Inductance and Capacitance as the Secondary Load

[Jimmy Lalani]

What are the effects of adding inductance and capacitance as secondary load of transformer? I am using a Ring /Toroid Transformer.

 

[Jimmy Lalani]

Hello hardy Sir,
I used a 100VA, 230V/ 2x24V, Secondary rated current Isc=2.08A ,L=36mh/2A/250V, C=58.8micF. I increased the current upto 2A of Inductor. Is this method correct?

 

[jim hardy]

I increased the current upto 2A of Inductor.

Well, what is the impedance of a 38 mh inductor at your line frequency?
What current do you expect per Ohm's Law?

 

[Jimmy Lalani]

Well, what is the impedance of a 38 mh inductor at your line frequency?
What current do you expect per Ohm's Law?

Hello Hardy Sir,
For 38mh inductor only the impedence is 11.938. If I use both inductor and capacitor the impedence is 42.196.
As there are two secondaries so two possible ways of connections are possible:
a) Connecting both in parallel so V=24, But Isec=4.16.
b) Connecting both in series so V=48, Isec= 2.08
The current expected should me near to the load current(rated current given).

 

[jim hardy]

Are you familiar with AC circiut analysis ? Phase angles, polar and rectangular notation and phasor notation ?

http://www.electronics-tutorials.ws/accircuits/complex-numbers.html
https://en.wikipedia.org/wiki/Phasor

Connecting your capacitor and inductor in parallel to a 24 volt winding, their currents will add.
24 volts / 11.938 ohms of inductive reactance = 2.01 amps . Do you know how to calculate its phase angle?
24 volts / 54.134 ohms of capacitive reactance = 0.443 amps at its phase angle .

Accounting for their different phase angles, when i add those two numbers I come up with 1.57 total.
What do you calculate?

 

[Jimmy Lalani]

Hello Hardy Sir,
With inductive reactance it is lagging so -90deg.
With capacitive reactance it is leading so +90
So the magnitude of currents subtract each other and I get 1.567.
Is it correct?

 

[jim hardy]

Yes !

 

[Jimmy Lalani]

Hello Hardy Sir,,
Thank you for your support.
If I want to measure different inductance values of a coil by varying the voltage and current values, then how do I measure it?
https://mail.google.com/mail/u/0/?tab=wm#inbox/15dbc557fa17e6b1?projector=1

 

[jim hardy]

I don't seem able to retrieve that image. Save it on your computer and use the "upload" button on the screen's bottom right.

You already know the answer to that.
You want inductance L ?

You know ohm's law, right ?
You have meters, right ?
You know frequency , right ? You used 50 hz earlier.

Measure volts V and amps I .
Impedance Z = V / I .

Measure DC resistance of your inductor, Ω .
Solve this for X_L
Z = √( Ω² + X_L²)

And you already know X_L = 2πfL
so L = X_L / 2πf

 

[Jimmy Lalani]

Hello Hardy Sir,
I used a signal generator and applied 300Hz frequency and amplitude 400mv. However, when I checked it on a multimeter it showed only 330mv. What can be the reason for this? What can I do about this?

 

[jim hardy]

What can be the reason for this?

I used a signal generator and applied 300Hz frequency and amplitude 400mv.

1. With what did you measure 400 mv ?
2. What changed between your 400mv measurement and your 330 mv measurement ?
for example..
(indent ) 2a. Were the readings taken one before and the other after you connected the load ? Do you think the signal generator might sag a little when loaded?
(indent ) 2b Were they taken with two different instruments ? Look in instruction manual for your multimeter - what is its frequency response ?

What can I do about this?

Make both measurements with the same instrument ?
Turn the knob up to get the output you want ?

.

 

[Jimmy Lalani]

Hello Hardy Sir,
I used HP 3325A signal generator. In this signal generator I entered both values-frequency and Amplitude.
However, when I connected the output of signal generator to a small multimeter, without any load, the multimeter showed 330mv instead of 400mv.

 

[jim hardy]

However, when I connected the output of signal generator to a small multimeter, without any load, the multimeter showed 330mv instead of 400mv.

Well then, one of them is mistaken.
What check can you run to determine which ?This is at 300 hz, correct ?

What kind of meter do you have? What is its frequency range and what is its AC accuracy?

Lastly , did you set the HP3325A synthesizer for sine wave output? It's capable of other functions.

 

[Jimmy Lalani]

Hello Hardy Sir,
Yes the frequency is 300Hz and I set the snthesizer for sine wave output.
I think the meter range is not so high. I connected the output of the generator to the oscilloscope and I get perfect sinusoidal wave with correct frequency and voltage.
In the generator the peak to peak voltage was 1.13 and in the oscilloscope it showed 1.24 V.

 

[jim hardy]

In the generator the peak to peak voltage was 1.13 and in the oscilloscope it showed 1.24 V.

Have you read the instruction manual ?
That HP instrument expects a 50 ohm load.
What load are you providing it?

https://www.scribd.com/document/122...nction-Generator-Operating-and-Service-Manual

 

[Jimmy Lalani]

Hello Hardy Sir,
You are right. I did this and I am getting the output. Thank you very much for your help.

 

[Jimmy Lalani]

Hello Hardy Sir,
With the same transformer, I changed the load to resistive and want to make a equivalent circuit of transformer. Practically, if I perform open circuit and short circuit test, I get all the parameters of the equivalent circuit.( R1,R2', X1, X2', Rc, Xc). However, If I want to calculate these parameters mathematically, than is it possible?

 

[jim hardy]

If I want to calculate these parameters mathematically, than is it possible?

Well anything is possible.

R1,R2

Do you know how to calculate resistance of a length of wire ?

X1, X2

Do you know how to estimate leakage reactance?

Rc, Xc

Do you know how to calculate inductance of a coil on a core ?
Do you know how to calculate core losses ? Look up Steinmetz equations.

 

[Jimmy Lalani]

Hello Hardy Sir,
For resistance Of a length of a wire R=pL/A. L=(4*pi(OD^2-ID^2))/H; Is this correct?
However, I am not ale to calculate the other parameters. If possible can you please explain me with an example?

 

[jim hardy]

For resistance Of a length of a wire R=pL/A. L=(4*pi(OD^2-ID^2))/H; Is this correct?

Wire is usually solid.

I am not ale to calculate the other parameters. If possible can you please explain me with an example?

I think you need to train your search engine and find tutorials appropriate for whatever is your level of familiarity, which i don't know.. Look for tutorials on transformer design. There are plenty in existence and i don't intend to write another.

Here's my first hit from "Inductance calculation"
http://www.rfcafe.com/references/electrical/inductance.htm

that should approximate leakage inductance.

You'll get the hang of digging out the basics. Search on keywords in the articles that you find .

You'll need to learn meaning of terms
magnetic flux , the Weber
magnetic flux density, the Tesla
magnetizing force, the amp-turn usually abbreviated MMF
magnetic reluctance , analogous to ohms , the ratio of magnetizing force and flux
permability , μ , ease with which flux passes through a medium

Magnetic circuits are like electric circuits with one exception. In electric circuits the current stays inside the wires. In magnetic circuits the flux can "leak" through the air bypassing parts of the intended circuit. Often you can ignore leakage, especially if there's a complete closed magnetic circuit of highly permeable material like a good steel transformer core..

Suppliers of Magnetic Materials have reference libraries. Investigate Coilcraft and Mag-Inc .

Good luck in your studies.

 

[Jimmy Lalani]

Hello Hardy Sir,
I have attached a picture containing the parameters calculated by measurement and theoretically. In the picture, I could not do the theoretical calculations of leakage reactances. The practical measurements are done by performing open circuit and short circuit test. The parameters are referred to primary.

Can you please look at the parameters and say if they are correct? I have doubt in Rc and Xc.

Regarding leakage inductances the formula which you provided me the value of relative permeability is coming a bit strange and so the inductance value and the reactance is also much far away from the measured value.
https://redirect.viglink.com/?forma...cafe.com/references/electrical/inductance.htm
I have B=1.57T, I have calculated H using H=(NI)/Le
Le is mean magnetic flux length given by
Le= (pi*(Rout+Rin))/2;
Then µ=B/H;
µr=µ/µ0;
Is this correct?

 

[jim hardy]

Can you please look at the parameters and say if they are correct?

?

I see two columns of figures there with no mention from where they came. How would i judge if they are correct? What did you measure ? How did you calculate them from your measurements ?
You'd have to show your work.

I have B=1.57T, I have calculated H using H=(NI)/Le
Le is mean magnetic flux length given by
Le= (pi*(Rout+Rin))/2;
Then µ=B/H;
µr=µ/µ0;
Is this correct?

That looks okay except for this:
Length of a circular path is pi X diameter and you seem to have calculated Le by pi X Radius ?

That's why one starts by defining all his abbreviations.

 

[Jimmy Lalani]

Hello Hardy Sir,
These calculations are performed as shown in the following figure.

From Measurement : Theoretical:
R1=R2'=25 R1=R2'=29.8
X1=X2'=0.72 X1=X2'=?
Rc= 62235 Rc=50865.38
Xc= 68293 Xc=107779.8
W0=1.04W

For magnetic path length Le, I found the formula from this website:
https://www.cwsbytemark.com/CatalogSheets/MPP PDF files/13.pdf.

 

[Jimmy Lalani]

Hello Hardy Sir,
I recently found this formula for leakage inductance.

page no:-9
https://www.mag-inc.com/getattachme...-Magnetics-Powder-Core-Catalog.pdf?lang=en-US

 

[jim hardy]

I recently found this formula for leakage inductance.

That catalog is for powdered cores intended for high frequency smps application. What kind of transformer do you have ?

....................................
What is it you are trying to accomplish ?

If it is to characterize a transformer from open and short circuit tests, i see no logic in your approach.
Use what you observe, apply what you know and build a complete picture as if working a jigsaw puzzzle.

Here's an equivalent circuit for a transformer, from Wikipedia.

At no load the only things dissipating power are Rp and Rc .
Did you measure the resistance of your winding Rp with an ohm meter ? What is it ?Your datasheet (this sure looks like homework ? A lab report due soon?)
tells you to ignore Rp for the open circuit test and ascribe all power to iron loss, Rc. See parentheses under "Wattmeter".
For the moment let us also ignore Xp , since it has only 5 ma through it the voltage drop can't be much.

Your observation tells you that no load power is 0.85 watt.

You know that P = E²/R, so R = E²/P
(230 volts)² / 0.85 watt is how much resistance ? That's Rc.

230 volts / 0.005 amps is how much impedance ? That's the parallel impedance of Rc and Xm, ignoring Xp..
What inductive reactance in parallel with Rc gives you the parallel impedance you just calculated ? That's Xm.


Now from short circuit test

teacher has again ascribed all power to one place, the copper.
That's not a bad assumption because flux in a short circuited transformer is quite low and losses are proportional to flux^^{(some power slightly greater than 1)}
Which means in the model above I am and Ic both drop to zero, or near enough zero to ignore.

So
12 watts = Ip² X Rp + Is² X Rs
and you can measure both Rp and Rc with an ohm meter.

By ohm's law we know what is voltage across Rp and Rs .
The rest of the applied voltage has to be dropped across the leakage reactances Xp and Xs . But you knew that already.
By ohm's law we also know what is impedance of the transformer when short circuited, 24.5 volts / 0.49 amps.
Call that Zsc for now.

What inductive reactance in series with sum of resistances Rp and Rs will give you Zsc ?
That's the sum of Xp and Xs in the above model.
How to separate each from the sum ?
Hmmm that's a good one...

I think that for purposes of this experiment
i would assume the windings are similar enough(and wound on same core) that their relative inductances will be in proportion to the squares of their respective numbers of turns.
After all. L = μμ₀N²A_rea/l_ength .http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indcur.html

So, what two X's meet this criterion--------
---They add to the sum of Xp and Xs just calculated, and have ratio (Np/Ns)² ---- ?

Now take your Xp back to the beginning and see if that 5 milliamps flowing through it it affects your calculations enough to notice. If so you'll have to iterate.

I'm sorry but i just can't follow your calculations from those images.

There's an approach. Try it to cross check your calculations.

old jim

 

[Jimmy Lalani]

Hello Hardy Sir,
Actually, the company in which I am doing my Master Thesis uses a software which on entering some basic input values and assuming some parameters gives the output of the transformer with electrical circuit of transformer and all other parameters.
I am trying to find some relation with the values provided by software with actual measurement of the transformer and also by doing calculations mathematically.

This photo is not some homework. I just downloaded it from net and put in the values measured to make you understand better.
The values from the measurement are determined by formula from the following link:
http://www.electronicshub.org/open-circuit-and-short-circuit-test-on-transformer/
The transformer which I have used as a reference is: (This pdf is data sheet of transformer in German Language, but easy to understand)
http://www.sedlbauer.de/media/ringkerntrafo_datenblatt_826029.pdf

 

[jim hardy]

I am trying to find some relation with the values provided by software with actual measurement of the transformer and also by doing calculations mathematically.

Ahh so that's the objective.

That electronics hub author you linked does not show the 'equivalent circuit' for a transformer . I don't see where he separates primary and secondary leakage reactances Xp and Xs. And he ignores primary leakage Xp reactance in OC test, as i did .

So his and my methods are nearly identical except
he calculates Rp and Rs by using the wattmeter and ammeter to determine Power Factor (PF) instead of using an ohm-meter tp measure them directly. Sounds roundabout to me.
I personally don't trust a wattmeter to have much precision. Especially with only 5 milliamps of current through it.
So i suggest you check your wattmeter/PF derived numbers for Rp and Rs against a decent ohm meter.
Be aware that inductance of a transformer likely will confuse a DMM , so measure each winding's resistance with the other winding shorted else your DMM may just flash meaningless random numbers.

If you study his and my method you'll find they are the same. Just he solves for R using a wattmeter&ammeter derived power factor and phase angle where i used ammeter&ohm meter , Pythagoras, and rectangular - polar impedance conversion .

I trust Mr Ammeter and Mr Ohm-Meter more than i do Mr Wattmeter. But that's a personal preference. You should determine resistance of windings both ways and see for yourself how closely they agree .
That author had trouble reading a wattmeter at such low power factor and small current. He even got a special watttmeter .

When the transformer is operating on no load, the current drawn by the shunt or parallel parameters is very small about 2 to 5 percent of the rated current. Thus, a low current will flow through the circuit during OC test. In order to be readable by the instruments, the measurements of voltage, current and power must be performed in the low voltage side.
And also, low range current coils and low range ammeter must be selected. The power factor of the transformer on no load is too low which is typically below 0.5 . So in order work with this low value, a LPF watt meter is selected. The equivalent circuit obtained by the OC test is shown below.

He ignored both Rp and Xp in his OC test which is a practical simplification that you can get away with because it doesn't cause much error.

note how closely it resembles this from the Wikipedia model

R and X represent core's iron loss and magnetizing current requirement.

Study that Wikipedia transformer equivalent circuit. Had your author used it in his paper it would have been a lot more easily understood.

I don't know what is your thesis topic
but i'd think you would want to understand the transformer model . It's a very nice tool to have in your "Bag of Tricks" .

 

[jim hardy]

That Hyperphysics page on Inductance will show you how to calculate Xm. Toroids have large Xm.

The Wikipedia transformer page is here https://en.wikipedia.org/wiki/Transformer

 

[jim hardy]

Actually i am very happy to see you try this practical experiment..

It is important to understand how things work so you know when a computer program is spouting gibberish.
Old saying "Garbage In, Gospel Out" - people tend to believe anything that's printed in orderly columns by a computer.

Don't be reliant on the computer , that's helplessness. Be instead the engineer programmers come to for advice.

old jim

 

[Jimmy Lalani]

Hello Hardy Sir,
Thank you very much for the compliment.
Now, With ohm meter I measure Rp=23.5ohm And Rs1=Rs2=O.7 ohm(two secondaries), The turns ratio =Vp/Vs= 230/24=9.5833.
With referred to primary and parallel equivalent resistance I get Rs'=32.12ohm. So the addition is Rp+Rs'= 55.62ohm.
from Short circuit test Zsc=Vsc/Isc= 24.5/0.49= 50ohm,
Rsc= P/i^2sc= 49.97ohm, so when referrd to primary I get Rp=Rs'=Rsc/2= 24.98ohm,
Xsc=1.73ohm, therefore Xp=Xs'=Xsc/2=0.86ohm

This is the problem I am facing when I actually measure the resistance with ohm meter as without short circuit test I cannot calculate Zsc and therefore Xp and Xs'.

 

[jim hardy]

[QUOTE="Jimmy Lalani, post: 5825807, member: 628145"T]he turns ratio =Vp/Vs= 230/24=9.5833.[/QUOTE]
Published numbers are for operation at rated load. They include voltage drop across Rp Xp Rs and Xs.

Turns ratio looks to me more like 230 / 27.39 = 8.397

That's why you run open circuit test, to eliminate or at least minimize the voltage drop across Rp Xp Rs and Xs

@Jimmy Lalani Try your calculations using observed turns ratio at open circuit
because that is when voltage drops across those impedances are negligible.. You'll come a lot closer to the actual turns ratio.

 

[Jimmy Lalani]

Hello Hardy Sir,
1) Is it possible to calculate the no load voltage mathematically? According to my observations it is always between 10-20% more than the rated voltage. However I am interested in calculating it mathematically.

2)For core losses I applied the following method:
The length of the medium flux path is: Lav = Pi*(ID+OD)/2 = 0.198 m.

The mass of the core is: M = Lav *(OD-ID)/2 *H* density =
= 19.8 cm * 1.7 cm * 3 cm * 7.55 g/cm3 = 762 g =0.762 kg
So the iron losses are: PFe = 1,1 W/kg * 0.762 kg = 0.84 W

3)And no load current= Pfe/230= 3.67mA (without considering the voltage drop across primary impedence)
Is it correct?Jimmy

 

[jim hardy]

1) Is it possible to calculate the no load voltage mathematically? According to my observations it is always between 10-20% more than the rated voltage. However I am interested in calculating it mathematically.

Sure it's possible if you know the actual turns ratio. The voltage ratio they give you is not the turns ratio it's the voltage you should expect at normal load.

2)For core losses I applied the following method:
The length of the medium flux path is: Lav = Pi*(ID+OD)/2 = 0.198 m.

Geometrically that looks fine and is probably close enough for experimenting at home..
Some folks correct for uneven flux density inside the core, see worked example here
http://pe2bz.philpem.me.uk/Comm01/- - Parts-NonActive/- Inductor/MakeCoils/terms.html
you'll find this approximation is used as well

The mass of the core is: M = Lav *(OD-ID)/2 *H* density =
= 19.8 cm * 1.7 cm * 3 cm * 7.55 g/cm3 = 762 g =0.762 kg
So the iron losses are: PFe = 1,1 W/kg * 0.762 kg = 0.84 W

The result of that will be as good as the 1.1 w/kg number. From where did that come ?

3)And no load current= Pfe/230= 3.67mA (without considering the voltage drop across primary impedence)
Is it correct?

?? i thought you measured no load current as 5 milliamps. See your 'observation ' table.
Which instrument do you prefer to believe , the wattmeter or the ammeter ? If the ammeter is digital and reading only on its least significant digit it's not very accurate.

 

[Jimmy Lalani]

Hello Hardy Sir,
1) I do not know the actual turns ratio. Is it still possible to calculate the No load voltage? Does the calculation differ for a frequency of 400Hz?
2) W/kg is an assumed value. However, I found that it does not work for all VA rating. I think this W/kg depends on the VA rating and the frequency considered, because when I performed the test for 200VA 50Hz I got core loss of around 2 W and for 600VA, 400Hz it was around 9W.
So, I searched on internet and I found that either I should find the W/kg for the given rating of transformer or the Core loss density at that frequency and Bm(Maximum Induction). How can I move on from here?

 

[jim hardy]

1) I do not know the actual turns ratio.

That's why you measure open circuit voltage ratio.

Is it still possible to calculate the No load voltage?

Why ? You measured it.. Your questions are circular . What information do you have about the transformer that you would use to calculate its open circuit voltage?

Does the calculation differ for a frequency of 400Hz?

? Open circuit voltage? To what accuracy? For all practical purposes, no the calculation does not differ. Of course core losses and leakage reactance will have more effect at higher frequency.. . but at no load they should still be small.
Have you studied the wikipedia article on transformer model? It's at
https://en.wikipedia.org/wiki/Transformer .

2) W/kg is an assumed value. However, I found that it does not work for all VA rating. I think this W/kg depends on the VA rating and the frequency considered, because when I performed the test for 200VA 50Hz I got core loss of around 2 W and for 600VA, 400Hz it was around 9W.

Look up Steinmetz formula for iron loss.

So, I searched on internet and I found that either I should find the W/kg for the given rating of transformer or the Core loss density at that frequency and Bm(Maximum Induction). How can I move on from here?

Take a look at this old old thread. It addresses similar questions.
https://www.physicsforums.com/threa...ses-hysteresis-eddy-current-constants.850475/