Re: Sam's HIGH POWER Ignition Coil Driver. Record pulses, driver dead:-(

> Before using the MOT I measured 2500V at 1.98A on the secondary
>output. As it seemed way too high, I measured the input: 220V, OFF
>SCALE on my 20A meter.
> Well, all seems to indicate that the MOT was running at 4500W (no
>exaggeration here! I really did measure these outputs. Those MOTs
>should not be considered shunted transformers).

Hello Sam.

What sort of load was your driver connected to when you made these
measurements on the secondary output?

With alternating currents, one must not (in general) calculate a power by
using the product VI. In the general case where:

V = Vo*cos*(w*t) and I = Io*cos*(w*t + p)

in a load, the average power is:

Pav = (1/sqrt(2))*Vo*Io*cos(p) = Vrms*Irms*cos(p)

where the term cos(p) is called the 'power factor' of the load, and is
usually quoted as a percentage.

Why is this important? Well, a common problem arises when large amounts of
power must be transmitted from a source to a load over a line of
appreciable resistance. Care must be taken to adjust the power factor close
to unity as the component of I that is an quadrature with V produces no
useful power in the load, but nonetheless gives rise to a power loss and to
a voltage drop in the line connecting the source to the load. 

As an example, the power factor of a resistor is cos(0) or 100%, and that
of a capacitor is cos(90) or 0%. Real inductors have both resistance and
inductance (if the frequency is quite high they become capacitive). If R =
w*L, then p = 45 degrees, and the power factor is cos(45) or 70.7%. 

Safe coiling,

Gavin Hubbard