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.