Thanks, Fuzzy, I had not seen that second shot before.

That is another one then that shows a strong steady +12 volts "on" voltage in the gate signal-- to Q1 presumably-- and yet no current is shown in the CVR during those "on" periods.

Hey Fuzzy.... in that schematic you just posted... is the text "Q2-Q5" in the same font as the other text labels in the schematic? Where exactly did that schematic come from?

Quote from: TinselKoala on May 11, 2012, 01:41:17 AM
Hey Fuzzy.... in that schematic you just posted... is the text "Q2-Q5" in the same font as the other text labels in the schematic? Where exactly did that schematic come from?

Hi TK,

The schematic was from one of Poynt99 postings after the incorrect schematic was found in the YouTube video ...

http://www.overunity.com/10407/rosemary-ainslie-circuit-demonstration-on-saturday-march-12th-2011/msg282318/#msg282318        Reply #708 on: April 20, 2011, 04:49:32 AM
http://www.overunity.com/10407/rosemary-ainslie-circuit-demonstration-on-saturday-march-12th-2011/dlattach/attach/52205/            Protoboard_schema_added.png

So the fonts may have been his doing 

Rosemary,

I have done my DC test now. I did set up the circuit as shown in the attached drawing.

Here is my theoretical predictions for this test:

Q2 will turn fully ON and conducting DC current by using a 12K resistor as a positive bias for Q2.
The current will flow from the 12 volt power supply through RLOAD, through Q2, through RSHUNT
and back to the power supply. The worst case RDS(on) for Q2 is 1.6 Ohm.

The circuit current will be:

Circuit total resistance = 10 + 1,6 + 0,25 = 11,85 Ohm. I = U / R according to Ohms law.
I = 12 / 11,85 = 1,0126582278481012658227848101266 Ampere.

Voltage over RLOAD will be:

U1 = R * I = 10 * 1,0126582278481012658227848101266 = 10,126582278481012658227848101266 Volt.

Voltage over RSHUNT will be:

U2 = R * I = 0.25 * 1,0126582278481012658227848101266 = 0,25316455696202531645569620253165 Volt.

Voltage over Q2 will be:

U3 = R * I = 0.25 * 1,0126582278481012658227848101266 = 1,6202531645569620253164556962025 Volt.

The three voltages above added = 11,999999999999999999999999999993 e.g. 12 Volt (by using a calculator).

Now, my digital volt meter can't display all those digits, I have only two digits after the comma.
So we must round all the values to:

I = 1,01A    U1 = 10,13V    U2 = 0,25V    U3 = 1,62V

The total Watt dissipated in the circuit will be:

P = (U * U) / R = (12 * 12) / 11,85 = 12,151898734177215189873417721519 Watt.

ACTUAL TEST:

Circuit was built as shown in the drawing. A power supply was adjusted to 12,00 Volt and connected
to the circuit. The voltage from the power supply was measured to be 12,29 volt. Q2 did turn ON
conducting current through the circuit and the 10 Ohm RLOAD resistor did heat up.

Measurement 1: The voltage over RSHUNT was 0,27 Volt.
Measurement 2: The voltage over RLOAD was 10,57 Volt.
Measurement 3: The voltage over Q2 was 1,45 Volt.

The current indicated by the power supply was 1,07 Ampere.

CONCLUSION

The above test describes what is happening in the Rosemary Ainslie circuit when the function generator
is at its positive pulse level. The Q1 will switch off and the Q2 will switch on and conducting DC current
for the duration of the function generator on pulse. So the Q2 will be operated as a DC switch
in this configuration. My measurements are in agreement with the predicted math and in the
usual +/-10% measurement range expected from a inexpensive digital volt meter. When using a
function generator then all we have to do is to compute the output Watt according to the duty cycle
length of the on pulse from the function generator.

We can actually use Ohms Law to compute the result when the function generator is at the positive pulse.
We can also use a inexpensive digital volt meter to measure the voltage over RSHUNT since the circuit
is in pure DC mode during the function generator on pulse.

Rosemary claims that the current flow is 320 mA at 62 Volt into the RLOAD (assuming 10 Ohm)
when the function generator is at the positive pulse.  This is impossible if the Q2 was biased fully on!

With a Q2 biased fully on the current through the circuit should be 62  / 12,25 = 5 ampere. The actual current
to through the circuit when we calculate in the 87.8% on time, is 4,39 Ampere. So the 320mA  looks like a
measurement error or a faulty Q2 MOSFET or a different RLOAD value than the 10 Ohm calculated here.

Rosemary, can you clarify what resistance your RLOAD was when doing the 62 Volt input run?
I have calculated all my math based on a 10 Ohm RLOAD. My test circuit also have a 10 Ohm RLOAD
and may have got the numbers wrong if you did use a different RLOAD on your circuit test run.

GL.

Thanks again, Fuzzy. Now you've got me reading again. And I found this bit of wisdom:

QuoteGuys - I need to explain something here.   And you really ned to understand it.  Proper measurement of electric energy is based on vi dt.  This requires the calculation - the product - of volts and amps in real time.  The minute one takes the average of the amperage over a certain period of time multiplied by the average of the voltage over a certain period of time - then one is actually only pointing to an average.  It may very well be a fair reflection of the fact - very much as you can average the global temperature - but it tells you NOTHING about the temperature at any precise point.  IF we measured electrical energy as an average - then it could be entirely misrepresented. 

Here's an example.  You can take the sum of the voltage across the shunt over - say - 1 minute - and then multiply it by the sum of the battery voltage over that same minute - and it will give a result that is only 'close' to but not 'representative of' the actual energy delivered or dissipated.  It will certainly HIDE the benefits in phase relationships between those voltages - and it may either enhance or fudge the actual facts.  Poynty is relying on this.  He is hoping that your own lack of knowledge of detailed measurements is such that he can carry his argument.  But pick up your phones.  Talk to your academics.  They're very approachable.  They'll explain how it is that any measurment based on an average - is absolutely NOT acceptable.  What is more - our own evidence shows that there are those relationships between the voltages that are ENTIRELY hidden by an average.  I've not even touched on that side of the evidence yet.  But Poynty et al - thay know of this.  And this NEED TO AVERAGE is therefore also MUCH NEEDED.  Else they'll have no argument.

We're here dealing with a really sophisticated effort to diffuse the efficacy of these forums.  And it's actually where the real energy pollution is. 

Regards,

And that was from
http://www.overunity.com/10407/rosemary-ainslie-circuit-demonstration-on-saturday-march-12th-2011/msg281955/#msg281955
Posted last April 18th, 2011.