how to undestand Power transformer specs

[pula58]

let's say a power tranny is spec'd to 600V, CT, @ 230mA. How do I use those numbers to calculate what the B+ voltage would be?

Let's assume solid state rectification. And, since the tranny has a center tap, the typical Twin reverb type of diode rectifier configuration.

also assume 4 12ax7 tubes biased at 1mA each, and two 6L6 tubes biased at 45mA each.

How do I calculate the expected B+ ?

[diagrammatiks]

if the transformer is 600vct that would mean 300-0-300.

using standard diodes and a capacitor input your expected unloaded b+ would be 1.4 times 300.

So 420 with a capacitor input.

IF you have 2 6l6s they will draw between 100-150 ma and each 12ax7 will be 8-15ma

The dropping resistor between the plate and the screens will drop somewhere around 100-120ma. The next one will drop the remaining current of the preamp and so on.

[pula58]

Then, the factor of 1.4 is really just using sqrt(2) to convert the RMS to peak (amplitude)?

How does the current listed in the transformer spec come into play?

[Phil_S]

In my experience, you often get a bit less than 1.4x. I'd figure 96% of that.

If I understand what you are saying correctly, the 230mA is what you can expect to safely draw from the high voltage primary. It is probably marginal for a pair of 6L6. You can't control, per se, the amount of current the tubes will draw. And often, the lower the plate voltage, the higher the current draw. You've also got to figure into it the screen current. Too many unknowns here.

It would be much easier to say if you had a particular schematic in mind.

For that PT, I'd be looking at four 6V6's with a tube rectifier, producing B+ in the mid 300's. That would be about 40W give or take, which is actually quite loud.

[David Root]

Old rule of thumb for us old guys with poor memories: Back into it!

Take the DC B+ voltage you want, add the rectifier tube voltage drop, if you're using one, divide by 0.9, multiply x 1.4 and divide by 2. The resultant number is the PT AC rating XXX-0-XXX.

350+50 (5U4GB)=400/.9=444X1.4=622/2=311-0-311

In practice, I have found if you have less than four 12AX7 in the mix, then the 0.9 factor comes out nearer 0.95.

This works for fixed bias. For cathode bias add ~20VAC to each side of the PT.

It is also at idle, not cranked, so you do need to know whether your PT is rated XXX-0-XXX at idle or at rated DC mA. If the latter, your DC voltage at idle will be 5-7% higher than the rule of thumb.

[pula58]

if the transformer is 600vct that would mean 300-0-300.

using standard diodes and a capacitor input your expected unloaded b+ would be 1.4 times 300.

So 420 with a capacitor input.

IF you have 2 6l6s they will draw between 100-150 ma and each 12ax7 will be 8-15ma

The dropping resistor between the plate and the screens will drop somewhere around 100-120ma. The next one will drop the remaining current of the preamp and so on.

Each 12ax7 drawing 8-15mA? That seems high wrt to what I have seen before (in the 1-2 mA range).

[pula58]

In my experience, you often get a bit less than 1.4x. I'd figure 96% of that.

If I understand what you are saying correctly, the 230mA is what you can expect to safely draw from the high voltage primary. It is probably marginal for a pair of 6L6. You can't control, per se, the amount of current the tubes will draw. And often, the lower the plate voltage, the higher the current draw. You've also got to figure into it the screen current. Too many unknowns here.

It would be much easier to say if you had a particular schematic in mind.

For that PT, I'd be looking at four 6V6's with a tube rectifier, producing B+ in the mid 300's. That would be about 40W give or take, which is actually quite loud.

What do you think the HT secondary should be rated for (current wise) for 2 6L6 and 5 12AX7 ?

I am trying to learn how to figure this stuff out, so, if you show equations and reasoning that would be great!

Thanks!

P.

[David Root]

If a 12AX7 draws 8 to 15mA it won't do it for very long at the plate voltages we use. The curves show about 3 mA per triode and that's pushing it. Perhaps Diagrammatiks meant 0.8 to 1.5mA.

Typical Fender values are 1mA per triode at 180V plate.

If you allow 2mA per 12AX7 you are in the ballpark, unless you have an unusual PI circuit.

[Phil_S]

This is just basic research. Here's how it's done. Look up each tube's spec sheet. Find them here: http://tubedata.itchurch.org/index.html or download the TDSL (very handy) here: http://www.duncanamps.com/tdslpe/

For the power tubes, find a reasonable match to plate voltage and screen voltage and go with a conservative number (you need to go on the high side if you aren't sure). I'll show you this one as an example, so you can see how it works. After this, you should be able to answer on your own.

I looked up the GE 6L6GC. Look for class AB push pull, where it says "values for two tubes." It gives you plate voltages of 360 and 450, so just average those if you assume 400V. There are actually graphs from which you can read the appropriate value. Since values are "typical" and each tube may deviate, it's really just an estimate anyway. So, at about 400V, maximum plate current is probably about 175mA + screen current about 17mA. Rounding up a bit, you'll need around 200mA.

Do the same for the 12AX7. The Sylvania sheet has 1.2mA per triode for typical operation. I usually figure 3mA per tube x 5 tubes = 15mA.

So, for your total, you are something north of 200mA and getting close to the 238mA rating. You might take a chance and it might work OK. The PT is going to run hot and it will probably sag pretty easily. This affects tone (good or bad depends on personal taste.)

Remember to also account for the filament requirements separately. The info for these is also on the data sheets. Filament windings are independent from the high voltage. In particular, you are asking for a pretty heavy filament load (you know where to go to figure it out) and that means a bigger PT.

Good luck with it.

[diagrammatiks]

I either moved a decimal or meant the total draw.

metric. what are ya gonna do.

I've been looking at some pt's and this one http://www.classictone.net/40-18005.pdf

is specced for 230 at 420vdc. Most of them seem around 200-230 for the fender style transformers so you should be fine.

Unless you're running 6 cathode followers and a reverb transformer 4 preamp tubes and 2 6l6s seem fine. Right in line with how fender did it.

[Hellhammer]

There's a problem when some manufacturers spec their transformers for DC. Most nowadays spec them for AC so you have to calculate for that. Here's a handy sheet. http://www.hammondmfg.com/pdf/5c007.pdf

In your case though, I AC=I DC if you're using a Full Wave rectifier and capacitor input load. Note that by using a choke input load, you get lower V DC but higher I DC from your PT.

[tubeswell]

What do you think the HT secondary should be rated for (current wise) for 2 6L6 and 5 12AX7 ?

Rule of Thumb for 2 x 6L6 in PP = about 140mA HT. Add 2mA for each 12AX7 and you have 150mA. The more current rating you allow for, the less likely it is to run hot. Most people add a margin of 10-20% in this regard, so the 150mA would become around 180mA. A tweed bassman 'stock' PT HT was 200mA.

SE is different. As a rule of thumb, you would want to allow around 100mA on the HT for 1 x 6L6 in SE, 200mA for 2 x 6L6 in parallel SE.

These are 'rough-as-guts' rules of thumb and there is a more precise method of calculating it which Merlin goes into in his 2nd book, but which is too long-winded to reel off here.

[Doctordog]

OK,

Not trying to contradict anyone, but thankful for the thread and replies, it helps me to try to get a grasp of this too. But, doesn't the OT primary impedance play into this? I was under the impression that it did, that the current draw would be different between say, a pair of 6L6's with a 4K primary and the same tubes with a 6K5 secondary.
I know that for efficiency's sake that a person would match the tubes impedance with the output's, but do these play into the power supply selection?

Thanks,

JB

[ampmike]

Hey Guys,Being on the topic of trannies,I found an old brand new in the box RCA PT.I would like to know if this thing would be useful on any kind of amp build.Here are the specs_Pri. 117 V., 60 cy. Plate 290.-0-292 V. 72 Ma DC Fil. #1 5 V. 2 amp Fil #2 6.35V 2.75 amp Thanks Guys,Mikey

[Phil_S]

I suppose the OT primary plays into the load of the power tubes. I would treat it as a fixed load and it is not very much of a load compared to the internal plate resistance. The load of the tube itself varies with the amount of heat and the signal being passed through it. I'd say that the tube itself, in respect of the OT primary is not modified very much by the OT for this particular determination. As I see it, this particular discussion concerns how to get a grip on the B+ supplied to the plate.

If you check the primary R of an output transformer, you'll find that it is not very much, maybe a few hundred Ohms.

Your question discusses reflected load and is a bit of a misdirect for this topic. Reflected load is totally different from the actual measurement of resistance on the coil itself. Reflected load is primarily a function of the turns ratio, where the square of the turns ratio is the impedance ratio. So, for example, in an OT that is 8K primary to 8 ohm secondary, we know the impedance ratio is 1000:1. From that we can determine that sqrt(1000) = ~31.5, which is the turns ratio (31.5:1). You don't actually have a DCR load of 8,000 Ohms on the primary. It's probably more like DCR = 200 Ohms. That's why I'm saying it isn't a significant thing to consider in estimating the B+.

I hope this makes sense.