PhysicsProf Steven E. Jones circuit shows 8x overunity ?

[hyiq]

Hi Hoppy,

Could you please explain this further? In my understanding, in Circuit V7, all power being consumed in the input side of the circuit should pass through the Input Shunt Resistor R2 giving a total calculation of the Input Current to the rest of the Circuit? Is this not correct? Have I missed something?

Thanks in advance.

All the best

  Chris

[Tudi]

Chris: the output is inversed as the input as i heard. Summing this in looped mode with the power source might still give a different result then feeding it directly to C2. You might need to invert the output of C1 before feeding it to C2

[hyiq]

Hi Tudi,

Yes, It is Inverted. It makes self Looping very difficult. Still working on it, no success yet.

Wont it be nice to build a small circuit that powers itself... Nice.

All the best

  Chris

[Hoppy]

Hi Hoppy,

Could you please explain this further? In my understanding, in Circuit V7, all power being consumed in the input side of the circuit should pass through the Input Shunt Resistor R2 giving a total calculation of the Input Current to the rest of the Circuit? Is this not correct? Have I missed something?

Thanks in advance.

All the best

  Chris

Chris,

The shunt resistor is just a convenient way of measuring the current in the complete circuit. Power does not pass through components, it is consumed by them, so each components dissipation is additive. Once you add the coil dissipation and dissipation in the transistor itself, then the total will be found to be much greater than the total power dissipation in the output circuit. Power in Watts over time in seconds = energy in Joules and its energy 'in' v 'out' that's important to get a full picture of efficiency.

Hoppy

[xee2]

@ hyiq

I think you should confirm the scope readings by replacing R4, D2, and VR2 in circuit V7 with a 1 K resistor and computing the output power by measuring the voltage across the 1 K resistor with a battery powered DVM. Watts = volts squared divided by resistance. Since the output is close to DC, this should give readings close to the scope readings. If not, the scope may not be giving accurate results.

EDIT:  using a 10 K resistor will produce higher voltage readings and may thus be more accurate.