Transformer tests good but questions..

[petemoore]

  Here's my problem, the DMM I have doesn't like to measure the tranny coil ohmages.
  But...'right' side has 5 wires
  two red, two orange, one black
  Left side has 4 wires
  Yellow and white
  two grey
  >which side is actually the primary?
  When 5.6vac is input to the left sides Y/W wires, I can get 10.2vac from the right side Red and Red wires.
  ...then slick found a rectifier circuit, 4 diodes, some caps and nice post type connectors and wire tabs...and easily connected the rectification circuit, perhaps I can find the link again [daft] I have it printed, was about to use parts off the board..then realized it's wired exactly like the schematic I have...
  and have a nice 11.4V*DC, which is working to 'quietly power an Lm386 testamp...powers it right up too !
  While I could use it at this voltage, or 2x this voltage as a supply, and a very nice one at that, I got to thinking about maybe the 'other' side is how it was originally wired.
  I would like to try transforming and rectifying AC117V from wall plug, and even though the wires on both sides of the transformer look quite hefty, and in fact I can find smaller AC line wire [sort of like a fuse], and I should use a fuse [and probably will when I can calculate it..after I figure out what I want DCV wise],
  So the question, what if I put 120VAC across the hefty wires of the secondary?
  And..this came out of a transistor amp, was the transformer used to step up or step down AC voltage, I get the idea maybe step down, all the amplifier circuits I can find [mosfet] are running at under 100v.
  The amps I have are mosfet source followers, ~10ohm source resistors, [gate biased], the transistors are rated at 100vdss, what would I gain by running them on a higher voltage power supply [they sound pretty good as low as 15v]?
  I tried step up and step down voltages, getting approximately 2x and 4x selectable both ways, starting with 5.6vac...I don't know where to get this type of info except here, and I want to work up the power supply to the best application for it, but...I don't want to blow a tranny or even a cap.
  So far all tests, and results have met the expectations, to the point of powering an amplifier !!
  Other than the DMM lead/testclip connection failing [and my cutting, then patching the PS which looked dead], then the testclip pinch connect I patched to the DMM lead failing...mahn...really gotta keep an eye on those DMM lead problems tonight...not too bad, I figured out it might be the wires about the time I started swapping wires from my other cheep DMM.
  Also I'm wondering about the pitfalls of operating two Mosfet Amps from one supply source...seems like all I'd have to do is add filter caps?
  Please excuse my "HV" talk, and accept my warning about working with higher voltages.

[mdh]

I would want to have a better idea of the winding ratios before I started applying 120VAC to any pair of wires.  I couldn't find the page I thought I had seen at GEO (your other transformer thread), but I did find this:  http://www.guitaramplifiermagazine.com/pub/2005/premier_issue/roccaforte_techtip.html.  I still think R.G. has written something up about this, but maybe it was just in a thread here.  Have you tested voltages on all other pairs of wires (not just red) with your 5.6V source on the Y/W wires?

One thing that is noted in the page that I linked above is that the heavy wires may be low voltage (high current) secondaries.  If you think the red wires may be the primary, a safer test would be to connect your 5.6V source to those leads and see what you get on other pairs of wires.  There also may be hints in the physical arrangement of wires, or at least a small one... is the one black wire positioned in between two wires of another color?  E.g., orange-black-orange?  If so, the black wire is probably a center tap.  That also may be a hint as to the identity of the primary windings... I would guess the 4 wire side, but there's no guarantee, and there's also no guarantee that you don't have the primary and a secondary coming out the same side of the transformer.  You didn't happen to do the salvage operation yourself and pay any attention to how the transformer was hooked up?

Sorry, this probably doesn't help much, but I think due diligence is really important when you could wind up dealing with high voltage.  I would tend to take things very slowly.  Good luck.

[petemoore]

  Well starting with a small voltage, I can double, quadruple, halve, or quarter it.
  A half way good indicator I guess might be that this came from a BJT output amplifier, and I would like to see a schematic of an SS amp, using four honking transistors in a 16'' x 4'' x 3'' heat sink. Actually I'd like to see a bunch of them or one that looks like it might be for the Onkyo this came out of...then maybe I could get an idea of whether the voltages were stepped up or down from 120VAC input...
  --------
Five wires from one side, two matching Red, two matching Orange [solid core], and a black odd one.
  ---------
  from the other side only four wires.
  two light colored, two grey colored [solid core].
--------------------------
  5.6vac across red wires makes 10.7vac appear across the Y/W wires
  5.6vac across the orange wires makes 22.3vac appear across the y/w
  --------------
  5.6vac across the y/w wires produces 1.1vac across the orange wires
  5.6vac across the y/w wires produces 2.6vac across the Red Wires.
  ---------------
  I need some more Mosfet Amp schematics, or to know like..what voltages..how to get alot out of an IRFP140, perhaps even Supertex's new HV mosfets...there's gotta be some amp designs for mosfet-guitar 'anti-fidelity' around somewhere..just a basic amp run conservatively or not so conservatively.
  BTW the gain drivers for the Mosfet Source followers [which don't really amplify gain, just drive the speakers] is on a 15vdc supply [I call it a 'preamp'] and I decided on 15v because it is less than the 20v thick sheet of glass in the mosfet gate, however, I'm reading on the irfp140 data sheet +/- 20vdc as max gate/source voltage...I guess if the source is following the gate I can apply ... alot more than 20vdc to the gate swing...as long as the source can follow, and the V+ rail isn't hit...that would require a larger source resistor to lift the source some...late, can't think right...daft !

[petemoore]

  So at this point I could apply 120vac and see what pops or measure the output...not exactly the type of damage the second 50 of 50/50 might cause.   
  I'd like to be able to figure out whether my design exceeds the capability limitations of this transformer while not exceeding them.
  The numbers on the tranny:
  NTP559D
  230154
  22IF
   The can ends are 2.25'' X 4''
   The outside of the leaves are 4.5'' x 4''
  I can figure I'll need to step down voltage for use with the Mosfets I have [100V Dssv max]..
  Data on the
  SylECG181
  A7913
  or the
  Hitachi 2SA756
  or the
  Motorola Q1060-7
  Would be helpful, these are the transistors which came out of the Onkyo Reciever amp.
  I typed into google a dozen things like transistor datasheet #'s and stuff like "Power BJT'...not much of any use I can fine floating around..
  There's gotta be a amp site about repairing or working with BJT amps, there are only about 10,000,000 of them around...I'm searching nearly every wrong place I guess.
  Actually any voltage used on Power BJT's or data sheets on a 'like' type BJT would help, I searched and searched...many sites, some sell used ones, data sheets are foriegn, or cryptic or unreadable, I just wanna know like...what's the E/C voltage applied to power BJT's in a typical 'high power' Hi-Fi setup...that might ballpark the output voltage derived from 120vac and 220vac which this transformer was assigned to produce.
  So I'm still searching and looking for a BJT transistor power amp schematic, something akin to what the Onkyo was, or anything utilizing the capability of these transistors in a Power Output design.
  What was a typical BJT for power amp transistor Voltage supply potential?...less than 100VDC or more than 100VDC ?
  Perhaps someone could throw a figure out, or a place I can research...anything to get me off posting HV questions on the SB Forum?
  I'd hate to have to go get a high Ma output, low voltage AC adapter, another transformer, used so I can use my transformer...looks like what it's boiling down to, either that or take the tranny I have outside, apply the 120VAC and find out that way if it pops.

[petemoore]

  It didn't pop.

[R.G.]

Maybe I can help a bit.

If it came out of an Onkyo amp, do you know the power rating of the amp? That would tell us a lot.

Here's the short story on solid state amps.
- the power supply voltage is nearly always bipolar, and each polarity is close to the peak voltage it can put out. The only difference in peak output voltage and supply voltage is the saturation voltage across the output devices, and that tends to be a couple of volts for bipolar, up to 5V for MOSFET.
- knowing the output power and impedance tells you the power supply voltage. If it puts out 100W into 8 ohms, that's really "100W of a single sine wave with no distortion" so the peak value of voltage is the value that gives 100W into 8 ohms. Since P = (Vpk^2)/2R, we can solve for Vpk as Vpk = SQRT(2*R*P). So 100W into 8 ohms is Vpk = SQRT(2*8*100) = SQRT(1600) = 40V pk. This would require a +/-40V power supply at minimum, and probably more like +/-44V. So the transformer winding would need to provide the AC that rectifies to 88V, or 88/1.414 = 62.5Vct.

That means that the transformer will have a beefy winding producing 60+Vct.

For 200W into 8 ohms, the voltage needed is Vpk = SQRT(3200) = 56.7V, so the transformer has to put out about 120Vct.
- if the Onkyo was a MOSFET amp, they often have auxilliary windings to make maybe 15Vdc to float on top of the main power to provide gate drive for the MOSFETs.
- many solid state amps provide +/-15V or 30Vdc, about, roughly, to run auxilliary circuits.

The power rating of a transformer is pretty much linear with MASS. If you want to find out how much power that thing will put out, find a known power rating transformer that's about the same size. That's real close to what this one will do just on a watts basis.

- transformers don't care what AC voltage you feed them as long as it's not too high or too slow. Get the voltage too high or frequency too low and the core saturates. A saturated core can no longer limit the magnetizing current from the wall socket, so the wall socket current pours through limited only by the wire resistances. Smoke and bad smell ensue.

- transformers DO care how much current you feed through the wires. Too much current overheats the wires, melts the insulation, and make smoke and bad smell.

[petemoore]

Maybe I can help a bit.
  Thank you for trying to help me.
  If it came out of an Onkyo amp, do you know the power rating of the amp? That would tell us a lot.
  I can only really guess, and then go down and look for remnants which may indicate the rating, 35w+...

  I'm going to try to study the figures you've provided...

[petemoore]

- knowing the output power and impedance tells you the power supply voltage.
  If it puts out 100W into 8 ohms, that's really "100W of a single sine wave with no distortion" so the peak value of voltage is the value that gives 100W into 8 ohms. 
  Here's a definition for 'peak foreward voltage':
  >The maximum instantaneous voltage applied to an electronic device in the direction of lesser resistance to current flow<.
  I see...has to do with headroom, so that the power supply rails + transistor voltage use is adequate to swing the transistor 'cleanly'.
  Since P = (Vpk^2)/2R, we can solve for Vpk as Vpk = SQRT(2*R*P). So 100W into 8 ohms is Vpk = SQRT(2*8*100) = SQRT(1600) = 40V pk. This would require a +/-40V power supply at minimum, and probably more like +/-44V.
  Not to pick it apart but to dissect it!
  'P=' means: Peak = what I guessed the definition is?
  P = (Vpk^2)/2R
  (.......) means work the equation inside the parenthesis first?
  V means voltage pk means peak, so Vpk means the swing of the signal + the transistor 'breathing room'
  '^'...this symbol means what?
  / means divide
  2R...means...an 'R' with a 2 before it which means I would ____ to work this symbol in the equation?
------
  Vpk = SQRT(2*R*P)
  Voltage peak = source transistor resistor threshold
  beyond me...sorry, I'll leave this here and try to come back to it. Thank you for trying to feed me, I'm sure the food is great, but a bit rich for my digestion.
  So the transformer winding would need to provide the AC that rectifies to 88V, or 88/1.414 = 62.5Vct.
  I need to take a look at a bipolar supply amp, the center tap of the output would probably make a fine supply.
  I get a glimpse of it, the voltage needs to be high enough to allow signal swings, the current needs to be adequate or the smoke comes out..and I can get the voltage thing figured out using approximate calculations supplemented with DMM readings...it is a 'real big' transformer, and was running 4 transistors and other stuff, I believe it'll run a couple mosfets...according to the much smaller WW's which didn't blow under the same load.
  I have gotten 2x, 4x, step up and step down voltages, also other voltages AC.
  I stepped up to 18vDC [after rectification] from 5.6vac, and ran an LM386 amp nicely and pretty quietly.
  However I tried 48VDC output from the rectifier and filter caps, and although it worked to power the Mosfet amp, the noise level was high, then got higher after a few seconds. And the transformer and PS circuit were in close proximity and not shielded from the Pre and amp..but it just 'felt' like direct coupling noise, as in not filtered enough.
  The rectifier circuit is 4 big diodes, and two caps 500uf and another one same size, I didn't pull it up to look at the print on it...perhaps there is adequate filtering applied with this...or is there additional filtering which I could add to try out?
  I found another transformer, this has a two lead primary and a two lead secondary [2 wires same color on each side], and produces about a 15:1 AC voltage conversion. I'll try to leave impedance matching speaker to transistor or plate through iron for another time, see if I can hack 1 'power output and impedance equates to supply voltage summing for the now.

[petemoore]

knowing the output power and impedance tells you the power supply voltage. If it puts out 100W into 8 ohms, that's really "100W of a single sine wave with no distortion" so the peak value of voltage is the value that gives 100W into 8 ohms. Since P = (Vpk^2)/2R, we can solve for Vpk as Vpk = SQRT(2*R*P). So 100W into 8 ohms is Vpk = SQRT(2*8*100) = SQRT(1600) = 40V pk. This would require a +/-40V power supply at minimum, and probably more like +/-44V. So the transformer winding would need to provide the AC that rectifies to 88V, or 88/1.414 = 62.5Vct.
  SQRT(2*8*100) = Vpk.
  Kinda Pretty simple, the '2' just kinda like...got in there...remembering that and...
  ^(2*X*Y)
  [X being speaker ohmage and Y being the wattage figure].
  ^(2*X*Y)...^(2*X*Y)...repeat...^(2*X*Y)...ok...^(2*X*Y)...I think I got it...wait...'the square root of '2' X ohmage X wattage'...yupp...lol !
  That's for just one side of the signal swing because then you've added the +/-...as in '+/- 40v' which equals total 80v of voltage swing.
  BUT...Youch..that's a bundle !
  Abstractions and takings for granted that's's...I'm beginning to see this is how it works, amazing that numbers can follow these algorythms..truly fantastic, reveals alot about the way things work..., as applied...
  Thank you for showing me these equation RG!
  and explaining it !
  Not that I can remember and re-create yet, ..fill in the knowns and work the 'unknowns to known' to find the answers to the equations without error...but it is very cool, super cool to see the equations work.
  Math actually works, and it is quite a feeling "Ahhh' HA' an relief when the values begin making sense...
  No wonder I didn't notice much difference with my 2w circuit into 8ohms when raising the supply voltage above about 15v.
  When you say 'Bipolar'...without a schematic reference...I'm not certain...this would be a push-pull amp? like one transistor working the - swings and another transistor working the + swings ?
  ^(2*X*Y)
  ok. for a 4ohm speaker @ 50w...
  ^(2*4*50)
    (2 x 4 x 50) = 400
  ^400 = 20.
  ..means I'd need +/-20V [plus whatever the transistor uses to operate, well give 5v to operating needs]...so +/- 25v supply voltage would produce 50w when presented to a 4 ohm load.
 
   
 
 

[petemoore]

 I typed "power supply filter" into google and found this from AMZ:
  This quick tip is about how to obtain cheap and effective power supply filter inductors. It's simple really; there are many inductors sold as noise filters for automobile sound systems. Even Radio Shack has a 4 amp inductor for $3.99 that can serve well in the power supply of your pedalboard (RS Part No. 270-030A) and MCM Electronics, Parts Express or other online suppliers have them as well.

The inductor (or choke) looks like a small transformer and is used in series with a capacitor filter to remove ignition noise or alternator whine from car stereo systems. The wire on the Radio Shack part is quite heavy to carry the 4 amp load and can easily handle the power requirements of effects circuits
  Seems like a choke might make a good addition to my amp power supply, I don't think I need more than 4 amps...
  Thank Jack Orman, and RS.
 

[petemoore]

  ^(2*X*Y)
  ok. for a 4ohm speaker @ 50w...
  ^(2*4*50)
    (2 x 4 x 50) = 400
  ^400 = 20.
   >>>this doesn't look right when I shoot from hip calculate...
  '100w @ 8 ohms...wouldn't that be about the same as 50w @ 4 ohms'
  ^(2*4*50]
  2x4=8
  8x50=400
  the ^400 = 20
  ----
  20x20=400
  Maybe I'm messing up the equation...still looks right to me in print though.
  I'm just trying to write an equation up using random, easy to work numbers to test my math when I'm trying to  solve: Vpk as Vpk = SQRT(2*R*P) with R being 8 [ohms] and P being 50 [watts].
  Ok...I left out the -/+ part, the signal swings -/+ 20v, so for 50w @8ohm, I'd want to start with ~40v '+' some for 'usages'.  Sounds about right.
 

[R.G.]

Pete - I typed one of my long, involved explanations, hit "post" and it both never appeared, and I can't get it back.

Let me shorten it up a bit. You're well on the way to understanding this. I was speaking in techno-math-shorthand and there's no reason you should understand it if you haven't been forced to sit through as much algebra as I have.

Some terms:
P= power (as I used it at least)
Vpk = peak voltage. Signals which are symmetrical aroung 0 have equal positive and negative peaks, so the
Vpk-pk = Vpp = peak to peak voltage, which is twice the peak voltage.
"^2" means "to the second power"
R = load resistance
* = multiply. Sometimes I leave this out, as 2R = 2*R
(....) = "do the calculations inside the parentheses first, then use the result of all that and do the calculations outside them

All power amplifiers are rated by their sine wave power - the output power they can produce when the output is a pure sine wave as big as the amp can make it, with very little distortion.

Computing power is easy for DC, it's just volts times amps. Four volts at two amps is eight watts. But what if you have a sine wave?

We compute sine wave power to produce equal HEATING to a DC voltage into the same resistance. In the above example,
4V*2A = 8W; this can only happen if the 4V is applied to a 2 ohm resistor.

If we want 4Vac to provide 8W into a 2 ohm resistor, we can't get it directly because a sine wave keeps changing value. The resistor is always being heated some, but how do we calculate it?

I'll spare you a bunch of integrals. It turns out that if we slice the sine wave up in to a zillion tiny time-slices, square the voltage at each time slice, average the squares over the whole AC cycle, then take the square root of that, we get a number that happens to be the same as a DC voltage would be that makes equal heating. We get this by taking the square ROOT of the Mean (or average) Square of the voltage. So 4Vrms is an AC voltage with the same power to heat a resistor as 4Vdc.

But how do we know how big the sine wave is? Again, sparing you some math, the peak of the sine wave is 1.414... times as big as the RMS number. So 4Vac RMS with a sine wave form is the same heating power as 4Vdc into a resistive load. Different waveforms have different multipliers for the RMS to peak relation. A square wave is obviously 1.000.

Going to speakers:
Let's say we want 10W into 8 ohms. If we used DC (which we can't, but if we could), then the power 10W = (V^2)/8.
or V^2 = 10W*8ohm = 80, so taking the square root to find V, then V= SQRT(80) = 8.944V.

We want to use AC voltages, so we need Vrms.
The peak value of this sine wave is 8.944 * 1.414 =12.65 V peak, and Vpeak-to-peak = 25.29Vpp

What that tells us is that we have to have a power supply of at least 25.29V to produce 10W into 8 ohms. Actually more, because the transistors we use to pull up and down on the load will not saturate to 0V.

What if we had an amp that would do this (10W into 8 ohms) and hooked a 4 ohm load up to it?

We'll assume for the purposes of this example that the amplifier can let enough current through to do this, which is not always true.

So the amp will produce 12.65V peak across 4 ohms. That is an RMS voltage of 8.944V, and P = (Vrms^2)/R = 80/4 = 20W.

That factor of 1.414 that is the ratio of the peak value of a sine wave to the RMS value of the sine wave comes out of the calculus used to compute it. It's actually the square root of two, in that 1.414... times 1.414... equals 2.000. The "..." at the end of that means that there is a never ending string of numbers going off into infinity, and that the "1.414" part is an approximation used as good enough.

We can actually calculate this stuff on peak values too.

Power P = (Vrms^2)/R  but we know that Vrms = Vpk/1.414... so we can substitute that in and get P = (Vpk/1.414)*(Vpk/1.414)/R.
If we simplify that down, the two square roots of two become a single 2 in the denominator, and
P = (Vpk^2)/2*R

And that's where the magic "2" came from in the calculation. I should have warned you about that.

So for your example, 4 ohms and 50W.

P = 50, R = 4

V = SQRT(P*R) = SQRT(50*4) = SQRT(200) = 14.14Vrms. Hey... that's just ten times the square root of two... yep.
That's 14.14...*1.414 = 20Vpeak, or 40V peak to peak. If you had perfect transistors, you could get 50W into 4 ohms out of a /-20V power supply.

For 100W into 8 ohms, P=100, R = 8, so Vrms = SQRT(800) = 28.28Vrms, or 32.24V peak and 64.5V peak to peak. So a 100W/8 ohm amplifier needs a power supply of at least +/-32V and actually more when losses are taken into account.

Since a power supply works from sine waves, we can calculate the transformer RMS voltage. In the last example, we need +/-32Vdc on the power supply (if our transistors are perfect) which is 64V peak, and that's a 64/1.414 = 45.25V center tapped (to get bipolar) transformer secondary.

Since we lose a couple of volts per transistor top and bottom, and 0.7V in the power supply diodes, a 48Vct transformer will just barely make 100W into 8 ohms under good conditions.

On the inductors. Don't do that. There is a difference between a capacitor input filter and an inductor input filter. The capacitor input filter charges to the peak of the AC sine wave from the transformer. An inductor input filter charges to much lower voltages on the output, and it's loading sensitive. An inductor filter works fine after the first rectifier caps, but it still needs some care in calculation. Don't go there until you have to.

Needing to learn something is the ideal situation for learning. You're going to be a whiz at this.

[petemoore]

  I typed one of my long, involved explanations, hit "post" and it both never appeared, and I can't get it back.
  Super-daft-it, nothing to like about when that happens, thanks immensely for retyping an excellent work.
  Needing to learn something is the ideal situation for learning. You're going to be a whiz at this.
  Awe, shuxxs, once it snaps and fits the first time, as long as I can remember how that went..letzee without looking...Vpk=^(2*R*P)...the majic '2' !
  I'm not sure what's up with the noise in my transformer AC step down/bridge rectifier PS setup, I suspect a few things.
  That the transformer was 1/2 adjacent and 6'' away from open signal wires and I didn't 'reorient' it point the field away from the wires...too tight, too HV etc.
  That the caps in the bridge rectifier aren't up to spec or specified/designed for the application [I observed the V ratings on the caps of course].
  Or that running 42vdc. on the circuits was somehow causing noise, perhaps the 10ohm source resistor on the mosfet driving the speaker, or the 2n7000 preamp using 1/4w resistors...
  Worked good stepping up 5.6Vac and rectifying to what was it...24vdc...I took so many voltage readings I forget what they all were.
  So I'm getting 2x and 4x step up and step down voltages, nice wide ranges, but still have details to work out before it's 'ready for prime time'.
  I would call the iron 'extra large', really old postal scale says 8.5 Lbs. [feels like more]
  the end cap protrusions are 2.25''x3.5'' [the 'inside square shape of sideview'
  The leaves are 4.75''/4''
  ---
  leaves thickness is 1.25''
  End cap width is 4''
  I salvaged two identical Onkyo Reciever/Amps, so if I get this PS details and application figured, I can try another...always good to have another.
  It certainly is a thrill when the realization of the fact that the math actually can work an ohmage/wattage/voltage...and RMS etc. equations is..hard to describe.
  The fact that it works at all is fantastic, add the 'curvy' math which can show this and 'super-fantastic' doesn't really do justice to how fantastic it is...
  There's alot going on, that math equations are 'allowed by mother nature' to follow and demonstrate and calculate all this...and that somehow humans figured all this out...how ?
  SomeoneS probably did extensive testing and documentation, somehow that got charted and looked like a curve that could be represented by algorythms...clear evidence of a plan IMO...
  Having great fun here, amps are working fine, I'm working on amps...lol.
  This is great !   
  I had to do some brain connection re-scrambling over a period of time before the light bulb came on! Thanks RG, this math is just too cool.

[R.G.]

You've probably heard the old saw that "magic may be defined as any technology you don't understand yet."

During the Dark Ages, alchemists and self-appointed wizards collected up bits of knowledge and spells into arcane books of magic. These "forbidden books" contained obscure descriptions of how to do spells that would cause effects that normal people could not do. Getting just the right amount of eye of newt and toe of dog into the cauldron and pouring the result over a dead cat at midnight at the crossroads was supposed to do ... magic.

The difference between that and math is - math works. You really can compute things that are impossibly opaque to most normal people. By the first definition, you're on your way to becoming a practicing wizard.