PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2022-01-18, 03:52:08
News: A feature is available which provides a place all members can chat, either publicly or privately.
There is also a "Shout" feature on each page. Only available to members.

Pages: 1 ... 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 [25] 26 27 28 29 30 31 32 33
Author Topic: Itsu's workbench / placeholder.  (Read 47755 times)
Full Member
***

Posts: 114


Buy me a drink
Hi Nelson,

I am fine thanks.   No problem with your activity, i still remember how it was when worked 10 hours  ;)

I can make some measurements on the base during loading off course, will that be compared to ground or to emitter or?

 
The base signal compared to ground looks very much like the blue base signal shown in the bottom screenshot here:
https://www.overunityresearch.com/index.php?topic=3691.msg80986#msg80986

Have a good night both.

Regards Itsu

Hi Itsu ,
Thanks by your availability to make such measures .
It will interest measure transistor base relative to D1/D2 Cathode with and without load , to see the type of fluctuation and if becomes negative during process of load in the output.
Many thanks Itsu .


---------------------------
Best Rewards
Nelson Rocha

" The goal is not to be successful, the goal is to be valuable.
Once you’re valuable, instead of chasing success,
it will attract itself to you. "
   

Group: Professor
Hero Member
*****

Posts: 2677
Finaly the scope with current probe showed that the DMM's, at least the one on the output, can not handle the high frequency of the current signals and is way off.
It is usual for DMM's error to increase with the crest-factor and with frequency.

Incandescent bulbs are very trustworthy as output power meters of the DUT because there they act as BOTH average current indicators AND the load.
Incandescent bulbs are NOT trustworthy as input power meters of the DUT because there they act ONLY as average current indicators.

Stepping up the output current at the expense of input voltage is not unusual.  It is what transformers do every day.
However, if the output current does not increase at the expense of the instantaneous product of the input voltage with input current, then it is an anomaly.  As usual, it is important not to multiply the average input current by the average input voltage to obtain the average input power, because multiplying averages disregards their relationship in time.
IAVG * VAVG == PAWG is true only for DC.
   

Group: Moderator
Hero Member
*****

Posts: 3332
verpies,

Quote
Incandescent bulbs are very trustworthy as output power meters of the DUT because there they act as BOTH average current indicators AND the load.
Incandescent bulbs are NOT trustworthy as input power meters of the DUT because there they act ONLY as average current indicators.

I think that very well could be the cause of the effect seen by Nelson his circuit.


Quote
However, if the output current does not increase at the expense of the instantaneous product of the input voltage with input current, then it is an anomaly.

So thinking about that i will calculate the output current via Iout = √ (Pout / R).
The "instantaneous product of the input voltage with input current" (Pin) i already have, so plotting this will show an anomaly if there is one?

I will do that tonight.

Itsu
   

Group: Moderator
Hero Member
*****

Posts: 3332

When i plot Iout versus Pin at certain loads i get the below data and graph.

Guess that downward line does not look promising.

Itsu
   

Group: Moderator
Hero Member
*****

Posts: 3332
Hi Itsu ,
Thanks by your availability to make such measures .
It will interest measure transistor base relative to D1/D2 Cathode with and without load , to see the type of fluctuation and if becomes negative during process of load in the output.
Many thanks Itsu .

Nelson,

Screenshot 1 shows the signal across base to cathode diodes with no ouput load (open).
Screenshot 2 shows the same signal with a 1K output load
Screenshot 3 shows the same signal with a shorted output (0 ohm load).


itsu
   
Full Member
***

Posts: 114


Buy me a drink
It is usual for DMM's error to increase with the crest-factor and with frequency.

Incandescent bulbs are very trustworthy as output power meters of the DUT because there they act as BOTH average current indicators AND the load.
Incandescent bulbs are NOT trustworthy as input power meters of the DUT because there they act ONLY as average current indicators.

Stepping up the output current at the expense of input voltage is not unusual.  It is what transformers do every day.
However, if the output current does not increase at the expense of the instantaneous product of the input voltage with input current, then it is an anomaly.  As usual, it is important not to multiply the average input current by the average input voltage to obtain the average input power, because multiplying averages disregards their relationship in time.
IAVG * VAVG == PAWG is true only for DC.



Hi Verpies,

Thank about your feedback .
I know bulbs are very trustworthy to validate current output , This was the main reason for i placing the lamp in series at the output of circuit , in order to have an easier visual perception of the current in the load by lack of equipment to measure ,  But I didn't know,  lamps, were not reliable reference when used the input showing because only show average current indicators, but we are all time learning; Thanks for the tip, because that aspect had never occurred to me sincerely .
Does this apply because the Non Linear Resistance of the bulb?

Verpies could this circuit be considered a current step down being the small cmc isolator transformer wired in 1:1 ?  maybe You could show other side that is escape me , because normally is winding ratio  that could define a step up or step down like happen in a normal operation transformer .
I will appreciate your answer if could contribute to improve my knowledge .
I really appreciate your opinion , many thanks .


---------------------------
Best Rewards
Nelson Rocha

" The goal is not to be successful, the goal is to be valuable.
Once you’re valuable, instead of chasing success,
it will attract itself to you. "
   
Full Member
***

Posts: 114


Buy me a drink
Nelson,

Screenshot 1 shows the signal across base to cathode diodes with no ouput load (open).
Screenshot 2 shows the same signal with a 1K output load
Screenshot 3 shows the same signal with a shorted output (0 ohm load).


itsu

Hi Itsu ,

Many thanks by the shots .
Seems a negative voltage signal exist when the circuit run in idle mode (without load)   , and when the circuit has a load the voltage increase to a more positive value, I’m understand right ?  Or am I misinterpreting the scope shots?
I remember when I did some measurements to Gyula , have the idea about that negative values, but not at this level :) If these negative values are confirmed, would it be interesting to consider the possible power factor coefficient calculation?
Just a thought.

Many thanks Itsu , need to think about that points and try made some tests this weekend  :) .


PS- I forget to say that merged frequencies seems very curious too ! :)



---------------------------
Best Rewards
Nelson Rocha

" The goal is not to be successful, the goal is to be valuable.
Once you’re valuable, instead of chasing success,
it will attract itself to you. "
   

Group: Moderator
Hero Member
*****

Posts: 3332
Hi Nelson,

yes, it seems there is a negative pulsing signal on the base compared to the diodes junction which changes towards less negative during loading.

Not sure what you mean by:
"If these negative values are confirmed, would it be interesting to consider the possible power factor coefficient calculation?"

I agree about the signal shape, looks curious.
The setting of the base resistor (a 500 ohm potmeter in my case) influences this shape considerable, as it does with the whole operation of this circuit.

I did some Pin measurements yesterday again now using my scope/math and i had quite a different outcome as before using a different base resistor setting.

So it might be there is a specific base resistance setting that increases the efficiency.
A needle in a haystack challenge.

Itsu
« Last Edit: 2020-05-16, 09:46:45 by Itsu »
   
Full Member
***

Posts: 114


Buy me a drink
Hi Itsu ,

Many thanks by the shots .
Seems a negative voltage signal exist when the circuit run in idle mode (without load)   , and when the circuit has a load the voltage increase to a more positive value, I’m understand right ?  Or am I misinterpreting the scope shots?
I remember when I did some measurements to Gyula , have the idea about that negative values, but not at this level :) If these negative values are confirmed, would it be interesting to consider the possible power factor coefficient calculation?
Just a thought.

Many thanks Itsu , need to think about that points and try made some tests this weekend  :) .


PS- I forget to say that merged frequencies seems very curious too ! :)


Good morning Itsu, Many thanks by your answer.

What I wanted to say is if the source of this negative pulsating signal can be related to a reactive nature state of the circuit when there is no load present at the output.
It would be interesting to put a block diode at input to prevent possible returns from possible reactive to main power source, as on different occasions it seemed to me that there was some kind of return to the input source  visible on the displays of my power supply just a thought .

Today i will Lift an oscilloscope I bought used, at  market auction.
It is an OWON HDS2062M-N a portable oscilloscope, but I couldn't resist the price 75 € :) it was stronger than me .....  I hope that all of their functions are 100% operational, because even that small money these days makes a difference to me.
Apparently from the photos and the feedback of seller it seemed to be in good condition, moreover during my participation in the project in Germany, I had identical equipment available, and despite being limited in some Math functions it is better than having nothing :) and for sure will help me . I'm flea! :)
I'll give you feedback later about the equipment .
I wish you a great day


---------------------------
Best Rewards
Nelson Rocha

" The goal is not to be successful, the goal is to be valuable.
Once you’re valuable, instead of chasing success,
it will attract itself to you. "
   

Group: Moderator
Hero Member
*****

Posts: 3332

Nelson,

ok about the "possible returns from possible reactive to main power source", i will check on that during my Pin measurements as it should be shown on the scope (negative Pin).

Good for the scope, my first scope was/is an OWON PDS8202 which i still have.
Nice piece of equipment, which worked fine, only had the rotary switches noise which caused the vertical and horizontal settings jump all over the place after a few years.
Had to add some 100nF caps across them to keep it stable.

i am sure you enjoy this one.

regards Itsu
   
Full Member
***

Posts: 114


Buy me a drink

Good morning Itsu, Many thanks by your answer.

What I wanted to say is if the source of this negative pulsating signal can be related to a reactive nature state of the circuit when there is no load present at the output.
It would be interesting to put a block diode at input to prevent possible returns from possible reactive to main power source, as on different occasions it seemed to me that there was some kind of return to the input source  visible on the displays of my power supply just a thought .

Today i will Lift an oscilloscope I bought used, at  market auction.
It is an OWON HDS2062M-N a portable oscilloscope, but I couldn't resist the price 75 € :) it was stronger than me .....  I hope that all of their functions are 100% operational, because even that small money these days makes a difference to me.
Apparently from the photos and the feedback of seller it seemed to be in good condition, moreover during my participation in the project in Germany, I had identical equipment available, and despite being limited in some Math functions it is better than having nothing :) and for sure will help me . I'm flea! :)
I'll give you feedback later about the equipment .
I wish you a great day


Hi Itsu ,
I already have the Owon Scope ,and he is in very good condition :) Seems is a reconditioned unit . I already make some validations and seems working fine .
Only the probes seem a little more worn, but apparently they are functional. Now need to read the manual  :D .
By the price of 75€ in think is a really nice unit , even for experiments abroad.
Some shots of the unit ;)




---------------------------
Best Rewards
Nelson Rocha

" The goal is not to be successful, the goal is to be valuable.
Once you’re valuable, instead of chasing success,
it will attract itself to you. "
   

Group: Moderator
Hero Member
*****

Posts: 3332

Looking good indeed Nelson, nice case, all stuff together there  O0 

I recognize the emitter signal in the middle picture.

congrats, Itsu
   
Full Member
***

Posts: 114


Buy me a drink
Looking good indeed Nelson, nice case, all stuff together there  O0 

I recognize the emitter signal in the middle picture.

congrats, Itsu

Hi Itsu ,
Yes is a shot of emitter signal in circuit :) good eye shot !
   I need to explore the basic operation and math functions , but in general I am happy that I can explore the circuit a little more :) with this oscilloscope .
Wish you a good end of day  Itsu .



---------------------------
Best Rewards
Nelson Rocha

" The goal is not to be successful, the goal is to be valuable.
Once you’re valuable, instead of chasing success,
it will attract itself to you. "
   

Group: Moderator
Hero Member
*****

Posts: 3332

I redid the input / output power measurements at different base resistor settings, but the outcome is that the efficiency stays at a max. of around 56% mostly in the 1K load area.

When i sweep/load the output with a 1M potmeter in series with a 100 Ohm resistor, no anomalies (negative input current / input power) are seen.
There is however a point around 72K where some sort of wild oscillations occure looking a the scope signals and the input bulb.

I will zoom in on that area later his weekend.

good evening all, regards Itsu
   
Group: Tech Wizard
Hero Member
*****

Posts: 948
Hi Nelson,

Your oscilloscope is a nice catch,  looks to be in good shape and hopefully the built-in Li-ion battery will serve you for a long time too. The 60 MHz bandwidth is also wide enough for most such pulsed circuits like this oscillator.   8)

Good night to you and all.

Gyula
 

   
Full Member
***

Posts: 114


Buy me a drink
Hi Nelson,

Your oscilloscope is a nice catch,  looks to be in good shape and hopefully the built-in Li-ion battery will serve you for a long time too. The 60 MHz bandwidth is also wide enough for most such pulsed circuits like this oscillator.   8)

Good night to you and all.

Gyula


Hi Gyula ,
I think it was a good buy, and is in very good condition, but battery condition I really don't know, only the next days will tell me if battery is in good condition.
The battery takes some time to fully charge, just over 3 hours, let's see what happens, but in general is a good tool piece even to help me sometimes in repair services in  CNC controllers and to help find fault encoders :) .
Gyula  and all have a nice Weekend !


---------------------------
Best Rewards
Nelson Rocha

" The goal is not to be successful, the goal is to be valuable.
Once you’re valuable, instead of chasing success,
it will attract itself to you. "
   

Group: Professor
Hero Member
*****

Posts: 2677
Does this apply because the Non Linear Resistance of the bulb?
No, it is because the voltage is also a factor in input power calculation. (next to the current), but when you connect an incandescent bulb as the SOLE load of the DUT, then the voltage stops to matter because the power becomes the function of only current flowing through it, according to P=i2R. This is all because a sole bulb is an output current indicator and the load in one package.

However, if you had a SEPARATE load connected in addition to the bulb, then the problem would reappear (just like at the input side) and the calculation P=i2R would not hold anymore. You'd have to multiply the output voltage by output current to get the output power.

...and since it is NOT true that the average of products is equal to the product of averages, you cannot just multiply an average voltage times average current to obtain the average power. Such operation loses the temporal relationship between the voltage and current and yields an invalid answer (except for DC). Unfortunately DMMs measure the mean (or RMS) voltage & currents and can't deal accurately with high frequencies and crest factors.
   

Group: Moderator
Hero Member
*****

Posts: 3332
I redid the input / output power measurements at different base resistor settings, but the outcome is that the efficiency stays at a max. of around 56% mostly in the 1K load area.

When i sweep/load the output with a 1M potmeter in series with a 100 Ohm resistor, no anomalies (negative input current / input power) are seen.
There is however a point around 72K where some sort of wild oscillations occure looking a the scope signals and the input bulb.

I will zoom in on that area later his weekend.


good evening all, regards Itsu

I zoomed in on his 72KhHz area by using a 100K potmeter as a load.

Now no wild oscillations are noticed, so i think the wild oscillations seen with the 1M potmeter were some kind of artifacts caused by a rapid load change with this potmeter.

Regards Itsu
   
Full Member
***

Posts: 114


Buy me a drink
No, it is because the voltage is also a factor in input power calculation. (next to the current), but when you connect an incandescent bulb as the SOLE load of the DUT, then the voltage stops to matter because the power becomes the function of only current flowing through it, according to P=i2R. This is all because a sole bulb is an output current indicator and the load in one package.

However, if you had a SEPARATE load connected in addition to the bulb, then the problem would reappear (just like at the input side) and the calculation P=i2R would not hold anymore. You'd have to multiply the output voltage by output current to get the output power.

...and since it is NOT true that the average of products is equal to the product of averages, you cannot just multiply an average voltage times average current to obtain the average power. Such operation loses the temporal relationship between the voltage and current and yields an invalid answer (except for DC). Unfortunately DMMs measure the mean (or RMS) voltage & currents and can't deal accurately with high frequencies and crest factors.


Hi Verpies many thanks by your answer.

After reading your answer, and analyzing it in depth, it seems logical and assertive to your reasoning.  This quote of yours was the key to I'm understanding the logic of your reasoning.
“and since it is NOT true that the average of products is equal to the product of averages,”
I will remove the small bulbs on input and output , seems better :) to analyze the circuit in a more assertive way.

I would be grateful if you could also give your personal opinion on the other question  I ask you .

Verpies could this circuit be considered a current step down being the small cmc isolator transformer wired in 1:1 ?  maybe You could show other side that is escape me , because normally is winding ratio  that could define a step up or step down like happen in a normal operation transformer .


Many thanks


---------------------------
Best Rewards
Nelson Rocha

" The goal is not to be successful, the goal is to be valuable.
Once you’re valuable, instead of chasing success,
it will attract itself to you. "
   
Full Member
***

Posts: 114


Buy me a drink
I zoomed in on his 72KhHz area by using a 100K potmeter as a load.

Now no wild oscillations are noticed, so i think the wild oscillations seen with the 1M potmeter were some kind of artifacts caused by a rapid load change with this potmeter.

Regards Itsu

Hi Itsu ,
did you mean that merged oscillations disappeared ? Or the ripple at output disappeared ?
Many thanks


---------------------------
Best Rewards
Nelson Rocha

" The goal is not to be successful, the goal is to be valuable.
Once you’re valuable, instead of chasing success,
it will attract itself to you. "
   

Group: Moderator
Hero Member
*****

Posts: 3332
I redid the input / output power measurements at different base resistor settings, but the outcome is that the efficiency stays at a max. of around 56% mostly in the 1K load area.

When i sweep/load the output with a 1M potmeter in series with a 100 Ohm resistor, no anomalies (negative input current / input power) are seen.
There is however a point around 72K where some sort of wild oscillations occure looking a the scope signals and the input bulb.

I will zoom in on that area later his weekend.

good evening all, regards Itsu

Hi Nelson,

what i mean is that the above mentioned wild oscillations seen when i sweep/load the output with a 1M potmeter in series with a 100 Ohm resistor are there not anymore
when  i sweep/load the output with a 100K potmeter (zoomed in).

The merged oscillations, as you call them, are still there as is the ripple at the ouput.

Regards Itsu   


 

   
Full Member
***

Posts: 114


Buy me a drink
Hi Nelson,

what i mean is that the above mentioned wild oscillations seen when i sweep/load the output with a 1M potmeter in series with a 100 Ohm resistor are there not anymore
when  i sweep/load the output with a 100K potmeter (zoomed in).

The merged oscillations, as you call them, are still there as is the ripple at the ouput.

Regards Itsu

Hi Itsu  hope you goes well ,
Now i understand your answer :)  I already asked Verpies this question, but I would like to know your opinion.
Can this circuit be considered a step down buck converter , with L2 / L3 being a 1: 1 transformer?
Many thanks Itsu






---------------------------
Best Rewards
Nelson Rocha

" The goal is not to be successful, the goal is to be valuable.
Once you’re valuable, instead of chasing success,
it will attract itself to you. "
   

Group: Moderator
Hero Member
*****

Posts: 3332
Hi Nelson,

doing well,   thanks.

This circuit "acts" more like a step up boost converter to me giving it steps up the 24V input voltage to 450V (my replication).

How this one works exactly hopefully can be explained by others, but in general, see:
https://en.wikipedia.org/wiki/Boost_converter.

Regards Itsu
« Last Edit: 2020-05-20, 08:45:41 by Itsu »
   

Group: Professor
Hero Member
*****

Posts: 2677
Verpies could this circuit be considered a current step down being the small cmc isolator transformer wired in 1:1
I will not answer this question because two windings on the same core act as a transformer only when the inducing and induced currents flow through them at the same time.
Is that happening in your circuit?
   

Group: Moderator
Hero Member
*****

Posts: 3332
It does happen in my replication:

green is the current, using my current probe, throught L2,
yellow is the voltage across a 1 Ohm 1% induction free csr in the L3 to bridge lead.

Output load was a 1K resistor, current probe was deskewed for 14KHz.

Itsu
   
Pages: 1 ... 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 [25] 26 27 28 29 30 31 32 33
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2022-01-18, 03:52:08