To be deleted

[Void]

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]

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]

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

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]

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]

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