Using Resistors to Change B+

[renshen1957]

I'm looking around for a new PT for my weird project amp. I want at least 140 mA available for idle current. The design calls for 550V on the plate windings (275-0-275).

I know about a couple of PTs that are very expensive and which fit the bill pretty well, but I located a cheaper one. Problem: it comes in at 600V.

http://www.classictone.net/40-18086.pdf

I assume I can just mess with the resistors on the power supply to get me down where I want to be. Correct? If I use a 5U4, I will be able to get ample current, and according to the PT specs, the voltage will be lower than a 5AR4 even before I change resistors.

Another question: the amps designs I've seen use the highest available voltages for B+, and then the preamp and so on are powered by lower voltages farther down the line. Is there any reason it has to work that way? If I need a higher voltage somewhere in the preamp, is there any reason I can't draw the power from the point closest to the PT and then draw the B+ from a lower voltage a cap or two away?

Hi New Steve.

Here are some suggestions based on your original post:

http://www.edcorusa.com/p/595/xpwr007_120 200mA 550V

You might want to check out the other power X-formers as Edcor USA

http://www.edcorusa.com/c/72/xpwrseries

XPWR004-120
Picture of XPWR004-120
Power transformer for a 120V, 60Hz. line to 550V with a 55V bias tap (275-55-0-275) at 325mA center tapped, two 6.3V at 4A and 5V at 3A.
$85.77

XPWR007-120
Picture of XPWR007-120
Power transformer for a 120V, 60Hz. line to 550V (275-0-275) at 200mA center tapped, 6.3V at 6A and 5V at 3A.
$57.33

XPWR009-120
Picture of XPWR009-120
Power transformer for a 120V, 60Hz. line to 550V (275-0-275) at 175mA center tapped, 6.3V at 4A and 5V at 3A.
$51.16

XPWR017-120
Picture of XPWR017-120
Power transformer for a 120V, 60Hz. line to 550V (275-0-275) at 250mA center tapped, 6.3V at 5A and 5V at 3A
$67.01

XPWR072-120
Picture of XPWR072-120
Power transformer for a 120V, 60Hz. line to 550V (275-0-275) at 250mA center tapped and 6.3V at 7A.
$57.33

XPWR163-120
Picture of XPWR163-120
Power transformer for a 120V, 60Hz. line to 550V (275-0-275) at 150mA center tapped, 6.3V (3.15-0-3.15) at 5A center tapped and 5V at 3A.
$55.03

XPWR210-120
Picture of XPWR210-120
Power transformer for a 120V, 60Hz. line to 600V and 550V (300-275-0-275-300) at 250mA center tapped, 6.3V (3.15-0-3.15) at 6A center tapped and 5V (2.5-0-2.5) at 3A center tapped.
$90.91


XPWR219-120
Picture of XPWR219-120
Power transformer for a 120V, 60Hz. line to 550V (275-0-275) at 150mA center tapped, 6.3V (3.15-0-3.15) at 4A center tapped and 5V (2.5-0-2.5) at 3A center tapped.


Best Regards

PS As a friendly suggestion from one Steve to another, invest in a copy of The Ultimate Tone volume 5 which discusses transformers in depth it's math isn't as complex as http://www.ampbooks.com books on the subject. I would also suggest The Ultimate Tone Volume 3, which discusses proper grounding, wiring, etc. It would cost about $140.00 for both but well worth the price of admission.

[diagrammatiks]

did you take voltage measurements with 1 ot and 2 tubes?

You're not leaking current anywhere...

I just built an amp with a 260vac winding at .09A

Unloaded b+ reads 364vdc while loaded b+ reads 300.

My transformer datasheet kinda implies that I'll lose about 15-20 vac under full load conditions...so when I draw current from just the plates the ac secondary goes to 240..

so I get 336vdc - then I lose some more across the actual cathodes, which in my case are biased at 16.5vdc with 470ohm resistors...

2 * 16.5 = 33. I should see 336 - 33 on my node and I get 300-305.

Like I said before, in the case of your amp...if the pt is fine then you get 290 unloaded...

You'll drop about 40 volts.

So it seems like your b+ is fine as long as your cathode voltages and everything else are fine.

[The New Steve H]

I'll tell you what. It's sad to have a physics degree and a year and a half of grad school and have people worrying that I won't understand math. I started getting back into differential equations and multivariable calculus a while back, and it was a whole lot easier than figuring out tube amps.

I really miss the days when my professors showed up several days a week and explained things in an organized, progressive manner! Picking this stuff up off the Internet in bits and pieces is not for everyone. I've been at this for a year, and I still don't know whether you can treat a tube like a resistor.

Thanks for the info.

[The New Steve H]

did you take voltage measurements with 1 ot and 2 tubes?

You're not leaking current anywhere...

I just built an amp with a 260vac winding at .09A

Unloaded b+ reads 364vdc while loaded b+ reads 300.

What am I missing here? My PT is 275-0-275, so I should have higher voltages than you, and I'm getting a LOWER B+ than you are, by fifty volts.

I'm sure I checked voltages with the OT out of the loop, but I haven't recorded everything in an organized way. I know the voltages did not improve.

Why would the B+ start out high, with the amp playing at normal volume, and THEN drop? If the load is making it drop, I don't see how it can ever get high enough to play in the first place.

[diagrammatiks]

beats me. I'm just going by the numbers that are in your schematic.

I cheat a lot. If I need to build an amp I usually go looking for a schematic of an amp that has the voltages on it that's close to what I want to build.

double the numbers to double the output tubes, halve the numbers to half the output tubes.

I don't really get what you mean when you say treating the tube like a resistor though?

There are 2 resistors already in the equivalent circuit.

the plate resistor and the cathode resistor.

The voltage drop across the plate resistor and the voltage drop across the cathode resistor should be nearly the same in terms of current.

Your number was close for the tube's internal plate resistance though.

A 12ax7 is anywhere between 80k to 56k depending on the b+.

did you take voltage measurements with 1 ot and 2 tubes?

You're not leaking current anywhere...

I just built an amp with a 260vac winding at .09A

Unloaded b+ reads 364vdc while loaded b+ reads 300.

What am I missing here? My PT is 275-0-275, so I should have higher voltages than you, and I'm getting a LOWER B+ than you are, by fifty volts.

just saw this.

I have a solid state rectifier and you have a tube rectifier. The tube rectifier will drop an additional 40-60 volts according to the 5y3 specs.

If I was using a 5y3 I'd expect to see about 240vdc on the plates.

If you look at a matchless dc30 - the schematics I have show a winding of 290-0-290.

That gives you a max possible voltage of 406vdc and the actual voltage is about 340. The rectifier used in that amp drops about 10-20 volts I think.

[The New Steve H]

Here's what I mean by treating the tube like a resistor.

I am using the 5F6A schematic all the way to the coupling caps past the PI. My PT provides lower voltages than a 5F6A PT, so the plate voltages on the first tube should be wrong. So how do I change that? I decided to use the known voltage drops in the chain, along with the current, to figure out the resistance between the plate and cathode. I treated this path as a resistor.

When I did the math, I got 8.428 times ten to the 4th power ohms, or ~85K. I put this back in the model, and in order to get 150 on the plates, I came up with 68K in place of the 100K resistors.

If I can't determine the resistance of the path the current takes through the tube, I don't see how I can determine the resistances needed above the plates without trial and error. It seems like my method worked okay. Maybe the resistance doesn't change much over a narrow B+ range?

[The New Steve H]

I stuck the 5AR4 back in there. This brought the B+ up to 290.

No improvement in performance. When I strum the guitar, the B+ drops over 20 volts.

[tubeswell]

If I can't determine the resistance of the path the current takes through the tube, I don't see how I can determine the resistances needed above the plates without trial and error. It seems like my method worked okay. Maybe the resistance doesn't change much over a narrow B+ range?

You need to start learning about load lines. A tube forms part of the load on the power supply because it completes the path of electron flow (assuming its the electrons that flow and not the +vely-charged holes that electrons 'leave behind'). But the load of a powered up tube is dynamic (altho under 'no signal' conditions, it assumes a relative state of quiescence) and it depends on how hot the cathode is, the propensity of the cathode material to 'release excited electrons, the size and voltage of the plate, and the distance of the plate from the cathode, as well as the bias voltage applied to the control grid, and the resistance of the cathode and plate resistors (which also form part of the load).

A load line is a graphic representation of the plate resistance of the tube (portrayed by the gradient of the grid curves), as well as a representation of how the plate current and voltage change depending on the plate resistor, and the cathode resistor (if there is one) or the bias voltage.

The less current there is flowing between the cathode and the plate, the less of a load the tube will make on the power supply - all other things being equal.

You can lower the tube current by either; lowering the power supply voltage, or by increasing the resistance of plate resistor or the cathode resistor, or by increasing the grid-to-cathode bias voltage. Each variable will have a different impact on the operating characteristics of the tube.

Lowering the power supply voltage will straight out decrease the plate-to-cathode voltage, lowering the overall gain.

Bigger plate and/or cathode resistors will also decrease the plate-to-cathode voltage, which will lower the gain, but a bigger plate resistor will mean there is more plate swing for a given quantum of tube current, which will counter the impact of decreased plate-to-cathode voltage.

Increasing the bias voltage will increase the plate-to-cathode voltage, because the plate voltage will increase more than the cathode voltage will increase (because there is less tube current to 'pull' the plate and cathode together), leaving less 'wiggle room' for the plate. If you keep increasing the bias voltage, you eventually get to a point where you virtually switch off the tube current altogether, and the plate will be at the same voltage as the power supply.

[diagrammatiks]

well I'm stumped if it's not giving you any output power even with only 2 tubes.

you should be able to pull at least 7 watts across an 8k output transformer.

what size did you make your shared cathode resistor?

and what did you bypass it with?

[Phil_S]

Steve, I don't think you are approaching the B+ the way it needs to be done. You are fiddling with the plate load resistor, which will affect gain and bias on the preamps, but not plate voltage in the way you desire. Changing the plate load from 100K to 68K will change the tone.

Try working with the dropping resistors on the B+ ladder.

The 6BM8's will have to accept what you get from the rectifier. You can tweak downstream. I'd work with idle voltage. You have 290 at the node. That means the screen node going to be about 285. The PI needs 230 (55v drop) and V2/V1 (last node) needs 150V (80V drop from the V3 node).

Figure 2ma per each of V1, V2. You have ~4ma at the last node and need to drop 80V. Ohm's law says you need a 20K resistor. Experiment until it's right. Tweak this first. Focus on voltage drop, not plate voltage.

The good news is that because it is already built, you can check the drop across the last dropping resistor, do the math (Ohms Law, you know the voltage drop and R), and determine the real mA draw for V1 and V2. Once you have that, replace 4mA with the real number and resulting resistor value will be darn close to what you need for a drop of 80V.

For the PI, you are working with 20mA and a drop of 55V. Ohm's law says R needs to be 2750. 2.7K is close enough. Tweak this one for plate voltage. When you have 230V here, V1 and V2 will fall into line. Again, you've already got it built. Check the drop across the resistor between the PI and the screen supply node. Figure the mA and replace it in the calculation. You should be darn close. (Actually, you should add the 4mA load from V1 and V2 because they are present downstream on the chain.)

You may have to go through a couple of rounds of tweaks as there is a certain circular logic here that isn't real strong.

How do I get 2mA for a 12AX7 or 20ma for a 12AY7? Look at the tube data sheet and the guess a little because the voltages are a little different.

Example, the 12AX7 says 1.2ma per triode at 250v and .5mA at 100V. Round to 1mA per and there are 4 sections, so it is 4mA. Close enough. In tweaking you will find the right dropping R, but your starting point is reasonably close.