1. ## Tube rectifier emulation

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

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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.

14. 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

15. 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

16. There's a thread on fenderforum.com which has some crossover to this
http://www.fenderforum.com/forum.htm...-10-2905:39:52
My contribution was to measure current draw with 6V6s and 6L6s with 5Y3 and GZ34, excert regarding 6V6s below:-

'6V6s, bias them to 19mA, B+ 458Vdc. OT can now supply a 8k or 4k load, depending which tap the 8ohm resistive load is on.
No signal - 60mA
8k Max sine - 160mA (14V into 8ohms)
8k Max square - 243mA
4k max sine - 167mA (11V into 8ohms)
4k max square - 257mA

Replace GZ34 with 5Y3, bias to 21mA, B+ 412Vdc.
No signal - 62mA
8k Max sine - 135mA (11V into 8ohms)
8k max square - 190mA
4k max sine - 137mA (9V into 8ohms)
4k max square - 198mA'

So, when you crank a pair of 6V6s, they can draw a lot of current and so you need to plan for 250mA to avoid your rectifier turning into a smoke machine. Peter.

17. Originally Posted by voxrules!
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 ).
Bob
Yep two diodes can suffice, and if they are 1000V 1A diodes, they probably will. But putting a couple in series like tx strat had shown does spread the load between the series diodes and offers backup if one of those diodes shorts (if that is a concern), and (FWIW) if you put a 600ish V film cap (like .01uf) in parallel with each diode in series, you can spread the load more evenly.

18. Originally Posted by tubeswell
putting a couple in series like tx strat had shown does spread the load between the series diodes and offers backup if one of those diodes shorts
Hi TW!Hi Peter!
I agree about the "backup" issue, OTOH using 1N4007s ( 1Amp continuous -1000V ) a diode failure seems a very unlikely event....

What I don't understand is how can this series configuration spread the load between the series diodes, as they're crossed by the same current, as soon as the first diode has cut half of the sine wave the second only adds its series Rd and its typical 0.6V drop to the equation.

The second "series" diode of each pair will work at conditions which are only a tiny bit better than the first, because voltage will be, say, 1-2 V less, but the current will be the same, so I still think it's not really needed.

Peter, thanks for adding your current measurements, surely using a 250 mAmps figure ( the one I used initially ) puts Matt on the safe side, so all that can happen is that the drop is less than expected if the current is less than that ( the "simulator" will "sag" less ) but there will be no safety issues.

Best regards

Bob

19. Hello all,

The reason for the two series diodes is that the required PIV in this circuit is equal to the sum of B+ and the transformer's peak output voltage. This works out roughly twice the peak output voltage when the power supply is lightly loaded.

So, if the B+ rail gets over 500V unloaded (say you turned the amp on and never used the standby, so the heaters are cold and the power tubes draw no current) then the diodes will see more than 1kV peak, hence 1N4007s may blow up.

When you put diodes in series, the PIVs add. They share voltage in series because each diode leaks a little current when its PIV is exceeded. So for 50 watt amps, we want two 1N4007s in series to get 2kV PIV.

Just to reiterate, the diodes are placed in series to share the reverse voltage. In the forward condition they both conduct and drop about 1V, and the 2V total drop across them is just as negligible compared to 450V as the drop of a single diode would have been.

20. Hi Steve,
I understand your point and I agree with you, having two diodes in series sums the peak voltage they can withstand, what I was questioning was that having two diodes was helpful in spreading the load between the diodes....the diodes are crossed by the same current and fed approximately with the same voltage.

I don' t think the extra diodes are a "must" in this particular design ( 6V6s work at lower voltages than 6L6s or EL34s ), but they cost very little so adding them would be advisable and would add "peace of mind" to the design.

Best regards

Bob

21. Peter,

I understand your post about the peak voltage and I'm a little concerned. I'm now thinking about using resistors with higher wattage. I didn't know that a peak could go up that far. My recently bought tansformer has a secondary of 2 x 330v CT at 120ma. Do you think it can deliver peaks at 250ma without blowing a fuse? I was thinking of a fuse at 1A slo blo.

Matt

22. What Steve said is what I meant by 'share the load', I was just too lazy to elaborate :-)

Sorry about the confusion it caused

23. Hi TW,
maybe I owe you some apologies too, I should have understood you were concerned about the PIV rather than the "load", but, as I told already, maybe my English is not as good as I thought, and this affects my understanding, ( or maybe I was too lazy to elaborate on that too ).

Hi Matt, a 1 amp slo-blo seems a little bit too much IMHO, I would start with a 500 mAmp slo-blo and maybe move to 630 mAmp if needed.

Well, I think we have analyzed Matt's problem pretty thoroughly, so now he should have the means to go ahead and come back to us with the results....

Good luck Matt!

Bob

24. Re the transformer winding, 120mA seems a bit low to me. Best to confirm with the transformer vendor that it cope in your circuit. BTW those currents aren't peak, but continuous steady state - when you're driving the output stage hard, those are the kind of currents you will be dealing with.
Regards the fuse rating, is that 1A for the primary or secondary? Secondary is implied, and I would go with along Bob's suggestion of 0.5AT.
Don't use resistors anywhere near their full power rating, as most all types get worryingly hot and smelly, get them at least 50% over your max calculated dissipation. Peter.

25. Peter,

the transformer is a Hammond replacement for the deluxe/deuxe reverb amp with original specs plus an additional 50v tap for the bias supply. You might have a look at the Hammond website - it's the BX290/BEX290 transformer.
They don't have the specs in their data sheets but on the transformer is a sticker with the values (660v CT @120ma, 5v@3A, 6.3v@3A).
I always thought 120ma is enough for a PP 6V6 amp.
The amp runs a GZ34 rectifier, two 6V6 and 3 12AX7. I believe it should work at least like the original (with all the disadvantages the original might've had).

I thought of putting a 1A slo blo in one of the primary taps and 0.5A slo blo in the B+ line right before the stand by switch.

Regarding the resistors: since there are not all kind of values (wattage) available I believe I'll go for 25W ones to be on the save side.

Matt

26. Matt, that should be fine then. The spec on the lable most likely means that with higher current draws it will lose regulation ie voltage will drop = power supply sag, but it should be plenty robust. The traditional / best place for the B+ fuse is in series with the B+ CT ground return.
Other best practice protection I like to fit is a fuse in the cathode ground return for each power tube, and a 1/4watt 47R resistor between the heater CT and ground. The good thing about the cathode fuse is that it can be quick blow, and as it is specific to each tube it can be smaller than the B+ fuse, so offering much better protection from the most common failure mode (tube shorts etc). The heater CT resistor will blow should there be a plate to heater short inside a power tube.
25 watt resistors tend to be the metal cased type which need heatsinking, might be better with 2 x 11 watt ceramics. Or just buy a non plug in copper cap from Weber? Peter.

27. Wow, that's a lot of fuses.

The cathode fuses and B+ fuse are fairly redundant, since the same current flows in the plate circuit as the cathode. Granted the cathode fuse can be half the rating, since the current divides in two. Or one-quarter for a four-tube amp.

The suggestion to fuse the heater CT is a little weird. If the heater ever did short to the plate, then the heater CT fuse would blow and the heaters would go up to B+, taking out the heater-cathode insulation in all of your tubes. So I think you want to keep the heater circuit solidly tied to ground, which would blow the HT fuse and minimize collateral damage.

I prefer to put the fusible resistor on the bias tap, since burning out the bias tap would trash an otherwise perfectly good transformer.

28. Originally Posted by pdf64
Or just buy a non plug in copper cap from Weber?
The more I think about it I tend to buy one and be satisfied.
On the other hand I was lately thinking of putting the diodes and resistors in parallel to the tube recto, the way I read about the Port Arthur rectifier. That would reduce the "stress" on the GZ34 and if the tube fails to rectify you could pull it and still play with the diode rectifier.

My idea of the fuse in the B+ line was not only for a short inside a tube but also for a cable short inside the chassis caused by vibration.

Matt

29. I built a Deluxe style amp (more like a 5E3) w/ a 120mA sec, and used a sag resistor.

I think there's a lot of variables so it's pretty hard to guesstimate its value.
In my case, I ended up w/ a 250 ohm 10W that dropped the B+ from 420V to 390V (IIRC, that was long ago, and I sold the amp since).

I think 100 ohm is a good start. Quite like a 5U4. I would try that and start from there.
I would also use the big 25W encased resistors, they're easy to screw on the chassis and can handle massive power.

30. Why not just go ahead and use the Weber? He's already done all this grunt work, plus you can pretty much order any voltage drop and current limit you want.
He uses a piece of copper pipe to house the unit.
I read last year a complaint from a guy who claims it overheated and dies, but he may well have been using the wrong part for the job...

31. Well,
don't want to start a philosophical debate, and don't want to question the usefulness of having a certain part readily available, but the reason why, IMHO, Matt should NOT buy a "ready to go" part is the very same that keeps all of us reading and posting on this forum.... the will to experiment, to learn, to discover how a certain thing work.....to find different solutions to a problem, to invent something different.....they call it evolution.

If Paul Bigsby, Les Paul and Leo Fender wouldn't have tried something different we would still be playing clumsy and prone-to-feedback guitars instead of the sleek, comfortable and efficient pieces of gear we're using today, and music's history ( and history itself ) would have been completely different. If Jim Marshall wouldn't have modified some Fender designs to his liking......If Seth Lover would have thought the single coil pickup to be the final stage of pickups' evolution.......If Floyd Rose would have thought the Fender bridge to be perfect.......and so on....and on......and on.......

Best regards

Bob

32. I can't argue with that, I hardly own anything I haven't tweaked, ruined or done the hard way just to see! It's always a balancing act between wanting to see "just how deep the rabbit hole goes", and just putting the iron down and Playing Already... a real tug of war...

33. That's what I'm thinking too. Sure I could go and buy one but this way's more fun too.
I decided to go for four diodes with two 220 ohm resistors at 11w put in parallel with the GZ34 tube (bought them today).
From the sheer calculating factor that would mean (assumed voltage of 420v) I have a voltage drop of 44v at 100ma (diode). While 6V6 have a max. current draw (typical) of 92ma that would be 40v and leaves me with 380v at the plates.
Slightly higher voltage with less current draw but thats to be expected. Even with more current draw the 11w should be enough.
Can tell you more when I finished the project and fired the amp up.

Matt

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