Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[poynt99]

So, if Q2-Q5 are 'on' the battery voltage (approx. 70V) is connected to the 0.5 ohm restricted input of the signal generator with just the heating element in series?
Must be quite a robust signal generator.....

The generator has a 50 Ohm output impedance, so there is some degree of protection there. But indeed, this is not something you really want to subject your FG to.

Also, there is no physical 0.5 Ohm resistor installed anywhere, that is just the resistance of the coaxial cable between the generator and the "Gate" input.

@TheCell,

What are you trying to say?

.99

[teslaalset]

The generator has a 50 Ohm output impedance, so there is some degree of protection there.

Also, there is no physical 0.5 Ohm resistor installed anywhere, that is just the resistance of the coaxial cable between the generator and the "Gate" input.

You're right in case one uses a signal generator instrument.
But be carefull if one is using e.g. a microcontroller like an Arduino.
Just a tip.

[TheCell]

First: I don't think, that it ever has been build up like that.

All considered during the off-phase.
If function generator C has negative voltage, than imagine a
2nd battery below ground potential B where it + is connected to B
and its - connected via the 0.5 Ohms Resistor to the SourceConnectors
of Q2-Q5.
The shunt 0.25 Ohm draws no current.
The whole voltage : 6*12V + UFunctionGen is connected to RL1 + 0,5Ohms at C.

.99 : Generator 50 Ohm output impedance , what the resistance of RL1 ?
The complete diagram is false, or the FunctionGenerator + battery drives the circuit during the off-phase.

[poynt99]

The circuit is built as shown. Verified from both the video analysis and from Rose herself.

The resistance of RL1 is about 11 Ohms. You are correct about Rshunt indicating 0 Amps for part of the cycle, and that the load is in fact being driven by Q2-Q5 through the function generator and BYPASSING the shunt. Rose has already admitted that they measured a significant power in the Gate circuit.

What a mess folks!

There is no way with this circuit configuration as it stands, can the INPUT power be accurately measured, when at least part of the time, the current measuring resistor is not even in the circuit. In fact, it could be argued that Q1 may not ever conduct any current and provide power to the load at all!

Stop the presses folks, this begs for a serious and close examination of what's really going on here! I think we have our answer as to why the shunt indicates a negative mean voltage and current.

BUSTED!

.99

[TheCell]

Still one thing to try :
Apply the function generator with a series resistor about 10 kOhms.

If it still oscillates during the off phase ...