Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: Willabe on January 18, 2011, 09:08:38 pm
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When you look on the box of a hammond PT they have a secondary rating for B+ and mA DC. For example hammond 272BX is listed as 300-0-300V, and 100mA DC, 95VA total, 5V 2A, 6.3V 3A. So if I'm understanding this, VA = V x A, so 5V x 2A = 10VA and 6.3V x 3A = 18.9. So 95VA - (10VA + 18.9) = 95VA - 28.9 = 66.1VA for B+ secondary.
So, VA / V = A = 66.1 / 300 = 220mA and 300 x 1.414 = 424.2V DC, and 66.1 / 424.2 = 156mA DC . Am I doing this right?
One more ? When tubenit and DaGeezer post schemo's and list a PT of say 275-0-275, 150mA, do they mean the 100mA DC rating thats listed on the box or are they refering to the 156mA DC after doing the math?
Thanks Brad :smiley:
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I think bringing up VA or even Watts, is not useful here, and only confuses the situation. See, http://www.powerstream.com/VA-Watts.htm
The current draw of the tubes is stated in mA. Remember power tubes have both plates and screens; and draw more current at full signal. These specs are stated on the tube sheets for various B+ voltages. Simply add-up the tube draw and choose your PT spec accordingly. (You can under-rate it if you want sag, e.g.)
Note that for an xxx-0-xxx PT, the current rating assumes that both sides of the secondary are being used. Ea side alone is only good for 1/2 the stated current.
For voltage, trannies are usually spec'd at the stated current draw (I think, or someone correct me). I.e., a 300-0-300 VAC @ 100mA, will give a 300V (RMS) reading with a 100mA draw. With no load, the reading will be higher; with a larger load the reading will be smaller (not to mention that a tranny could diverge widely from its stated specs). Of course, after rectification to DC, the RMS issue disappears, and the DC reading will be "magically" 1.414X higher than the RMS AC reading, less the impedance of a tube rectifier.
But again the tubes are spec'd for DC voltage, so it never becomes necssary to consider the watts or VA aspects.
For heaters, the draw is specified in Amps or mA, so so, again the watts or VA issue is unnecessary.
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Ea side alone is only good for 1/2 the stated current.
no, both halves of the secondary of PT will run at the rated current - think like both halves are 2 elements in series, in that V drops are = across the 2 like elements, but I stays same.
ex. if you have a hammond rated 200-0-200 @ 100ma and used with a FW cap loaded; IDC = IAC. you can use 1/2 primary 200-0 @ 100mA with a FWB & attain roughly the same Vo, however, IDC will be .62xIAC. or 62mA. hammond simplified things for you by calculating and printing the IDC rating of their transformers calculated into a FW cap loaded recto. so, respectfully, the statement above is not entirely correct.
again, the rectifier arrangement and load you choose to use will have differing current requirements. link to document below will help to clarify.
http://www.hammondmfg.com/pdf/5c007.pdf
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The VA rating of each secondary is important when you want to calculate the mains fuse size on the primary side. You add up the sec VAs to get a combined sec VA, which you then divide by the pri Voltage to get the current going into the primary winding - and then you double that number for the fuse rating (just so that your fuse doesn't pop all the time unnecessarily).
But in terms of calculating appropriate current ratings for the secondaries, that depends on what tubes you will be running. There is a chart for current draw of common guitar amp tubes in the tech section here: http://www.el34world.com/Forum/index.php?topic=10569.0
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The VA rating of each secondary is important when you want to calculate the mains fuse size on the primary side. Good to know! Probably also important for secondary fuse sizes.
no, both halves of the secondary of PT will run at the rated current - think like both halves are 2 elements in series, in that V drops are = across the 2 like elements, but I stays same. . . link to document I'm not getting this. The Hammond illustration is measuring across the outer legs of the PT. So, a 300-0-300 PT 100mA, per the Hammond document, would read 600VAC RMS @ 100mA. (After rectification you still have the benefit of the current handling capacity of both sides of the secondary.) But, unlike the illustration, if you measure from the center tap to one outer leg, you would read 300VAC RMS with presumably only the current handling capacity of that one side of the secondary winding. I do not understand why this is wrong.
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I do not understand why this is wrong. Nevermind, I think I understand. The 2 secondaries (like any 2 matched, stand alone trannies) can be combined to either: a) double the voltage at the same current rating; or b) double the current rating at the same voltage. With the xxx-0-xxx PT, we have choice "a)" hard-wired for us. Is this right?
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a) double the voltage at the same current rating; or b) double the current rating at the same voltage.
yes, windings are a) in series, b) in II.
:smiley:
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a) double the voltage at the same current rating; or b) double the current rating at the same voltage.
yes, windings are a) in series, b) in II.
:smiley:
Is "b) in II" supposed to mean in parallel???
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a) double the voltage at the same current rating; or b) double the current rating at the same voltage.
yes, windings are a) in series, b) in II.
:smiley:
Is "b) in II" supposed to mean in parallel???
yes.
:smiley:
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For resistive AC loads, heaters, the VA is as you say.
For _DC_ loads, with rectifiers, especially with capacitor-input filters, the AC VA is much higher than the DC Watts. The narrow pulses are hard on (heat-up) the windings. Your simple subtraction is not valid until you correct the DC Watts into AC VA.
There is an additional factor. The Hammond Tube-series Center-tapped transformers are rated in a mixed system. The 600VCT is AC, but the 100mA is specifically for 2-diode rectifier and capacitor input filter.
Each side of the CT winding will take rated current HALF the time. That's how the 2-diode CT rectifier works. It is inefficient of copper, but very thrifty with cathode material and insulation. Back when vacuum diodes were the only choice, a little excess cost in the PT saved a ton of cost in rectifiers and their heat. (Silicon diodes change the picture, but not enough to bother re-inventing and re-desiging.)
600V CT at 100mA will give 280V-420V DC (depending on diode loss) at 100mA. It "could" maybe instead give 600V-800V DC at 50mA (though the insulation is not sure to be good enough for high voltage winding to frame).
> 272BX is listed as 300-0-300V, and 100mA DC, 95VA total,
I can't make Hammond's numbers work out, but I won't question them. There are several limitations on transformer ratings. Heat, resistance, etc. They did the math, I don't care to repeat that (I don't have all the information).
> 272BX is listed as 300-0-300V, and 100mA DC, 95VA total,
280V-420V DC (depending on diode loss) at 100mA plus the heater windings.
It is very likely you can pull more. It may sag. It may heat. Life may decline from 100,000 hours to 50,000 hours... may not matter for a stage-amp.
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Thanks, PRR, and ALL, for your words.
For _DC_ loads, with rectifiers, especially with capacitor-input filters, the AC VA is much higher than the DC Watts.
If the pri. winding is rated at 95 VA total, and I subtract the two heaters Watts that leaves 66.1VA. I don't uderstand how the AC VA can be any/much higher. Where is the extra current coming from? I must not be thinking of this right. Your simple subtraction is not valid until you correct the DC Watts into AC VA.
How do you do that? The can of worms is open, might as well take the lid all the way off? 280V-420V DC (depending on diode loss) at 100mA plus the heater windings.
So, this PT will support a champ 6L6GC type amp to a full 10w output, or a BF Deluxe 6V6 type amp to 20w's or so, and PP pair of 6L6's to 20w's (class AB1) ?
I'm just trying to be able to figure out how to pick PT's for plate current.
Thanks Brad :smiley:
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..If the pri. winding is rated at 95 VA total, and I subtract the two heaters Watts that leaves 66.1VA. I don't uderstand how the AC VA can be any/much higher. Where is the extra current coming from? I must not be thinking of this right.
The AC VA rating is probably for the primaries. Don't forget about efficiency of the transformer. You won't get 100% from any transformer.
I'm just trying to be able to figure out how to pick PT's for plate current.
A good place to get an idea is this web site:
http://www.dreamtone.org/Calculate_Current_Form.htm (http://www.dreamtone.org/Calculate_Current_Form.htm)
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a PS design tutorial link below - given are some examples of selection from their product line, but a decent primer nonetheless.
http://www.mcitransformer.com/i_notes.html
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> AC VA rating is probably for the primaries.
Probably the core.
A 76VA core with "easy" winding may show 3 Watts iron loss and 11 Watts copper loss.
The double-high-voltage winding used for the 2-diode rectifier is not "easy". Much of the space for the secondary is full of shellac and paper insulation. Copper losses will be higher than with the same core working at more reasonable voltages.
> I don't uderstand how the AC VA can be any/much higher. Where is the extra current coming from?
Any transformer can deliver much more current than it is rated; but it will sag and run hot.
RATED current may be limited by sag or by heat.
_IF_ the current wave were sine or DC, no problem.
But the cap-input filter takes NARROW SPIKES of current.
Below is a hasty sim. The DC output current is 90mA. The PT current is zero most of the time with narrow spikes to 678mA. The RMS (heating effect) of this spike-wave is 155mA.
Even if heat were not an issue, the cap charges to the peak PT voltage minus copper loss times the PEAK current. Since peak may be 6 times the DC current, the sag is 6 times worse.
> The can of worms is open, might as well take the lid all the way off?
This topic takes 20 pages in Grossner, and does not come to a firm answer for the cap-input filter case. It Really Depends.
> I'm just trying to be able to figure out how to pick PT's for plate current.
I think you are trying to be too clever.
When you look on the box of a hammond PT they have a secondary rating for B+ and mA DC. For example hammond 272BX is listed as 300-0-300V, and 100mA DC.
Hammond TELLS you the _DC_ current rating.
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I'm just trying to be able to figure out how to pick PT's for plate current.
I think you are trying to be too clever.
Thanks PRR for your input. Maybe this will explaine my thick headedness, then again maybe not. :laugh: But, it's not my math, I got it from a book. :rolleyes: There's a chapter on "Transformer Specs" which has text on figuring out how the "older" ratings of Hammonds "Classic" series relate to the "newer" ratings. The whole line is listed once converted, and are for capacitive input filter and half bridge/full bridge, full-wave SS rectification. Here's two examples;
272BX 95VA 6V3_18.9VA 5V_10VA 600V CT_ 66VA (100mA DC-rated) 5.5 lbs
3A 2A 300V_220mA or 600V_110mA
424V DC_156mA DC 848V DC_78mA DC
272JX 236VA 6V3_50.4VA 5V_20VA 600V CT_165.6VA (250mA DC rated) 10.5 lbs
8A 4A 300V_552mA or 600V_276mA
424V DC_ 390mA DC 848V DC_195mA DC
From text; "if we use the usual factor of 1.414 to convert from AC to DC, then our DC voltage based on 300V AC becomes 424Vdc, and the current becomes 390mAdc. How does the 250mAdc original rating figure into this? The meters we use to monitar or measure voltage and current are designed to average out variaions. (ect. ect.) We end up with "average" and "RMS" readings and ratings, depending on the method of averaging the parameter. (ect. ect.) A DC ammeter in the high-voltage load line might measure an average value lower than the DC value we calculated."
Then latter; "If you were to install an ammeter in series with the output stage, you would see a value that is much lower than the calculated peak currents. Again, the meter is averaging the variation to provide a steady reading. Our 88mA SE amp might look like it is drawing as little as 62mAdc, while the 124mApk for the PP amp might look like 87mAdc."
"Using these values, the DCmA rating for the "Classic" transformers seem to make some sense."
I did noticed in the sim. the trans was puting out 150mA, _but_ only 89mA was delivered to the load. I would think it's what's delivered to the load is the most important value here. Which seems to show Hammonds figures are correct and not the math from the book.
Thanks guys for all the links, they are very helpfull to me. I have saved them to my favorites . :wink:
Thanks Brad :smiley:
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I'm WAY behind everyone here on this and I don't want to interrupt the discussion for too long so just a couple quick, dumb questions.
I don't even know what 275-0-275 means. Would 275 be the voltage present on the 2 high voltage secondary outer windings and the 0 is the center tap voltage?
This is the PT I'm using on my current build: http://www.magneticcomponents.net/40-18016_Rev_A_Stock.pdf
Would that be a 330-50-330, 120mA transformer?
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I don't even know what 275-0-275 means. Would 275 be the voltage present on the 2 high voltage secondary outer windings and the 0 is the center tap voltage?
yes
Would that be a 330-50-330, 120mA transformer?
Not quite. It is a 330-0-330 @ 120ma with a 50VAC tap. Measuring AC volts, if you put one meter lead on the 0v lead (center tap, red/yel) you will measure 330vac to the two red leads and 50vac to the blue (bias) lead.
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Got it - thank you.
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I'm WAY behind everyone here on this and I don't want to interrupt the discussion for too long so just a couple quick, dumb questions.
samato, your not way behind, everyone has to start somewhere, I know how that feels, just read some of my post's. :rolleyes: Remember what the "Jimi Man" wrote/sang "stand'in next to a moutain, chop it down with the edge of my hand", it's gonna take awhile. There's a lot to learn, and that's good. I don't think your "interrupting", your on topic, were talking about PT's and there ratings. Your questions are not dumb, how can any of us learn and grow if we don't ask questions? :wink:
And Hey, you got your photo gallery idea open for all of us! Nice. :toothy9:
Brad :smiley:
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Would that be a 330-50-330, 120mA transformer?
clarification - with a CT and a 50V tap then would it not be written as 330-50-0-330?
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330-50-0-330 Methinks this causes a disturbance in the Force. I urge it to be called 330-0-330 with a 50V bias tap, assuming that's what it is. (And if so, the 50V tap is not good far any real current.)
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> I got it from a book.
Care to share your bibliography? (Or is it TUT?)
In case you didn't get "Grossner", here's a full citation:
Transformers for Electronic Circuits (ISBN: 0070249792 / 0-07-024979-2) Grossner,Nathan R
If you want a not-as-good freebie:
Electronic Transformers and Circuits, Reuben Lee
http://www.tubebooks.org/technical_books_online.htm
Schade's graph on page 66 is, still, the best practical guide. Schade covers all cases, real parts are a subset. nwCR must 10 to 100 for good filtering economics. %Rs/Rl will normally be between 10% and 2%.
> "if we use the usual factor of 1.414 to convert from AC to DC, then our DC voltage based on 300V AC becomes 424Vdc, and the current becomes 390mAdc.
Correct for voltage, NOT for current.
Schade's graph gives the current conversion.
> I would think it's what's delivered to the load is the most important value here.
We are asking for a transformer rating. We look at what happens IN the transformer.
I'm guessing the hard truth of "RMS" needs refreshing. RMS may be defined as "heating value".
1 Amp in 1 Ohm is 1 Volt and 1 Watt.
2 Amps in 1 Ohm is 2 Volts and 4 Watts.
Here we have peaks SIX TIMES higher than the DC current. While that current is flowing, the winding-heat is thirty-six times higher than the DC current alone would be. True, this peak current flows only part of the time. Using too-crude eyeball integration, I get about 1/20th of the total time. 36/20 is like 1.8 times the heat that the steady DC current would be. That shows how violently the cap-input rectifier filter abuses the transformer.
> "If you were to install an ammeter in series with the output stage, you would see a value that is much lower than the calculated peak currents.
Yes, that is how amplifiers work.
> Our 88mA SE amp might look like it is drawing as little as 62mAdc
That seems to be a bad or mis-stated example. The Ideal SE amp can pull 50mA DC with peaks to 100mA and zero mA. A Champ with 40mA DC should do near 75mA and 10mA.
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> the 50V tap is not good far any real current
Not specified.
In fact, it is "usually" the SAME winding as the HV. 100mA or more. Any current drawn from 50V does add to current in the HV winding.... but only 1/7th of the whole winding. Heat and sag will be relatively minor. If the HV taps are only loaded 80% of rating (say 80mA on a 100mA rating), I'm sure that an added 50% (50mA on the 50V tap) does no harm.
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In fact, it is "usually" the SAME winding as the HV. 100mA or more. Live & learn! :book1:
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Methinks this causes a disturbance in the Force. I urge it to be called 330-0-330
not in the winders force: 330 (start of winding) - 50 (tap) - 0 (center tap) - 330 (end of winding) meaning that the 50V tap is between the start side of the winding and the center tap.
make sense now? :smiley:
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> I got it from a book.
Care to share your bibliography? (Or is it TUT?)
Yep.
Look I'm greatful to learn anything from any one I can, and I am just trying to sort it out,
(clever, hah, altho my mom and dad would've like that) I'm also not so clear on what can be quoted from someones book and what can't. If I go to the author, their going to give me answers according to their thinking, which leaves a guy like me in a place of only trust, that they are right in their thinking, so how do you put it to the test? Plus sometimes people are getting at/saying the same thing, just coming from/starting from a different angle, and ends up adding up to the same thing. Then there's the "ME" factor. :laugh: Sometimes good, sometimes not so good! I seem to remember you saying something like "read everything you can, just don't belive everything you read", which has been said before many times thru history. I'm trying to figure out whats valide and whats not. Even Einestine said that he "cooked" the math to get "what he wanted" at one point, _ then_ when he came to terms with what and why he did it, he went back and re-did the math, and the rest is history. Need I say Ba'Boom!
Yet, I don't think he ever swung a hammer in heat or cold. hee hee!
As a carpenter, to prove my work, the "facts/resultes" of what I have built/laided my hand to, seem to be much simpler to conferm. I don't need a scope, meter, ect. all I need is,_ HAS_ the building weathered the storm of time in the long run, and _HAS_ it weathered the weather, over time. I can see that, so can the home owners.
I've talked to Doug, Gerald, and KOC, on the phone several times, and all were great guys. And I'm greatful to them all, they have all helped to bring all of this about on the WWW.
Any way.....
Thanks Brad :smiley:
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Einestine said that he "cooked" the math to get "what he wanted" at one point, _ then_ when he came to terms with what and why he did it, he went back and re-did the math This is the infamous Fudge Factor, dressed up as the Cosmological Constant. To get his theories to match the supposed ideal of a stationary universe, Einstein found that he had to multiply his answer by a certain factor, much to his consternation. Later Hubble discovered that the unverse was not constant but rather expanding. So Einstein scrapped his Fudge Factor, but never came up with a proper equation. With great irony, the fudge factor is now back in the form of supposed dark matter and dark energy. The existence of dark matter is posited to provide "gravitational glue" to explain why local clusters of galaxies remain clustered. Meanwhile it has been discovered that more distant galaxies are not merely moving away from each other at a constant velocity, which could be explained by the Big Bang. Rather they are accelerating. There is no explanation for this other than to fudge and posit the existence of dark energy which is supplying the force necesaary to produce this acceleration.
Meanwhile relativity theory, in trerms of frames of reference, is our friend inside the amp; e.g.: "the grid is more negative than the cathode". To the outside observer the grid is at -0- VDC. But if the cathode is at +25VDC, -0- is more negative than that. From the frame of reference of the cathode, the grid is at (-25)VDC.
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... is now back in the form of supposed dark matter and dark energy. The existence of dark matter is posited to provide "gravitational glue" to explain why local clusters of galaxies remain clustered. Meanwhile it has been discovered that more distant galaxies are not merely moving away from each other at a constant velocity, which could be explained by the Big Bang. Rather they are accelerating. There is no explanation for this other than to fudge and posit the existence of dark energy which is supplying the force necesaary to produce this acceleration.
jjasilli = darth vadar! Use the force Luke! :laugh:
(jjasilli I'm only kidding)
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When I was taking electronics in College my Instructor would say "Voltage positive or Negative only depends on where you stand when you spit. :grin:
Tony