Building EL-84 push/pull amp out of Zenith Radio Parts

[DRH1958]

Pretty sure Gonz is building a head not a combo, so no worries about that part. I have two pairs of the 6P14P-EV which is exteded life version=E, mechanically rugged= V. Agree with Chuck, the K version is for vibration service. They also have the thickest glass of any tube I own. Here is a listing for them, scroll down for the data. They are only rated for 14 Watts no matter the voltage. One thing that I think is overlooked is filament voltage. They are rated for 5.7-7 volts. Nice! I think it is a safety factor since they were made for the military. See if you can get the ones made in the '70s. They were made in the Reflector plant. Same as RFT? They sound great. I think they might have gotten a bad rap for fragility the same way JJs have recently gotten one for their octal tubes not holding up. Newer tubes just aren't the same as vintage tubes.6P14P EV EL84M 6BQ5 Matched Pair Gold Grid Strong Russian Tubes | eBay
Agree with Chuck, no real reason to push them to the edge for an extra watt or two. You won't hear the difference anyway and will save on tube life. From 16w to 18w is only .5dB louder. 320-340v is the max on all my EL84 amps. I run the screens about 5-8v lower than the plate.

[DrGonz78]

So I am researching away on the different ways to drop B+ using zener diodes. I came across the affordable option on fleabay of course and it looks like it is from china. If the price is low it must be too good to be true. However, I have to ask if anyone has used these types of 50 watt zener diodes from these sources. Should I just stay away from such stuff or should I take a chance on something like this?

1N3338B Zener Diode 50 Watt 82 V | eBay

So there is the version of just daisy chaining 4-6 5watt zeners in series from CT to ground. That version makes me very nervous that the zeners are going to get so hot that they desolder themselves from the circuit. I read more on the idea of using pennies as heat sinks too. Then I read about the mosfet with two resistors and a zener connected in series form CT to ground on the negative connection to the first filter cap. <<< See attached pic of something I drew up as an example.

So If I don't go with the big 50 watt version then yes it looks like mosfet time! I got plenty of heats sinks lying around here to make that one work out too.

[DRH1958]

Obviously, there are many ways to implement the zener method. I like the string of 5 watt zeners best because you have flexibility in adding more later(or removing). If you place them in series with the B+ rail, it seems to be a better way. If you put them in series with the PT secondary CT and you take your bias supply form one of the legs of the main winding, it will affect the raw AC voltage going to your bias supply and will change your bias voltage.
To put to rest the wattage/heat factor worries, let's assume you draw 125mA, a reasonable amount for your amp. Let's use six, 10v, 5 watt zeners instead of 12v like in the posted circuit. 10v X.125= 1.25W. In the posted circuit, you also have to consider ripple current factor, so add in another 50%, so 1.875 watts. Even adding an additional 50% for safety, you get 2.8 watts per diode, a little over half the 5 watt rating. You have dropped 60 volts and left the option to change it if you want. If you want to only drop plate voltage, just put them in series with the OT CT. Even less dissipation, so you could use less higher voltage zeners. See this listing, a dollar for two: http://www.ebay.com/itm/151695488519...%3AMEBIDX%3AIT

[Chuck H]

Obviously, there are many ways to implement the zener method. I like the string of 5 watt zeners best because you have flexibility in adding more later(or removing). If you place them in series with the B+ rail, it seems to be a better way. If you put them in series with the PT secondary CT and you take your bias supply form one of the legs of the main winding, it will affect the raw AC voltage going to your bias supply and will change your bias voltage.
To put to rest the wattage/heat factor worries, let's assume you draw 125mA, a reasonable amount for your amp. Let's use six, 10v, 5 watt zeners instead of 12v like in the posted circuit. 10v X.125= 1.25W. In the posted circuit, you also have to consider ripple current factor, so add in another 50%, so 1.875 watts. Even adding an additional 50% for safety, you get 2.8 watts per diode, a little over half the 5 watt rating. You have dropped 60 volts and left the option to change it if you want. If you want to only drop plate voltage, just put them in series with the OT CT. Even less dissipation, so you could use less higher voltage zeners. See this listing, a dollar for two: 2 Pcs 1N5347B 10V 5W Zener Diode by on Semi Lead Free | eBay

Just to add... Using my proposed circuit idea above you can let some cheap resistors do the heavy lifting (eminently more able to take the heat reliably). An added benefit to this is that even if a zener opens up you'll still get an approximation of your voltage drop and the amp will finish a gig!

[DRH1958]

I just re-read my post #18 and the last line about the zeners in series with the OT CT a more clear description would be "use fewer but higher voltage zeners", instead of "less higher voltage zeners".
+1 for Chuck's idea.

[DrGonz78]

So I ordered some 5 watt 10v and some 5 watt 15v zener diodes to get this ball rolling. I am willing to implement the idea of putting a resistor in parallel with the zener diode. So we are saying that I could make a series string of these same zener diodes and put a resistor in parallel to each zener, is that correct? Which value of resistor should I use in conjunction with each zener diode?

Edit: So let's say I draw 125ma and B+ is 390v. First target voltage drop is 380v, drop in 10v increment and I would put an 80 ohm 5 watt dropping resistor. Target voltage 375, drop in 15v increment and I would put a 120 ohm 5 watt resistor. Does that sound feasible? I only say 5 watt as 1.8 watt (or 2.8 watts as DRH stated for safety) is the minimum required wattage for the first voltage drop and as the voltage drops so will the wattage, right? I will stop there to see if I am even on the right track, thanks again ya guys.

[Chuck H]

A couple of things to consider are:

1) The resistor will drop more voltage under max current.

And

2) You probably don't want the diodes switching during operation of the circuit. They should always be dropping some volts.

So the resistor should drop almost all of the voltage under max current, but not more than 10V. If the resistor drops more than 10V the diode will be switching and that may add noise to the line (not sure of this). There would also be the effect of the resistance acting as a dynamic voltage drop, but to a fixed shelf when the diode switches. That would probably feel/sound weird. So we need the resistor to drop a tad under ten volts at max current.

It's surprising just how much current can be momentarily running through a tube amp. In my trials I've measured el84's (whole tube, plate and screen) drawing as much as 18W! This would be an A/B circuit where the tube would spend an equal time drawing only, say, 5 or 6 watts. So, using this figure as a likely max current I ran some simulations on PSUD2.

The results with a 5u4 rectifier are a little different than the results with a 5y3 so I figured for a value that would work with either. The zeners will be doing a tad more "work" with the 5y3. It's close enough that it shouldn't matter. A difference of about a volt, so seriously NBD.

Since you already bought the diodes it was necessary to value the resistors to the diodes action. But resistors (the inexpensive and easy to get ones) are only available in certain common values. Further complicating this, you'll want a certain type of resistor. Metal oxide. Since these are typically rated to handle the high voltages in the HV rail. You got lucky as in my simulations it looks like a 47ohm resistor will drop about 9.5 volts max in your proposed circuit. The resistor will be dissipating something like 2watts so buy a 5watt rating.

Soooo... To work with the 10V zeners you want 47ohm/5watt metal oxide resistors. One resistor parallel with one diode for each incremental step down of 10V B+.

Things do change a little as more voltage is dropped but not enough to matter with respect to the goal. You won't be putting ten of these units in series or anything.

EDIT: IGNORE THIS THREAD. I WRONGLY CONCEPTUALIZED THE ZENER/RESISTOR ACTION IN REVERSE. NEW INFORMATION TO FOLLOW!

The only reason I didn't erase it is to avoid confusion around my, and Dave's interaction below.

[DRH1958]

All of this is cool idea but even with the safety factors built in, the zeners are only around half their wattage rating. I have seen very few designs implement using both components. And without resistors, there is no possibility of switching noise. They are always conducting in reverse bias. You also need to make room for six, 5 watt resistors. Depending on how much room you have in your chassis, this could be a problem. The whole reason for getting 5 watt zeners is so they will handle the wattage. But it's your amp and Chuck's idea will certainly work too.

[Chuck H]

All of this is cool idea but even with the safety factors built in, the zeners are only around half their wattage rating. I have seen very few designs implement using both components. And without resistors, there is no possibility of switching noise. They are always conducting in reverse bias. You also need to make room for six, 5 watt resistors. Depending on how much room you have in your chassis, this could be a problem. The whole reason for getting 5 watt zeners is so they will handle the wattage. But it's your amp and Chuck's idea will certainly work too.

I get the point. Why trouble with the resistors, figuring for them and making room in the chassis if the circuit has already been managed by the use of series diodes? And I honestly only have two reasons...

Historically and IMHE resistors might be a tad more reliable. And since they would be doing most of the work, they offer a more reliable component to take the load off the diodes.

With the resistors in there, even if a diode goes open the amp will continue to work. I can think of a couple of gigs I've done where this sort of feature on a circuit would have been a great help.

So... Pro's and con's.

[Dave H]

So the resistor should drop almost all of the voltage under max current, but not more than 10V.

10V is the zener voltage isn't it? If the resistor isn't a high enough value to drop more than the zener voltage at max current then the zener will never conduct and could be removed as all the voltage drop will be resistive. If you don't want any resistive 'sag' you'd have to set it up so that the resistor drops the zener voltage (or more) at the min (idle) current. Perhaps it would behave more like a tube rectifier if the resistor was sized to drop the zener voltage at some current between the min and max current?

Edit.
Or you could just use a resistor without a zener. I used a 220R resistor to drop the voltage from 375V to 330V. It reduced the power from 20W to 15W but it sounds better and doesn't cook the tubes.

[DRH1958]

Yes, you make good points. Always good to explore pros and cons since this is the essence of the amp designing process. There always seems to be a trade off. There seems to be no absolutely best way to do any of it. More power vs. longer tube life, fixed vs. cathode bias, etc. Always fun to learn new things and there is ALWAYS something to learn if only we are willing! I guess it's up to Gonz.

EDIT: Dave H and I were posting at the same time and what he said was bouncing around in my head also.

[Chuck H]

10V is the zener voltage isn't it? If the resistor isn't a high enough value to drop more than the zener voltage at max current then the zener will never conduct and could be removed as all the voltage drop will be resistive. If you don't want any resistive 'sag' you'd have to set it up so that the resistor drops the zener voltage (or more) at the min (idle) current.

The circuit in question is a resistor parallel to a reverse zener in line with the HV rail. Not a load. Think about it... Suppose a 10V zener. The zener will see to it that there are 10V less on the other side of it. If a resistor is parallel to that, but only drops 9V, there will be no sag or zener switching. If the resistor drops more than ten volts at any given time the zener will switch and the resistor will do any work, sag an all, for voltage above the zener voltage.

[Dave H]

... Suppose a 10V zener. The zener will see to it that there are 10V less on the other side of it. If a resistor is parallel to that, but only drops 9V, there will be no sag or zener switching.

There won't be 10V less on the other side of the zener unless the current through the resistor multiplied by the resistor value comes to more than 10V. When the resistor has 9V across it so does the zener (they are in parallel). At 9V the zener is non conducting which means all the voltage drop is down to the resistor so there will be resistive sag. When the zener starts to conduct the voltage across the resistor is clamped to 10V so there will be no more resistive sag.

[Dave H]

Perhaps it would behave more like a tube rectifier if the resistor was sized to drop the zener voltage at some current between the min and max current?

Talking to yourself is a sign of ness isn't it?

I was thinking that a resistor added in series with the zener could make the circuit into a two slope approximation of the way a tube rectifier's slope resistance reduces with increasing current. (See Chuck's sag thread)

[DRH1958]

There won't be 10V less on the other side of the zener unless the current through the resistor multiplied by the resistor value comes to more than 10V. When the resistor has 9V across it so does the zener (they are in parallel). At 9V the zener is non conducting which means all the voltage drop is down to the resistor so there will be resistive sag. When the zener starts to conduct the voltage across the resistor is clamped to 10V so there will be no more resistive sag.

This is the correct explanation of the operation. Unless there is 10v or more across the zener, the reverse breakdown voltage will not be reached and it cannot perform the function as intended and be an open circuit, forcing all the current through the resistor. I actually have never seen a design with resistors in parallel now that I think about it and this is why.
The zener method I proposed is because there will already be sag from the tube rectifier. No more is needed. This is why I didn't propose a large resistor. The large resistor is good for SS rectifiers where sag isn't a concern but you don't mind having it or you want it.