary is surely a good thing where a vintage Xfmr is concerned. I await Ed's comment with interest.

[SurreyNick]

Morning Nick.

It doesn't matter which Diagram you use but your Diodes are backwards. 'Conventional Current' flows from the Anodes to the Filament so they should 'Point' in that direction.
Transformer Windings don't like being left floating so it's best to give them a bit of Load.

Alan

Thank you Alan.
Is this (pic) correct now?
I'm of a mind that if a load will do no harm then it's inclusion is worthwhile as a belt-and-braces measure. But we will see what others have to say on the matter.
The question that still confuses me is which of the two options 'A' or 'B' to adopt. You say either, but Brian mentioned in post #3 that I must examine the wiring to see which heater pin to use because it must be the pin to which the reservoir capacitor is connected. Brian can you help me out on this with a little more explanation please and how I go about identifying it?
Thanks
Nick
   

[Brianc]

First, identify the mains Xfmr winding wires which go to the heater pins (3 & 4). One pin will have an extra wire on it which leads either to the reservoir cap (or to an input choke). It makes no difference which as the heater pin with the extra wire will be the HT +ve feed. It is to this pin to which the diode cathodes MUST be connected.
I see that you have drawn a 100R resistor from pin 3 to 4. This will dissipate nearly 0.75 Watt so you must use at least a 2W resistor as it is in a confined space - I really don't think that that is a good idea. Lets wait to see what Ed says.

[pwdrive]

100 ohms, 4 volts....

Lawrence.

[pwdrive]

If it were me I would just rig up the valve base as shown in the picture having first established which heater connection is the HT +ve out.

Use a 4 ohm 7 or 10 watt resistor in place of the 100 ohm resistor shown in the picture, that will simulate the heater current draw of the U10 rectifier valve that Nick says he has, once it can be established that there are no HT shorts etc or no serious heater winding/connection leakage to HT-ve in the PSU then he can then try the U10 rectifier.

Lawrence.

[Nowhere-Man]

In another (wanted) thread seeking a full wave rectifying valve to replace a duff 1821 ppppenguin suggested hiding a couple of 1N4007 and a limiting resistor in the base of the duff valve so it could still be used. I have been giving this some thought and I'd like to have a go. Someone even suggested hiding a tiny 6v bulb in the base too.

Although I have acquired a U10 I don't want to just pop it into the Mullard PM Unit and hope for the best, because it would reduce me to tears if it went 'Pop'!

I have never attempted anything like this and so would appreciate some instruction.

Do I remove the valve base from the balloon of the 1821, and if so how do I do this without breaking the glass? They seem well attached to each other.
I'm assuming I will then have bare wires projecting from the valve. Do I snip these off?
How do I then wire up the diodes and resistor to the pins of the B4 base? I am assuming I'll need 4 diodes because the valve was a full-wave rectifier and that's also what the HT Unit takes.

I am very conscious all I ever seem to do is ask for help on this forum and contribute very little but I promise I did have a look on Google first to see if I could find an instructable but turned up nothing.

Nick

Apologies in advance that I simply don't have time to go through all the responses to this thread (this public place closes at 17.45. So, given time is short, I'll just add a few ideas about rectifiers.

Firstly I'm currently working with a U10 so am familiar with the tube itself. I spent some time setting up silicon rectifier alternatives so here's how..

O.K. here goes and very precisely. For some this may be a bit simple:

It's always a very good idea to remove all your tubes first. You stated you fear something catastrophic could be upsetting which is why by removing all the tubes you can test your rectified D.C. in peace and quiet. Output tubes can be stubborn to remove but the rest are much easier. I put my fingers dead on the base of the tube while pushing down with my knuckles on the chassis. You have to ratchet up the force with some care and out it will come.

A U10 has 4 pins. At any rate, it's always wise to be absolutely clear where are your transformer anode wires and where your cathode wire is. The rectifier tube cathode wire is easy to locate - it leads straight from a tube pin to the big reservoir can As a point of interest, if you should test the cathode wire with a digital multimeter when a rectifier tube is working you should get 0 volts A.C. but around 380 or around D.C. (I did this test last week).

Confirm where the transformer anode wires are (the cathode wire leads to the HT  can). Both anode wires from the transformer will go to the silicon diode anodes. There are two diodes so for full wave rectification the two transformer wires will lead to both of the diode anodes.

Output will be now at the joined cathodes (bridged to make one output) and you can measure voltage if you wish. It ought to be 1.414 times the A.C. voltage of the transformer.

At this point I put in parallel with each diode a 10 Nanofarad capacitor between anode and cathode of the diode each side. That is also 0.01 Microfarad if you wish. This reduces the high voltage at the cathode somewhat through capacitive and inductive reactance.

Finally the choice of series resistor. No harm starting with 1 K to see what the voltage will be. It must be at least 5 watts. If voltage is too low, reduce the Ohms. I found 220 Ohms worked just right.

Check also the grid bias resistor to avoid any excess positive bias on the grid.

Locate correct output valve anode voltage in schematic and use that as a guide. When all seems well you can connect the H.T. to the H.T. can. Then put tubes back in.

Output will be at the cathode. These leads are indicated by a stripe.

CAUTION: All of this work involves very high voltages and none of it should be undertaken till absolutely certain. I only outline the mechanics of it all but how you approach it requires a personal understanding. Matters of load on the system I left out of it.

[Nowhere-Man]

Morning Nick.

It doesn't matter which Diagram you use but your Diodes are backwards. 'Conventional Current' flows from the Anodes to the Filament so they should 'Point' in that direction.
Transformer Windings don't like being left floating so it's best to give them a bit of Load.

Alan

Good point that, Alan. Funny enough I'd done my own post before I then noticed the matter of load has been raised by you here. Personally I did all my own testing with no load on the transformer although, yes, it does make sense to consider the load factor. One thing I find a bit unavoidable is when, on testing a new set for a trial, you may wind up with lit filaments but no H.T. or anode load (because the H.T. supply had a fault). That worried me when it happened as I feared it would have damaged my output tube. Luckily it didn't but probably didn't do it any good either.

The golden rule I learned on rectifier flow:

AC supply braided wire from transformer to tube anode blocks the negative cycles. Only + cycles pass and appear as a voltage (pulse) at the cathode. That is, cathode is the + pulse. To clarify I tested this on both A.C. setting on a digital meter and on D.C. Only D.C. showed the high cathode voltage with reference to chassis.

This cathode wire will go to the H.T. can +terminal.

Works the same with silicon. Transformer leads connect to anodes and output in a DC ripple at the cathode.





Note, I refer to voltage as the electron current always flows from cathode to anode (heated filament free electron cloud attracted to anode H.T.)

[Nowhere-Man]

Final point, on starting up it's normal for the voltage to drop on the H.T. can + terminal as the output tube "kicks in".

[Brianc]

100 ohms, 4 volts....

Lawrence.

I don't understand what you mean, Laurence. Perhaps I am being stupid or something!
A 100R resistor across 4V will dissipate .16W (not the value I stated, very simple maths too ).  However, a 4R resistor, 4Watt. This is a lot of energy in a very confined space inside the valve base and it (and the diodes) will get very hot.

Another way of looking at it:
If leaving a winding o/c on a normal mains transformer is bad, how do these general purpose Xfmrs with multi-windings survive in use with only some of the windings used? They do.
If an ordinary mains Xfmr primary is connected to the mains with no loads on its secondaries, does it overheat or fail? They don't.

These are the questions to which the answers suggest that no "load" resistor is necessary.