PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2024-11-27, 07:40:23
News: Forum TIP:
The SHOUT BOX deletes messages after 3 hours. It is NOT meant to have lengthy conversations in. Use the Chat feature instead.

Pages: 1 2 [3] 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ... 24
Author Topic: Investigating "anomalies" in Bifilar coils  (Read 221036 times)
Group: Guest
Brad, does your scope have cursors? If so can you read the values of the Math trace using the cursors?
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4728


Buy me some coffee
Your Atten can't do trace math? ISTR something about that. Well, I don't really trust the Rigol's trace math very much either, really, and I didn't include RMS values in those shots up above because the scope's values are a little smaller than they should be for pure sine waves... wait, maybe that's telling me they aren't really pure sine waves and the "p-p x 0.3535" calculation doesn't apply.

For the FG pumping out the power, it's not hard at all to cobble together a simple powered transistor amplifier that can be driven by the FG and put the same sinus signal at greater voltage into your coils.

As far as a 4ch scope with isolated channel references... that's a toughie. I've used some 2 channel scopes, portables, with isolated references like the Fluke 199 ScopeMeter and the Tek THS720P (has power quality analysis built in!) but I can't think of any quality 4ch units with isolated references. However... the cure for that is Differential Voltage Probes !! And real Current Probes !! They can be costly too though. I wish I had one of each myself.  But there are pitfalls.... we are dealing with a situation where inductance can have a great effect on measurements and that's not even including phase shift/power factor.

Remember this blast from the past?
http://www.youtube.com/edit?o=U&video_id=KWDfrzBIxoQ

Yes,the atten can produce a math trace,but dose not give you the value.
Been through this with many people,including Poynt via skype video,and even he couldnt work it out.

The youtube link you posted,is my youtube channel -wrong link perhap's ?

Im thinking that you may be referring to that BPC circuit i came up with a few years back,where you replicated it,and had the brightest LEDs you'd seen-without blowing--for an !apparent! 0 watt input lol.


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4728


Buy me some coffee
Brad, does your scope have cursors? If so can you read the values of the Math trace using the cursors?

Oh there's been plenty of cursing,trying to work out the math on that thing lol

Yes,it has cursors
Now your going to ask if i know how to use them-right? lol

Brad


---------------------------
Never let your schooling get in the way of your education.
   
Group: Guest
Argh, I got the wrong link, I blame YT changing things.
Here's the one I meant:

http://www.youtube.com/watch?v=KWDfrzBIxoQ

I'll also edit my original post.
   
Group: Guest
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4728


Buy me some coffee
Argh, I got the wrong link, I blame YT changing things.
Here's the one I meant:

http://www.youtube.com/watch?v=KWDfrzBIxoQ

I'll also edit my original post.

Wow
That was a big difference across a short piece of wire.


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4728


Buy me some coffee
Yes, this one was fun:

http://www.youtube.com/watch?v=hvf9Uo7UVx0

Yes-thats the one.
Brightest LEDs ever-without blowing lol.
Must be negative current  :D


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Group: Guest
Yes-thats the one.
Brightest LEDs ever-without blowing lol.
Must be negative current  :D


Brad

Yep, you saw it right there... on both meters at the same time! Negative current!
 
I love those Cen-Tech meters. They are nearly as accurate as my Fluke 87-III and if I break one I can just toss it. I have about six of them, my friends get them for free or a dollar or two at Harbour Freight sales and give them to me.
   

Group: Professor
Hero Member
*****

Posts: 3499
I would appreciate it if somebody could confirm or deny my suspicions that this is a distributed, parallel LC network,
Yes, it is a coiled distributed LC network, that exhibits all of the behaviors of a transmission line when its length approaches or exceeds the wavelength of the signal applied to it.  Like in this video:

https://www.youtube.com/watch?v=ozeYaikI11g
   

Group: Professor
Hero Member
*****

Posts: 3499
EDIT: Do we need to make any corrections for possible phase shift at this point?
No, because the current and voltage are always in phase in a resistor.

But is this really correct? We know that the pure resistors are dissipative loads that give off power as heat, lost to the system.
Yes

But what about the coils? Neglecting their DC resistance of course... is the inductive load actually dissipating power like the resistors do ... or is it _storing_ it?
A pure inductive load can only store energy and give all of it back.  It is lossless in that respect.

The current and voltage across an ideal inductor are not in phase so the formula for average dissipated power PAVG=VRMS*IRMS does not hold (is false in that case).
Actually the power dissipated in an inductor is zero over entire cycles, because the power factor for a pure inductor is always zero, too.  This is because cos(90º)=0.
The same goes for capacitors because cos(-90º)=0, too

EM radiative losses are another story.
« Last Edit: 2017-04-28, 15:22:08 by verpies »
   
Group: Experimentalist
Hero Member
*****

Posts: 1808
I am thankful to see that Verpies has answered the reactive power question as no one seemed to read or believe what I had previously posted.  I had already prepared this before I read Verpies response but I will post this anyway as there is one thing to add and that is the direction of currents in the discussed TBP coil as I had also previously pointed out. 

Attached is a sim plot of TK's TBP that shows three separate power measurements over an even number of cycles.  The red trace shows the average power drawn from the input while the green and blue traces show the average power in the L1-L23 winding relative to the currents in R1 and R2(see previous schematic).  Note that the average powers in the winding nearly cancel which will be very close to TK's actual bench circuit measurements if taken in this manner.

PM
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4728


Buy me some coffee
No, because the current and voltage are always in phase in a resistor.
Yes
A pure inductive load can only store energy and give all of it back.  It is lossless in that respect.

The current and voltage across an ideal inductor are not in phase so the formula for average dissipated power PAVG=VRMS*IRMS does not hold (is false in that case).
Actually the power dissipated in an inductor is zero over entire cycles, because the power factor for a pure inductor is always zero, too.  This is because cos(90º)=0.
The same goes for capacitors because cos(-90º)=0, too

EM radiative losses are another story.

Verpies

Could you take a look at post 40,and let us know what you think.


Cheers


---------------------------
Never let your schooling get in the way of your education.
   

Group: Professor
Hero Member
*****

Posts: 3499
Could you take a look at post 40,and let us know what you think.
I agree with TK's calculations

2-Dissipated power of R1+L1
3-Dissipated power of R2+L2
With respect to this, I must add that these questions do not have a clear answer because L1 and L2 do not dissipate any energy - they just store it, only to give it up later. 
Ideal inductors and capacitors are non-dissipative (lossless).

The resistors R1 and R2 will dissipate electric energy as heat, of course ...just like TK has calculated.


Perhaps a better question to ask would be: "What is the ratio of stored to dissipated energy in R1+L1 ?," etc...
   
Sr. Member
****

Posts: 286
Yes, it is a coiled distributed LC network, that exhibits all of the behaviors of a transmission line when its length approaches or exceeds the wavelength of the signal applied to it.  Like in this video:

https://www.youtube.com/watch?v=ozeYaikI11g

Is there any data regarding the lengths of wire used? I see that upwards of 1MHz has been used in these tests.  That would be a wavelength of 300 meters. None of those BP coils look to me like they are within that range... please feel free to correct me if I'm wrong

Dave
   
Sr. Member
****

Posts: 472
How about placing resistor outside the pancake coil and measuring the difference of temperatures between those resistors ? If there is COP 2 then resistor connected to the center tap should be twice hotter
   

Group: Professor
Hero Member
*****

Posts: 3499
That would be a wavelength of 300 meters. None of those BP coils look to me like they are within that range... please feel free to correct me if I'm wrong
Yes, the physical length is too small but the electrical length might not be because of the additional mutual inductance and capacitance resulting in a small velocity factor.
Lately, Itsu had a video demonstrating a small VF like that. Ask him for a link.
   
Group: Experimentalist
Hero Member
*****

Posts: 1808
Regarding the velocity of TK's TBP using his overall inductance of 680uH and distributed capacitance of 2.8nfd with the formula v=1/√LC, we get v=724.71e3 m/s.  The velocity factor would then be Vf=1/c√LC=2.416e-3 where c is the speed of light or 300e6 m/s.

With the simulation first dip at 277kHz, the wavelength using the above velocity would be 2.61 meters or 1.31 meters for half wavelength.

However, this may not tell the accurate story due to the cross coupling of the distributed capacitance.  This is easier to visualize when looking at the attached schematic with the lumped equivalent TBP sections redrawn.  This is not a conventional transmission line.

PM 
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4728


Buy me some coffee
I agree with TK's calculations
With respect to this, I must add that these questions do not have a clear answer because L1 and L2 do not dissipate any energy - they just store it, only to give it up later. 


The resistors R1 and R2 will dissipate electric energy as heat, of course ...just like TK has calculated.




Quote
Ideal inductors and capacitors are non-dissipative (lossless).

Yes,but we are far from ideal.

Quote
Perhaps a better question to ask would be: "What is the ratio of stored to dissipated energy in R1+L1 ?," etc...

Well,as we know what the dissipated energy is in both R1 and R2,the better question would be-what is the dissipated energy in L1 and L2.

Due to the fact that i am using a very small amount of energy,and the coil are of a very large size wire,measuring heat dissipation is out of the question. So how to calculate the dissipated energy by L1 and L2 ?.

Is it just automatic that we say that it's stored energy,because if it's not,then we have an OU occurrence--and we couldnt have that now-could we lol.

If the energy is being stored in the inductor,then the total P/in should be equal to the power being dissipated by the resistor only.
If in my case,R2 is removed,the the Pin should go down by the same amount that R2 was dissipating ,providing nothing else is changed during the test.

If an inductor/coil is radiating a magnetic and electric field,is it not dissipating !some! power?.


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4728


Buy me some coffee
I agree with TK's calculations
With respect to this, I must add that these questions do not have a clear answer because L1 and L2 do not dissipate any energy - they just store it, only to give it up later.
Ideal inductors and capacitors are non-dissipative (lossless).

The resistors R1 and R2 will dissipate electric energy as heat, of course ...just like TK has calculated.


Perhaps a better question to ask would be: "What is the ratio of stored to dissipated energy in R1+L1 ?," etc...

It is my understanding that it takes energy to produce a magnetic and electric field that changes in time.
This being the case,then L1 and L2 must be dissipating some energy at all times.


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Professor
Hero Member
*****

Posts: 3499
Yes, but we are far from ideal.
In that case they should be treated as ideal inductors in series with a parasitic resistor R0.
So, instead of writing L1+R1 you'd write L1+R0+R1.  The R0+R1 can be added together and called e.g. Rx, thus simplifying the entire expresion and analysis to L1+Rx.

Well,as we know what the dissipated energy is in both R1 and R2,the better question would be-what is the dissipated energy in L1 and L2.
None, but there will be energy dissipated in R0...and in Rx.

So how to calculate the dissipated energy by L1 and L2 ?.
You don't.  L1 and L2 do not dissipate energy by the very definition of pure lumped inductance in a circuit.
You measure/estimate the parasitic R0 and calculate its power dissipation according to i2*R0

If an inductor/coil is radiating a magnetic and electric field,is it not dissipating !some! power?
It is my understanding that it takes energy to produce a magnetic and electric field that changes in time.
This being the case,then L1 and L2 must be dissipating some energy at all times.
Yes, but EM radiation from inductors is a different story and a minor effect in most cases.

P.S.
Ferromagnetic hysteresis losses (core losses) and Eddy currents losses can be significant.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4728


Buy me some coffee
Regarding the velocity of TK's TBP using his overall inductance of 680uH and distributed capacitance of 2.8nfd with the formula v=1/√LC, we get v=724.71e3 m/s.  The velocity factor would then be Vf=1/c√LC=2.416e-3 where c is the speed of light or 300e6 m/s.

With the simulation first dip at 277kHz, the wavelength using the above velocity would be 2.61 meters or 1.31 meters for half wavelength.

However, this may not tell the accurate story due to the cross coupling of the distributed capacitance.  This is easier to visualize when looking at the attached schematic with the lumped equivalent TBP sections redrawn.  This is not a conventional transmission line.

PM

PM

Forgive me for not making some comments on your post,but i am still in the process of learning about transmission lines,and not much point making comments on something i know very little about ATM.


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Group: Ambassador
Hero Member
*****

Posts: 4045
Tinman
I hope some of us[Me] can learn right along side you, and honestly it would seem intuition does Tug at this with the "whatifs"??

thanks

Chet
   
Group: Guest
Regarding the velocity of TK's TBP using his overall inductance of 680uH and distributed capacitance of 2.8nfd with the formula v=1/√LC, we get v=724.71e3 m/s.  The velocity factor would then be Vf=1/c√LC=2.416e-3 where c is the speed of light or 300e6 m/s.

With the simulation first dip at 277kHz, the wavelength using the above velocity would be 2.61 meters or 1.31 meters for half wavelength.

However, this may not tell the accurate story due to the cross coupling of the distributed capacitance.  This is easier to visualize when looking at the attached schematic with the lumped equivalent TBP sections redrawn.  This is not a conventional transmission line.

PM

Or about 0.65 meter for the 1/4 wavelength.

I've been trying to relate those wavelengths to the physical dimensions of the TBF coil.  According to this calculator
http://www.giangrandi.ch/soft/spiral/spiral.shtml
my "unwrapped" bifilar length is about 11.74 meters, which seems about right.
   
Group: Ambassador
Hero Member
*****

Posts: 4045
Russ is exploring here also [Thx to Matt Watts]

https://www.youtube.com/watch?v=-HDwOwfIHns
   
Group: Guest
His current probes cost six times what my car is worth! 

But does he have a set of these? This is the kind of thing you really need for this work.

http://www.aaronia.com/products/antennas/RF-Field-Probes-PBS1/

They are only 1500 Euro for the set, or 2500 Euro with a 40dB preamplifier setup.

   
Pages: 1 2 [3] 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ... 24
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-11-27, 07:40:23