Okay, now the result.

The input is about 10 Volts x 80 mA= 0.8 Watts.

The 80 mA can be seen, by folding the square wave
ontop the DC level to a steady value, so the voltage
gives about 8 Volts, if you integrate over all areas
of the cycles..
8 Volts / 100 Ohm is 80 mA.
You already heat via the input at the 100 Ohm Resistor
a Wattage of (8 Volts )^2 / 100 Ohm = 0.64 Watts.


Now output power at R2= 1 Ohm is
about (0.4 Volts)^2 / 1 Ohm= 0.16 Watts,
when you fold again all the areas to a steady DC level.

So the output is much lower than the input.

You can try different resistor values,
but without a transformer it would probably
always be output < input.

Regards, Stefan.

Dr. Spark,
just try it again the other way around:
R1= 1 Ohm
and
R2= 100 Ohm.

This way it will make more sense,
as the 1 Ohm at R1 will only be a shunt to measure the
input current and not all input energy  is already heated
there.

Regards, Stefan.

Thanks Stephan,

The bulb thru the 100ohms and the 2 10 ohm resisters one burning hot, sure had me fooled!

Ill still try the transformer circuit. Soon as parts arrive...

drspark

Stephan

How does this explain the burning resister?

drSpark

Quote from: drspark on March 21, 2007, 05:39:48 PM
Stephan

How does this explain the burning resister?

drSpark

You seem to have used 0.25 Watts types resistors.
If you have used more power in one resistor,
it surely gets hot...