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Transformer efficiency

T

Tony Williams

Jan 1, 1970
0
Tom Bruhns said:
OK, so I've been thinking well outside your box here, and I
readily admit it, but partly that's been in an effort to find out
just what the box is supposed to be in this case. I don't think
it's completely defined yet, but at least a few of the sides are
in place.

I don't think the box is too unreasonable Tom. Certainly not in
power applications, where the efficiency can be important.

The main assumptions are a fixed primary voltage, (so that the
iron loss can be said to be reasonably constant), and it is a
reasonably decent tranformer. ie, Where the shunt current is
small compared to the load current, and with reasonably decent
coupling (so that it is ok to lump all resistive losses over onto
one side).

Output Vo*Io
Efficiency = -------- = -----------------
Input Vo*Io + Pcu + Pfe.

2
Pcu = Copper Loss = Icu*R. Where R is the total resistance.

Pfe = Iron Loss, which is assumed to be a design constant.

The need is to find out if/when (Pcu + Pfe) has a minimum.

There's one of those obscure mathematical thingy's that come
in handy for this. It says...... If we have (A+B), and can
show that A*B is always a constant, then (A+B) can be shown
to be a minimum when A=B.


Vo*Io Vo
Efficiency = ----------------- = ----------------------
Vo*Io + Pcu + Pfe. Vo + (Icu*R + Pfe/Icu)

(The assumption is that Io is Icu.)

Multiply Icu*R by Pfe/Icu. Result is R*Pfe, which is a constant,
as required by the (A+B) thingy.

So (Icu*R + Pfe/Icu) is a minimum when Icu*R = Pfe/Icu, which is
the same as saying when Pcu = Pfe.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Tony Williams
) about 'Transformer efficiency', on Thu, 23 Oct 2003:
There's one of those obscure mathematical thingy's that come
in handy for this. It says...... If we have (A+B), and can
show that A*B is always a constant, then (A+B) can be shown
to be a minimum when A=B.

You can also obtain the result by simple differentiation. Since it's a
text-book result, I don't propose to post the mathematics here.
 
M

Martin Riddle

Jan 1, 1970
0
Roy McCammon said:
I don't think that you can make it zero, but you reduce it
with more turns (assuming a voltage transformer). If your
copper loss is not zero, then more turns means more copper
loss. So one goes up as the other goes down, it is plausible
that the minimum would include a balance between the core loss
and copper loss. Its not obvious to me that it would be one to one.
Perhaps its a fluke of typical core material constants.

Yes, the chart I have shows a function of 1/(Acw - Anp), and 1/Anp.
Acw - crosssectional area of the core
Anp - primary winding area.
Where these two functins meet at 1/2 Acw, yields the lowest combination of losses.
The actual power loss is related to the core material. Thus if you change material you need to
reoptimize your design to change the Pcu losses.

Cheers
 
G

Glen Walpert

Jan 1, 1970
0
I read in sci.electronics.design that Tony Williams


You can also obtain the result by simple differentiation. Since it's a
text-book result, I don't propose to post the mathematics here.

Cutting and pasting from your previous post of the mathematics:
-------------
Well, it's an example of a general principle, 'fixed loss' = 'variable
loss' is the condition for highest efficiency. For a fixed core size
and induction, the 'iron' loss is fixed, and the copper loss is
variable.

You can do it the other way round, but you get the same answer. You
can find it in Section 121 of Fundamentals of Electrical Engineering,
by E Hughes, Longmans, London 1955, but I'm sure it's also in many
better-known textbooks.

In Hughes' treatment, the only assumption is that the copper loss is
small compared with the load power.

The output power is Vs*Is*cos[phi].

The input power is Vs*Is*cos[phi] + P(iron) + Is^2*Res.

phi = phase angle of load (it doesn't affect the result),

Res = Rs + Rp*n^2,

n = turns ratio.

So the efficiency is:

Vs*cos[phi]/{Vs*cos[phi] + P(iron)/Is + IsRes},

which is a maximum when the denominator is a minimum, if Vs is nearly
constant, i.e. copper loss is small).

d/dIs(denom) = 0 if P(iron)/Is^2 = Res,

i.e. P(iron) = Is^2*Res.

Q.E.D.

----------------
and I replied:

Thanks, John. I think that one additional assumption is required to
extend the observation that a transformers efficiency is maximized by
the above equations when core and copper losses are equal, to the
statement that a transformer should be designed for equal losses: that
the 'cost' of copper efficiency is equal to the 'cost' of core
efficiency. Where 'cost' is whatever combination of volume, weight,
price etc. is important to your application.

Since the 'cost' of copper and core efficiency is generally not equal,
given any set of total cost limits, the efficiency/cost optimized
design having generally unequal losses will be more efficient than an
equal loss design.

An example of true optimization is presented in Smith, "Magnetic
Components" for the case of volume optimization (lowest volume for a
given efficiency):

Expressing core loss as C1 * B^X,
C1 = constant (actually a function of frequency and B, but constant
under design conditions)
B = Flux Density
X = Flux Density Exponent, typically around 2.5
(where C1 and X are derived from core mfgr's data sheet)

Then for various values of Flux Density Exponent, the optimum % core
loss (from the minimum volume optimization equations) is:

X core loss

3 40%
2.5 44% (most typical)
2 50%
1.5 57%
1 67%

Optimizing for something other than efficiency per unit volume would
no doubt result in different results.

But I will conceed that the equal loss rule is a reasonable first
order approximation, even though Smith correctly points out that it is
not in any way useful for the design of truly optimized transformers.
 
B

booboo

Jan 1, 1970
0
I read in sci.electronics.design that Glen Walpert <[email protected]>


The OP asked for 'maximum efficiency' in a *high-school* context. Not
for a fully-optimized design, based on post-graduate concepts.

Equal losses DOES give maximum efficiency. It doesn't necessarily give
lowest product cost, although for small transformers, the differences
are very small.



Well alrighty then !!
If I've got a transformer with 10 Watts of copper loss, and
only 5 Watts of core loss using M-6 core material... I can just
use M-20 or M50 to get more iron loss and the losses will be equal and
then I will have higher efficiency !!!

And it will be cheaper !!

booboo
 
J

John Woodgate

Jan 1, 1970
0
(in said:
If I've got a transformer with 10 Watts of copper loss, and
only 5 Watts of core loss using M-6 core material... I can just
use M-20 or M50 to get more iron loss and the losses will be equal and
then I will have higher efficiency !!!

Yeah, you carry on thinking that.
 
B

booboo

Jan 1, 1970
0
Yeah, you carry on thinking that.


But serioulsy, the efficiency of a transformer will be highest when
the ESR is the lowest. That's the big factor, despite pri/sec
balancing or winding density or anything else. (excluding $$ cost)

Have a day !
:)
booboo
 
J

John Woodgate

Jan 1, 1970
0
But serioulsy, the efficiency of a transformer will be highest when the
ESR is the lowest. That's the big factor, despite pri/sec balancing or
winding density or anything else. (excluding $$ cost)

No. Core loss doesn't appear in ESR. You see it only on the primary
side. You can minimise ESR, only to find your core loss is HUGE.
 
B

booboo

Jan 1, 1970
0
No. Core loss doesn't appear in ESR. You see it only on the primary
side. You can minimise ESR, only to find your core loss is HUGE.

Well, that would depend on which side you consider as the primary. I
can get core loss or measure ESR and drive either side of a
transformer. Correct ?

booboo
 
B

booboo

Jan 1, 1970
0
No. Core loss doesn't appear in ESR. You see it only on the primary
side. You can minimise ESR, only to find your core loss is HUGE.



Oh, and yes, you're absolutely right about the core loss... After
that's taken care of the ESR is the place to optimize...


booboo
 
B

booboo

Jan 1, 1970
0
Oh, and yes, you're absolutely right about the core loss... After
that's taken care of the ESR is the place to optimize...

Yeah, I know... I talk to much...! I better stop posting without
thinking first !
ESR would be the important thing at full power where core loss is the
lowest.

booboo
 
J

John Woodgate

Jan 1, 1970
0
(in said:
Well, that would depend on which side you consider as the primary. I
can get core loss or measure ESR and drive either side of a
transformer. Correct ?
I think you are trolling. The primary side is the side that is connected
to the supply. You can't measure ESR on one side because both (or 'all'
if you come back with a facetious remark about multiple windings)
winding resistances contribute to ESR.
 
B

booboo

Jan 1, 1970
0
I think you are trolling.

No, sorry if it looked like that... (I had to look up "trolling")


The primary side is the side that is connected
to the supply. You can't measure ESR on one side because both (or 'all'
if you come back with a facetious remark about multiple windings)
winding resistances contribute to ESR.

Nahhh.... One primary and one secondary assumed here...

Have a day !
 
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