Author Topic: OU/COP>1 switched cap PS cct like Tesla's 'charge siphoning' (Read 145112 times)

poynt99 · « **Reply #240 on:** January 01, 2009, 05:00:55 PM »

Hi Sandy.

Quote from: nul-points on December 31, 2008, 12:00:08 PM

hi Poynt

yes, removing R1 does increase max stored voltage achieved - did a trial run, couple months back i think, to see if worth following up & decided it will be

think efficiency was high but dissipation was focussed in DC res of L and therefore inaccessible, so just parked that for future test & continued with full RLC branches

It makes sense that there will be some losses in the inductor itself, although removing R1 will shift the total circuit impedance largely back to a reactive one so most of the energy should be transferred into C2 from C1. Hopefully most (if any) excess energy tapped in the process will end up in C2 as well.

In my sim, C1,C2 settle at a voltage of 5V with R1 present. Without R1, they settle at about 5.4V. In both cases, the presence of the inductor is bringing the transfer efficiency above 50%, because without it, the final C1,C2 voltage would be 4V.

Quote

original tests (using my handdrawn schematic you just re-posted) were showing efficiencies round 125% - BUT - this was on the assumption that the external energy converted in R whilst charging was equivalent to the final stored energy in the output cap, as claimed by textbook treatment

my subsequent tests showed this relationship was not constant - though for my particular tests it didn't reduce the overall efficiencies below 100%

latest switching control uses PC parallel port o/p thro' SFH618 opto-isolators, re-shaping test-side signals thro' CMOS Schmitt devices, driving straight into MOSFET Gates

opto i/p LEDs powered by PC port drive; all isolated test circuitry powered by C1, as you noted from earlier tests

most recent change to test circuit is replacement of P-chan MOSFETS (caution: partnos. from memory!) FDN304P replaced with IRF5305(?) because circuit chewed up the P-chan every month or so & wiped out my stock (i'd salvaged a few from scrap I/O boards)

all the best
s.

Thanks for the info and the circuit. I'll maybe draw something up showing the whole or most of the complete circuit.

Regards,
.99

AbbaRue · « **Reply #241 on:** January 01, 2009, 09:09:18 PM »

One point that causes confusion when dealing with capacitor charge and discharge times.
A capacitor charges to two thirds the input voltage in the first time constant.
Then it charges to two thirds the remaining voltage in the second time constant.
eg. If a capacitor takes 5 seconds to charge to 99.6 volts from a 100 volt source
It will charge to 66.6 volts in the 1st second
Then to 88.8 Volts in the 2nd second.
Then to 96.2 Volts in the 3rd second. (This is considered fully charge for most circuit applications.)

And when you discharge that same capacitor it follows the same scheme.
It will discharge two thirds it's stored voltage within the first time constant.
So a capacitor charged to 100 volts will discharge to 33.3 volts within the 1st time constant.
Then to 11.1 Volts in the 2nd time constant.
Then to 3.7 Volts in the 3rd time constant.

nul-points · « **Reply #242 on:** January 02, 2009, 05:00:07 PM »

thanks for the TC info, Abba - useful for measuring the cap values in these circuits - the switching operation between caps in these circuits is mostly taking place at a very small fraction of the circuit TC

hi Poynt

Quote from: poynt99 on January 01, 2009, 05:00:55 PM

In my sim, C1,C2 settle at a voltage of 5V with R1 present. Without R1, they settle at about 5.4V. In both cases, the presence of the inductor is bringing the transfer efficiency above 50%, because without it, the final C1,C2 voltage would be 4V.

i found i had a copy of the reduced Rload test run i mentioned - not exactly fitting your suggested variation but possibly enough of a pointer to see if it's going in the right direction - i can try a closer test when i get access to my setup again

input power was being provided by my original C1 (0.31F) so the test run covered approx 430 charge/discharge cycles of C2 (196uF) to discharge C1 from 8V to 7V

i need to confirm this following aspect when i can retest, but i believe the trace shows the voltages on Rload (1ohm) and C2 at the point of mid-energy input (approx 7.53V on C1)

the max stored voltage achieved on C2 is 5.33V

the results show slightly higher Rload energy dissipated in charging C2 than in discharging it - DC res of L is 0.5ohm so the total dissipated energy is at least 50% higher than that recorded on Rload

hope this is useful

all the best
s.

poynt99 · « **Reply #243 on:** January 02, 2009, 05:45:45 PM »

Perform the testing you would like Sandy.

I recommend however simplifying as much as possible, which is why I suggested ( http://www.overunity.com/index.php?topic=4419.msg146734#msg146734 ) the simple C1 to C2 discharge/charge test through L1, and sticking with your 190u and 196u capacitors. The circuit is operated until VC1=VC2, and the voltage noted.

I don't think it can get much easier or simpler. If you get higher than about 5.6V then chances are you got it

.99

nul-points · « **Reply #244 on:** January 03, 2009, 11:01:10 AM »

Quote from: poynt99 on January 02, 2009, 05:45:45 PM

Perform the testing you would like Sandy.

I recommend however simplifying as much as possible, which is why I suggested ( http://www.overunity.com/index.php?topic=4419.msg146734#msg146734 ) the simple C1 to C2 discharge/charge test through L1, and sticking with your 190u and 196u capacitors. The circuit is operated until VC1=VC2, and the voltage noted.

I don't think it can get much easier or simpler. If you get higher than about 5.6V then chances are you got it

hi Poynt

yes, i was just mentioning some values from that previous test since i'm not able to try your suggested test for a few days 'til i get back at my setup and my previous test was close to your suggestion of removing the Rload

i thought this test might give you some idea if it's in the right ballpark - it appears that the average input energy was approx 5.4mJ per charge/discharge pulse train and the associated total output energy converted was around 6mJ on the 1ohm Rload (with around 1.6mJ extra charging energy lost on DC res of L)

i'll let you know as soon as i've been able to run your suggested test

all the best
s.

nul-points · « **Reply #245 on:** January 04, 2009, 12:29:12 AM »

hi Poynt

what level of energy dissipation will you be expecting to occur in the DC resistance (mostly sum of ESRs for L, Cs & D) of the circuit when i run your proposed test?

thanks
s.

poynt99 · « **Reply #246 on:** January 04, 2009, 02:13:05 AM »

Quote from: nul-points on January 04, 2009, 12:29:12 AM

hi Poynt

what level of energy dissipation will you be expecting to occur in the DC resistance (mostly sum of ESRs for L, Cs & D) of the circuit when i run your proposed test?

thanks
s.

Sandy,

I ran a sim to get some numbers.

First it is difficult to find numbers for capacitor ESR so I used 100m Ohms total for both. This seems to jive with data I was able to find on the web. I also added 50m Ohms for the switch resistance. Total circuit R then I used is 0.65 Ohms, which includes 0.5 Ohms for the inductor. The 10 Ohm R1 is removed.

From the scope shot you'll see that the diode appears to continually use energy throughout the switching process and uses 268uJ after 100ms (blue). The energy lost in the total circuit resistance of 0.65 Ohms levels off quickly at 208uJ (yellow).

The total energy loss after 100ms run time is about 476uJ.

C1 (red), C2 (green) final voltage is about 5.39V.

After about 80ms, the capacitor voltage begins to slowly decrease due to the continual switching loss in D1. From this the optimal run time would be about 80ms, which is about where the two capacitor voltages appear to level off. The total Joule loss in this case decreases a little from 476uJ down to 460uJ.

Regards,
.99

nul-points · « **Reply #247 on:** January 05, 2009, 02:20:44 AM »

thanks, Poynt, that's an interesting prediction from the sim

in the past test where i removed the 10ohm Rload and replaced it with a 1ohm current sensing resistance the measured results showed a charging current which resulted in an energy dissipation in the inductor and sensing resistance around 2x that stored in C2, 196uF (which was 2.8mJ approx)

this division of output energy between stored and dissipated has been a common feature of the results for these tests, as Dave (CTG Labs) mentioned previously

so, on a previous test, the effective removal of Rload didn't cause a greater proportion of the output energy to accumulate in the o/p cap

it reduced the effective load resistance on C1, which increased the peak pulse current, but the dissipated energy was still several times greater than that stored in C2

so unfortunately i don't think this proposed test is going to provide the simple, single-value measure we'd hoped for

i can still go ahead with the test but obviously the dissipation will also have to be monitored, we can't just note the final stored energy

nailing jell-o to a tree must be a breeze

all the best
s.

poynt99 · « **Reply #248 on:** January 05, 2009, 07:14:17 AM »

Sandy.

I'm not sure what else I can say.

Removing R1 will significantly increase the transfer efficiency of energy from C1 to C2. I don't think you would disagree with that. If you want to do the same test but with the 10 Ohm load present, then go ahead. It makes no difference really because you will account for all the energies in the circuit anyway. All that will happen with R1 present is more of the energy will get wasted as heat instead of being transferred to C2. The final capacitor voltage will be lower with R1 present.

At the end of the day you would measure the voltage on the two capacitors when they level off (they will be the same voltage) and add up all the losses and see how it all comes out. All the energies should equal (or better) that which started in C1.

.99

nul-points · « **Reply #249 on:** January 05, 2009, 02:24:46 PM »

hi Poynt

i was just following-up on your original intention for a simple test which relied on reading only the single final voltage on the two caps:

> "The circuit is operated until VC1=VC2, and the voltage noted.
> I don't think it can get much easier or simpler."

i agree this test idea is easy & simple, i'm just noting it also appears from previous results that this single measurement won't be sufficient on its own:

ie. we'll still need to try and get a measure of any dissipation - that doesn't mean we need to use a 10ohm load to do it - and hopefully my previous post doesn't suggest that we should (i mentioned using a 1ohm current-sensing resistance previously for this purpose)

all the best
s.

Grumpy · « **Reply #250 on:** January 05, 2009, 03:08:27 PM »

Quote from: poynt99 on January 05, 2009, 07:14:17 AM

I'm not sure what else I can say.

.99

Hey .99, question for you: will a capacitor store a displacement current?

poynt99 · « **Reply #251 on:** January 05, 2009, 03:27:42 PM »

Grumpy.

I don't know. This is one subject that gets debated from time to time.

Do you mean "flow through" vs. 'store"?

The DC theory does make sense, but I do not know if it has ever been proven one way or the other.

Why do you ask?

.99

Grumpy · « **Reply #252 on:** January 05, 2009, 03:51:03 PM »

Quote from: poynt99 on January 05, 2009, 03:27:42 PM

Grumpy.

I don't know. This is one subject that gets debated from time to time.

Do you mean "flow through" vs. 'store"?

The DC theory does make sense, but I do not know if it has ever been proven one way or the other.

Why do you ask?

.99

Let's say it can. What would this imply? What might one be able to do if this is the case?

Come on .99, have a drink with ol' Grumpy!

poynt99 · « **Reply #253 on:** January 05, 2009, 09:53:51 PM »

I can only guess at what you may be thinking, so my guess is that if a capacitor can store displacement current, it does so in the form of electric charge (or flux), whereas our beloved inductor stores conduction current in the form of magnetic flux.

This might imply that the capacitor can be treated as an exact inverse of the inductor, such that a collapsing electric flux results in the reversal of built up displacement current, just as the collapsing of the magnetic field (built up current) around an inductor results in the reversal of the voltage across it.

Not sure if this is the parallel you were looking for...probably not.

.99

Grumpy · « **Reply #254 on:** January 05, 2009, 10:50:49 PM »

Quote from: poynt99 on January 05, 2009, 09:53:51 PM

I can only guess at what you may be thinking, so my guess is that if a capacitor can store displacement current, it does so in the form of electric charge (or flux), whereas our beloved inductor stores conduction current in the form of magnetic flux.

.99

Come on .99 - work with me here.

If a capacitor can store displacement current, when you send a very sudden impulse to a capacitor, you are applying the miniscule current in the wire plus the enormous displacement current produced by the sudden change.

Might this result in greater energy stored in the capacitor than you initially applied?

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Author Topic: OU/COP>1 switched cap PS cct like Tesla's 'charge siphoning' (Read 145112 times)

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