F670 T103/203 transformer question

[CJ]

Area Product  Y-6431 >  Ap = 1.786 cm^4


what  can we do with this number?  (to see how we derived it, see above)

all kinds of cool stuff,

Volume of XFMR  = Kv * Ap^3/4      where Kv = a constant for lams of 19.7

first, lets use the Y^x button on the calculator to get rid of that 3/4 power: 1.786 ^ 3/4 = 1.545

so from the area product, we can easily get the volume of copper and iron:

Vol = 19.7 * 1.786 ^3/4 = 30.43 cm^3

hey, this is easy, what about weight, you say?

Weight = Kw * Ap^3/4

for lams, the Kw constant is 68.2

Wt = 68.2 * 1.545 = 105 grams

lets go for Surface Area >

Surface Area = Ks * Ap^ 1/2

1.786 ^1/2 (square root) = 1.336,    the constant Ks of lams is 41.3 , so

Surface Area = 41.3 * 1.336 = 55.2 cm^2

one more, Current Density - J

J = Kj  * Ap^ -0.125

or J =Kj / Ap ^ .125    if negative exponents are not your thing,  :-X

the constant for Kj for lams is 366 for a temp rise at 25 degrees

Current Density = 366 * 0.930 = 340 Amps / cm^2

so these transformer engineers have it easy, the Ap numbers for each lam are in the catalog,

all they do is plug in a constant to get all the data they need, bastards! 

so your formulas and results for the UTC Y-6431 



Vol = Kv * Ap^3/4
Weight = Kw * Ap^3/4
Surface Area = Ks * Ap^ 1/2
J = Kj  * Ap^ -0.125


Vol = 19.7 * 1.786 ^3/4 = 30.43 cm^3
Wt = 68.2 * 1.545 = 105 grams
Surface Area = 41.3 * 1.336 = 55.2 cm^2
Current Density = 366 * 0.930 = 340 Amps / cm^2

 

[Moby]

ok, i am doing my part, who is going to handle the 670 pad calculation?

Are you talking about resistors R145,146,147,148,149,150 in front of T103?

 

[CJ]

no, i was talkin about R245, 246, 247, 248, 249 and 250. 

lets get on the same page here, 


well, lets have a closer look...

ok,




Go!    ;D ;D ;D ;D

 

[CJ]

while you are working on the pad, check this out:

a power xfmr has to support>

a coil/core on the pri,  and a coil/core on the sec

lets say we want to build an iso transformer, 120 in, 120 out,

we need 1 amp to drive a big ol Tektronix vac tube scope,

1 amp at 120 is 120 watts, are you with me so far?  :-\

because if you are, this next part will blow your mind>

while the sec is keeping us happy with 120 at 1 amp, but guess what?

the pri has to do the same thing.

this ain't no werid pulse transformer, or switcher, this is sine wave in, sine wave out,

so, this has the revealing notion that >

the core has to really handle twice the power as the customer wants out the ass end.

well blow me over with a sand storm, that is a good one, i would like to see the math?

ok,

Apparent Power = Power In + Power Out

now we have losses in transformers, but they still get  95% most of the time,

so use 0.95 and we can modify this for the real world,

Power In = Power out / n , where n is efficiency, so

P in = P out / 0.95  -------    (eq 1)

P t (apparent power) = P in + P out, and

we can sub in (eq1) to simplify:

P t = P out / 0.95 + P out ,      see that common factor of P out?  remove it:

P t = P out (1 / 0.95 + 1)    , now we are approaching the desert counter:

Apparent Power  = 2.05 * Power Out

this is the number the engineers would use to design with,

so the transformer, with losses has to handle 2.05 time the actual power you want.

 

[CJ]

coming up, we have a  useful application for Area Product, finally!

there is a formula that looks real dangerous, but it is very easy to tame,

Area Product  = Apparent Power * 10^4 ^1.16 /( K  B max  f  Ku  Kj)

translated into english, this really says that:

if we know the power, we know the lam,

let us build a custom equation for input transformers,

use Supermalloy at 20 hertz and we will have a very simple formula,

Constants:

K = wave coef. = 4.44 for sine wave,  4.00 for  hendrix  ;D

B max = 0.5 Tesla for Supermalloy,

f = Frequency = 20 hz

K u = Window Utilization  = 0.4 ( how much copper actually gets through the core)

K j = Current Density = 366 for lams ( 322 for C cores if  interested )

so plug that into the whole enchilada and pray to the calculator gods,

Area Product = (P t * 10^4)^1.16 /( K * B * f * Ku * Kj)

Ap = (10,000 Pt)^1.16 / (4.44 * 0.5 T * 20 hz * 0.4 Wu * 366 Kj)

Ap = (10,000 Pt / 6500)^1.16

Ap = (1.538^1.16)

Ap = 1.648 * Pt

what this says is that for any Supermalloy input xfmr we want to build,

1.648 times the apparent power will equal the lam area product.

an example is worth a thousand paragraphs, so here we go...

lets us the Y-6431 since we already know it is Supermalloy, we want it down to 20 hz,

and we already calculated the area product.

For UTC Y-6431, the Area Product for the 31 UI lam and 15/32" =  1.786 cm^4

So if Ap = 1.648 *  Pt, then

1.786 cm^4 = 1.648 * Pt , solving for Pt,

Pt = 1.786 cm^4 / 1.648 = 1.0837 ,

since Apparent Power = 2.05 Power out,  that lam stack will yield:

1.0837 / 2.05 = 0.5286 watts of input power max, at 20 hz

so the UTC Y-6431 handles about 500 milliwatts at 20 hertz, totally maxed out.

is that a good number or not? use the db scale to find out:

RDH4 says 500 millwatts = about 19db    6 mw ref into 600 ohms

UTC catalog says 18 db for the HA 100X

so that ain't too shabby for such a huge formula with all those hand derived constants

now you could reverse this process:

"I want a supermalloy input that will handle 30 db at 20 hz, and i want it now! 

which lam?

easy, from our formula  Area Product = 1.648 * Apparent Power ,

and RDH4 saying 30 db = 60 volts and 6 watts, ( i like nice, big  round ones!  ;D ;D ;D ;D)

6 watts doubled is 12, thats what the core will see, (apparent power) so,

Area Product =  1.648 * 12 watts = 19.776 cm^4

we could use a lookup  table for finding the lam that has an Ap of  19.776 cm^4

those are really hard to find online, but  Grossner says he has a plan:

the Ap of a square stack E Lam is:

Ap = 1.5T * 0.5T * T^2    (T stands for Tongue)  transposing,

T = (Ap/0.75)^1/4 , insert the Ap we already have up top,

but watch out! > mathematical bear trap ahead, remember the Hubble?

Grossner uses inches, McLymann is metric, so we need the Ap in inches cubed,

1 centimeter^4 = 0.0240250961 in^4

19.776 cm^4 = 0.475 in^4

0.475/0.75 = 0.6333

0.6333^1/4 = 0.89 inches

So we need a square stack of Supermalloy with a tongue width of about 0.892> 0.875

0.875 = 7/8 inch

This is a common lam, we call it the type 87 EI

so a 30 db Supermalloy input xfmr would be bigger than a Neve output, I like it!



ok, any questions?
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good..
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because i am all transfomered out and shot to h*ll!

;D

next up, enuff theory, lets go to.............The Lab!

 

[CJ]

just got done taking the Y-6431 down,

32 lams, not 33, 

49 Ni, not 80 Supermalloy, so this thing should take quite a bit more level than a HA100-X

3 internal cans, exactly like the HA100-X,

has a shield between the pri and sec,

the sec is not wound with two pies side by side like the A-10-HA100-X,

the sec is wound as one pie, 230 turns per layer,

the shield wire was connected to all 3 internal cans, as well as the core and  case,

 

[abbey road d enfer]

the core has to really handle twice the power as the customer wants out the ass end.

well blow me over with a sand storm, that is a good one, i would like to see the math?

Not quite, CJ, because the flux created by the secondary is opposite to the primary flux (the age-old action-reaction law). Even if there was no secondary, or disconnected, the core would still be at whatever induction you've chosen to operate it. The algebraic sum of primary + secondary induction is a constant (almost).

 

[vari-mu]

Thank CJ , you are transformers Encyclopedia !!!

 

[CJ]

yeah, abbey, i did not explain that very well,

it sounds like i am saying that you need twice the power to power a load,

or that the flux is doubled, or something like that, it relates to the transformer itself,

this is a good read:

"The concept of a resistive load or working power is fairly easy to understand. For example, a light bulb that consumes 100W of power generates 100W worth of heat and light. That is a pure resistive load. An inductive load, on the other hand, is a little harder to understand. Think about a transformer, which has coil windings to generate an electromagnetic field and then induce current in another set of windings. A certain amount of power is required to saturate the windings and generate the magnetic field, even though no work is being done. A power transformer that is not connected to anything is a perfect example of a pure inductive load. An apparent power draw exists to generate the fields, but no working power exists because no actual work is being done.

When the transformer is connected to a load, it uses both working power and reactive power. In other words, power is consumed to do work (for example, if the transformer is powering a light bulb), and apparent power is used to maintain the electromagnetic field in the transformer windings."

so we are talking about the power needed to make the transformer function.

luckily for us, this whole confusing concept has been avoided:

the lam makers already figure this into the catalog specs, so you never see the math. 

 

[CJ]

it is fun to do an approximation on what you think is going on,

then actually find out in real time if it's real or not, 

so i took apart the Y-6431 over in the lab and this is what we have for input level,

the stack is 7/16 in * 5/16 in = 0.1367 in^2 , 1 square in = 6.4516 square centimeters,

Answer: .13671875 in² = 0.882054 cm²

i also got to count the exact turns on the primaries,

total primary turns = 480 * 2 coils = 960, so we had that right,

so we have B = 100,000,000 / (4 stacking) (0.8821 cm^2 lams)(20 hz)(960 turns)

= 100,000,000 / 67745 so for the Y-6431,

B max = 1476 E,  where E is Volts - rms,

i also found out that the lams are of 49% nickel alloy, so we can boost the input level:

Permalloy 49 : B max 12,000 gauss, so set that equal to our core:

12,000 gauss = 1476 E

E = 8.13 Volts rms = 18 db (1 mw ref)




you can do this: 960 turns / 8.13 volts = 118 turns per volt

so for this core, if you wanted 10 volts instead of 8.13, to would have use:

10 volts * 118 turns / volt = 1180 turns instead of 960 turns.

if you only wanted 1 volt, like a low level mic, you would only need:

1 volt * 118 tuns /volt = 118 turns ,

once you get the core constant for that lam and that stack, the rest is duck soup.

so, the more volts you want, the more turns you need in order not to saturtate.