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Author Topic: To be deleted  (Read 44818 times)

nul-points

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Re: To be deleted
« Reply #90 on: December 11, 2018, 03:45:42 PM »
as i said, everyone is entitled to their own opinion...  using bad maths to try & justify it, however, is not acceptable

Turbo

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Re: To be deleted
« Reply #91 on: December 11, 2018, 06:28:02 PM »
It's all bad math based on theory and assumption and we tend to stick to the formulas and models that work best but the truth of the matter is that we don't have a clue about what is really going on.


nul-points

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Re: To be deleted
« Reply #92 on: December 11, 2018, 07:25:55 PM »

the operating cycle for the circuit, based on a pnp blocking oscillator, is as follows:-

 -  drain 'half-cycle':
at power on,  the transistor switches on, the primary voltage rises and its current starts to charge the elcap via the coil and also to drive the main LEDs;

the antiphase secondary voltage drives the transistor harder with its low-level output;

when the primary winding saturates, the secondary voltage rises sharply, and turns off the transistor;

 - feedback 'half-cycle':
the resulting output current pulse from the coil-field collapse is fed via the feedback LED and elcap, back into the supply battery;

after several cycles, the elcap charges to a steady DC voltage across the main LEDs, so that those LEDs are continuously lit throughout the cycle, not just pulsed once each cycle;

the elcap receives a current pulse via the primary each cycle, to restore the charge which is passed back to the battery during the feedback pulse;

although the feedback current is approx. 20% of the drain current, there is imperceptible ripple on the elcap, compared to its DC voltage


due to the feedback current opposing the drain current, the net supply current is lower than the drain current by approx 20% (in this circuit -YMMV!)


hopefully this is a factual and non-contentious description of the circuit operation

this is a simple circuit - you don't need to believe either the claims, or the counter-claims for its performance - i'm sure you're all well able to build, test & decide for yourselves


np
 

itsu

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Re: To be deleted
« Reply #93 on: December 11, 2018, 09:33:13 PM »

NP,

thanks for your description of your circuit.

I am terrible at math so i won't go there, but about this statement above:

Quote
due to the feedback current opposing the drain current, the net supply current is lower than the drain current
by approx 20% (in this circuit -YMMV!)

Are you sure the feedback current is opposing the drain current?
I mean, yes its of opposite direction, but it is shifted in time, see the screenshot of the supply current R1
in my post #81 here:
https://overunity.com/16384/flyback-data/msg527979/#msg527979

To me it seems that the supply current = drain current and then after the transistor is off, the supply current
stops, followed by the FeedBack current in the opposite direction charging the battery.

This FB current is part of the supply/drain current but time shifted, so one should not subtract it from the
supply current to get a new net supply current.

just my 2 cents,  

Regards Itsu
 

nul-points

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Re: To be deleted
« Reply #94 on: December 11, 2018, 09:52:41 PM »
hi Itsu


thanks for your question, i understand what you're saying - we don't get a lot of experience of AC current with battery sources  :)


think of current as an amount of charge which could flow (in either direction) - if, say, two lots of the same value of charge flowed out of, and back into, a battery then technically it doesn't matter if those 2 flows of charge occurred in 2 sequential periods of 30 minutes (eg. driving a motor, then recharging the battery), or 2000 sequential mS periods of alternate flow direction

the important thing in our calculations here is to choose a regular period which contains the same number of each flow direction (it could be 1 of each, for one cycle; or many cycles each containing 1 of each flow direction). where each flow direction occurs within each cycle is not important

our 2 flow directions here have different average current value, but if you could make them equal then the net charge which has flowed would be zero (regardless of the total time period for the exchange) and the battery would be in the same state of charge after the test as it was before (except maybe temperature!)


i hope this helps - i don't think i've explained it well !


np

nul-points

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Re: To be deleted
« Reply #95 on: December 11, 2018, 09:54:55 PM »
[Edit: i'm going to remove this post comparing battery run-times until i can confirm that i'm comparing use of the same battery and not 2 from the same batch]

i'll repost later


np
« Last Edit: December 12, 2018, 09:20:07 AM by nul-points »

nul-points

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Re: To be deleted
« Reply #96 on: December 12, 2018, 12:46:17 AM »

[...]

This FB current is part of the supply/drain current but time shifted, so one should not subtract it from the
supply current to get a new net supply current.

[...]

hi Itsu

i don't think i fully addressed your question earlier - i see your concern that the feedback current should only be considered as part of the 'drain' current, Iin

yes, you're right - all charge must have come from the initial drain current, Iin, for each cycle

but if, say, 2 Coulombs of 'charge' flow into the circuit, as Iin, and 0.5 Coulombs of charge flows back out, as Ifb, in the same cycle, then a net 'charge' of 1.5 Coulombs has flowed into the circuit, and the effective current to achieve that is (Iin - Ifb), regardless of the sequential timing of those 2 different charge-transfer directions (within the same cycle)

i hope this answers your question


np

itsu

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Re: To be deleted
« Reply #97 on: December 12, 2018, 10:59:29 AM »

NP,

thanks for trying to explain  :)

i understand what you are saying, but in my mind there must be a flaw somewhere otherwise we would have excess energy in this
circuit and the battery would not drain down.

Not sure though what this flaw is, perhaps its NOT in the same cycle (supply and feedback).

Anyway, the duration tests could shed some light on that i think.

Today i will finish my last two measurements for a total of 10 so i should have a decent average.

Then i will remove the feedback led connection and do some measurements to see how this new situation influence
the severall current / energy flows.

Itsu

nul-points

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Re: To be deleted
« Reply #98 on: December 12, 2018, 08:52:59 PM »

hi Itsu

i understand your doubts about the drain and feedback currents alternating in direction within each cycle, but the green LED wouldn't illuminate if there wasn't a flow of current from the circuit to the battery, and that current can only be generated by the coil-field collapse current (which occurs once at the end of each cycle, as i described in post #92)

the value of the current returning to the battery is only about 20% of the current that the battery supplied to the circuit, so all that will happen is that the circuit runtime will be extended.  we can't create extra energy (because energy is conserved) - but what we can do is store & then convert some of the original energy again to do some extra work (because work is not conserved)

each cycle, the circuit stores some energy in the coil field and in the elcap; the coil field collapse causes some current to flow in the secondary and this current can draw on energy stored in the elcap to flow through the feedback LED and battery (this is the extra work done)

the only way for the battery to not drain down is if we could return more current to the battery than the battery supplies to the circuit (it would need to be more because the battery recharge efficiency is only about 70%)

we're only returning about 20% of Iin, so we're nowhere near a self-runner (and i guess we never will be); but we are increasing the runtime of the circuit for the same initial energy stored in the supply

i look forward to seeing the results from your battery rundown tests


np

nul-points

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Re: To be deleted
« Reply #99 on: December 12, 2018, 09:03:00 PM »
[Edit: i'm going to remove this post comparing battery run-times until i can confirm that i'm comparing use of the same battery and not 2 from the same batch]
i'll repost later
np


i've supplied my circuit from a 1F capacitor to do some comparative run tests

i charged the cap to 4+ V and started the run as the voltage slowly approached 4V

i stopped the run as the cap voltage just crossed the 3.4V

i measured the time between the crossing at the 4V & 3.4 cursors



the runtime for the circuit WITH NO feedback is approx 125s

the runtime for the circuit WITH LED feedback is approx 140 s

(i also measured the runtime for a circuit variant WITH SCHOTTKY DIODE feedback at approx 224 s
- but i think the nearly doubled time was mostly due to a change in drive conditions)


np

itsu

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Re: To be deleted
« Reply #100 on: December 12, 2018, 10:20:39 PM »

NP,

thanks for the run down graphs, how did you make those?

Good idea to use a supercap, else run down times would be endless.
So the feedback (led) provides for about 15s longer runtime which is 18.75% (your 20% feedback current?)

Anyway, i have completed my 10 runs for the normal circuit, see below spreadsheet top part.
The 10x average is on the bottom of this first colored part underscored.

Then i quickly today ran 4 runs without the feedback (led) attached, to show that we have a limited increased power
drawn from the battery 15.28mW v 13.65mW (+1.63mW).
See lower part of the spreadsheet, again 4x average on the bottom of this lower part underscored.
Guess some power is lost somewhere as the ratio "pwr users / input" drops from average 94% to 85%.


Will start with some run down tests now,               looking for my super caps......


Itsu

nul-points

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Re: To be deleted
« Reply #101 on: December 12, 2018, 11:19:55 PM »

hi Itsu

you've been BUSY!!!!

those power user results look pretty comprehensive


its a shame that you haven't been able to replicate the true net supply current value (ie net flow of charge into & out of the circuit) - your scope traces show the 2 parts, so they are there (and your Green LED current value shows that charge is flowing back to your supply)


yes, large value caps seem to be the way to go when you have many tests to do

my 1F cap is physically large as well as large value - i think they were designed for crazy-loud car audio systems, so presumably very low ESR

i just replaced the battery with the 1F cap, charged it slightly above my required start voltage, then watched the voltage approach the cursor and started the circuit when the voltage was slightly above. i stopped logging just after the end cursor was crossed and subtracted the end time from the start time


good luck with your next tests
np

nul-points

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Re: To be deleted
« Reply #102 on: December 14, 2018, 12:47:14 PM »
Then i quickly today ran 4 runs without the feedback (led) attached, to show that we have a limited increased power drawn from the battery 15.28mW v 13.65mW (+1.63mW).

hi Itsu

i'm not sure about your input power calcs...

on your spreadsheet, you make them:
15.28mW (with feedback)
13.65mW (no feedback)

from your data, i make them:
3.75V * 6.97mA = 26.14mW (with feedback)
3.75V * 6.45mA= 24.19mW (no feedback)


thanks
np

itsu

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Re: To be deleted
« Reply #103 on: December 14, 2018, 01:13:28 PM »

Hi NP,

indeed, you make them:
3.75V * 6.97mA = 26.14mW (with feedback)
3.75V * 6.45mA= 24.19mW (no feedback)


but i did not make them, i had the Tektronix TDS3054B scope calculate them to be:
15.28mW (with feedback)
13.65mW (no feedback)

Your formula is for DC signals only, while we are dealing with a complex AC like current waveform.
The scope is able to deal with those complex AC like waveforms using its elaborate math functions.

Itsu


AlienGrey

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Re: To be deleted
« Reply #104 on: December 14, 2018, 02:21:27 PM »
I don't know its just another blocking oscillator like all the rest but its upside down!
Sling the ferrox core out and replace it with as NANOCRYSTAL one and tune for best output.

Not trying to be clever but that stuff the nanoparticles stay charged and if you just pulse the device
you're bound to get again.