Volt Amp Turn Relationship

Discussion in 'Electronic Design' started by Ian Stirling, Apr 22, 2004.

1. Ian StirlingGuest

What do you mean by secondary amperages are so high.
The current will only be that high if you put a small resistance across
Don't do that.

2. Randy GrossGuest

Greetings,

I'm having a little trouble developing a power supply to deliver 14 vac
@ 10 amps. I modified a MOT to deliver 14 volts but, the primary (5
amps) and secondary (41 amp) amperages are so high, I had to put a large
Iron cored inductor in series with the sec. to reduce the 41 amps down
to 10. The power supply dwarfs the original design perimeter. I need to
design a tranny to deliver what I need in a smaller package. What I need
to determine is the primary to secondary turns and core size to deliver
14 vac @ 10 amps.

I have formulas for determining/modifying the volt amps in an existing
design and, I understand these. What I don't know is how to target a
specific set of values and how the core is used to achieve them. If you
know of a source of information, please direct me.

I realize I could find and purchase a supply to fit this need but, what
would I learn?

Randy

3. Robert BaerGuest

14V times 10A = 140W; for a 150W transformer, here are the numbers for
E-I laminated core:
212 turns for 110V primary, 1.96 turns per volt (all windings), 3.5
square inches cross-sectional area (which is 2 times center core area).
For wire size, i use the wire of cross-sectional area in circular mils
equal to or greater than the milliamps drawn at full load.
I never get copper losses and always have space to spare for the
windings.

4. John WoodgateGuest

I read in sci.electronics.design that Robert Baer
Where can I get this room-temperature superconducting wire?
So how hot does it get? Oh, of course, you have no copper loss, so the
iron loss is the only cause of heating. Or have you eliminated that as
well?

6. John PopelishGuest

(snip)

I am not at all concerned if you dot your i's and cross your t's. The
problem you described did not narrow down the possibilities enough for
anyone ot comment on sensibly.

Was the primary current you mentioned what you got with the secondary
side open, loaded or shorted? Was the secondary current what you
measured with the secondary driving some load or short circuited?

You simply didn't describe what you were measuring with enough detail
for us to imagine what was happening.

7. Ian StirlingGuest

What's wrong with a fuse?
Or if you want to get fancy, a circuit breaker?

8. John PopelishGuest

THis is very high for a 140 watt design. Sounds like the core is
coming close to saturation and needs some extra turns on the primary.
You might be surprised how many percent this current will go down with
1% more turns.
Is that open circuit (no load) voltage?
Obviously such low resistances are hard ot measure accurately with an
ohm meter. But if the voltage applied was really 14 V and the current
really was 41 A, then the actual resistance was 14/41=0.34 ohms.
The voltage was 1.5 volts less than when you loaded it to 41 amps?
http://www.micrometals.com/material/pc_coreloss_txt.html

N= number of turns, E = RMS sine wave volts, F= hertz, A = cross
sectional area in square cm, B = peak flux in gauss
For silicon iron, I think you need to keep the peak flux down to
something like half of the 15,000 gauss saturation flux.

So...

N=120V * 10^8/(4.44*60Hz*7.14cm^2*7500*gauss = 842 turns

But I can't believe that is right for a core that big. It seems like
way too many turns. Even if you allow the flux to swing all the way
to saturation in each direction you still get 421 turns. How many
turns are typically on the primary of one of your microwave oven
transformers?

Wait a minute. 2.8125 in^2 = 2.8125*2.54^2 = 18.1 cm^2

so if the area in square inches is right,

N=120V * 10^8/(4.44*60Hz*18.1cm^2*7500*gauss = 332 turns

That's a bit better.

I'll bet your primary is somewhere between that and half that.

9. Randy GrossGuest

Sometimes I forget that I'm corresponding with learned professionals who
dot the i's and cross the t's and, assume little.

The first stage of my goal is a dedicated, unregulated 12vdc supply,
rectified and the ripples smoothed. The device this supply is dedicated
to is a dc motor: shunt wound, 2800 rpm, 10 amp with an internal
resistance of 2 ohms. The load placed on this motor will not vary, once
started the rpms will remain constant until it is stopped.

My concern arose when I examined the potential of the potential I chose
to employ. This is much more power than the device will ever need and,
in the event of mishap, has the potential to completely destroy my
efforts, not to mention expense. I would be much more comfortable if I
brought the maximum to just above need.

Randy

10. Robert BaerGuest

I use plain ordinary copper wire in the transformers; using the
equations for winding, the transformers barely warm up, and using the
"circular mils = current in mils" rule of thumb, one cannot see the IR
drop in the windings (it is that low in all cases).
One of the transformers i wound was for a regulated 3.3v 100A power
supply.
The rectifiers and the pass transistors got rather warm, but the
transformer kept cool (not cold as if off) to the touch.

11. Randy GrossGuest

Greetings John,

Disrespect was not my intention. More often than not, I assume that
others know where I am headed with a quest when really it was ambiguous.
I attempted an apology.

The test I used (and still do) was the one you gave me back in 2000 when
I needed a high current load to test modified transformers. Your method
entailed immersing approx. 15 feet of 18 ga. wire in cool water and
running the output current through it to dissipate the heat. It works
perfectly up to about 80 amps.

Details: Primary current 5 amps, no load on secondary.

Secondary was rewound from 2000vac down to 14vac.

Using the wire test above, as a high current load, the secondary
produced 41 amps at 14vac. Resistance of the load measures 1/2 ohm.

The second test I performed was to use a 2 ohm resistance as the load to
match the internal resistance of the motor. The current drawn was 7.1
amps and some major heat dissipation. Voltage drop across the resistor
was 12.5vac.

The cross sectional area of the core measures 2.8125 sq. in. or 7.14 sq.
cm.. This core can easily be converted from 700VA to a 150VA transformer
and bring the output closer to the needs of my design. I was just
unclear as to how to calculate the turns of the primary to produce the
goal of 14vac at 10 amps in the secondary.

I ran across these formulas from a Google search:

N=(E*10^8)/(4.44FAB) or N=(k*E)/A , k=6.256 @ 60HZ

I just wanted to confirm that I am on the right track.

One more word about dotting the i's and crossing the t's, In the future
I'll try to be less mental and more descriptive. My wife says I talk in
riddles. I'm beginning to understand.

Randy

12. John WoodgateGuest

I read in sci.electronics.design that Robert Baer
So you would rate 19 AWG at 1.288 A. I use 3000 A/sq.in, and 19 AWG has
an area of 0.001012 sq. in. so I would rate it at 3 A. I believe modern
transformers use even higher current densities.
Then your transformer is larger, heavier and more costly than it need
be. If you don't mind about that, you're home and dry.

13. Robert BaerGuest

First, you are incorrect on my current rating method.
Number 28 wire has a cross-sectional area of 159.8 circular mils, so i
use it for currents not exceeding 160mA.
Number 24 wire has a cross-sectional area of 404.0 circular mils, so i
use it for currents not exceeding 400mA.
Number 20 wire has a cross-sectional area of 1022 circular mils, so i
use it for currents not exceeding 1A.
Number 18 wire has a cross-sectional area of 1624 circular mils, so i
use it for currents not exceeding 1.6A.
Number 19 wire is not available except on special order, which i have
found to be the case for *all* odd-number size copper wire.

Second, concerning the transformer sizes, my equations, back in the
1950s to 1980s, resulted in zero physical size changes to handle a given
power.
More recently, the silicon steel alloys used for power transformers
have improved, resulting in smaller commercially made transformers.
Here, one could either accept the implied de-rating and require a
larger core - which would be the same size as those made in the 1940s
(or so), OR wind according to the present core capability -which would
give no size penalty as mentioned.

14. John WoodgateGuest

I read in sci.electronics.design that Robert Baer
Nothing incorrect: I just chose 19 AWG because its area give easy
arithmetic. The fact that it's not freely available is utterly
irrelevant.

The rest of your text simply doesn't make sense.

15. R.LeggGuest

How does your iteration compare (in size and weight) to a commercial
product rated for a 50degC rise, with the same ratings?

While commercial products do tend to be rather cost conscious,
intermediate wire guages where necessary, to provide a smaller part
with a ~ lighter shipping weight.

While size and temperature rise may not play an important role in
one-off prototyping, we are all using roughly the same tools and
materials in 50/60Hz isolation devices.

This results in a situation where the 'winner' may have the insulation
system with the higher permissible operating temperature, smallest
package or lowest assembly/shipping cost. Every application has it's
own priorities, and every sale has it's clinching salient hook - one
that may have nothing to do with 60Hz magnetics performance at all.

This shouldn't prevent a designer from tackling the goal of optimizing
designs within the operating environmental constraints of the power
level addressed. In order to do this, it is important to refer to
specific material grades, lamination sizes / stack heights and
temperatures predicted/recorded (in degrees of a recognizable scale),
when describing an iteration.

I'm not sure what formulas you were refering to earlier, unless you
meant the original poster's web reference, ie:
( members.tripod.com/~schematics/xform/xformer20.htm ) If this is the
case, then core size is being selected by means of a curve-fitting
guideline that does not reference either target magnetic material
grade or design temperature rise. While these obviously entered into
the equation originally at somepoint in time, being related to power
loss at the throughput levels charted and flux density proposed in a
square lamination, they should be stated.

Unlisted insulation systems are usually restricted to applications
where rises are demonstrated not to exceed 30degrees C above the
ambient temperature. Listed insulation systems for higher operating
temperatures are offered for use, free of charge or reassignment, by
many varnish and insulation material manufacturers.

http://www.pdgeorge.com/
http://www.dolphs.com/
http://www.pleo.com/index.html
http://www.e-insulationsys.com/english/select.lasso

RL

16. Robert BaerGuest

In reference to this particular thread, i have not disclosed my
formulae (ie: plural); on at least two previous occasions, i had
responded with the equation set.
While i have never measured the temperature rise of any of the
hundreds of transformers that i have rewound, as long as the load was
within specifications, the transformers, at worst, got slightly warm.
If i use laminations from a modern transformer, and apply "my"
equations according to the core size, then i get a transformer with a
lower volt-ampere rating than the OEM rating.
Using laminations from an "old" transformer results in no "penalties".

17. Robert BaerGuest

In reference to this particular thread, i have not disclosed my
formulae (ie: plural); on at least two previous occasions, i had
responded with the equation set.
While i have never measured the temperature rise of any of the
hundreds of transformers that i have rewound, as long as the load was
within specifications, the transformers, at worst, got slightly warm.
If i use laminations from a modern transformer, and apply "my"
equations according to the core size, then i get a transformer with a
lower volt-ampere rating than the OEM rating.
Using laminations from an "old" transformer results in no "penalties".

18. R.LeggGuest

I'm afraid that without a more specific reference or link, I'm unable
to intelligently comment on the formulas in question.
I suggest that this is not an advantage, except that it allows
reliable replacement with a commercial jellybean sporting the better
VA rating in the same real estate, if required at a later date.
Winding low frequency magnetics is both time consuming and labour
intensive. There has to be a reason to do it. Achieving superior
performance is a minimum requirement before I would ever bother to do
so.

http://members.shaw.ca/legg/1985b.html
Though these do not age, their guage and material grade are fixed and
possibly unidentifiable. The penalty will be in having to assume a
worst case material grade and in being restricted to the original
guage used. The more handling that an individual lamination gets, the
more unruly it behaves in the stack. I would only attempt it if the
original set were unvarnished, or when attempting to restore/repair an
accurate antique using original materials.

RL