To be deleted

[nul-points]

[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

[nul-points]

[...]

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]

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]

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]

[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