Author Topic: To be deleted (Read 45406 times)

Turbo · « **Reply #75 on:** December 10, 2018, 12:17:28 PM »

There is a function called Sound_Play() in mikroC library and i have had success with that.
The Prototype is void Sound_Play(unsigned freq_in_hz, unsigned duration_ms);
https://download.mikroe.com/documents/compilers/mikroc/pic/help/sound_library.htm
So you can easily pass frequency and duration as parameters and so you can roll your own function that incorporates the mark duration, as well as delay for the space duration.

If you need higher frequency's you have to switch to classic PWM.
But who does not remember being a child and hearing the 20K whistle coming from the CRT Flyback tv when we was young ? _{and the resulting crackle from the electrostatic field on the tube when touched...}
Of course now i'm old and i can't hear that high no more.

nul-points · « **Reply #76 on:** December 10, 2018, 12:27:01 PM »

time to recap on the original circuit we're looking at here (and which Itsu is kindly looking at replicating)

i'm attaching an overview of the circuit - it has 1 supply branch and effectively 2 branches (ignoring a higher impedance biasing branch with an effective impedance of approx 10k ohm at 130kHz)

one branch is entirely a current drain, Iin, on the supply and the other is entirely feeding current, Ifb, back to the supply

the net supply current, Isupply, is Iin - Ifb

np

Void · « **Reply #77 on:** December 10, 2018, 03:01:03 PM »

Quote from: itsu on December 10, 2018, 11:31:38 AM

Hi Void,
on this wiki there is some info on rms values / waveform:
https://en.wikipedia.org/wiki/Crest_factor
So every waveform needs its specific calculation to arrive at the correct rms value.
I can not see how a scope does that, so probably there is some generic way.
But i agree, (power) measurements are no easy task and marginal cop > 1 values should be treated with causion.
Itsu

Hi Itsu. I think the way a scope does it may be as follows:
The way a scope calculates RMS values is to square all the measured sample values taken over a period of time,
then calculate the mean of those squared values, and then take the square root of the calculated mean.
It's been a long time since I studied such things, but I think this may be how it is done in scopes (or similar).
As my scope's manual states: "RMS: Root mean square of all data values."

Your measurements so far are showing a fair bit of variation, so not so super reliable, possibly because
the circuit is not so stable, but likely at least in part due to the current probe not giving highly
accurate measurements at those low current levels. Still it seems pretty clear already that there is no
indication of OU in that circuit arrangement when measurements are done properly.

Void · « **Reply #78 on:** December 10, 2018, 03:18:12 PM »

Quote from: nul-points on December 10, 2018, 12:27:01 PM

time to recap on the original circuit we're looking at here (and which Itsu is kindly looking at replicating)
i'm attaching an overview of the circuit - it has 1 supply branch and effectively 2 branches (ignoring a higher impedance biasing branch with an effective impedance of approx 10k ohm at 130kHz)
one branch is entirely a current drain, Iin, on the supply and the other is entirely feeding current, Ifb, back to the supply
the net supply current, Isupply, is Iin - Ifb
np

Hi Nul-points. Although Itsu's measurements are not showing as super consistent/accurate,
which may possibly be partly due to circuit instability and very possibly/likely an accuracy limitation of his
current probe at low currents, the measurement method Itsu is using appears to be correct (and it passes sanity checks),
so Itsu's measurements should be close enough for general analysis purposes, so it should be pretty clear by now
that there are so far no signs of OU in your circuit arrangement.

My recent comments have been to demonstrate how tricky it can be to make really accurate measurements
on circuits that contain complex AC waveforms, and how important it therefore is to question and examine everything
very closely to see where a person might possibly be going wrong or overlooking something in their measurements.
Nul-points, you seem to have ignored this, and also ignored that your previous measurements and calculations
are way off compared to what Itsu has been measuring using an (apparently) correct measurement method (although
apparently not so super accurate with Itsu's current probe). Nul-points, what do you think about this? Do you acknowledge
that Itsu's measurement approach is correct, even if it may not be super accurate/consistent when using his current probe?
If you don't think Itsu's measurement approach is correct, please explain why.

P.S. Regarding your latest drawing posted above, efficiency is calculated as a ratio of average power out to average power in,
so you taking a ratio of currents is completely wrong. You also seem to be trying to analyze the circuit as if it is some
sort of DC circuit or simple AC circuit, which is not correct, as the circuit produces complex AC waveforms, and I have already
gone into much detail in this thread to demonstrate and explain what to watch out for and how to make proper measurements
in that type of circuit, and showed possible ways to do a sanity check on measurements to make sure the measurements are
reasonably in the ballpark of what should be expected. As best as I can tell, Itsu's measurement approach is correct, and is not
showing any signs of OU.

Void · « **Reply #79 on:** December 10, 2018, 05:16:29 PM »

Quote from: itsu on December 10, 2018, 11:31:38 AM

But i agree, (power) measurements are no easy task and marginal cop > 1 values should be treated with causion.

This bears repeating.

At low power levels such as in the low mW or lower,
a difference in measurement of even several mW can be well within the margin of measurement error,
so, yes, I would say jumping to conclusions when measuring on AC circuits with complex waveforms
at low power levels is not at all a good idea. There is just too much room for error or oversights there.

Now, if a person can scale up their circuit to much higher power levels and still measure a higher average
output power than average input power with a difference in the order of many Watts, then IMO that would bear a
much closer look, assuming the measurement methods stand up to close examination. The easiest of all is
just to self-loop a circuit and see if you can get it to self-sustain, as that way no measurements are required.

itsu · « **Reply #80 on:** December 10, 2018, 10:21:24 PM »

I added some more measurements onto the spreadsheet.
I also removed the very first one as that one seems wrong as i probably did not follow my measurement protocol.

But i don't think that like Void mentions it is inaccurate.
We are talking about milli-volts and micro-amps deviations, so pretty accurate if you ask me.

_{Anyway, here it is:}

Itsu

itsu · « **Reply #81 on:** December 10, 2018, 10:32:40 PM »

Quote from: nul-points on December 10, 2018, 12:27:01 PM

time to recap on the original circuit we're looking at here (and which Itsu is kindly looking at replicating)

i'm attaching an overview of the circuit - it has 1 supply branch and effectively 2 branches (ignoring a higher impedance biasing branch with an effective impedance of approx 10k ohm at 130kHz)

one branch is entirely a current drain, Iin, on the supply and the other is entirely feeding current, Ifb, back to the supply

the net supply current, Isupply, is Iin - Ifb

np

NP,

thanks for the recap, i see what you are trying to say, but i think it is somewhat more complex.

The below screenshot shows the 3 currents:

Isupply grey/white R1
Iin White R2
Ifb green

The yellow trace is the voltage across the transistor to show when it turns on and off.

R1 is Isupply and includes the positive cycle Iin and the negative Ifb totaling 7.17mA
R2 is Iin and only has the positive cycle as during the negative Ifb the transistor is off, see yellow trace.
Green is Ifb and can only go left (transistor off) into the supply.

I have overlayed green (and inverted) on R2 to show it is similar as the Isupply waveform R1.

The rms value of Isupply is higher (7.17mA) with the negative cycle, compared to Iin without
the negative cycle (6.5mA), but this is due to how AC rms voltage/current works, it is equivalent to the same
DC value, but it uses both the positive and negative parts to do work (if you can call it that here).

Itsu

Void · « **Reply #82 on:** December 10, 2018, 11:04:04 PM »

Quote from: itsu on December 10, 2018, 10:21:24 PM

But i don't think that like Void mentions it is inaccurate.
We are talking about milli-volts and micro-amps deviations, so pretty accurate if you ask me.

Hi Itsu. It's all relative. You have now removed one measurement entry that had wide deviation,
so with that entry removed the measurement deviation seems not so wide now. Looking
at the load LED power measurements, you have one measurement at 10.3 mW and one
measurement at 11.27 mW. That is a variation of about 1mW, or about 9% variation,
so not so small a deviation in measurements. Part of the problem could be due to circuit
instability however, as mentioned previously.

There is no point at all in comparing input and output currents in regards to trying to analyze
efficiency of the circuit. As I have already explained, efficiency of a circuit is determined from the ratio
of average output power to average input power. Comparing currents alone gives no indication whatsoever
on the efficiency of a circuit. I have also explained and demonstrated why looking at RMS readings alone in
circuits of this type with complex AC waveforms will very likely lead a person completely astray.

At any rate, it should be clear already that this circuit configuration gives no indication at all of OU.

itsu · « **Reply #83 on:** December 11, 2018, 10:10:39 AM »

Void,

thanks, we get it now.

I have to agree, all the power is almost being accounted for (95%), the rest is very probably consumed like Hoppy mentioned in the potmeter, transformer, cap esr, etc.

Left is now how to correctly interpret the data and the energy flows.

Itsu

AlienGrey · « **Reply #84 on:** December 11, 2018, 10:19:41 AM »

Oh yes very nice triangle waveforms of loss, the problem with AC and magnetics is you get loss as your magnetising a transformer
core and then you destroy that magnetic flux by reversing it and doing it again in the other direction!
What you really need to do is recover that energy before it's lost in heat.
If you used only positive or negatively charged pulses that didn't reverse in polarity and ironed out the back 'ringing' in your magnetic drive
the results you obtain might turn out far better, also better magnetic devices might produce better results.

nul-points · « **Reply #85 on:** December 11, 2018, 11:35:51 AM »

Quote from: Void on December 10, 2018, 11:04:04 PM

[...]

There is no point at all in comparing input and output currents in regards to trying to analyze efficiency of the circuit.
As I have already explained, efficiency of a circuit is determined from the ratio of average output power to average input power.
Comparing currents alone gives no indication whatsoever on the efficiency of a circuit

[...]

the supply and the 2 significant current branches share a common voltage (let's call that voltage Vsupply)

the generic equation for a measure of Energy = Volts * Amps * Time

say we measure our energy for a period of 20 cycles (let's call the total time t1)

Our energy supplied for this period is

Esupply = (Vsupply * Isupply * t1)

Our energy converted for this period is:

Ein = (Vsupply * Iin * t1)

therefore
Efficiency, n, = Ein / Esupply = (Vs * Iin * t1) / (Vs * Isupply * t1)

the voltages cancel (because they are common to Numerator and Denominator)

so now Efficiency, n, = (Iin * t1) / (Isupply * t1)

the time values cancel (because they are common to Numerator and Denominator)

and now Efficiency, n, = (Iin) / (Isupply)

this is the calculation i show in my overview diagram (and i also used in some previous posts)

so - we ARE able to calculate Efficiency using only current values WHEN the current branches share a common voltage (which they do) AND all the current readings are averaged over a common measurement period (which they are)

in my example above (overview diagram), Vsupply = 3.7V, t1 = 20 cycles (approx 60uS) - these cancel and leave us with:

Efficiency, n, = (6.93 mA) / (5.7 mA) = 1.2 (= 120%)

Quote from: Void on December 10, 2018, 11:04:04 PM

[/size]
[...]

At any rate, it should be clear already that this circuit configuration gives no indication at all of OU.

lol

np

Void · « **Reply #86 on:** December 11, 2018, 12:10:44 PM »

Sigh...

itsu · « **Reply #87 on:** December 11, 2018, 02:30:02 PM »

Come on guys, lets agree to disagree for the time being.

I need to make some 4 more measurements to get a decent average and then i can make some duration tests.
See how long it takes to drain the battery.

I can also change the frequency to see if there are major changes in the severall power users, perhaps there is a sweet spot somewhere.

_{But step by step.}

Thanks, itsu

nul-points · « **Reply #88 on:** December 11, 2018, 03:00:57 PM »

hi Itsu

everybody is entitled to an opinion (that includes all of the posters to this thread) - i hope that no opinions have been restricted on this thread?

maths is maths however, and if two sets of values, eg. Current and Energy (under the limits explained), are proportional then you can certainly use one set as validly as another when calculating a ratio/proportion or percentage

if i see someone claiming otherwise** as justification for denying a possibility then i am going to call it out

...and i would hope that ANY other well-intentioned and informed person would do the same

(**does anyone here think that we should not flag up maths issues like this?)

the duration tests sound good - as one part of my investigation with these tests i've been logging battery terminal voltage vs time - it''ll be interesting to compare our results and try to get a better idea of the characteristics of this circuit

np

Void · « **Reply #89 on:** December 11, 2018, 03:30:59 PM »

Quote from: itsu on December 11, 2018, 02:30:02 PM

Come on guys, lets agree to disagree for the time being.

Hi Itsu. Anyone is free to do as they wish. It is already very clear that the circuit
is not producing OU however, and that is fair and honest assessment. All the best.

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