Tube rectifier emulation

[txstrat]

My new (6G3) built will be equipped with a GZ34 tube rectifier I've already ordered. Being scared of a gig with a failing tube and no replacement I thought of a solid state rectifier as a spare. Since there is less voltage drop I fear it would blow the filter caps.
Recently I've read about a replacement with solid state diodes and a resistor in series with each of them to emulate the tube recto.
Could anybody help me out with the values of the diodes and resistors for a GZ34 replacement and what wattage the resistores should have?
Is this the same way a copper cap or the Port Arthur rectifier is working?

Thanks

Matt

[Robert M. Martinelli]

Hi Matt,
I never opened a Weber copper cap or the like, but I think all you need is to take a look to the datasheets : you need a total Rd ( differential resistance ) much the same as a rectifier tube, so you' ll need to add an external resistor in series with the diodes ( 1N4007s will be ok ), as to the value, take a look to the 5AR4 ( or 5U4, or the tube you want to "emulate" ) datasheet and examine the "Voltage vs. Current" curve, then choose the appropriate resistor value according to Ohm's law; This value will be the sum of the diodes' Rd plus the external resistor you're adding.

As to the resistor's power rating, once the current and the maximum voltage drop across it have been stated, multiply the drop by the current ( which will be in the 200-300 milliAmperes range ).

This kind of arrangement should cause the amp to "sag" like a "tube rectified" one.

Hope this helps

Best regards

Bob

[txstrat]

Hi Bob,

I'm not very expierienced in reading those data sheets but I believe I found the right value for the resistor(s). The max. dc current of the GZ34 is 250ma at 350v according to the data sheet.

Input voltage to the rectifier will be 330v = rectified around 465v. I go for 390v at the plates that would mean I have to add a 300 ohms resistor for a voltage drop of 75v (= 18.75w at 250ma). Those are theoretical values (rectifier Vin * 1.41 = Vout) with no load.

1. I believe with the tubes in, the voltages will drop more than above stated (I think of 422v rectified) = voltage drop of around 32v to get to 390v. That would need a 120 ohms resistor at 8w
2. Do I have to calculate the wattage of the resistor(s) with the full amount of current? Since the PP 6V6 would never draw that much.
3.Could I also put two resistors of the half wattage in series with each diode?

If so I would decide for 56 ohms at 10 or 15 watts each.


Matt

[Robert M. Martinelli]

1- yep, the voltage will probably be somewhat lower, but that' s not the point, the point is you want the amp to "sag" ( =voltage to drop ) under dynamic conditions ( = while playing ).
2-yep, that will put you on the "safe" side, because the resistors will never work at their rated wattage.
3-yep, provided the voltage drop on one diode is half of the desired total, so the series resistor's value will have to be halved as well.

As to reading the datasheet, resistance is the slope of the line describing V/C. Given the equation V=R*I and translating it into y=mx ( line equation in a XY plot ) R can be considered as the "m" coefficient.

Hope this helps

Best regards

Bob

[txstrat]

Hi Bob,

would you mind having a look at the pic I posted.
Paralleling two resistors would mean half the value. I don't know if it matters here, cause both are coming from different diodes.
I wonder if the second rectifier wwould have the same effect on voltage drop and compression like the first. (the values are just for the understanding)

thanks

Matt

[Robert M. Martinelli]

Hi Matt,
the voltage drop would be the same in both cases, only I don't understand why you put four diodes instead of two, as they're shown wired in series....this way the second diode of each series does nothing but adding its Rd to the equation as the first diode already rectifies the incoming sine wave....Only two diodes can suffice with the configuration you have drawn, simply connect the anodes to the HT secondary lugs and the cathodes together ( one resistor ) or each cathode to its dropping resistor and the resistors together on the other side ( two resistors ).

Oh....about the values..... I think the real resistor(s) values will be more likely in the 50 to 100 Ohm range ( depending on the desired drop ), not 50-100K...

Lemme know how it works.....

Best regards

Bob

[txstrat]

Thanks Bob,

would you allow one more question? According to ohms law I can calculate the voltage drop by resistance*current. Do I have to calculate with the current the tubes and grids are actually pulling (how would I know this value) or with the current the transformer is able to deliver?

Matt

[Robert M. Martinelli]

Hi again Matt,
well, your actual +B current will vary from a minimum in quiescent conditions=not playing ( this current will be the sum of the bias currents for all the tubes, but, since the preamp tubes have bias currents in the 1 mAmp range, most of the quiescent current will be flowing through the output tubes ) to a maximum ( playing at maximum volume ) - you should use this last value, and not the current the tranny could deliver. Being the Ohm's law linear, playing at higher volumes will cause more drop and "sag" ( the tubes will "saturate" earlier due to the lower +B ).

If you have a multimeter with "min-max memory", you could easily measure this maximum current, and add the proper dropping resistor to suit your needs ( = create the exact amount of drop and "sag" you desire ).

Best regards

Bob

[Robert M. Martinelli]

Oh, I was forgetting....

about the layout....did you really intend to use two diodes in series or was it a ( mis-drawn ) Graetz bridge rectifier? I' m asking this because I see you use the sqrroot of two ( 1.41 ) in your formulas....

If so be careful because using a Graetz bridge the CT of the PT cannot go to GND ( like shown in your layout ) if the negative of the bridge is grounded.....( it would be much like shunting half of the PT to ground through half of the bridge, this would result in the bridge rectifier to be destroyed ).

Regards

Bob

[txstrat]

Well, because of the center tap I'm about to use two diodes (like in the pic - only one on each side). Isn't the fomula of 1.41 the same with bridge recto or two diodes as recto? If not, what would the formula be for two diodes?

Matt

[Robert M. Martinelli]

Hi Matt,
I took a look to a GZ34/5AR4 datasheet and I think 82 Ohm would be good as a starting point for your resistor ( or one 39 Ohm resistor in series with each diode ). You' ll have some 20V dropping at 250 mAmps, so you' ll need a 5 Watt resistor ( or two 3-watters ). For your peace of mind you could even use 7 or 10 watt resistors.

If you actually have the center tap and are willing to use two diodes as a full wave rectifier the 1.41 still applies, I simply was not sure about the layout you were willing to build....

Let me know how it works ( and sounds ) - I might build something similar on ( on a socket ) and try it on my 1964 AC30 TB

Best regards

Bob

[txstrat]

Actually I'm planning to build it on an old tube socket as well. Do you already have an idea how to manage that? I don't wanna leave the socket open, due to the high voltages but haven't had a clue yet how to cover the parts in the socket.

Matt

[Robert M. Martinelli]

I would think to fit a cylinder to the socket, to insulate the thing both thermally ( the resistor(s) will heat up somehow ) and electrically. Teflon would be great as it can withstand temperatures up to 300 deg F, and maybe something more.

[txstrat]

Great idea, only thing I don't know where to get teflon cylinders.

BTW I thought of two 120 ohms resistors. Due to the current for two 6V6 tubes a figure of 250ma appers to high IMHO. I was calculating with 120ma (still high for 6V6 ?). A resistance of 240 ohms would then cause a voltage drop of 28.8v. Assumed I have a rectified voltage of around 422v under load, the voltage drop would leave me with 393v and that's just what I'm going for.
The (2) resistors would need 3.5 watts. I think I would be save with 5 watts or maybe 10 to make sure.
I didn't calculate with the GZ34 values cause that tube is also able to deliver current for two 6L6. Too much for my built, practically.

Matt

[Robert M. Martinelli]

Well, I used the GZ34/5AR4 differential resistance because a 5AR4 is used in the original design, the 250 peak mAmp were only a figure ( or maybe I was already thinking about modifying my Vox....powers of the unconscious ) , your amp will more likely be in the 100 mAmp range, so your values seem a good starting point.

Have fun and come back with the results....

Best regards

Bob