setting the B+

[cooldude666]

Are you there forums? It's me, Margaret.

I'm confused on how to achieve desired B+ voltages in my circuit. I haven't been able to find much literature on the subject either. Most resources describe how to design a stage with a given B+, but not how the designer dropped the B+ of a previous stage to the desired value. I know a resistor is used, but how to calculate that resistor value is eluding me. I am feeling like it's ohms law. Desired voltage / Anode current = R. Is it that simple? Is the hard part in working out the draw for your tube? Right now I'm working on figuring out a long-tail phase inverter that will be driving a pair of 6550s. My 6550s are at 480v. I have chosen to use a 12ay7 as my PI tube. So according to my data sheet, 12ay7 at 250v with a 47k anode resistor biased at -2.5v will draw approx 2.25 mA. Since this is the PI we then double this to see draw. 4.5mA. Now I need to choose a tail voltage? 60v is about 1/4 B+ so should provide enough swing to drive my output tubes. So my total B+ needed is then 310. So if I'm correct about ohms law being my guiding voice here- a 68k 2 watt resistor should supply me? With 310v at idle. Am I right? Does this all look sound? Thank you kindly for advice!

[Amp Kat]

Are you there forums? It's me, Margaret.

I'm confused on how to achieve desired B+ voltages in my circuit. I haven't been able to find much literature on the subject either. Most resources describe how to design a stage with a given B+, but not how the designer dropped the B+ of a previous stage to the desired value. I know a resistor is used, but how to calculate that resistor value is eluding me. I am feeling like it's ohms law. Desired voltage / Anode current = R. Is it that simple? Is the hard part in working out the draw for your tube? Right now I'm working on figuring out a long-tail phase inverter that will be driving a pair of 6550s. My 6550s are at 480v. I have chosen to use a 12ay7 as my PI tube. So according to my data sheet, 12ay7 at 250v with a 47k anode resistor biased at -2.5v will draw approx 2.25 mA. Since this is the PI we then double this to see draw. 4.5mA. Now I need to choose a tail voltage? 60v is about 1/4 B+ so should provide enough swing to drive my output tubes. So my total B+ needed is then 310. So if I'm correct about ohms law being my guiding voice here- a 68k 2 watt resistor should supply me? With 310v at idle. Am I right? Does this all look sound? Thank you kindly for advice!

There will be other tubes downstream from there tugging on the supply also so maybe if there were no other tubes it would work out that easy but that isn't the case. Not to be un calculas but I just plop resistors in there till I find the right combination. Did this with my last build and couldn't be happier with all of the clean and distortion channel voltages. 68K seems a little high to me I'm thinking about 20 to 30k sounds more like it when all said and done.

[defaced]

If you can estimate the draw of each tube, the Duncan Power Supply simulator does a good job of simulating the power supply. But as to be expected, the info out of the simulator is only as good as the info put into it. I'd probably use it as a good "place to start", then season to taste.

[cooldude666]

Big bummer that is windows only.

[ReadyTeddy]

If your name is Margaret, shouldn't your screen name be cooldudette666?

[cooldude666]

Unlike tube technology, gender roles did not stop evolving in the 1950s :-)

[loudthud]

Back to the subject of this thread, the problem is that of a simultaneous equasion with three variables. You don't know that the tube will draw unless you know the B+ and you also don't know the voltage on the supply side of the dropping resistor because it will drop when you connect the resistor. Find some version of SPICE that will run on your computer or use the old method called "Cut And Try It".

[trobbins]

Ohm's law is the correct method, along with Kirchoffs law I recall. Each stage can be given an estimated operating current and voltage. A power supply dropping resistor can be connected between each stage. Each stages operating current sums together as more sections are added - starting from the input stage and progressing to the output stage.

If your preamp stages sum up to draw 2mA total, then you add that to the PI's 4.5mA to get 6.5mA passing through your first dropping resistor from 480V to 310V - so that dropping resistor needs to be (480-310)/6.5 = 26k. Then move one stage back and do the same calculation but with the relevant currents for that stage - etc till you get to the input stage. Quite logical, and easy to do a quick design sketch.

You can use datasheet operating levels of V and I for each stage, or draw an anode loadline with cathode bias line to estimate an actual operating current for a stage. Actual operating V and I are likely to be within +/- 10%, and resistors come in set value ranges (or whatever you have in your parts bin), so the pre-design allows a good guess at what you'll need to initially solder in. Measure actual supply, anode and cathode voltages at idle to confirm that each stage is operating as you would expect from your pre-design. I use this technique, and use a MS Word doc with datasheet curves inserted in to a drawing, and loadlines added - it makes it easy to move loadlines around graphically for each stage - and to cut/paste/copy for other stages or amps.

[jazbo8]

Try this simple simulation that I made, you can adjust any of the components and see the resultant node voltages.

[R.G.]

Back in the day when tubes were the only active components, amp designers relied on the fact that most tube circuits are class A and pull a fairly constant current, and even more that preamp style tubes were used with almost-fixed sets of circuits, varying only a little from design to design. They'd sketch out what they wanted, guess the plate and screen currents based on experience or manufacturers' published literature, then adjust the series resistors in the plate supply as needed. This was helped by the fact that especially cathode biased triodes are very, very tolerant of varying plate voltage.

You can do the same.

Or you can take a step sideways. You can force the voltages to be what you want independent of the currents, using the modern devices we now have. A single power MOSFET can be used for a source-follower regulator. You put a voltage on its gate, and the source voltage follows at that voltage minus the few volts of threshold voltage, which can usually be ignored in the hundreds of volts in a tube amp. So you could make B+ and a single zener-regulated reference string. From there, hook any number of MOSFET followers between the B+ and their output voltages. You would then get the voltage you wanted, fixed very solidly for a tube circuit.

But you may want proportional sag. OK, ditch the zener, and use just a resistor string from B+ for each section. Now the voltages follow B+ around proportionately.

Notice that this is a nice companion to star grounding, start power distribution.

There are so many ways to skin a cat that it's a wonder that there are any cats left.

[cooldude666]

Is the same technique used for selecting a resistor to set the screen voltage with? Tap off the B+ to the screen based on current draw? Datasheet says zero signal draw is 9mA but full signal draw is 38mA! Quite a difference. I think my biggest fear is burning up those resistors when thing is dimed. Most of the 6550 schematics I look up are ultralinear so they just dodge the question. Is this another game of "pick one, see what happens"? There's gotta be a way to estimate it...

[Enzo]

Are you talking about the individual screen grid resistors? Or the dropping resistor between plate and screen nodes? In either case, use Ohm's Law. If you know current and resistance, you can easily calculate dissipation. 38ma through a 1k resistor? I get about 1.44 watts, so the common 5 watt resistor is fine for that.

[cooldude666]

I figured a 5 watt is what was necessary but how did you arrive at a 1k resistor? I guess my entire question in this post is how to I achieve a desired voltage drop. My plates are at 480, I want my screen at 310. Datasheet tells me that at maximum signal I'm drawing 38mA. So my ohms law equation goes like this?
480 - 310 = 170. I want to drop 170v across this resistor. V/I = R; 170v/38mA = 4473 but a 3.9k is the closest value. But 170v * 38mA is more than 5 watts. I know I'm doing something wrong, I just don't understand what. Is it that at maximum signal I won't be dropping 170v across Rg2?

[Jazz P Bass]

Where did you come up with that screen voltage target?
You cannot 'drop' a voltage without upsetting all the other voltages.
Voltage is not dropped, it is consumed.

[cooldude666]

datasheet. The 6550 datasheet (http://www.drtube.com/datasheets/6550a-ge1972.pdf) says that for 2 6550s push-pull in fixed bias to set the screen to 310v.