Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[Rosemary Ainslie]

And guys, this is that circuit that I referred to some time back.  I have drawn it with a switch.  It needs a transistor - ideally a MOSFET - but I have no idea how this is to be done - what supply to use - or where to establish the common ground for those switches. 

But if anyone here is clever enough to put this together - then here's the thing.  It would be a far better way to prove the principle that I'm trying to bring into focus.  Effectively, in this arrangement - the one battery will always discharge to recharge either itself or the other battery. 

If, as is claimed by the thesis, that charge is entirely retained, in other words, if current flow ONLY ever returns to the source and IF there is no material loss at that supply - then this would unarguably be a closed system. 

And as a reminder.  The idea is that in the transfer of energy through a circuit NO material is lost or gained by the suppy.  In other words, current flow belongs to its source and only ever returns there.  In the same way - current flow that is induced from the circuit material ALSO ony returns back to that material.  The heat that is dissipated on circuit components is a different discussion.  I'll get there.

Be very glad if someone can resolve the principles in the circuit.  It's been something that I've been toying with for about 2 years. 

Kindest regards,
Rosemary

Hope it's clear.  I can't seem to resize my pictures any more.

I'm trying this again.  I think I've resized it.   

[nul-points]

this is that circuit that I referred to some time back.  I have drawn it with a switch.  It needs a transistor - ideally a MOSFET - but I have no idea how this is to be done
...
It would be a far better way to prove the principle that I'm trying to bring into focus.  Effectively, in this arrangement - the one battery will always discharge to recharge either itself or the other battery. 
...
Be very glad if someone can resolve the principles in the circuit.  It's been something that I've been toying with for about 2 years. 
...
Rosemary

hi Rosemary

i don't think that your new circuit will operate as you expect

because the circuit is circular, the Diode/Switch pairs and the remaining individual components can be moved around the ring to any position and the circuit will still be exactly equivalent

therefore your circuit (on the left, below) can be arranged, for example, as the other two circuits below

it might be easier to see from the arrangement on the right:-

a) if the switches always operate in anti-phase, then the only current which will ever flow is due to the combined voltage of the two cells (or batteries) flowing through the combined series resistance (R1 + R2 + reverse impedance of either D1 or D2) <<depending which switch is closed

ie. B1 & B2 will take a long time to discharge because basically the circuit is not doing a great deal (eg. the cells are merely discharging via diode leakage current, regardless of duty cycle or switch rate)


b) if the switches should ever be allowed to operate in-phase, then when both switches are ON the current will be (V1+V2)/(R1+R2);
  and when both switches are OFF the current will be similar to (a) (except the diode leakage will be due to two diodes in series)

ie. the discharge behaviour of B1 & B2 will depend on the duty cycle and the switch rate


** ignoring milliohm resistance in the switch on-states in both (a) & (b)


hope this helps
np

http://docsfreelunch.blogspot.com

[Rosemary Ainslie]

Hi Nul-Points.  Many thanks for that - but I was hoping you or someone would design it with a transistor - something appropriate to allow that dual path.  I agree it won't work with a reed switch or somesuch unless one can get a difference in the two battery supply voltages. 

Can I impose on you to try a design on this principle?  Is it even possible?  I just don't know enough about switches to answer this myself.

But either way - many thanks for this work.  I'll get back here and try and explain the point of it all - but it will be later today.

Kindest regards,
Rosemary

[nul-points]

Hi Nul-Points.  Many thanks for that - but I was hoping you or someone would design it with a transistor - something appropriate to allow that dual path. 
...
I agree it won't work with a reed switch or somesuch unless one can get a difference in the two battery supply voltages. 
...
Can I impose on you to try a design on this principle?  Is it even possible?  I just don't know enough about switches to answer this myself.
...
Rosemary

hi Rosemary

here's one possibility below (RA-2c)

i'm not sure why you think there will be any effective difference in behaviour between RA-2b and RA-2c ?


if i correctly understand the weight of all your previous explanations, so far,  then the diodes provide the '2nd path' in your 'dual path' expectations of switched current

if that is so, then for your purposes, circuits RA-2b and RA-2c should effectively provide the same action

the net effect is the same - both B1 & B2 would discharge over time, regardless of what was happening at S1 & S2 - or Sig 1 & Sig 2


a "difference in the two battery supply voltages" is not relevant to the operation of your circuit (RA-2a), or the equivalents which i've shown (RA-2b & RA-2c), since the two cells ("batteries"), as you can see, effectively form one true battery of voltage (V1 + V2)


hope this helps
np

http://docsfreelunch.blogspot.com

[Rosemary Ainslie]

Hi again nul-points.  You've sort of put two batteries in series.   And then given two loads with two switches.  Is that all I've managed in that design?  LOL.  If so, let me try this again.  I'll get back here when I've redrawn it.

Many thanks for your efforts there by the way.  But I'm still trying to point to something.  I think I must make a small adjustment there.  I need to explain it more fully by getting it workable with those reed switches.  Then - possibly you'll be able to put in the required transistors.

Hang ten.  It takes me ages to draw - photo - then down load. 

Kindest as ever,
Rosie