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### Author Topic: To be deleted  (Read 24452 times)

#### itsu

• Hero Member
• Posts: 1836
##### Re: To be deleted
« Reply #105 on: December 14, 2018, 02:30:53 PM »

Finaly got some drain duration results.

Using 3x 10F caps in series = 3.3F

Charged till 4.15V, activate the circuit, at 4V start the timer, at 3.4V stop timer.

Circuit with feedback      shows drain time from 4V => 3.4V to be average (10X) 10 minutes 14 seconds = 614s
Circuit with no feedback shows drain time from 4V => 3.4V to be average (10x)   9 minutes                    = 540s

So the remove of the feedback shorts the draining time by 1m14s = 74s = 12%.

The graph shows the both situations:

Regards Itsu

#### Free Energy | searching for free energy and discussing free energy

##### Re: To be deleted
« Reply #105 on: December 14, 2018, 02:30:53 PM »

#### nul-points

• Hero Member
• Posts: 995
##### Re: To be deleted
« Reply #106 on: December 14, 2018, 03:34:30 PM »
thanks Itsu, your cap draw-down results seem comparable with mine

here is a way to visualise what is occurring with this circuit in terms of average quantity for the transfer of charge (ie. current flow):

say you have 3 watertight containers (2 larger ones, Cb & Cc, and 1 smaller cup Ct) and a timer

(NO FEEDBACK)
after using Ct to fill Cb with a quantity of 20 cupfuls of water,
start the timer, and then at every new minute use Ct to transfer a quantity of 5 cupfuls of water from Cb to Cc;
(so a quantity of water is transferred in one direction only, each minute, from Cb to Cc)

Q1) after the 4th minute of water transfers, how many cupfuls of water remain in Cb?
A1) 0 cupfuls

Q2) what is the average quantity of water transferred per minute?,
A2) (5 cupfuls * 4 transfers) / 4 minutes = 5 cupfuls per minute

now we repeat the test with a slight difference to the water transfer procedure:

(WITH FEEDBACK)
after refilling Cb with 20 cupfuls of water,
start the timer, and then every minute use Ct to transfer 5 cupfuls of water from Cb to Cc, and also transfer 1 cupful back from Cc to Cb
(so 2 different quantities of water are transferred, one in each direction, each minute, between Cb and Cc)

Q3) after the 4th minute of water transfers , how many cupfuls of water remain in Cb?
A3) 4 cupfuls

Q4) what is the average quantity of water transferred per minute?
A4) ((5-1) cupfuls * 4 transfers) / 4 minutes = 4 cupfuls per minute

note that you still have 4 cupfuls of water left in Cb - it would take another minute's transfer event to empty Cb. (this extra runtime behaviour matches what we're seeing with our capacitor draw-down tests)

so, in the posted circuit, if the voltage waveform across a CSR (or equivalent sensor) at the battery terminal shows 2 pulses of different polarity, then 1 pulse is caused by current, Iin, flowing INTO the circuit from the battery to power the components, and the other is caused by current, Ifb flowing OUT of the circuit back into the battery, recharging it.  the average supply current, Isupply = (Iin - Ifb)

in my circuit, the average supply current, Isupply = (Iin - Ifb) = (6.5mA - 1.2mA) = 5.3 mA. (DMM reading was 5.4mA)

np

#### nul-points

• Hero Member
• Posts: 995
##### Re: To be deleted
« Reply #107 on: December 14, 2018, 03:48:01 PM »
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.

ah, ok

my 'formula' Vsupply * Isupply is the standard calculation for input power calculated from the average Voltage & Current supplied to the circuit

i understand that you're reporting the values provided by the scope, but i thought that those values would have some similar meaning (ie.  that you could multiply the voltage reading by the current reading and that the result would produce some meaningful value for the input power)

in that case, it's not clear to me what is the relationship between the data provided by the scope, and what is the meaning of any individual reading in terms of standard power calculations - eg. is there any reading on the spreadsheet which represents 'average input current' in the conventional meaning of that phrase?

thanks
np

#### Free Energy | searching for free energy and discussing free energy

##### Re: To be deleted
« Reply #107 on: December 14, 2018, 03:48:01 PM »

#### itsu

• Hero Member
• Posts: 1836
##### Re: To be deleted
« Reply #108 on: December 14, 2018, 04:15:34 PM »

NP,

as said, the scope uses its elaborate math function to compute the data found on the traces it is processing
(here CH1, yellow, voltage times CH4, green, current).
The red math trace presents this data averaged over many samples.

Besides what the math function is doing with the, in this case, data found on CH1 and CH4, i can display the
data on those channels to my likings, like in rms, mean etc.

Normally i put the data on those channels in rms, but if i would put the data in mean, nothing would change on
the math trace as it is still doing its thing with the data.

In post #56 of this thread:
https://overunity.com/16384/flyback-data/msg527867/#msg527867

you can see in the first screenshot that i have the input power being calculated and present the (DC) voltage
yellow in rms and the complex AC current green in both rms and mean (together with the math red calculation in mean).

If you think that the complex AC current in green presented in mean would be more meaningfull, then i could use
that for next calculations, but i still consider the calculated power via the red math trace more reliable.

Itsu

#### nul-points

• Hero Member
• Posts: 995
##### Re: To be deleted
« Reply #109 on: December 14, 2018, 04:32:57 PM »
ok, i understand what you're doing

thanks
np

#### Free Energy | searching for free energy and discussing free energy

##### Re: To be deleted
« Reply #109 on: December 14, 2018, 04:32:57 PM »

#### itsu

• Hero Member
• Posts: 1836
##### Re: To be deleted
« Reply #110 on: December 15, 2018, 11:30:04 AM »

As the percentage "power users versus input" drops from 94% with feedback to 85% without feedback (see post #100),
i was trying to find out where that missing 15% in the no feedback situation went.

I was doing that by monitoring the temperature of some components like the transformer, the potmeter and the
electrolytical cap using my laser temp. meter.

But it seems that this method is too rough for these small currents as i do see some temp. changes, but in a
irregular way, so probably due to room temp changes.

Guess i have to scope the current paths to these components to get a more accurate view.

Itsu

#### nul-points

• Hero Member
• Posts: 995
##### Re: To be deleted
« Reply #111 on: December 15, 2018, 12:50:42 PM »
that's a shame about the resolution issue of a low-powered circuit, Itsu - it would have been a very interesting view of what's going on where

i've repeated the 1F (nominal) cap draw-down test swapping out the original coil with a larger coil (both turns and ferrite, also a toroid rather than solenoidal)

the average runtimes (for same draw-down voltage range) are as follows:-

NO FEEDBACK: t = 68.4s

LED FEEDBACK: t = 89.6s

ratio = 89.6 / 68.4 = 1.31

thanks
np

#### Free Energy | searching for free energy and discussing free energy

##### Re: To be deleted
« Reply #111 on: December 15, 2018, 12:50:42 PM »

#### itsu

• Hero Member
• Posts: 1836
##### Re: To be deleted
« Reply #112 on: December 15, 2018, 09:21:07 PM »

I started measuring the current through the potmeter with and without feedback
and the current to the electrolytical cap with and without feedback.

We are dealing with minute currents through the potmeter, but we see a big increase of the current when feedback
is removed from 402uA with, to 636uA without,  see screenshot 1.
White is the current through the potmeter with feedback, green without.
Also to notice is that the frequency went up when feedback is removed.

The current to the elco is much bigger, but behaves the other way around, less without feedback, from 6.22mA with, to
5.04mA without, see screenshot 2.

I don't think though that this will account for the missing power in the system with feedback removed.

Itsu

#### nul-points

• Hero Member
• Posts: 995
##### Re: To be deleted
« Reply #113 on: December 16, 2018, 02:11:49 AM »
yes, good call Itsu

the frequency does increase with 'no feedback' (and i think this is because in the feedback case, the f/b pulse increases the cycle duration whilst the coil collapse energy performs work, charging the battery somewhat & lighting the feedback LED - in the 'no feedback' case the coil collapse energy dissipates wherever it can)

interesting about the resistor pot - after i used the pot to select a suitable operating drive level i replaced the pot with a fixed resistor

the elco behaviour seems reasonable -

with feedback, the elco is drained in 2 ways:
1) buffering the LEDs
2) completing the coil collapse current circuit back thro' the Feedback LED

without feedback, the elco is only drained as in 1) above

thanks
np

#### Free Energy | searching for free energy and discussing free energy

##### Re: To be deleted
« Reply #113 on: December 16, 2018, 02:11:49 AM »

#### nul-points

• Hero Member
• Posts: 995
##### Re: To be deleted
« Reply #114 on: December 16, 2018, 02:15:06 AM »
QUESTION:  Is the average supply current, Isupply, equal to the difference between the current drawn, Iin, and the feedback current, Ifb?

ANSWER:  Yes it is

SCOPE TEST:  measure Isupply at a mid-range voltage (3.75V)

(With Feedback)
signal cycle = 70.4kHz, full cycle = 14.1us

DC average, across 1 ohm CSR, using scope math from traces shown below

Isupply: 8.0mA
(full cycle)

Ifb: -5.24mA  *   (4.7us / 14.1us) = -1.67mA  (averaged over full cycle)

Iin: 14.6mA * (9.4us / 14.1us) = 9.7mA  (averaged over full cycle)

Difference between Iin & Ifb = (9.7 - 1.67) = 8.03mA

#### itsu

• Hero Member
• Posts: 1836
##### Re: To be deleted
« Reply #115 on: December 16, 2018, 11:39:58 AM »
yes, good call Itsu

the frequency does increase with 'no feedback' (and i think this is because in the feedback case, the f/b pulse increases the cycle duration whilst the coil collapse energy
performs work, charging the battery somewhat & lighting the feedback LED - in the 'no feedback' case the coil collapse energy dissipates wherever it can)
interesting about the resistor pot - after i used the pot to select a suitable operating drive level i replaced the pot with a fixed resistor

the elco behaviour seems reasonable -

with feedback, the elco is drained in 2 ways:
1) buffering the LEDs
2) completing the coil collapse current circuit back thro' the Feedback LED

without feedback, the elco is only drained as in 1) above

thanks
np

NP,

you mention: "in the 'no feedback' case the coil collapse energy dissipates wherever it can".

So that i should be able to measure, but up till now (temperature tests, current tests) i was not able to detect.
Perhaps i should check with the Spectrum Analyzer if the spectrum is more polluted without feedback.

Itsu

#### Free Energy | searching for free energy and discussing free energy

##### Re: To be deleted
« Reply #115 on: December 16, 2018, 11:39:58 AM »

#### itsu

• Hero Member
• Posts: 1836
##### Re: To be deleted
« Reply #116 on: December 16, 2018, 11:58:54 AM »
QUESTION:  Is the average supply current, Isupply, equal to the difference between the current drawn, Iin, and the feedback current, Ifb?

ANSWER:  Yes it is

SCOPE TEST:  measure Isupply at a mid-range voltage (3.75V)

(With Feedback)
signal cycle = 70.4kHz, full cycle = 14.1us

DC average, across 1 ohm CSR, using scope math from traces shown below

Isupply: 8.0mA
(full cycle)

Ifb: -5.24mA  *   (4.7us / 14.1us) = -1.67mA  (averaged over full cycle)

Iin: 14.6mA * (9.4us / 14.1us) = 9.7mA  (averaged over full cycle)

Difference between Iin & Ifb = (9.7 - 1.67) = 8.03mA

NP,

i cannot confirm your above question as i did not record the average (mean) current.
My voltages and currents are in rms, and then you see that Isupply is almost Iin while Ifeedback is way more then the difference.

Itsu

#### nul-points

• Hero Member
• Posts: 995
##### Re: To be deleted
« Reply #117 on: December 16, 2018, 02:09:55 PM »
you mention: "in the 'no feedback' case the coil collapse energy dissipates wherever it can".

So that i should be able to measure, but up till now (temperature tests, current tests) i was not able to detect.
Perhaps i should check with the Spectrum Analyzer if the spectrum is more polluted without feedback.

Itsu

hi Itsu
yes, i was thinking energy radiated away from the circuit - the waveform across a supply CSR with no feedback shows a burst of HF and also some LF oscillations

thanks
np

#### nul-points

• Hero Member
• Posts: 995
##### Re: To be deleted
« Reply #118 on: December 16, 2018, 02:22:32 PM »
i cannot confirm your above question as i did not record the average (mean) current.
My voltages and currents are in rms, and then you see that Isupply is almost Iin while Ifeedback is way more then the difference.

Itsu

that's ok, thanks Itsu - my general post about the average supply current was just provided as supporting evidence relating to the results from my circuit, for other interested viewers of the thread

thanks
np

#### itsu

• Hero Member
• Posts: 1836
##### Re: To be deleted
« Reply #119 on: December 16, 2018, 02:42:02 PM »

NP,

i just redid the basic power measurements now including the average (mean) current through the severall paths.

Picture is probably to wide for the thread, but decreasing it would make it unreadable i guess.

I looked at the Spectrum with my SA, but only a very slight increase in overall bottom level in the first 33MHz
is noticed, no specific peaks or noise bands, both with or without feedback.

By the way, just to confirm, i remove the green led cathode from the plus rail to measure without feedback.

Itsu