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Author Topic: To be deleted  (Read 44820 times)

itsu

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Re: To be deleted
« Reply #30 on: December 05, 2018, 07:11:17 PM »

While waiting for the correct BC327 transistor i setup the circuit on a breadboard using a 2N5401 transistor instead.

Smallest pot core i had using 0.45mm dual bonded magnet wire (green/red) 5 turns measuring 184uH each and when shorting
the other windings i have 0.84uH left (for leakage inductance calculations).
The capacitance inbetween the 2 windings is 96pF.

Using a 3.7V lion battery pack.

Using the 50K pot i set the frequency to 130Khz.

Input current was measured in 3 ways,
1 using a normal DMM in mA DC (10mA),
a 10 Ohm 1% inductionfree resistor in the return line measured with the yellow probe 97.mV / 10 = 9.7mA and
finally my current probe showing 9.6mA (green trace).

With a 1 Ohm csr it was hard to correctly measure the current.
 
So i am fairly confident that the DC input current is around 10mA and i will use the DMM for continue monitoring.

Video here: https://www.youtube.com/watch?v=22IEuVtAIIk&t=24s


Itsu

itsu

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Re: To be deleted
« Reply #31 on: December 05, 2018, 10:02:10 PM »

From the above video we see that in this setup the input power is 10mA x 3.88V = 38.8mW

The green led is pulling 5.7mA rms at 2.52V rms for a power of 3.985mW see screenshot 1

The red (load) led is pulling 11.22mA rms at 2.056V rms for a power of 23.07mW see screenshot 2

The transistor path (main led branch??) is pulling 16.08mA rms at 3.912V rms for a power of 43.38mW see screenshot 3


Not sure yet where to take additional measurements to get to the same conclusion as NP did, but will figure that out.

What i notice is that the lion battery pack voltage is raised somewhat (3.88 v 3.91), not sure where that comes from, probably some chemical reaction.

Yellow is the voltage across the led / transistor
green is the current (current probe) through that led / transistor
red is the math trace yellow x green (voltage x current)
blue is the voltage across the lion battery pack.
Itsu 

nul-points

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Re: To be deleted
« Reply #32 on: December 05, 2018, 10:34:05 PM »

thanks for the info, itsu, a very thorough inspection as always! - it looks like you have the whole circuit covered for measurements!

i agree about the 1 ohm CSR - i only used the CSR with the scope, to make a ballpark check on the supply current duty cycle and relative magnitude & current direction (for which it performed acceptably and supported the DMM current readings); so most written readings taken using DMM,  no CSRs, no supply capacitor (so far)


i also checked the supply to the drive branch (ie. the non-feedback branch) and confirmed that the current there is unidirectional from the supply to the transistor/primary/main LEDs, and also that the average of that current is higher than the average supply current by approx 20%

i see that i forgot to mention that the 2 main LEDs are 'cold white' type (hence 2.7VDC across the electrolytic cap //main LEDs); and just for info the bias pre-set resistor value is approx 20k Ohm

i hope this additional helps confirm the circuit i'm using

thanks

itsu

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Re: To be deleted
« Reply #33 on: December 05, 2018, 10:45:02 PM »

NP,

thanks for this additional info,  good to know, i will make some led changes then.
I measured my potmeter to be at 8K for the 133Khz

My measurements are just some praktice runs untill i get the BC327.



During the few hours of testing, the DMM monitoring the input current went from 10mA to 10.7mA and
the voltage across the battery pack (not monitored from the beginning  :-[ ) slowly drops using a Fluke 179 meter
from 3.791V to half an hour later 3.788V.

Itsu


nul-points

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Re: To be deleted
« Reply #34 on: December 05, 2018, 11:04:12 PM »
that all sounds pretty much nominal  - your cct is drawing approx 2x current as mine using a bias resistor value about half of mine

my battery is 60 or 70 mAh and a full charge takes approx 10-11hours to deplete, and i believe your battery is approx 840 mAh, so a full charge on your battery should deplete in approx 84 hours


thanks

nul-points

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Re: To be deleted
« Reply #35 on: December 06, 2018, 05:17:40 PM »

some observations about this circuit SO FAR:-
(plugging in example measurements displayed for itsu's preliminary practice run)

a) the current between the supply and the circuit is bi-directional
     (confirmed using scope probe on CSR in supply lead)

b) the CSR can be considered to be resistive
     test example:  10 Ohm non-reactive, negligible phase-shift component

c) the 'supply' current is the average of the RMS current values for the 2 half-cycles
     test example: 10mA

d) Input voltage is a DC voltage, essentially constant, negligible phase shift
     test example: 3.88V DC

e) the Power In (per cycle) is the product of the static supply voltage and the RMS 'supply' current
    = 3.88 × 10 = 38.8mW

f) the positive half-cycle RMS current enters the circuit into the transistor branch
    test example: 16.08 mA

g) the total dissipative load across supply rails (per cycle) is the product of the static supply voltage, as above, and the positve half-cycle RMS 'supply' current
     = 3.88 × 16.08 = 62.4 mW

h)  the Efficiency n, for driving this load is    dissipative load / power in
      = 62.4 / 38.8  =  1.6   (or 160%)


(additional energy is dissipated in the feedback LED which, when included in the results, increases the efficiency of the circuit

Void

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Re: To be deleted
« Reply #36 on: December 06, 2018, 05:21:53 PM »
Hi Itsu. Nice test. I would guess your LED power consumption measurements/calculations
are reasonably close to actual since your Tectronix scope RMS readings are probably
reasonably close, and since the current and voltage spikes across the LEDs
appear to be reasonably in phase.

However, your transistor power calculation looks quite suspect to me.
Ignoring the base current, the power consumption for the transistor should roughly be the
RMS collector to emitter voltage times the RMS collector current (assuming they are
reasonably in phase), when the transistor is conducting. The collector to emitter voltage should be quite low
when the transistor is conducting. When the transistor is not conducting, power consumption should be near 0.
Maybe you already explained it in your video, I can't watch it right now, but where are you getting that
3.912 V from? That doesn't look right to me at all.
All the best.

itsu

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Re: To be deleted
« Reply #37 on: December 06, 2018, 07:34:42 PM »

Thanks NP, Void,

for Void, yes i was wondering the same about that measurement, but its how NP measured it, so i did the same, but i too don't think it is the correct way.
The 3.912V i mention at the transistor is the voltage across the whole 'main led branch', so including the load led voltage etc.


In my below new setup i have changed that to as what i think is correct (voltage across E / C).



New setup using 2x 10mm white leds.
Also the 10 Ohm csr is removed.

The frequency had to be adjusted again to 130Khz by adjusting the 50K pot, its now set at 3.8K.

All measurements are done by the scope to be able to compare (backed up by the DMM's).

Current probe and voltage probes are deskewed as much as possible to rule out unwanted phase shift (important in AC like signals), but i am left with a 8ns delay (phase shift) between current and voltage probes.
This has no significant influence as shown by comparing with the 10 Ohm csr.

Below the circuit with some measurements.

The combined power users (leds and transistor) make up for a total of 21 + 1.4 + 2.3 = 24.7mW .
The input power is 28.4mW, so i think we have accounted for most of all the power into the circuit.

There are severall current measurements done, and with some i have a problem, like in the 'main led branch'.
The 9.2mA at the transistor comes short for supplying the follow on current through the load led (7.2mA) and to the feedback link to the elco (5.4mA)

Anyway, these are the results up till now:



Regards Itsu
« Last Edit: December 06, 2018, 09:54:32 PM by itsu »

Void

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Re: To be deleted
« Reply #38 on: December 06, 2018, 08:09:00 PM »
for Void, yes i was wondering the same about that measurement, but its how NP measured it, so i did the same, but i too don't think it is the correct way.
The 3.912V i mention at the transistor is the voltage across the whole 'main led branch', so including the load led voltage etc.

Hi Itsu. I see. Thanks for the clarification.
Yes, the transistor power dissipation should be relatively small, but I think it will be tricky
to measure it accurately with those pulse waveforms. 
How did you arrive at the 1.5V and 2.3 mW for the transistor?


itsu

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Re: To be deleted
« Reply #39 on: December 06, 2018, 08:57:27 PM »

Hi Void,

nice to have you take a look at this.

I let my scope do the hard work, by calculating the power from the voltage across the C/E and the current through this transistor.

In the above diagram, i have my yellow probe ground lead on the Emitter and the probe tip on the Collector (channel inverted on).

The current probe is at the arrow above the Emitter (also channel inverted on as my current probe is pointing the wrong way (up)).
 
The values and the math trace (yellow x green) is in the screenshot below.

Itsu

Void

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Re: To be deleted
« Reply #40 on: December 06, 2018, 09:13:34 PM »
Hi Itsu. Thanks for the details.
Yes, looks good to me for your transistor measurements.
Around 2mW transistor power dissipation looks reasonable.
Having that scope current probe and that math function is very handy. :)


itsu

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Re: To be deleted
« Reply #41 on: December 06, 2018, 09:30:15 PM »
some observations about this circuit SO FAR:-
(plugging in example measurements displayed for itsu's preliminary practice run)

a) the current between the supply and the circuit is bi-directional
     (confirmed using scope probe on CSR in supply lead)

b) the CSR can be considered to be resistive
     test example:  10 Ohm non-reactive, negligible phase-shift component

c) the 'supply' current is the average of the RMS current values for the 2 half-cycles
     test example: 10mA

d) Input voltage is a DC voltage, essentially constant, negligible phase shift
     test example: 3.88V DC

e) the Power In (per cycle) is the product of the static supply voltage and the RMS 'supply' current
    = 3.88 × 10 = 38.8mW

f) the positive half-cycle RMS current enters the circuit into the transistor branch
    test example: 16.08 mA

g) the total dissipative load across supply rails (per cycle) is the product of the static supply voltage, as above, and the positve half-cycle RMS 'supply' current
     = 3.88 × 16.08 = 62.4 mW

h)  the Efficiency n, for driving this load is    dissipative load / power in
      = 62.4 / 38.8  =  1.6   (or 160%)


(additional energy is dissipated in the feedback LED which, when included in the results, increases the efficiency of the circuit

My reaction below each statement

a) the current between the supply and the circuit is bi-directional
     (confirmed using scope probe on CSR in supply lead)

   Not what i see, the supply current is a DC current of 10mA out of the battery pack.
   I do have a smoothing cap of 2200uF and a decoupling cap of 0.1uF on the supply rail.
   Without them, i see an all positive pulsing current.

 
b) the CSR can be considered to be resistive
     test example:  10 Ohm non-reactive, negligible phase-shift component

   agreed.


c) the 'supply' current is the average of the RMS current values for the 2 half-cycles
     test example: 10mA
 
   Not sure what you mean here.

 
d) Input voltage is a DC voltage, essentially constant, negligible phase shift
     test example: 3.88V DC

   Agreed.


e) the Power In (per cycle) is the product of the static supply voltage and the RMS 'supply' current
    = 3.88 × 10 = 38.8mW

   Both voltage and current here are DC, so no cycle, further agreed.


f) the positive half-cycle RMS current enters the circuit into the transistor branch
    test example: 16.08 mA
   
   Not sure what you mean here, the supply current (10mA) together with the feedback current (5.7mA) enters
   into the transistor branch (16.08mA).

   
g) the total dissipative load across supply rails (per cycle) is the product of the static supply voltage, as above, and the positve half-cycle RMS 'supply' current
     = 3.88 × 16.08 = 62.4 mW

   This is where i have problems with as the transistor only consumes little power. See also Voids remark
   
h)  the Efficiency n, for driving this load is    dissipative load / power in
      = 62.4 / 38.8  =  1.6   (or 160%)

    Also here i have a problem as the former statement g) is not correct in my mind.


Itsu

itsu

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Re: To be deleted
« Reply #42 on: December 06, 2018, 10:20:48 PM »

I did some tests on statement a) above and have to adjust my reaction there already  :D

When removing both the 2200uF elco and 0.1uF decoupling cap i have on the rail, i get the below shown current in green.

The inline DMM in mA setting still shows 6.85mA, but the scope now shows 9.164mA rms and 6.871mA mean current.

So it seems the DMM is averaging the current.

On the total input it makes little difference as the input power still calculates around 27.7mW

Itsu

Void

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Re: To be deleted
« Reply #43 on: December 06, 2018, 10:33:53 PM »
So it seems the DMM is averaging the current.

Hi Itsu. Yes, that makes sense as ordinary DMMs will measure roughly the average AC voltage and current,
whereas true RMS meters will indicate the RMS values. I say 'roughly' because many cheaper DMM's are not
so accurate and most DMM's and true RMS meters are not spec'd to measure AC current and AC voltage at
higher frequencies such as 130 kHz, so their accuracy at such frequencies can't be counted on.


nul-points

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Re: To be deleted
« Reply #44 on: December 07, 2018, 03:23:55 AM »

thanks for the update Itsu, i think we're making progress - apologies for the delayed reply, i was out at band practice until late this evening (Thurs)

changing the Red LED to White has made the circuits more similar, that's good - my feedback LED is Green, as in my schematic, but if your circuit uses a White that should be ok

(if your feedback LED is White and it's illuminated by a 2mA current  however, then i question that the voltage across it is only 1.54V, so that measurement might need checking (did you use DMM or scope?)

yes, as i mentioned in post #32 the circuit doesn't have a supply capacitor, and CSRs are not included either (only added to take certain scope shots), so that was important that you removed the 2200uF elcap (& 0.1uF too?) to also make the circuits more alike

ok, so removing the supply caps changes your circuit's supply behaviour, but you're still only seeing positive current flow** into the circuit - so this is a major difference between our circuits (and the reason why steps (c) & (f) don't mean anything to you) - when i observe the supply current using a scope & 1 Ohm CSR (inserted inline with the positive supply) the current is bidirectional: the feedback current, Ifb, flows into the supply, and the main drive current, Iin, flows into the circuit as you'd expect

these 2 different current flows show as opposite polarity voltages across the CSR (scope trace shown below hopefully).  the duty cycle of these 2 approx. triangular pulses within each cycle is around 1.5:8 and the average current indicated by their relative areas and proportion of the cycle time gives reasonable confirmation of the values measured by DMM

the different current flow direction, or polarity, (Iin, current into main drive path; Ifb current out of feedback LED path) is confirmed (on my circuit, at least) with a DMM

hopefully this info will help you see what steps (c) and (f) achieve (assuming your circuit can achieve bidirectional current flow in the two parts of each cycle)

[Edit: ** actually, i believe your circuit does have bidirectional current flow in the supply lines: your latest scope shots in post #42, from the supply line probe (after removing supply caps) appears to show a narrow elongated negative going triangle pulse followed by a larger flattened somewhat triangular pulse of opposite polarity
   i also note the same approximate relationship between Isupply, Iin & Ifb in your circuit and in mine:
   Iin = Isupply + Ifb ]

the steps (a) to (h) of my result overview posted above in post #35 only use 3 current measurements - these are the 'combined' supply current and the 2 separate current flows, Iin & Ifb, discussed here in this post


i didn't attempt to measure the transistor power dissipation - i measured Iin and calculated the total power dissipation for the whole of that path from +ve supply line to -ve supply line and subtracted main LED dissipation to arrive at the dissipation in the transistor and transformer windings - you can see this measurement in post #19

step (g) doesn't mention the transistor (because i thought that you mentioned that your reading of 43.38mW in post #31 was for that whole branch, between +ve & -ve supply lines, not just the transistor)

so step (g) refers to "the total dissipative load across supply rails (per cycle)" ie. the whole current path between supply +ve & -ve lines, entering the circuit into the Tr. emitter - this would include the transistor 'on' state, transformer winding losses, & main LEDs (ignoring the much smaller values of bias losses, elcap leakage etc.)


so, i see that your circuit has been updated with new readings - let's plug them into steps (c) to (h):-

c) the 'supply' current is [the average of] the RMS current [values for the 2 half-cycles]
     test example: 7.4mA

d) Input voltage is a DC voltage, essentially constant, negligible phase shift
     test example: 3.76V

e) the Power In (per cycle) is the product of the static supply voltage and the [RMS] 'supply' current [the mean current, as per your updated measurement after removing supply caps]
    = 3.76 × 7.37 = 27.3 mW

f) the [positive half-cycle] RMS current enters the circuit into the transistor branch
    test example: 9.2 mA

g) the total dissipative load across supply rails (per cycle) is the product of the static supply voltage, as above, and the positve half-cycle RMS 'supply' current
     = 3.76 × 9.2 = 34.6 mW

h)  the Efficiency n, for driving this load level is   ( dissipative load / power in)
      = 34.6 / 27.3  =  1.27   (or 127%)

[unit time, common to both terms to produce units of energy, cancels out]

(additional energy is dissipated in the feedback LED which, when included in the results, increases the efficiency of the circuit)


i hope this helps
« Last Edit: December 07, 2018, 10:18:36 AM by nul-points »