Why the efficiency of air core transformer changes (up and down)

[UtsavRaj]

The graph below shows the efficiency of an AIR CORE transformer at different voltages (from MAINS). At low voltages, it has high efficiency and gets lowers because of high magnetising current causing heat loss. But why is it higher at some points again? I do not understand this pattern.

Please give me a detailed explanation with scientific terminology.
Thank you.

Graph : https://postimg.org/image/vd50yniox/

Table of current and voltage: https://postimg.org/image/cojojmvhd/

Efficiency and voltage:
https://postimg.org/image/di6wj9jlj/

The setup for air-core: the coil were put on top of each other with cardboard in between for support .(
Coil used: http://int.frederiksen.eu/shop/product/coil-f--student-transformer--400-turns

 

[CWatters]

At low voltages, it has high efficiency

Looking at your graph it appears that at _one_ data point it has "high" efficiency (0.088%). Either side of that data point the efficiency appears more consistent with the rest of your graph. Is that data point correct?

 

[UtsavRaj]

The graph looks like this if you zoom near the end parts. Link:https://postimg.org/image/t2y1eqhgh/
I want to understand why it has a bouncing ball type graph (as it goes up and down and less up and down and so on)
I am sorry but I do not understand what you are talking about.
Thank you for your time

 

[CWatters]

I am sorry but I do not understand what you are talking about.

Ok I will try to explain.

You said..

At low voltages, it has high efficiency

So I looked at this graph..
https://postimg.org/image/vd50yniox/

At low voltages (<1V) there are four data points.

0.2V approx 0%
0.4V approx. 0.09%
0.6V approx. 0%
1V approx. 0%

Only one of these data points has "high" efficiency. Is it possible you made a mistake when recording the data?

The graph looks like this if you zoom near the end parts. Link:https://postimg.org/image/t2y1eqhgh/
I want to understand why it has a bouncing ball type graph (as it goes up and down and less up and down and so on)

Is this...
https://postimg.org/image/cojojmvhd/
_all_ of the data from the experiment?

If that is all of the data you have then the "bouncing ball type graph" is probably due to the curve fitting program you used. You do not appear to have enough data points to get a nice smooth curve that bounces up and down.

 

[UtsavRaj]

Only one of these data points has "high" efficiency. Is it possible you made a mistake when recording the data

I didn't make any mistakes. because i checked 3 times The problem is that I want to understand why it has the "up-down-up-down" relationship and why the air core transformer behaves that way.

Is this...
https://postimg.org/image/cojojmvhd/
_all_ of the data from the experiment?

The data i used to plot is the one below:

Efficiency and voltage:
https://postimg.org/image/di6wj9jlj/

I don't get why it rises again at the specific voltages and believe that people here can help me. I used the bouncing ball idea as it was easier to convey but it really does show a similar pattern although i agree that the data is a little low to truly call it that. But i am interested in the explanation of the rise-drop than the trend itself.
Thank you for your time

 

[UtsavRaj]

At low voltages (<1V) there are four data points.

0.2V approx 0%
0.4V approx. 0.09%
0.6V approx. 0%
1V approx. 0%

I apologise as I framed it wrongly. What i meant was that at specific voltages the there an increase in efficiency but as you from lower to higher voltages, the increase became smaller and smaller.

Hope this helps you understand what i want meant.

 

[CWatters]

What was the turns ratio of the transformer? I tried calculating it using the primary and secondary voltages and get either 3:1 (3 data points) or about 27-30:1 (8 data points).

 

[CWatters]

How were you measuring the secondary current? What was the secondary load? A resistor?

 

[mfb]

I didn't make any mistakes. because i checked 3 times The problem is that I want to understand why it has the "up-down-up-down" relationship and why the air core transformer behaves that way.

I don't think it is an effect of the transformer operation itself. I would expect some problems with data-taking, or maybe mechanical issues like a variable coil position (did you just put the other coil on top, or did you fix its location somehow?).

You have three groups of efficiencies: 0.00033% (most values), 0.003% to 0.013% (three values) and then this one outlier of 0.088%. That could indicate three different setups, or some changes between them. The small differences within the low-efficiency group can come from measurement uncertainties, I wouldn't worry about those too much. The massive difference between the groups needs an explanation first.

It would help to know more about your experiment, as CWatters already commented. A picture of the whole setup could be interesting as well.
Which setup and which device(s) did you use to measure voltage and current? Did you change the range of the instruments?

 

[jim hardy]

How did you determine efficiency? Power in vs power out ?

How did you measure AC microamps to six significant figures the in presence of a line frequency magnetic field?

What is physical arrangement of the coils? The meters and their leads?

Be aware that in an air core transformer the flux is not constrained to an iron core, instead it's free to meander
so as secondary current increases it will make flux find another path back to primary coil

I think that a lot of what you measured is EMI.and you would have been better off to leave the secondary open circuited , using it as a flux detector, that way it wouldn't affect flux distribution
Lastly

PLEASE A DETAILED EXPLANATION WITH SCIENTIFIC TERMINOLOGY .

When in a nice restaurant, if you shout demands at the servers you're apt to get soup spilled in your lap.

old jim

 

[UtsavRaj]

First of all, I would like to thank you all for commenting as i have been constantly bugged by this bizarre results and seeing people help me out makes my day. :)

When in a nice restaurant, if you shout demands at the servers you're apt to get soup spilled in your lap.

old jim

I totally agree sir but this is actually something i wrote for yahoo answers where people give minimal expression. I sincerely apologise if it came out rude or as shouting but i need the scientific explanation because i want to understand it. Also, I really like this how you made me understand sweetly.

What was the turns ratio of the transformer?

It was 400:400 which gives 1:1 ratio

The setup for air-core: the coil were put on top of each other with cardboard in between for support .(
Coil used: http://int.frederiksen.eu/shop/product/coil-f--student-transformer--400-turns

I used this coil for BOTH primary and secondary coil.

How were you measuring the secondary current? What was the secondary load? A resistor?

Which setup and which device(s) did you use to measure voltage and current? Did you change the range of the instruments?

I was using a bulb as a load (I think a load is an electrical appliance which uses up energy, please correct me if i am wrong).
As for data collection, Digital ammeter and Voltmeter were used which had the option of A, mA and uA (the reason i could measure even small currents).

How did you determine efficiency? Power in vs power out ?

Yes. It is efficiency so the ratio is made into a percentage.

I don't think it is an effect of the transformer operation itself. I would expect some problems with data-taking, or maybe mechanical issues like a variable coil position (did you just put the other coil on top, or did you fix its location somehow?).

The coils have a square gap which you can see. So i put cardboard to hold it still.

I think that a lot of what you measured is EMI

I believe so that i measured the EMI.

Be aware that in an air core transformer the flux is not constrained to an iron core, instead it's free to meander
so as secondary current increases it will make flux find another path back to primary coil

Could you please elaborate on this?

It would help to know more about your experiment, as CWatters already commented. A picture of the whole setup could be interesting as well.

Actually, I am a grade 11 student working on this as my essay and i will credit the people who help me here. The problem is that if i give my images here then it will come as plagiarism. Therefore, I will give you a short summary and some other pictures- if any further problems then, I can provide everything i have (I sincerely apologize).

The method went like this:

• Powerpack attached to rheostat
• rheostat attached to digital ammeter
• Digital ammeter attached to Transformer coil (primary winding)
• Transformer coil attached back to powerpack.
• The digital voltmeter attached in parallel to the primary coil.
• The secondary winding was attached to another Digital ammeter.
• The digital ammeter attached to a bulb.
• Bulb attached back to the secondary winding.
• Another digital voltmeter attached in parallel to secondary winding.
• All wires were copper wires.
• All the voltage and current readings were taken from the respective devices and recorded.

As for the setup of Air-core transformer
The image below will help:
Note: Coil were resting on the table and no gap between the coils.

IMPORTANT NOTE
Interesting observation i forgot to mention: At higher voltages, the primary coil got heated to pretty high temperature. The time taken to increase in temperature decreased with Higher voltages. I didn't find significant heating in the secondary coil.

Thank you for your time and guidance.

 

[jim hardy]

I totally agree sir but this is actually something i wrote for yahoo answers where people give minimal expression. I sincerely apologise if it came out rude or as shouting but i need the scientific explanation because i want to understand it. Also, I really like this how you made me understand sweetly.

I too should apologize , you are likely not accustomed to local customs where i live
so if you'll accept my apology and excuse this "grumpy old man" 's outburst we'll get along fine.

Actually, I am a grade 11 student working on this as my essay

That is impressive ! Glad to hear there are still practical courses in schools .

Could you please elaborate on this?

I work in thought experiments... so,
Fire up that young imagination !

Your coil makes a "solenoid", try a search on magnetic field of a solenoid
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html
In yourimagination
think of the magnetic field from your primary winding
it might look like this

of course reversing itself at line frequency

Magnetic flux travels in closed loops just like current, so the analogy of "Kirchoff's Flux Law" applies.
The picture just isn't big enough to show the bigger loops of flux
but with no iron around flux does spread out into the surrounding space

Now stack an identical coil on top of your solenoid but don't connect the lamp load yet .
Some of the flux from your bottom solenoid, quite a bit of it in fact, will continue up and exit through the upper coil linking ts turns and inducing voltage there.
I drew that flux in blue, please excuse my lack of draftsmanship. It's "MSPaint" after all...

An open circuited coil in a magnetic field makes a simple flux detector - voltage is rate of change of flux through the coil , and whatever flux goes through it is changing at line frequency.

Current is what pushes flux, and the bottom coil pushes some of its flux through the top one. Since the top coil has no current it doesn't push back against the flux from bottom one.

Now connect a load to the top coil and let current flow in it
Now the top coil can push flux of its own
and it will push against the flux from bottonm coil because of something called "Lenz's Law"
giving you in effect opposing electromagnets
distorting the flux field
so some of the flux from bottom coil won't go up into top coil it'll detour out into air, lowering the efficiency of your transformer.

You could demonstrate that using paper iron filings and DC current through your coils
i didnt draw the re-routed flux on this one , i think you're savvy enough to have figured it out by now ...
observe opposing fields

so any current flowing in your secondary pushes flux down against the upward flux from primary, lowering secondary voltage
i think if you open circuit secondary you'd find secondary voltage linear with primary current
but as soon as you allow current to flow it shape shifts your flux field .

Now iron conducts magnetic flux hundreds or thousands of times better than does air
so with ah iron core almost all the flux stays in the iron
as shown in this picture from the datasheet for your coil that you linked
http://int.frederiksen.eu/shop/product/coil-f--student-transformer--400-turns , click on pdf

There's a small amount of flux that goes through the air
indicated by the solitary black lines going there in that picture
but a huge amount of flux going through the iron.The flux that shortcuts through air is a necessary evil in transformer design, it's called "Leakage Flux"
and precision transformers arrange the windings carefully so as to minimize it.

Your arrangement has a lot of it.

Magnetics is rather fun
i hope you get interested
try searching on some of the terms
and get a feel for the two basic units
nowadays everybody uses SI
The magnetism that flows around in loops is called Flux it's a diffuse continuous field but we draw it as lines to show intensity, SI unit is Weber
and you'll find plenty of sites where intensity is shown by color
and a few that still use old cgs units of Maxwells or Lines instead of Webers
the force that pushes magnetic flux around loops is MagnetoMotive Force (MMF) , SI unit is Amp-Turns,

most folks work in flux density Webers per square meter, called Teslas , that way their formulas apply to any size coil.

I hope this gets you started.

Now a practical thought
When you connect a meter in the presence of an airborne magnetic field,
any flux encircled by your meter leads creates voltage in them
so the voltage arriving at your meter is in error by that induced voltage.
It'll be small for your case because your 400 turn primary encircles a lot more flux than does the one turn loop made by your meter
but i wonder how much it affected those feeble secondary current readings ?
To reduce that induced test lead error you minimize the area they encircle by twisting them together. That's why you see so much "Twisted pair" wiring in audio work.

It looks like you are in an exceptional school. I had the good fortune to be in a good one during my high school years.
Have fun, learn a lot , think simple

old jim

 

[mfb]

Some current flows through the voltmeter. Knowing the internal resistance of the voltmeter would help, especially if you switched ranges in between (this changes the internal resistance and therefore influences different measurements in different ways).

I think that a lot of what you measured is EMI

I believe so that i measured the EMI.

Quite sure jim meant electromagnetic interference. Noise from the environment. Then everything will influence the measured values, including the current location of the cables in the setup and even your own body position.

 

[UtsavRaj]

Quite sure jim meant electromagnetic interference. Noise from the environment. Then everything will influence the measured values, including the current location of the cables in the setup and even your own body position.

I thought he meant electromotive force (EMF). My mistake.
If you don't mind, can you elaborate on this

 

[UtsavRaj]

Some current flows through the voltmeter. Knowing the internal resistance of the voltmeter would help, especially if you switched ranges in between (this changes the internal resistance and therefore influences different measurements in different ways).

I am sorry but the resistance of the voltmeter would be hard to obtain because my school is closed.

 

[UtsavRaj]

I too should apologize , you are likely not accustomed to local customs where i live
so if you'll accept my apology and excuse this "grumpy old man" 's outburst we'll get along fine.

It's my pleasure to be guided by smart and intelligent people by you.

I have some question regarding your explanation but i will most likely post them later because it is 3:32AM here.

I am very happy to have commented on this forum because not only everyone has helped me but also i have learned a lot of new things;So, do not feel annoyed if i ask a lot of questions.

Thank you for your precious time everyone

 

[UtsavRaj]

Sir,
Your explanation is really great but still does not answer why there is a up-down-up-down graph than for eg. 1/x relationship
Questions hoarding you later

 

[mfb]

I thought he meant electromotive force (EMF). My mistake.
If you don't mind, can you elaborate on this

Alternating currents, e. g. from the power grid and all connected cables in the room, lead to changing electromagnetic fields around them. Your secondary circuit might pick those up, independently of the transformer coupling. It works a bit like a radio: you can receive radio transmission without connecting the radio to anything. You just need an antenna - and your circuit acts as one.
Sensitive AC equipment never operates at line frequency for this reason (60 or 50 Hz depending on the region), unless it absolutely has to. Too much noise.

 

[jim hardy]

Your explanation is really great but still does not answer why there is a up-down-up-down graph than for eg. 1/x relationship

It'll take some more investigation
Using your data

could you plot secondary voltage on vertical versus primary amps horizontal ?
The closer that is to a straight line the less your upsy-downsy was due to secondary coil's MMF pushing backward against primary coil's MMF
Since the ratio of currents was like 10,000 to one i expect a fairly straight line.

Primary voltage is divided between I X R drop in the winding and voltage required to push magnetic flux , called "induced voltage" or "counter EMF".

Have you an idea about the resistance of your bulb load ? Plot secondary volts versus amps and see how straight is the line.
Then you'll begin to get a feel for what your transformer and load were doing.

One observation is worth a thousand expert opinions. See what you can glean from your observations.

Have you had complex arithmetic yet ? Are you accustomed to polar and rectangular representation of volts and current ?

 

[UtsavRaj]

Alternating currents, e. g. from the power grid and all connected cables in the room, lead to changing electromagnetic fields around them. Your secondary circuit might pick those up, independently of the transformer coupling. It works a bit like a radio: you can receive radio transmission without connecting the radio to anything. You just need an antenna - and your circuit acts as one.
Sensitive AC equipment never operates at line frequency for this reason (60 or 50 Hz depending on the region), unless it absolutely has to. Too much noise

A really simple to understand explanation. Thank you very much.
But can elaborate a little about how my equipment could have read this at the 50 Hz region (Singapore)? Is it because of going very low for current readings or something like that?

 

[UtsavRaj]

could you plot secondary voltage on vertical versus primary amps horizontal ?

The graph:

Can you tell me how this relationship tell you about the

The closer that is to a straight line the less your upsy-downsy was due to secondary coil's MMF pushing backward against primary coil's MMF

?

I find this graph very intriguing and most likely will be used in my Essay. Can everyone tell some more interesting graph like this?

Important note
I actually looking at the efficiency of Iron core and Air core transformer at same voltages and trying to explain the difference in it. These graphs can help me a lot or other information can help me a lot.

Have you an idea about the resistance of your bulb load ?

I believe it to be between 2 Ohms to 4 Ohms. But most likely 2 Ohms.

Have you had complex arithmetic yet ? Are you accustomed to polar and rectangular representation of volts and current ?

I don't know this but I am eager to learn.

Question regarding the previous explanation:

but with no iron around flux does spread out into the surrounding space

Does this mean that they are not closed or just very big loops?

so some of the flux from bottom coil won't go up into top coil it'll detour out into air, lowering the efficiency of your transformer.

Due to the pushing of the top coil to the bottom. Will it go into the air because of that only?

There's a small amount of flux that goes through the air

A little bit explanation would be nice here. Is it due to saturation?

Thank you everyone for time

 

[Baluncore]

Maybe someone has trouble with reading or positioning decimal points.

Looking at the graph at the top of post #21, I see three points out of line. For the two points in the top left corner 0.35V and 0.4V, I suspect the decimal point has been misplaced in voltage. If the values are reduced by a factor of ten they become 0.035V and 0.04V which then fall on the line.

The point on the LHS axis should be closer to zero volts which will be in the local EMI noise floor.
So maybe 0.12V is also out by a factor of ten making a corrected voltage reading of 0.012V.

 

[UtsavRaj]

Thank you Baluncore. I will check again

 

[UtsavRaj]

Interesting graphs

Can you tell me how this relationship?

Also, I can find the resistance by taking out the gradient of the straight line. What resistance would i be taking about then?

Important note
I actually looking at the efficiency of Iron core and Air core transformer at same voltages and trying to explain the difference in it. These graphs can help me a lot or other information can help me a lot.

Saturation for my Iron core.
NOTE: Not my graph.
https://openenergymonitor.org/emon/sites/default/files/yhdcreport/yhdc05.gif
by plotting Secondary coil current vs Primary coil current- I can see whether any saturation has occurred or not?

 

[jim hardy]

Can you tell me how this relationship tell you about the

Thank you for doing that.
It'd be really nice if Baluncore is right and there's a decimal point error or interference that pushed those three points off the line.
The reason i asked is this

1. A coil hanging in space that's occupied by a changing magnetic field will experience induced voltage.
That voltage will be the number of turns X rate of change of magnetic flux going through the coil
if no current is allowed to flow in the coil, it is then acting like a flux detector . A simple coil and voltmeter will do the job.

2. A coil hanging in space that has current forced through it will produce a local magnetic field with intensity that is directly proportional to the coil's current.
That is shown in the hyperphysics link i gave for field of a solenoid...
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html
Since your two coils have the same number of turns
and your secondary current is thousands of times smaller than your primary current
i figured the the magnetic field in the region surrounding your two coils comes >99.9% from primary and <0.1% from secondary
which means your secondary makes a decent if imperfect flux detector, certainly close enough for plotting a graph.
It'll measure the flux in the area

What i wanted to see is -
Is measured flux linear with current ?

Secondary voltage measures flux
Primary current causes flux
if the ratio of those two is constant you'll get a straight line, ie constant slope

which would confirm that the primary flux is behaving as predicted by that hyperphysics link, which said
B = μnI
flux density B is directly proportional to current I, because μ is a constant of free space an n is 400 for your coil
giving one great faith in his data

If you find that those three outlying points were errors of transcription and they do indeed lie on the line
then your "efficiency" plot might straighten out.

I like to plot raw data for sanity checks.

For a hint at bulb behavior, plot secondary voltage vertical versus secondary current horizontal.

by plotting Secondary coil current vs Primary coil current- I can see whether any saturation has occurred or not?

To study saturation you'd do much better to plot secondary voltage measured with zero secondary current, again just use coil as a flux detector. (No load except voltmeter.)
After all that's what saturation is, too much flux for the iron.

 

[jim hardy]

Also, I can find the resistance by taking out the gradient of the straight line. What resistance would i be taking about then?

That gives you the "impedance" which is slightly more than resistance.
Have you studied inductance yet ?

 

[jim hardy]

Important note
I actually looking at the efficiency of Iron core and Air core transformer at same voltages and trying to explain the difference in it. These graphs can help me a lot or other information can help me a lot.

Make the same plot, secondary volts(unloaded, no bulb) versus primary current with iron core.
I doubt you'll be able to push even an amp through primary.

 

[mfb]

A really simple to understand explanation. Thank you very much.
But can elaborate a little about how my equipment could have read this at the 50 Hz region (Singapore)? Is it because of going very low for current readings or something like that?

Right. The noise will lead to some current readings that do not depend on the primary coil. But the misplaced decimal point explanation fits better.

Do you also have a plot (secondary voltage) versus (secondary current)? It should also be a linear relation, deviations would point to problems.

 

[UtsavRaj]

I will be going through the piles of data to check the voltages and currents again.

To study saturation you'd do much better to plot secondary voltage measured with zero secondary current, again just use coil as a flux detector. (No load except voltmeter.)
After all that's what saturation is, too much flux for the iron.

The problem is my school is closed. Although i can do this after my school opens.
But the graph i used can be used as a substitute?

Do you also have a plot (secondary voltage) versus (secondary current)? It should also be a linear relation, deviations would point to problems.

For a hint at bulb behavior, plot secondary voltage vertical versus secondary current horizontal.

 

[UtsavRaj]

That gives you the "impedance" which is slightly more than resistance.
Have you studied inductance yet ?

I have not in school but i have been studying articles. But if you don't mind, A simple explanation would be good.

 

[UtsavRaj]

Oh my god, I am really embarrassed by my mistake.

Looking at the graph at the top of post #21, I see three points out of line. For the two points in the top left corner 0.35V and 0.4V, I suspect the decimal point has been misplaced in voltage. If the values are reduced by a factor of ten they become 0.035V and 0.04V which then fall on the line.

The point on the LHS axis should be closer to zero volts which will be in the local EMI noise floor.
So maybe 0.12V is also out by a factor of ten making a corrected voltage reading of 0.012V.

Baluncore, you are absolutely right. It's actually what you predicted.

I am really sorry to everyone because of mistakes. Therefore, I will check all my work before I post all the changed graph.

Thank you to all who have been helping me.

 

[jim hardy]

But the graph i used can be used as a substitute?

Be sure to note that it was somebody else's data.
For studying core saturation It would be better to have secondary open circuit voltage, look how much more current you have.

For studying power transfer it'll be okay
When you said bulb for load i just asuned you mean a light bulb, incandescent ?
just be aware your load is not constant because incandescent lamps are nonlinear. Cold current is ~10X hot current.
Look at the scatter in your secondary volts vs current plot. Volts divided by current is load resistance. Constant resistance would give a straight line.

 

[UtsavRaj]

When you said bulb for load i just asuned you mean a light bulb, incandescent ?

Yes

Be sure to note that it was somebody else's data.

I am going to use my data of secondary and primary current.

Look at the scatter in your secondary volts vs current plot. Volts divided by current is load resistance. Constant resistance would give a straight line.

As soon as i check everything, I will plot it.

 

[mfb]

just be aware your load is not constant because incandescent lamps are nonlinear. Cold current is ~10X hot current.
Look at the scatter in your secondary volts vs current plot. Volts divided by current is load resistance. Constant resistance would give a straight line.

Apart from the 4 outliers we found already, it looks quite linear. We are talking about ~10 µW of power.

 

[jim hardy]

~10 µW of power.

Well that surely shouldn't heat things up very much at all.

I have not in school but i have been studying articles. But if you don't mind, A simple explanation[of inductance - jh] would be good.

Impedance is the property of a circuit that opposes flow of current.
"Impede" is the memory aid (mnemonic _{look up that word} )
Impedance is measured in Ohms
It is what you get by applying ohm's law Volts = Amps X Ohms
or Ohms = Volts / Amps

1 volt will push 1 amp through 1 ohm

so your transformer's apparent primary side impedance is primary volts divided by primary amps
when you get data squared away plot that, primary volts vertical primary amps horizontal
then plot on same graph a second line of the ratio (primary volts/primary amps) and see if it's a straight line
that second line is primary side impedance

What is impedance of your transformer from primary side V and I ? __________ ohms
............Impedance comes in two basic flavors
Resistance or Reactance
both are measured in ohms
The difference between them is this
when current flows through resistance it dissipates energy into heat
when current flows through reactance it does not make any heat, the energy goes instead into an electric or magnetic field.

we will consider reactance only for AC because you are working with AC and the math is straightforward
Impedance is the sum of resistance plus reactance
but they don't simply add like regular numbers do
they add like the two sides of a right triangle do, Pythagoras (you knew there was some reason for trig, didn't you ?)
so the impedance is √(resistance² + reactance²)

We usually denote Resistance by letter R
and Reactance by letter X
and Impedance by letter Z (because I is already taken by Current)
so Z = √(R² +X²)

and you will know what is impedance from your Vpri/Ipri ratio plot - what is Zprimary ? ___________ ohms

Next you will take your ohm meter and measure resistance R of your primary coil Rprimary = ___________ ohms
and you will calculate what is X by pythagoras' theorem X = √(Z² - R²)

Now reactance comes in two sub-flavors, Inductive and Capacitive
Inductive reactance is a property of coils, the magnetic field surrounding them holds the energy from current that i mentioned above
Capacitive reactance is a property of plates separated by a small distance, the electric field between them holds the energy from current i mentioned up above

and since you are working with coils - surely you can guess which sub flavor of of reactance it is that you just calculated?

We designate capacitive reactance by Xc (uppercase X sbuscript C as in capacitance)
We designate inductive reactance by Xl (uppercase X subscript L as in lag)

So how many ohms of Xl did you come up with for your transformer primary ? _____________________

Inductive reactance is a characteristic of inductors, which have the physical property " inductance." and it's an important concept for any budding EE to grasp.
Inductance is named after a scientist Henry and we measure inductance in Henries. But we designate Henries not by H but L ( for some reason.)
Inductive reactance that you calculated from your measurement is the product of your coil's inductance and the frequency at which you are operating it.
Are you 50 hz or 60 ?

Xl = 2 X π X f_requency X L_henries
so L = Xl / 2πf

What is the inductance L of your primary winding ? _______________________
The physics definition of Inductance is (Flux Φ per Ampere I ) X #of turns N
L = NΦ/I

Now think back - that's why i asked you to plot secondary volts(proportional to flux) versus primary current. It tells you what is inductance of your coil.
And an air core coil will have constant inductance. But an iron cored one will not, because of saturation.

You specify inductors by their inductance L because they might be used at any frequency.

That's a too-basic of introductions to inductance, but it's tailored to your experiment. Now that you have successfully measured it, you'll want to read up on it .

old jim

 

[Tom.G]

pythagoras' theorem X = √Z2 - R2)

Did you mean Z² - R²) ?

 

[jim hardy]

Did you mean Z² - R²) ?

Yes musta hit subscript instead of super?

found a few others...

Thanks Tom ! i appreciate the help .

old jim

 

[The Electrician]

UtsavRaj, be careful not damage your primary coil. You mentioned in an earlier post that the coil was getting hot.

There is a warning in: http://int.frederiksen.eu/Admin/Pub...iles/exp/137000/137710-EN-The-transformer.pdf

It says:
"When the core is not in place in the primary coil, the
current can get so high that the coil overheats and the
plastic coil former is damaged. The coils used have
these maximum currents:
462510 200 windings Max. 2 A
462520 400 windings Max. 1 A
462525 800 windings Max. 0.5 A"

Be sure to make your measurements at those high currents quickly and then turn the voltage down.

 

[jim hardy]

There is a warning

Good Catch !

 

[The Electrician]

the impedance is √(resistance² + reactance²)

I think the OP will have difficulty making this measurement on his "air" core setup. The reactance of his primary coil will be so small compared with the resistance, that (with the equipment he has) he won't be able to see the difference between Z and R.

I have an old Radio Shack filament transformer that I keep around because the laminations weren't varnished so they can be removed. This makes it handy for measurements of various parameters with and without an iron core:

'

You can see the green rubber bands holding everything together.

With the core in place, the primary impedance measured at the primary @ 60 Hz with .5 volt excitation is shown in 3 different formats herewith:

Here the reactance is the first (top) value, and resistance is the second value.

Here's the impedance and phase angle.

In the second analyzer image, we see that the reactance and resistance are the same order of magnitude.

 

[The Electrician]

Now here's the same transformer with the core removed:

And here are the same 3 images showing the primary impedance @60 Hz with an "air" core:

Notice the phase angle is nowhere near 90 degrees. The primary impedance is nearly a pure resistance.

In the second analyzer image, we see that the reactance is much smaller than the resistance. Also, note that the resistance without the core is quite different than it was with the core. That's because the core loss becomes part of the measured AC resistance.

My transformer has primary and secondary wound on the same bobbin, but the principle will apply to the OP's setup. I don't think the OP will be able to see the difference between Z and R when the core is absent.

The coefficient of coupling in the OP's setup is abysmal as is the efficiency. He's going to get a good lesson in the importance of an iron core in transformers!

 

[jim hardy]

He's going to get a good lesson in the importance of an iron core in transformers!

By hands- on ! That's wonderful.

@UtsRavj - where is your school? I'm delighted to hear somebody still teaches hands-on electronics .

 

[UtsavRaj]

UtsavRaj, be careful not damage your primary coil. You mentioned in an earlier post that the coil was getting hot.

There is a warning in: http://int.frederiksen.eu/Admin/Public/DWSDownload.aspx?File=/Files/Files/exp/137000/137710-EN-The-transformer.pdf

It says:
"When the core is not in place in the primary coil, the
current can get so high that the coil overheats and the
plastic coil former is damaged. The coils used have
these maximum currents:
462510 200 windings Max. 2 A
462520 400 windings Max. 1 A
462525 800 windings Max. 0.5 A"

Be sure to make your measurements at those high currents quickly and then turn the voltage down.

That's the first concern me and my teacher had when it got really hot - the insulation and wire could melt that's why i did the experiment really quickly and with one minute break (so the temp doesn't affect my experiment.) Thank you for your advice.

My transformer has primary and secondary wound on the same bobbin, but the principle will apply to the OP's setup. I don't think the OP will be able to see the difference between Z and R when the core is absent.

So are you saying that Impendance would be equal to resistance (mathematically speaking)

@UtsRavj - where is your school? I'm delighted to hear somebody still teaches hands-on electronics .

It is in Singapore. I chose this topic as part of my essay.

UPDATE ABOUT THE GRAPHS
I will most likely post all the updated information tomorrow afternoon. (SGT)

Thank you for your time everyone.

 

[The Electrician]

So are you saying that Impendance would be equal to resistance (mathematically speaking)

Yes, for the case where there is no iron core.

The current in the primary of a transformer has two main components. One part of the current is due to the secondary current being "reflected" to the primary. This is the current that supplies power to the load on the secondary; it's the "useful" part of the primary current. The other part of the primary current is the "magnetizing" current. It is only there because the inductance of the primary requires it; laws of physics and all that. But since the magnetizing current must pass through the resistance of the copper wire of the primary, it causes losses (heating mostly) in the wire. These losses reduce the efficiency of the transformer. If the magnetizing current could be reduced, the losses would be reduced and the efficiency increased.

Since the magnetizing current is only there to energize the primary inductance, it would be reduced if the inductance of the primary were increased. That's what the iron core does.

Looking at the images of the measured impedance of my transformer, you can see that when the iron core is present the inductance of the primary is about 1.59 henries. When the core was removed, the inductance was only .012 henries, more than 100 times less. This means that the magnetizing current would be 100 times less when the core is in place.

The main purpose of an iron core is to reduce magnetizing current. It also increases the coupling between primary and secondary, especially in an arrangement like your transformer where the primary and secondary are not wound on the same bobbin.

 

[jim hardy]

Primary counter-emf could hardly be less than emf induced in an open circuit secondary since it's the same flux
well, same flux on a bifilar winding anyway where primary and secondary wires are so very close that hardly any flux fails to link both .
He has a lot more leakage though

His data shows maybe 30 to 1 ratio primary to secondary voltage
so i expect detectable inductive reactance even if just a few % with air core

old jim

 

[The Electrician]

Consider the OP's table of primary and secondary voltages. The OP is keeping way too many digits for a lot of his values. I make Baluncore's suggested changes to the first 3 secondary voltages. Then calculate the ratio of primary to secondary voltages for all 12 pairs. It appears that the 6th and last are different enough from 30 that they could reasonably be considered outliers, so I delete them. I'm left with 10 pairs and calculating the ratio of primary to secondary voltage I get these 10 values:

{33.3, 30.8, 28.9, 29.9, 31.3, 27.7, 27.6, 27.7, 27.3, 27.9}

Now subtract 30 from each value, divide by 30 and multiply by 100 to get a percent deviation from 30. I get a list like this:

{11.0%, 2.67%, -3.67%, -.33%, 4.33%, -7.67%, -8.0%, -7.67%, -9.0%, -7.0%}

That's quite a bit of variance. The OP might just barely be able to claim an accuracy of around 10%.

Let's say we applied 1 amp to the primary and expected about 1 volt across it due to IR drop in the primary and 1/30 volt due to inductive reactance.

If the resistive voltage and reactive voltage added arithmetically, we would expect a drop due to 1 amp of 1 + 1/30 or 1.033 volts, compared to just 1 volt due to the resistive drop. Given the variance in the OP's measurements, stray voltages due to placement of his voltmeter leads around the magnetic fields present, and other error sources, I wouldn't bet on his ability to see the difference.

But resistive and reactive voltages don't add like that. They add like SQRT(1^2 + .0333^2) which would give 1.00055 volts. I'm fairly certain the OP can't distinguish that from 1.00000 volts with the equipment and setup he has.

 

[jim hardy]

Ahh the beauty of graph paper !
Plot deviation vs current - is it random, IR drop, or heating ?

As you said earlier - when he gets iron i there he'll be amazed

if we can give him a feel for inductance at this early stage in his education, what a boost it'll be .

 

[The Electrician]

Once he gets the iron in, measuring the inductance should be possible even with his less than ideal setup.

UtsavRaj, would you please measure the resistance of your coils with a DVM; also the resistance of your bulb load, and post the results here.

Do you know what the rated voltage of the bulb is?

 

[UtsavRaj]

First of all, i am very sorry that i did not reply for this long. It was a medical emergency with one of my close relatives which have subdued for now.

UtsavRaj, would you please measure the resistance of your coils with a DVM; also the resistance of your bulb load, and post the results here.
Do you know what the rated voltage of the bulb is?

Make the same plot, secondary volts(unloaded, no bulb) versus primary current with iron core.
I doubt you'll be able to push even an amp through primary.

The result would only come after 11th August because by then my school will open.

Plot deviation vs current - is it random, IR drop, or heating ?

I do not get this.

UPDATE ON GRAPH
I will definitely tomorrow.

IRON CORE

Once he gets the iron in, measuring the inductance should be possible even with his less than ideal setup.

I have already done that.

The methods is

The method went like this:

• Powerpack attached to rheostat
• rheostat attached to digital ammeter
• Digital ammeter attached to Transformer coil (primary winding)
• Transformer coil attached back to powerpack.
• The digital voltmeter attached in parallel to the primary coil.
• The secondary winding was attached to another Digital ammeter.
• The digital ammeter attached to a bulb.
• Bulb attached back to the secondary winding.
• Another digital voltmeter attached in parallel to secondary winding.
• All wires were copper wires.
• All the voltage and current readings were taken from the respective devices and recorded.

The picture of the transformer:

More Details

Explanation for trend:

Sudden increase:

• The induced magnetic field becomes stronger due to permeability of iron core which provides magnetic flux densities 10000 times that of air.high permeability means that the most of magnetic field is concentrated in the iron core and therefore allows a better inductance.n iron core is often used to provide a low-reluctance path for the magnetic flux
• The resistance of a given piece of wire depends on three factors: the length of the wire, the cross-sectional area of the wire, and the resistivity of the material composing the wire. All of the factors remain the same. Therefore, coil have a steady resistance.
• Copper losses are low as resistance is the same and the current is low.
• The iron core Magnetic Stray losses at minimum although it increases with negligible with higher voltage.
• At this point, the hysteresis and eddy currents are minimal (also because of insulated Iron core).
• As weak magnetic field produces weak eddy currents and hysteresis.
• Magnetostriction is low as well
• Electric hum low due to less magnetic stray.

At higher voltage the increase becomes less.

• More consistent value:
• Magnetostriction increases due to a strong magnetic field.
• Copper losses increases because of increase in current (A big source of energy loss).
• Eddy Current becomes a bigger problem because of increase in magnetic field that causes bigger current being made in the iron core.
• Hysterosis also increases. (Don’t exactly know)
• Magnetic stray loss is still negligible but increases.
• The induced magnetic field becomes stronger.
• Electric hum a little bit due to more magnetic stray (which causes friction and therefore energy loss).

Is this explanation of trend right or am i wrong or less somewhere?Some questions i have:

• Does Hysteresis increases with increase in voltage and current?
• How does magnetising current affect iron core in efficiency?
• "To maximize flux linkage with the secondary circuit, an iron core is often used to provide a low-reluctance path for the magnetic flux. The polarity of the windings describes the direction in which the coils were wound onto the core. Polarity determines whether the flux produced by one winding is additive or subtractive with respect to the flux produced by another winding" Can this explain anything about the Iron as well as Air core efficiency?

• "In lower frequencies (mains) the efficiency is a lot worse, because an air core will have an enormous magnetizing current that will cause losses by heat on the wires." Why high magnetizing current for air core?

As always, thank you for your time.

 

[UtsavRaj]

Explanation for trend:

Sudden increase:

• The induced magnetic field becomes stronger due to permeability of iron core which provides magnetic flux densities 10000 times that of air.high permeability means that the most of magnetic field is concentrated in the iron core and therefore allows a better inductance.n iron core is often used to provide a low-reluctance path for the magnetic flux
• The resistance of a given piece of wire depends on three factors: the length of the wire, the cross-sectional area of the wire, and the resistivity of the material composing the wire. All of the factors remain the same. Therefore, coil have a steady resistance.
• Copper losses are low as resistance is the same and the current is low.
• The iron core Magnetic Stray losses at minimum although it increases with negligible with higher voltage.
• At this point, the hysteresis and eddy currents are minimal (also because of insulated Iron core).
• As weak magnetic field produces weak eddy currents and hysteresis.
• Magnetostriction is low as well
• Electric hum low due to less magnetic stray.

At higher voltage the increase becomes less.

• More consistent value:
• Magnetostriction increases due to a strong magnetic field.
• Copper losses increases because of increase in current (A big source of energy loss).
• Eddy Current becomes a bigger problem because of increase in magnetic field that causes bigger current being made in the iron core.
• Hysterosis also increases. (Don’t exactly know)
• Magnetic stray loss is still negligible but increases.
• The induced magnetic field becomes stronger.
• Electric hum a little bit due to more magnetic stray (which causes friction and therefore energy loss).

Is this explanation of trend right or am i wrong or less somewhere?

This is for the graph above.