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#1
03-03-2006, 07:31 PM
 achieveforce New Member Join Date: Mar 2006 Posts: 2

Hello,

I have a question regarding the description of how the transformer works in "Lessons In Electric Circuits -- Volume II Chapter 9 TRANSFORMERS".

The chapter explains how the current on the secondary side does not make any additional magnetic flux. Here is the paragraph that has the explanation:

"No current will exist in the secondary coil, since it is open-circuited. However, if we connect a load resistor to it, an alternating current will go through the coil, in phase with the induced voltage (because the voltage across a resistor and the current through it are always in phase with each other).

At first, one might expect this secondary coil current to cause additional magnetic flux in the core. In fact, it does not. If more flux were induced in the core, it would cause more voltage to be induced voltage in the primary coil (remember that e = dΦ/dt). This cannot happen, because the primary coil's induced voltage must remain at the same magnitude and phase in order to balance with the applied voltage, in accordance with Kirchhoff's voltage law. Consequently, the magnetic flux in the core cannot be affected by secondary coil current. However, what does change is the amount of mmf in the magnetic circuit."

I am thinking that if the voltage that applied to the primary is a square wave. The current appeared on the secondary side will be a square wave too (if the load on the secondary is a resistor).

Let's assume this current DOES introduce additional flux in the core. Since it is square wave (not sine shape), it will just OFFSET, NOT CHANGE the shape, or dΦ/dt, of the exciting flux. In other words, the additional flux caused by the secondary current should be allowed.

If that is true, then, how does the secondary load current force the primary to generate a current to ballance the secondary current as stated in the paragraph?

I have been having this question for quite a while. I really hope the author can explain it to me. I really appreciate it!

-achieveforce
#2
03-03-2006, 10:06 PM
 windoze killa Senior Member Join Date: Feb 2006 Location: Melbourne Australia Posts: 605

Quote:
 Originally posted by achieveforce@Mar 4 2006, 07:31 AM Hello, I have a question regarding the description of how the transformer works in "Lessons In Electric Circuits -- Volume II Chapter 9 TRANSFORMERS". The chapter explains how the current on the secondary side does not make any additional magnetic flux. Here is the paragraph that has the explanation: "No current will exist in the secondary coil, since it is open-circuited. However, if we connect a load resistor to it, an alternating current will go through the coil, in phase with the induced voltage (because the voltage across a resistor and the current through it are always in phase with each other). At first, one might expect this secondary coil current to cause additional magnetic flux in the core. In fact, it does not. If more flux were induced in the core, it would cause more voltage to be induced voltage in the primary coil (remember that e = dΦ/dt). This cannot happen, because the primary coil's induced voltage must remain at the same magnitude and phase in order to balance with the applied voltage, in accordance with Kirchhoff's voltage law. Consequently, the magnetic flux in the core cannot be affected by secondary coil current. However, what does change is the amount of mmf in the magnetic circuit." I am thinking that if the voltage that applied to the primary is a square wave. The current appeared on the secondary side will be a square wave too (if the load on the secondary is a resistor). Let's assume this current DOES introduce additional flux in the core. Since it is square wave (not sine shape), it will just OFFSET, NOT CHANGE the shape, or dΦ/dt, of the exciting flux. In other words, the additional flux caused by the secondary current should be allowed. If that is true, then, how does the secondary load current force the primary to generate a current to ballance the secondary current as stated in the paragraph? I have been having this question for quite a while. I really hope the author can explain it to me. I really appreciate it! -achieveforce Quoted post
I am not sure if I can answer this totally but I will give it a try. Lets start with the square wave signal. A transformer is an inductor and as such the basic understanding of an inductor should tell you that it will resistor a change in current flow. So although you apply a square wave signal to the transformer you will not get a square wave of current flowing. Depending on the inductance and the frequency the current flow in the secondary will start to resemmble a sine wave.

Next you say "Let's assume this current DOES introduce additional flux in the core" Well you can't assume this because it doesn't happen. What the text is trying to say is that the voltage in the primary CAN NOT be greater than the applied voltage. ie. If you apply 100VAC to the primary then that is all that can appear at the primary.

If you consider what you were trying to say then if the secondary current flow induced further votage into the primary then this increase in primary current flow would inturn increase the current flow in the secondary and again would increase the current in the primary. Can you see where this is going??? A very hot transformer exploding with over current. Fortunately this doesn't occur because it doesn't induce any current into the primary.

Hope that helps a little. And I hope I got that right. I am sure someone will point out if I didn't.
__________________
Eddy Current is a little dizzy over Milli Volt.
#3
03-05-2006, 05:08 AM
 thingmaker3 Super Moderator Join Date: May 2005 Location: Rural, Oregon GMT -8 Posts: 5,072 Blog Entries: 6

__________________
"I want to establish in your mind very clearly that you must not think I deny all that I do not admit. On the contrary, I think there are many things which may be true, and which I shall receive as such hereafter, though I do not as yet receive them; but that is not because there is any proof to the contrary, but that the proof in the affirmative is not yet sufficient for me"
#4
03-08-2006, 10:50 PM
 dougp01 Junior Member Join Date: Dec 2005 Location: Colorado Posts: 25

Actually, the suggestion that the secondary load would cause additional flux in the core is worth considering, if only for the educational purpose. In my response I will consider the transformer as an ideal and not take into consideration the inductive effects. Generally speaking you may leave out the inductive effects if you design within the core saturation limits and do not gap the core material.

Often folks think of the transformer as simply capable transforming the Voltage Current ratio into another Voltage Current ratio based on the turns on the primary and secondary. This is a correct assumption. That is, as 120V, 1A primary converted to a 12V, 10A secondary (neglecting losses).

When considering the flux in the core, we realize this is really a two-step process. The primary EMF is first converted to the core MMF and then the core MMF is converted to secondary EMF. The Magneto-motive force is often stated simply as flux. There is flux intensity and flux density which can be plotted. This is often referred to as the BH curve. This is probably deviating from the concern but suffice it to say that these things are available for learning.

If you change the load conditions, say from open to short or some point in between and assume a constant voltage on the primary, the current in the secondary and the primary will also change correspondingly.

If you place a square-wave on the primary, this is not different as long as the transformer is designed for higher frequencies. A square-wave is really comprised of a series of sine-waves, all odd harmonics of the fundamental. Correctly designed, a transformer will happily transform a square-wave to the secondary an all the flux can be accounted for.

Bmax= (E x 10E8)/4.44NAF (for sine waves)
Bmax= (E x 10E8)/4NAF (for square waves)

Where:

Bmax = Maximum saturation density of a core material in Gauss
E = Primary Voltage
N = Number of primary turns
A = cross sectional area of the core in CM^2
F = the highest frequency of interest in Hz

Using simple algebra, you can solve for frequency. As you can see a transformer designed for 50 or 60 Hz is not going to do well at higher frequencies. For example in audio applications.

I am generally old school so I am most comfortable using Gauss, most people now use Tesla's in their calculations and the equation is adjusted accordingly.

Lloyd Dixon of Texas Instruments has a number of online papers that can help.

Here is one: http://focus.ti.com/lit/ml/slup205/slup205.pdf

I'm afraid I may have deviated quite a distance from your initial question. Hope, this helps a little.

-doug
#5
03-09-2006, 11:54 AM
 aac Junior Member Join Date: Jun 2005 Posts: 35

Quote:
 Originally posted by achieveforce@Mar 3 2006, 03:31 PM Hello, I have a question regarding the description of how the transformer works in "Lessons In Electric Circuits -- Volume II Chapter 9 TRANSFORMERS". The chapter explains how the current on the secondary side does not make any additional magnetic flux. Here is the paragraph that has the explanation: "No current will exist in the secondary coil, since it is open-circuited. However, if we connect a load resistor to it, an alternating current will go through the coil, in phase with the induced voltage (because the voltage across a resistor and the current through it are always in phase with each other). At first, one might expect this secondary coil current to cause additional magnetic flux in the core. In fact, it does not. If more flux were induced in the core, it would cause more voltage to be induced voltage in the primary coil (remember that e = dΦ/dt). This cannot happen, because the primary coil's induced voltage must remain at the same magnitude and phase in order to balance with the applied voltage, in accordance with Kirchhoff's voltage law. Consequently, the magnetic flux in the core cannot be affected by secondary coil current. However, what does change is the amount of mmf in the magnetic circuit." I am thinking that if the voltage that applied to the primary is a square wave. The current appeared on the secondary side will be a square wave too (if the load on the secondary is a resistor). Let's assume this current DOES introduce additional flux in the core. Since it is square wave (not sine shape), it will just OFFSET, NOT CHANGE the shape, or dΦ/dt, of the exciting flux. In other words, the additional flux caused by the secondary current should be allowed. If that is true, then, how does the secondary load current force the primary to generate a current to ballance the secondary current as stated in the paragraph? I have been having this question for quite a while. I really hope the author can explain it to me. I really appreciate it! -achieveforce Quoted post

I understand your why you are uncomfortable with the no additional flux caused by the secondary and agree with you. Anytime there is a current in a wire there is a magnetic field and therefore flux. You should look up Lenz's law. What you'll find is that the additional flux is always in opposition to the change that caused it. So in the transformer, the flux caused by the secondary circuit is in the oppsite direction to the flux in the primary. No problem with creating more voltage in the primary since this would reduce the flux. What it does look like on the primary side, however, is that the inductance is smaller. Inductive reactance happens because the inductor is returning enercy to the circuit. If some of the energy is stolen by a secondary coil, then less energy is returned to the primary making it look like a smaller inductor. The means for the same voltage, there will be larger current because the reactance looks smaller. Square or sine doesn't really matter but I think you've got this pretty well figured out in your thought experiment.
#6
03-14-2006, 03:14 AM
 goutham249 New Member Join Date: Mar 2006 Posts: 2

Quote:
 Originally posted by achieveforce@Mar 4 2006, 02:01 AM Hello, I have a question regarding the description of how the transformer works in "Lessons In Electric Circuits -- Volume II Chapter 9 TRANSFORMERS". The chapter explains how the current on the secondary side does not make any additional magnetic flux. Here is the paragraph that has the explanation: "No current will exist in the secondary coil, since it is open-circuited. However, if we connect a load resistor to it, an alternating current will go through the coil, in phase with the induced voltage (because the voltage across a resistor and the current through it are always in phase with each other). At first, one might expect this secondary coil current to cause additional magnetic flux in the core. In fact, it does not. If more flux were induced in the core, it would cause more voltage to be induced voltage in the primary coil (remember that e = dΦ/dt). This cannot happen, because the primary coil's induced voltage must remain at the same magnitude and phase in order to balance with the applied voltage, in accordance with Kirchhoff's voltage law. Consequently, the magnetic flux in the core cannot be affected by secondary coil current. However, what does change is the amount of mmf in the magnetic circuit." I am thinking that if the voltage that applied to the primary is a square wave. The current appeared on the secondary side will be a square wave too (if the load on the secondary is a resistor). Let's assume this current DOES introduce additional flux in the core. Since it is square wave (not sine shape), it will just OFFSET, NOT CHANGE the shape, or dΦ/dt, of the exciting flux. In other words, the additional flux caused by the secondary current should be allowed. If that is true, then, how does the secondary load current force the primary to generate a current to ballance the secondary current as stated in the paragraph? I have been having this question for quite a while. I really hope the author can explain it to me. I really appreciate it! -achieveforce Quoted post

hello,
I really didn't get your doubt but it seems you made a small mistake by considering output for a square wave input to a transformer is same as the input
but it is not so . The output would be impulse voltage(on no load condition).For a
resistive load current would be same as the voltage i,e impulsive.One more thing
is secondary current will never introduce additional flux, according to lenz'law
however it always produces flux in opposite direction to that produced by exciting
current.If net flux decreases emf induced in primary coil decreases .Because of
difference in (d Φ /dt)and applied voltage V.more exciting current is drawn and previous flux is reestablished.Thus operating fulx in a transformer is always constant at least for ideal transformer.
#7
04-24-2007, 06:13 PM
 recca02 Senior Member Join Date: Apr 2007 Location: India-22.35° N 82.68° E. Posts: 1,211

well this post is really cold now,but since i chanced upon this one i am obliged
to post my views.
i do not know what wud happen with a square wave.probably an impulse voltage,
but i agree with aac the primary voltage can not increase instead it decreases back emf
due to lenz's law or an analysis of the phase of magnetic flux created by secondary reveals it is in opposition to flux due to primary. so a back emf
is induced in primary tends to reduce as now flux in the mag ckt has reduced.
thus back emf decreases with respect to back emf and an additional current
in primary is caused. this additional current again increase the flux to its normal value and thus flux in transformer remains constant no matter at what load it is working ,even at short ckt on secondary.
this is why transformer is referred to as constant flux machine.
the phase values of the flux change depending on whether load is inductive or capacitive.

Last edited by recca02; 05-08-2007 at 08:47 AM.

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