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### Network # Less is More??

R

#### Randy

Jan 1, 1970
0
I'm wanting to set up a transformer to deliver 50% of its' power by
applying a tap at the 7th turn of a 14 turn secondary. There are two in the
circuit but, only one will be modified. The only data I have on the
transformers is from my own calculations: (ignoring losses)

Power: .7kVA (each) Pri.: 120 turns Sec.: 14 turns Ratio:
116

The formula I used to get I is:
Isec = Ipri x Tpri /
Tsec

Using this formula, I get 14 volts @ 49.97 amps. Again, using this same
formula for the tap, I get 7 volts @ 99.94 amps. Something has to be
wrong with my approach. I'm tapping for less but, according to this, I'm
getting more. What did I miss or, am I thinking in reverse?

J

#### John Popelish

Jan 1, 1970
0
Randy said:
I'm wanting to set up a transformer to deliver 50% of its' power by
applying a tap at the 7th turn of a 14 turn secondary. There are two in the
circuit but, only one will be modified. The only data I have on the
transformers is from my own calculations: (ignoring losses)

Power: .7kVA (each) Pri.: 120 turns Sec.: 14 turns Ratio: 116

I don't know what ratio refers to, here. If it was turns ratio, it
would be 120/14=8.57
The formula I used to get I is:
Isec = Ipri x Tpri / Tsec

What is Ipri?
Using this formula, I get 14 volts @ 49.97 amps. Again, using this same
formula for the tap, I get 7 volts @ 99.94 amps. Something has to be
wrong with my approach. I'm tapping for less but, according to this, I'm
getting more. What did I miss or, am I thinking in reverse?

The power the primary may be able to supply is not necessarily the
same as the secondary can deliver. If the transformer was designed to
supply 700 watts at 14 volts, that implies 700/14=50 amps. But this
also implies that the gauge of the secondary was chosen to handle up
to that much current, and not much more. If you use only half of the
turns, the remaining turns can handle a bit more than rated current,
but not twice as much. A better estimate would be about 1.4 times the
full winding current. This causes the half secondary to produce the
same heat as the full secondary at rated current (I^2*R, with R being
half of the full winding resistance). This concentration of heat is
offset by the fact that the primary will be running derated at only
71% of rated current, and producing only half the heat it was designed
for.

All that said, there is no reason to take this much current from the
transformer. That will be controlled by the load resistance. If you
have a load that draws 25 amps from whatever voltage the half
secondary delivers, that is all the transformer will deliver.

R

#### Randy

Jan 1, 1970
0
: >Power: .7kVA (each) Pri.: 120 turns Sec.: 14 turns Ratio: 116
:
: I don't know what ratio refers to, here. If it was turns ratio, it
: would be 120/14=8.57

It should have been .116, I won't use that one again.

: > The formula I used to get I is:
: > Isec = Ipri x Tpri / Tsec
:
: What is Ipri?

It's supposed to be the current of the primary, no subscript.

: The power the primary may be able to supply is not necessarily the
: same as the secondary can deliver. If the transformer was designed to
: supply 700 watts at 14 volts, that implies 700/14=50 amps. But this
: also implies that the gauge of the secondary was chosen to handle up
: to that much current, and not much more. If you use only half of the
: turns, the remaining turns can handle a bit more than rated current,
: but not twice as much. A better estimate would be about 1.4 times the
: full winding current. This causes the half secondary to produce the
: same heat as the full secondary at rated current (I^2*R, with R being
: half of the full winding resistance). This concentration of heat is
: offset by the fact that the primary will be running derated at only
: 71% of rated current, and producing only half the heat it was designed
: for.

This is what I was trying to avoid, more current! I see now that I can't
reduce the current internally without adding more turns and changing the
configuration, which is not control. I don't want to chop up the sine
either. That leaves an adjustable primary.

Thanks again John.

R

#### Randy

Jan 1, 1970
0
<[email protected]>...
: On Mon, 14 Jul 2003 23:11:22 +0100, John Popelish wrote:
:
: > I don't know what ratio refers to, here. If it was turns ratio, it
: > would be 120/14=8.57
:
: 14.8/120 is within walking distance of 0.116
:
: What a silly way of stating turns ratio!

Sorry for the confusion. As a good friend of mine is so fond of saying: I'
ve slept since then Randy Gross
:
: --
: Then there's duct tape ...
: (Garrison Keillor)
: [email protected]
:
:

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