how can i determine what voltage this PT will give me?

[daz]

I'm thinking about a lower voltage PT to put under 400v at the el34 plates, preferably around 380. Currently theres 440 on them. I was considering this hammond to replace mine who' model i don't have handy at the moment. But can i look at what the specs are on my current PT and determine what i would get on the plates with this one?

[big_teee]

Your AC wall voltage has a lot to do with what you will get out of the PT.
Here mine runs 125-127V AC.
With that I came up with a 1.44 multiplier at my house.
My 345-0-345 PT, gives me 497 V out.
I'm sure this is not that accurate, but a good rule of thumb for my location.
You're talking TrainWreck voltages.
http://www.classictone.net/40-18065.html
http://www.classictone.net/40-18065.pdf
T

[daz]

Your AC wall voltage has a lot to do with what you will get out of the PT.
Here mine runs 125-127V AC.
With that I came up with a 1.44 multiplier at my house.
My 345-0-345 PT, gives me 497 V out.
I'm sure this is not that accurate, but a good rule of thumb for my location.
You're talking TrainWreck voltages.
30W TW Express, Liverpool & Rocket Style Power Transformer, 120V / 240V
http://www.classictone.net/40-18065.pdf
T

Hmmm.....which secondary do they use on those TW PT's and what does each get out?

[big_teee]

At my house that calculates to 375v & 432v.
One good thing, you could try both and gives you more variables.
T

[daz]

Doesn't seem to give a current rating one those. Do you know what it is? The marshall classictone i had was 150ma, which from what i read led to very hot running PT's. But my hammond is 250ma.

[Shane]

Its never as simplle as a straight forward 'multiplier' factor like 1.414 etc because it depends on loading, source impedence etc.
Have a look here:


The Valve Wizard

Cheers
Shane

[big_teee]

http://www.classictone.net/ClassicTo...e_Database.xls
Shows it to be 300ma.
That 1.44 is the loaded idle voltage here.
Like I said it is just a rule of thumb to get me started when I order transformers.
The High Voltage all the power companies run, messes everything up on old voltage specs.
A good example of this is the 350-0-350 for a 5E3 amp.
Don't know how many threads where they exceed the 450v Cap specs, because of the High Secondary.
T

[daz]

Its never as simplle as a straight forward 'multiplier' factor like 1.414 etc because it depends on loading, source impedence etc.
Have a look here:


The Valve Wizard

Cheers
Shane

well, thats why i asked if i can determine it by comparing my current PT's specs to a new one.

[R.G.]

ACK!!! Transformer output voltages!!!

It's predictable - ish!

Transformers not only don't have impedances, they don't have voltages. What they have is ratios. They are variable see-saws, including the bump at the bottom.

When someone says a transformer is 120V to 600Vct at 100ma DC (just pulling numbers out of my hat, for example) here's what that really means.
(1) DC output:
They have told you an AC output at a DC output current. That makes no sense at all, except that they have assumed that you're rectifying this to DC somehow, then filtering it, and finally loading it down so there's 100ma of current at DC. This is handy for them as they don't tell you the rectifier/filter setup, they assume some kind of circuit.

For tube amp transformers with CT windings, the rectifier is nearly always a full wave center tap setup with only two diodes, not a bridge. This will be important later.

(2) AC output voltage:
With the implicit rectifier and loading, what they have not told you and just assumed that you know is that the windings have wire resistance, the diodes have forward voltages, and the filter caps affect the pulse currents feeding the filter caps. The resistances of the wire coils lowers the output voltage when there is current flowing through them. Transformer makers tell you a voltage at some current. If the current is not the max load, but zero, there is no voltage lost in the wire resistances, so the open circuit voltage on a transformer is higher than the specified output voltage by some amount. They're guaranteeing you a minimum - you'll get at least this much AC voltage in the assumed circuit at this DC current.

(3) Rectifier and filter cap losses:
They don't mention rectifier losses or filter cap losses at all. Solid state rectifiers start to conduct at about half a volt, and may go up to a volt or so at big currents. Vacuum rectifiers start to conduct at about 0V, but have a really high resistance which decreases a bit, then stays at a lower, but still high level. A decent rule of thumb is that vacuum rectifiers drop 30-50V at working currents in a tube amp power supply.

Here's the real story. The AC power line is (nominally!) a sine wave with the AC rms voltage. Power companies used to put out 110, then 112, then 115, 117, 120, 125... 130Vac is not all that uncommon. Transformers are locked to a ratio. A transformer designed for 110 to 600 will put out 682Vrms at 125V in. This is why things like the Vintage Voltage adapter are good for older amps with transformers from half a century ago.

A sine wave has a fixed value of the peak of the AC wave to the RMS value. Power companies report the RMS value, as that's the true value of AC for delivering power. But diodes conduct any time they're forward biased, and if you put a sine wave into a diode and then into a cap, the cap charges up and the diode stays on until the peak of the AC wave, then turns off when the AC wave gets lower than the peak. So a diode/cap is a peak detector. The cap can't get any higher than the peak of the AC wave, minus one diode forward drop. The peak of an AC wave is 1.414... times the RMS value. So a 120Vac rms sine wave has a peak value of 120*1.414 = 168.68V. If this goes into a diode then a cap, the DC voltage on the cap will be 168.68V minus the diode drop. And that's about 0.7V to 1V for silicon, and 0 to 50 for a vacuum rectifier. This is one huge source of the variation in DC output for vacuum rectifier power supplies.

But the actual voltage on the cap also depends on how much voltage is lost to wiring resistance in the transformer. And it's worse than most people know, because of the peak-voltage thing. A DC supply has the caps hold up the DC current between the diodes conducting. The caps run down between peaks, and the diodes then conduct as soon as the AC wave feeding them gets higher than one diode drop above wherever the cap has run down to. This run-down/charge up is the ripple voltage. It's almost entirely determined by the capacitor value and the DC current sucked out of the cap. Since caps are nearly always chosen to give small ripple voltages compared to the DC output voltage, the diodes only get to conduct for a very small part of the time between AC voltage peaks. When they do conduct, they conduct enough current to supply the entire circuit for the next half cycle til they can turn on again. So the diode current comes in pulses of typically about 1/20 (i.e. 5%) of the time between pulses and correspondingly about 20 times the DC current going out.

So the current into the diodes and cap are very much bigger than the DC current out, and this current, the momentary peak current, is what drops the voltage coming out of the transformer windings. Also, since the current in the transformer is big, isolated peaks, it has a different RMS heating value than the DC output power. For full wave center tap rectifiers, this is about 1.2-1.4 times the DC current. For full wave bridge rectifiers, it's 1.6 to 1.8 times.

So when a transformer maker says "120 to 600CT at 100ma DC" they're not telling you enough to do a decent design, only to make some guesses, and that's why it's hard. On top of that, the variation in the AC power line voltage wobbles the answer by a proportional amount.

What you need to do an accurate design is the open circuit voltage ratio of the transformer, the winding resistances or a specification of the voltage ratio at some load, which is the same thing. This last is often quoted as a "regulation" number, which is always a percent under 100, and generally under 20%. It's not an active regulation the way we think of it today, it's a passive how-bad-does-it-NOT-regulate.

All this uncertainty adds up. It's calculable, and I spent some years having to do the calculations. But it leads to wide variations that are not easily hand waved with ratios unless a lot of stuff is specified. Like business accounting, it's precise(ish), but hard for non-pros to understand until they sit down and do some calculations, possibly with help.

[daz]

But RG, as i suggested above, while you are citing differences in caps and rectifiers etc etc, shouldn't i be able to estimate is very closely when replacing a PT with ALL ELSE staying the same in the amp? EX: if i have a 300-300 PT now and it's making 440VDC, shouldn't i be able to estimate a 260-260 to give around the same percentage in respect to the spec? In other words, with a 300-300 PT, if i take 300 and multiply that times 1.47 and i get 439, about what i have now. So if i have a 260-260 PT and multiply 260 times 1.47 i get 382 VDC. Would that that be accurate with all else in the amp being the same as i had it with the 300-300 hammond??

[g1]

Daz, I think what you are proposing should work or give a ballpark figure if 2 conditions are met
1) The 2 transformers are spec'd for similar output current capability
2) You compensate for the difference in AC line voltage, as the Hammond is spec'd at 115V input and I think you will be putting in more than that, so you will have to add a % increase.

[daz]

Thats true, forgot about that. Guess i'm gonna pass on trying this rather than buy a PT i end up having no use for. My math skills are almost non-existant and i'm not gonna be able to estimate this close enough to risk the $.

[R.G.]

But RG, as i suggested above, while you are citing differences in caps and rectifiers etc etc, shouldn't i be able to estimate is very closely when replacing a PT with ALL ELSE staying the same in the amp? EX: if i have a 300-300 PT now and it's making 440VDC, shouldn't i be able to estimate a 260-260 to give around the same percentage in respect to the spec? In other words, with a 300-300 PT, if i take 300 and multiply that times 1.47 and i get 439, about what i have now. So if i have a 260-260 PT and multiply 260 times 1.47 i get 382 VDC. Would that that be accurate with all else in the amp being the same as i had it with the 300-300 hammond??

If you can measure the winding resistances for the primary and HV windings on the two transformers, you can estimate it closely. I'd be happy to help with the math if you can get that, plus the no-load to some-measured-load sag on your existing supply. Each bit of stuff you have to guess at gets you a less accurate answer.

[loudthud]

Hammond publishes a data sheet, it can be found here: http://www.hammondmfg.com/pdf/EDB270HX.pdf

It's a little tricky but PSUD2 can be used to get output voltage and ripple.

Have you tried the Hammond 270DAZ

[daz]

If you can measure the winding resistances for the primary and HV windings on the two transformers, you can estimate it closely. I'd be happy to help with the math if you can get that, plus the no-load to some-measured-load sag on your existing supply. Each bit of stuff you have to guess at gets you a less accurate answer.

Thanks, but I'd have to buy it to do that which would defeat the purpose of trying to find the voltage.

Hammond publishes a data sheet, it can be found here: http://www.hammondmfg.com/pdf/EDB270HX.pdf

It's a little tricky but PSUD2 can be used to get output voltage and ripple.

Have you tried the Hammond 270DAZ

Isn't that the one they don't publish specs for because they can't figure it out?