Quote from: l0stf0x on October 11, 2016, 05:53:35 AM
....
I made some measures..

Direct the load to the battery: 3.70V --- 0.58mA  ---> 0.2146 Watt
Through the circuit:                4.90V --- 93.2mA  ---> 4.5668 Watt

That's 2028% difference..

The loal direct to battery is very very dimmed..
Through the circuit is so bright that you can't see it directly.. near to burn I guess.

Hi,

Unfortunately, both calculations are wrong, if you do the math correctly, you get 0.002146 Watt i.e 2.146 mW the LED load consumes directly from the battery, 
and when you see the 93.2 mA current, then the power is 4.9 V * 0.0932 A = 0.45668 Watt, ok? Now the further question is where did you get the 4.9 V?
Across the LED load when it was brightly lit in the circuit? Even so, you cannot compare the two, see below why?

Quote from: kEhYo77 on October 11, 2016, 10:24:54 AM
...
I'd like to know why are you using the LED in measuring the input on the input side. It is not clear right now.
Why the LED is not bright when you measure the input current?
...

When he connects the LED directly across the battery, his LED is not bright because his battery voltage is lower than the LEDs forward voltage.

So it is not the input current to the circuit but only to the LEDs, (the circuit is not involved), ok?

If the goal is to get circuit efficiency, either with the LEDs or a DC motor as the load, then the input current to the circuit should be measured
while the load is running.  There is no sense to run the load directly from the battery and check the input current and voltage to it and then run the same load
from the circuit and check the output current and voltage. 
The two power levels has no any connection: the 2.1 mW to LEDs directly has nothing to do with the 0.47 W LED power received from the circuit. 

Gyula

Quote from: l0stf0x on October 11, 2016, 01:18:45 PM
Not yet and I am away from my workplace to do it now.. Tomorrow I ll do that, and other tests..

I will stick to the dc motor and forget about the leds.. there is something strange going on with this LED pcb. Maybe its the bridge rectifier on pcb .. I really don't have an idea.

I have to test the battery discharge times for both cases.

and to check if I can recharge a second battery or the same battery in closed loop.

After all that .. I will move to more advanced coils setups. I guess the voltage/current will be more if you add copper turns to secondaries. We ll see

Hi l0stf0x - Thanks for reportng your great work!

Sometime back, we did a Battery charge test using the same circuit. This was another of Wistiti's works, orriginally from SkyWatcher I think?

Using standard Battery formula's and proceedures, we did get some good results!

So, This is worth investigating!!! See below Image.

Gyula, please correct me if my calculations are wrong! We did do several experiments and got slightly different results at different stages of these works.

   Chris Sykes
       hyiq.org


P.S: Digital Multimeters are very unreliable at High Frequency and should not be trusted %100, but all in all, good work!!!

Quote from: EMJunkie on October 11, 2016, 05:40:01 PM

....
Gyula, please correct me if my calculations are wrong! We did do several experiments and got slightly different results at different stages of these works.
....
P.S: Digital Multimeters are very unreliable at High Frequency and should not be trusted %100, but all in all, good work!!!

Hi Chris,

No offense intended but I am not here to check whether someone can correctly multiply some numbers.
I did an exception for member l0stf0x because he seems to be new here. He obviously made a wrong comparison between
his input and output measurements. This was my point in writing to him at all. His errors in multiplying the numbers was a secondary issue.
I hope you understand.

Gyula

Quote from: gyulasun on October 11, 2016, 06:41:31 PM
Hi Chris,

No offense intended but I am not here to check whether someone can correctly multiply some numbers.
I did an exception for member l0stf0x because he seems to be new here. He obviously made a wrong comparison between
his input and output measurements. This was my point in writing to him at all. His errors in multiplying the numbers was a secondary issue.
I hope you understand.

Gyula

Of course, Understand.

It is very much easier to make incorrect measurements than correct ones, but this is no reason to suspect or expect that there is not anything important to ones work!

   Chris Sykes
       hyiq.org

Thanks Gyula.. Yes you are absolutely right with calculations, Its mA.. Sorry about that

Also you are right about the wrong way of measuring.. I should measure input output voltage and current with circuit in play.
Its due to these leds.. They trick me and confuse me.
Sorry guys 


Anyway I will continue through the day and post with proper measurements.. if possible.. because as EMJunkie said, you can't trust digital meters in high freq.

Quote from: gyulasun on October 11, 2016, 01:53:07 PM
Hi,

Unfortunately, both calculations are wrong, if you do the math correctly, you get 0.002146 Watt i.e 2.146 mW the LED load consumes directly from the battery, 
and when you see the 93.2 mA current, then the power is 4.9 V * 0.0932 A = 0.45668 Watt, ok? Now the further question is where did you get the 4.9 V?
Across the LED load when it was brightly lit in the circuit? Even so, you cannot compare the two, see below why?

When he connects the LED directly across the battery, his LED is not bright because his battery voltage is lower than the LEDs forward voltage.

So it is not the input current to the circuit but only to the LEDs, (the circuit is not involved), ok?

If the goal is to get circuit efficiency, either with the LEDs or a DC motor as the load, then the input current to the circuit should be measured
while the load is running.  There is no sense to run the load directly from the battery and check the input current and voltage to it and then run the same load
from the circuit and check the output current and voltage. 
The two power levels has no any connection: the 2.1 mW to LEDs directly has nothing to do with the 0.47 W LED power received from the circuit. 

Gyula