5 Watts in 60 Watts out ...

[DeepCut]

Ooops !

Ye that should have been :

OUTPUT VOLTAGE : 138 Volts DC

OUTPUT CURRENT : (138/388) 0.3556701030927835 (Calculated).

OUTPUT POWER : 48 Watts.

**EDIT** and i've not divided by 66 but only by the coil resistance which is 388 **EDIT**

My load resistor is across the rectifier output and i'm measuring from there.

I just tried to self run it, it stops immediately and the safety neon flashes very brightly for an instant.

I will try it with a puffer cap in a bit but i need a cup of tea and a smoke !

brb

Gary.

[void109]

I still think you're measuring the output current incorrectly - you place the load on the output of the rectifier, probably with a big smoothing cap to get an accurate reading with your digital meter, then you place the ground and hot lead of the meter on either side of the load to get the voltage.  Then you divide that voltage by the resistance of the load to yield the current.  Shoot me if I'm wrong, I'm new too, but that's my understanding of the situation.

I'll see if I can draw a picture in paint or something.

[DeepCut]

Thanks void.

Ye that's exactly what i'm doing.

I'm just looking for a suitable cap right now.

[void109]

haha excuse my paint skills!

[penno64]

Hi Gaz,

From one Gaz to another,

Power at the output has nothing to do with the resistance of the output coil.

What voidy is trying to tell you is -
Measure the voltage across your load resistor (66ohm)
Then use E = I x R like this
I = E/R

So if you read for example 24v -

I = 24v / 66 ohms
so I = .36 amps
and

Power = I x V
so
P = .36 x 24
8.73 Watts

Hope that helps.

Regards, Penno