Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[poynt99]

Poynty - I can no longer see your waveforms and i would LOVE to see them.  There are others of us who also cannot open those files.  WOULD YOU PLEASE POST A PICTURE and just size it that it fits this thread.

Kindest regards,
Rosie

ADDED
This is copied over from your forum - written by you Poynty Point.

'Indeed, I am having difficulty figuring out why her shunt trace is at zero. Mine is showing about 1.5V or so, and hey, 0.25/11 x 72 = about 1.6V. It would seem what I'm showing is about right.'

I think what you meant is 0.25/11 x 72 = about 1.6 watts, NOT VOLTS.  In which case can you then explain the temperature over the load which, typically, is at 6 watts or greater at 72 volts applied.

Perhaps someone can unzip the files and resize them. It's a shame that Stefan does not fix the problem so that this is not necessary. I offered him a solution already, as this works fine at OUR.

Regarding the calculation, no I meant Voltage. I am calculating the approximate voltage that should be across the shunt when the FET is ON, and under ideal conditions, those being that the ON resistance of the FET is much much smaller than your CSR of 0.25 Ohms. It is a simple voltage divider between the load resistance (11 Ohms) and the CSR resistance (0.25 Ohms) if we do not include the inductances.

So, 0.25/11 Ohms, times 60 Volts ~ 1.36V. [I used 72 volts last time assuming you had 6 batteries.]

.99

[Rosemary Ainslie]

Perhaps someone can unzip the files and resize them. It's a shame that Stefan does not fix the problem so that this is not necessary. I offered him a solution already, as this works fine at OUR.

Regarding the calculation, no I meant Voltage. I am calculating the approximate voltage that should be across the shunt when the FET is ON, and under ideal conditions, those being that the ON resistance of the FET is much much smaller than your CSR of 0.25 Ohms. It is a simple voltage divider between the load resistance (11 Ohms) and the CSR resistance (0.25 Ohms) if we do not include the inductances.

So, 0.25/11 Ohms, times 60 Volts ~ 1.36V. [I used 72 volts last time assuming you had 6 batteries.]

.99

Poynt - we all know that you can do that resize.  Is there a reason you won't?  Are you keeping this hidden for a reason?  Have you not managed to show that waveform afterall?  Come on Poynty.  How about it?  It's a lame claim if there's no evidence.  And right now it's hidden from view.

Still don't understand your sum.  Where does the 11 Ohm's come from? 

Rosemary

[poynt99]

Still don't understand your sum.  Where does the 11 Ohm's come from? 

Rosemary

What is the resistance value of your load resistor then?

.99

[Rosemary Ainslie]

What is the resistance value of your load resistor then?

.99

11 Ohms is close.  But why are you dividing the shunt value by the ohms value of the resistor?  Are you saying that there's a 1.3 volt across the FET?  Still don't get it Poynty.  The FET voltge is much higher than this during the 'on' period. 

added

[Magluvin]

Hey Rose and Woopy

At lunch so Ill make it short. And sweet.  ;]

Here is a circuit with 2 mods. 

1  a resistor is added to simulate resistance in the inductor.

2 added the diode across the batt/switch for purposes as described last night.  ;]  =]

I ran it on the sim and it works!     

But in the real world the circuit will need more additional circuitry to compare when the cap reaches 5v to cut the switch.

What I did to enable myself to cut the switch manually was slowed down the sim so that I could come very very close to cutting a 5v, and yep that diode allows current in the forward direction to keep flowing into the cap and I was getting over 6v into the cap.
It varied as I could not hit the switch to get a perfect 5v cutoff.

This should be a cop>1.  =]

Woopy  I will look at ways to enable you to automate the cutoff in your setup to do the test tonight.  ;]

Today im not even tired, on 2hours sleep, Im all uppity!  =]  I wonder why?  ;]

Mags