OU/COP>1 switched cap PS cct like Tesla's 'charge siphoning'

[nul-points]

[EDIT - see 1st post above for recent developments]

[NerzhDishual]

Hi Nul-Points,

I'm still on line...
Your results sound promissing.

I have printed out your posts and your pictures. I'm just trying to figure out
what it is all about. I'm far from an electronics specialist and I must confess
that I'm lost notably with you acronyms...

What do you mean by CCT? CirCuiT, I guess and not "Cortical Collecting Tubule"
or "Canadian College of Teachers".      (http://www.acronymfinder.com/)

Your (HEF?)4093B IC (Integrated Circuit) is a "Quadruple 2-input NAND Schmitt trigger"?

More to come ASP.

Best

[Feynman]

@nul-points

You have some serious math skills!  This will take some time to understand.

[nul-points]

thanks for the encouragement, guys

sorry about the acronyms - i'm a software engineer like you ND (not yet retired tho') but i trained in Elec Eng many decades ago so i got used to writing cct for 'circuit'

yes, 4093 is quad Schmidt NAND, two gates used as astable oscillator with two outputs, regular & inverted - one switches 1st Darlington pair Trs to charge 4u7 switching cap from input cap (or battery), then second switches 2nd Darlington pair Trs to discharge 4u7 cap into load

i think my maths has improved as i've got older, Feynman - which just shows how bad it must have been when i was a student!

ok, the bad news is that my 15R series measurement resistor definitely limits the charging of the switched cap in the cct - so, although the scoped input appeared to confirm the meter reading,  the output would definitely not still have been providing the power previously measured

i've rerun the tests using just the input 0.25F cap stack starting at 8V, stopping at 7V - this supplies a total of 1.875 Joules (= 1875mWseconds)

so i divide that by the time it takes to discharge from 8 to 7V and that gives me the average Pin in mW for each test

i've measured the load resistance - in these latest tests it's 468R - and i've smoothed the output with a 1000uF cap so i can scope the load volts and then calc the load mA & mW

the reactance-free circuit handled 18.9mW av. & the input was 34.2 av. so the actual COP was 0.55 - close to the predicted 50% loss

so i repeated the test with a coil of copper wire, as supplied on its reel, as an inductor feeding the 4u7 cap, and this enabled the circuit to retain about 30% more energy - a COP of 0.82

it seemed to me that the output smoothing cap was producing a Vout value lower than the area under the graph would - i might try comparing them sometime - it's easier & much faster.to read the smoothed value direct with a meter, but if its altering the cct efficiency it would be better to go back to the Excel method

i guess the next step for the cct is to see if the additional inductance(s) need to be tuned to the value of switching cap used & also see if the losses can be reduced - some more thinking to do there

@ND
i tried that idea you mentioned in Tesla Switch thread, of charging 2nd cap thro a dc motor in series:
-  if i connected input cap stack at 8V thro a CD load-tray motor direct to empty 2nd cap stack & discharged input to 7V then the motor ran and also the 2nd cap charged to 1V - charge was conserved but we get the motor running too

  - if i used my switching cct, same start volts on caps, then motor runs but this time 2nd cap charges to 1.2V - charge conservation has been violated, as you and i have found before, and we still get to power the motor too!

i'll get back here when i have some new results - or to follow up any comments or questions
sandy

[nul-points]

 i was abit over optimistic in thinking that i could account for all relevant losses just by taking an offload power reading - the onload currents cause significant volts drops within the circuit

the true COP value of the circuit should be greatly improved by including the onload losses in the charging branch, the load branch, the switch drive resistors and the switch circuit itself

i've run a new test, again using a charged cap stack as supply, and taking the extra readings and doing the calcs, i've measured 53mW supplied as input, 30mW drawn by load, and a total of 36mW 'losses' (losses only in the sense that they are not output to the load - they represent useful power in making the circuit work!)

COP = (30 + 36)/53 = 1.24

so, the full picture now shows that the switch-charged capacitor does indeed handle more power than is supplied!

could this be related to the fact that charge is not conserved in this type of circut, as i reported in the Tesla Switch thread?

i used the front-end of this circuit to discharge one cap into a 2nd cap: initial charge 2 Coulombs; final charge approx 2.9 Coulombs - an  increase in charge of over 1.4 and a clear violation of Conservation of Charge(checkout Wikipedia for the implications of breaking this 'law')

considered comments - and verifications!! - welcome
sandy

(measurements & calcs shown below...)


0.25F supply cap;
discharged from 9V to 8V
time taken = 40.04s
9V 10.125Joules;
8V 8.0J

Input supplied:
Ein 2.125J  = 2125mWs over 40.04s
Pin 53.07mW av.
Vin 8.5V av.
Iin 6.24mA av.

Output to load:
Vout 3.77V av.
Iout 8.06mA av.
Pout 30.37mW av.

Losses:
  Switch cct
Vin 8.5V av.
Is 0.35mA av.
Ps 3mW av.

  Switch drive
duty = 1/9 (0.1 charge; 0.9 load)

  12k R drive-charge
Vin 8.5V av.
Irc 0.71mA x 0.1 av.
Prc 0.64mW av

  12k R drive-load
Vrl 4.82V av.
Irl 0.40mA x 0.9 av.
Prl 1.74mW av.

  charge-branch
Iin (6.24mA - (0.35+0.007)) = 5.88 mA av.
Vdrop 8.5-4.82 = 3.68V av.
Ploss 21.64mW av.

  load-branch
Vdrop 4.82-3.77 = 1.05V av.
Iout 8.06mA av.
Ploss 8.46mW av.

Totals:
Power in 53.1mW av.
Load power out 30.4mW av.
Losses 21.6 + 8.5 + 2.4 + 3 = 35.5mW av.

COP = (power handled/power supplied)
   = 65.9/53.1 = 1.24