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

[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.

 

[The Electrician]

In your chart check the primary voltage of 6.59 volts. The corresponding primary current appears to be about 10 times too large. Same thing for primary voltage of 11.11 volts. These two points on the graph appear to be outliers.

If you fix those two values your graph will look much better; although the very first two voltage values look questionable too.

 

[UtsavRaj]

In your chart check the primary voltage of 6.59 volts. The corresponding primary current appears to be about 10 times too large. Same thing for primary voltage of 11.11 volts. These two points on the graph appear to be outliers.

If you fix those two values your graph will look much better; although the very first two voltage values look questionable too.

Why does it have to be a straight line again?

 

[mfb]

All currents and voltages should be proportional to each other if we can neglect nonlinear effects (e. g. heating-induced things).

 

[jim hardy]

Why does it have to be a straight line again?

Primary current is the MMF that pushes flux.
Secondary voltage is a measure of the resulting flux.

What is the formula that relates flux to current ? Which terms in that formula might be nonlinear ?

 

[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?

Or do we need more info?
Thank you

Primary current is the MMF that pushes flux.
Secondary voltage is a measure of the resulting flux.

What is the formula that relates flux to current ? Which terms in that formula might be nonlinear ?

Are you asking me or mfbd?

 

[jim hardy]

Why does it have to be a straight line again?
Are you asking me or mfbd?

I thought you were asking me about the graph in post 49.
Perhaps i responded to the wrong question.

old jim

 

[UtsavRaj]

I thought you were asking me about the graph in post 49.
Perhaps i responded to the wrong question.

old jim

The question i was asking you sir was the explanation in post 55

Thank you for your time everyone

 

[UtsavRaj]

In your chart check the primary voltage of 6.59 volts. The corresponding primary current appears to be about 10 times too large. Same thing for primary voltage of 11.11 volts. These two points on the graph appear to be outliers.

If you fix those two values your graph will look much better; although the very first two voltage values look questionable too.

Sorry, the graph is actually this one

 

[UtsavRaj]

can please everyone tell me whether the explanation i gave at post 55 is right or wrong or not enough info?

 

[UtsavRaj]

One part of the current is due to the secondary current being "reflected" to the primary.

I do not get this

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

This tells us whether it behaves like a solenoid or not?

Thank you for your time