High B+ using SS rectifier

[expat]

I believe its 2.69ma for all three - 12ax7 is 1.2ma, 6v6 is 1ma, 6at6 is .49ma.

 

[Cqwet Dbdfte]

If the 6V6 is used to drive a speaker thru a transformer, it would draw more than 1mA under load.
I'm guessing 45mA no signal, with 12.5V on the cathode.
https://frank.pocnet.net/sheets/127/6/6V6.pdf

 

[expat]

Ok, scratch that. What I seem to be hearing from AI is that these tubes would draw 1.24amps under load. That's the best I got.

 

[Cqwet Dbdfte]

That AI stuff cannot be trusted. Look at the datasheet above.
You can add a switch to the power transformer center tap and turn it on after tubes are warmed up, the internal resistance of the power supply works is series with the tubes (warm up) to form a voltage divider. Say hello to Mr. Ohm's law.
The unloaded powersupply will show a higher voltage.

 

[expat]

I say hello to him all the time, he just sneers at me. I'm just not at the level yet where it helps me much. Like - not knowing the current draw under load. I know Ai can't be fully trusted, but it is showing the math and how it arrives at the results. If I fully trusted it, I wouldn't be on a forum!

 

[expat]

Here's what it reported:

To accurately calculate the voltage drop across the series resistor with all three tubes (6AT6, 12AX7, and 6V6) in place, we need to consider the total current drawn by all these tubes combined.

Let's sum up the currents we estimated earlier for each tube:

• 6AT6: 0.15A
• 12AX7: 0.3A
• 6V6: 0.79A

Total current draw total=0.15A+0.3A+0.79A=1.24A
Now, we can use this total current draw to calculate the voltage drop across the series resistor.
Given:
Initial voltage: 403.2V
Desired voltage: 350V

Voltage drop =403.2 V−350 V=53.2 V drop

Now, using Ohm's Law you would need a resistor with a value close to 42.9 ohms.

 

[Cqwet Dbdfte]

...and 1.24A x 350V = 434W with a 5W speaker?
Realistic?
Looks like heater currents.
Plate current I'd say 1 or 2mA each for the AX7, 1mA for 6AT6, these have a 10K series dropping resistor, 6V6 24mA? The third grid is floating, this grid is normally connected to the cathode. (nevermind its internally connected).
The schematic already has a 100R in the PSU output.

 

[expat]

Oh man, if I only knew! Just trying to get this sucker rollin' and learn in the process!

 

[Cqwet Dbdfte]

You could power up the heater transformer separately on 120ACV and use the variac to bring up the +B.

 

[expat]

Not sure I follow you. It's all one PT and has 5v and the 6v heater taps on it. I know on my schematic they seem separate, but its because I use an old version of Circuit Maker.

 

[Cqwet Dbdfte]

Can you get another 6V supply?

 

[expat]

I reckon I could make one or something, but I guess I'm not quite following what you mean.

 

[Gus]

Look around GEOFEX.com.
There is a section that shows how to use a Zener at the center tap to lower B+ voltage

 

[Cqwet Dbdfte]

once the tubes are heated up you can power up the +B slowly with the variac to desired +B voltage, then measure cathode voltages in your circuit, which are the currents of each tube, and the 100R dropper would also give the total value.

 

[Cqwet Dbdfte]

The tubes characteristics are related to heater temperature, so a low heater voltage will not give a correct answer...

 

[expat]

I'm hoping there might be a good guess on a starting resistor value to try out in series and see where that lands me voltage-wise before building a 6v PS. ...and my buddy didn't leave me the variac!

 

[amplexus]

Here's what it reported:

To accurately calculate the voltage drop across the series resistor with all three tubes (6AT6, 12AX7, and 6V6) in place, we need to consider the total current drawn by all these tubes combined.

Let's sum up the currents we estimated earlier for each tube:

• 6AT6: 0.15A
• 12AX7: 0.3A
• 6V6: 0.79A

Total current draw total=0.15A+0.3A+0.79A=1.24A
Now, we can use this total current draw to calculate the voltage drop across the series resistor.
Given:
Initial voltage: 403.2V
Desired voltage: 350V

Voltage drop =403.2 V−350 V=53.2 V drop

Now, using Ohm's Law you would need a resistor with a value close to 42.9 ohms.

This is why you can’t trust AI for anything technical.

That AI can’t differentiate between heater currents and plate currents, and it certainly can’t tell you to consider the difference between quiescent current and current drawn when signal is applied.

Stop using AI and put a small amount of time and effort into understanding how to use ohm’s law and the relationships it describes, how to read a datasheet- esp for tubes, and a basic understanding of the parts of a tube and how current flows in it (ie: so you know basics like the scale of the current flow of plate versus filament versus screen and don’t get fooled into huge mistakes by a dumb AI algorithm)

Incidentally most old 6V6 tubes would tick away quite happily at 400+ volts for a good long time as long as plate AND screen dissipation were carefully considered in the circuit setup. Many current production tubes will also function similarly tho at a shortened lifespan. The current production JJ 6V6 is spec’d at 500VDC on the datasheet and I can tell you from a significant amount of experience that it’ll take 450-475vdc all day long for a very reasonable lifespan. Of course it’s constructed quite a bit more like a 6L6 in a smaller bottle- but still.

Consider that almost every classic Fender circuit and all the modern reissues that use 6V6’s are running between 390-440VDC loaded on the plates- and these amps are produced by the thousands and are generally running the cheapest chinese or russian tubes Fender can rebrand. And they all function reasonably well with regards to tube life. (Of course they’re biased cold as ice from the factory, but that’s easily corrected and they still work fine)

If you used a current production JJ tube in your amp as-is, biased it respecting dissipation and keeping it to 55-65%, AND throw some reasonably sized screen grid resistors on there (~1kOhm) following a decent dropper for the screen node- you’ll probably make 25-30W and the tubes should go for a good long time.

If you wanted to use any current production tube- esp something that sounds a little more like a classic 6V6, then sure, knock your B+ down to 425-450 with a big ass zener, do all of the above, and call it a day.

This is a guitar amp, trust me when i say not only do you not need regulation- but adding a regulator circuit in this instance is adding more circuit complexity that you don’t need and more things to fail. And when things fail at 450V they tend to make other things fail along the way. (Ahem: every modern traynor amp i’ve ever fixed with their stupid over complicated power supply design) and that kind of tight regulation will work against most of what you want _tonally_ to happen in a guitar amp. (Unless you play speed metal)

ALSO- unlike the capacitors in circuit- which DO care about a static voltage difference across them exceeding ratings at all times- you could put 1000V on the plate of a cold tube and it wont care (ok, a slight exaggeration but i digress). There is no current flowing in a cold tube so the tube is not dissipating any power. If your no-load voltage with the tubes out is 500V, but drops to 425v when under static quiescent load conditions then the 425V is the only plate voltage you need to realistically worry about in real world conditions for the TUBE PLATES (again- capacitors need to be rated a healthy margin over the peak B+ at any point if you want them to last)

In practical terms- clip your dmm between the plate and cathode (this is the _real_ plate voltage the tube sees and is less than full B+ to ground) and see what happens when the tube warms up. I promise any not already failed 6V6 isn’t going to explode or catastrophically fail because it sees a few seconds of higher plate voltage than spec.

If it settles in to anything under 450, I say slap some screen grid resistors on, keep screen volts lower, bias to 60% and plug in your guitar.

 

[expat]

Stop using AI and put a small amount of time and effort into understanding how to use ohm’s law and the relationships it describes, how to read a datasheet- esp for tubes, and a basic understanding of the parts of a tube and how current flows in it (ie: so you know basics like the scale of the current flow of plate versus filament versus screen and don’t get fooled into huge mistakes by a dumb AI algorithm)

Oh gosh, I build Ai Chatbots and Ai website tools for a living - I definitely know about their limitations and hallucinations. Sometimes its great, sometimes its not. Just like every reply on a forum from a real human. But its all in an effort to learn (and also hopefully to get this amp running one day!)

You say 'spend a small amount of time' to learn all that stuff?! Well, i'm definitely trying, and with the great input from guys like you, I'm hopefully getting better, albeit verrrrry slowly. Thanks for everyone's patience.

So, you said slap some screen grid resistors on there - what is that about - are mine not high enough in value?

Does this voltage need to be reduced using a zener or will series resistance do it, and what might that resistor value need to be?

If I felt safe to figure this stuff out on my own, I wouldn't ask, so thanks for everyone's input.

 

[Cqwet Dbdfte]

The screen grids are effectively another "plate" , but they also control the plate current, but with much lower gain. Keeping the screen voltage within spec is a must, unless you have an infinite supply of tubes. A (bypassed) HV Zener makes sense.

 

[abbey road d enfer]

So, you said slap some screen grid resistors on there - what is that about - are mine not high enough in value?

I don't knw why Amplexus mentioned that, because the oneon your schemo seems adequate to me.

Does this voltage need to be reduced using a zener or will series resistance do it, and what might that resistor value need to be?

It could be either.
A zener will give a tighter response than a resistor.
You need to drop about 50V with an idle current of about 50mA; that computes to a a 100 ohm resistor. It will dissipate an average of 2-3W, I would put a 10W. This will give significant sag.
You should order a selection of resistors of 22, 47 and 100r so you can adjust on test.