# how transformer work

Discussion in 'Homework Help' started by TAKYMOUNIR, Aug 27, 2013.

1. ### TAKYMOUNIR Thread Starter Active Member

Jun 23, 2008
351
1
how a transformer works. Why does a power transformer have high in rush current? When are the losses in the core of a power transformer the highest, at no load or full load and why?

2. ### MaxHeadRoom Expert

Jul 18, 2013
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Induced Current principle.
At power on the transformer presents a resistance equivalent to the DC resistance of the winding, just a few Ohms.
Until the inductive reactance effect takes place it then resorts to whatever the Impedance of the primary presents.
When a load is placed on the induced current of the secondary, this load current is reflected back to the primary.
Eddy currents are a source of losses, hence the laminated core in order to suppress this effect.
Max.

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3. ### The Electrician AAC Fanatic!

Oct 9, 2007
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This is not correct.

Transformer cores are made of a material which doesn't retain much flux when the exciting current is absent. So, when voltage is first applied, the flux is nearly zero (that is, the core isn't magnetized), but this doesn't mean that the inductance is low. The inductance is proportional to the slope of the B-H curve, and when the core isn't saturated, that slope is not (approximately) horizontal, as it is when the core is saturated. This means that the inductance of the primary is high when the core isn't magnetized, not the other way around.

The "inductive reactance effect" takes place immediately.

Saturation is a result of the core becoming as magnetized as it can, and it occurs only after some time has passed when starting out with a core that isn't already magnetized. It's as a result of saturation that the inductance decreases precipitously; the inductance of the coil becomes nearly the same as if the core weren't there.

So we see that the reason transformers of all types exhibit an inrush current (separate from the inrush due to charging capacitors on the secondary) is not because the magnetic field is building up--the surge in current occurs when the magnetic field has stopped building up due to saturation of the core. This is also the reason that the surge that occurs is dependent on just where the grid sine wave is when the switch is closed. If the grid sine is just crossing zero, then the surge will be a maximum. If the switch is closed when the grid sine is at a peak, then there won't be any significant surge at all.

I have a linear power supply with a 400 VA toroidal transformer, and I put a current shunt in series with the primary winding (120 VAC here in the U.S.), and disconnected the secondaries from the rest of the circuitry. I set up an oscilloscope to capture the current in the primary and then applied line voltage. I turned the supply on and off a lot of times until I had captured the current surge when the primary voltage was applied just as the sine was crossing zero volts. The peak surge current was about 180 amps.

The grid voltage is applied to the primary of the transformer at about 2.2 cm (green). The current pulse (purple) doesn't even begin until about 5 mS later, at about 4.2 cm. This is because it takes that long for the flux to reach saturation (the flux is proportional to the integral of the applied voltage).

If your explanation were correct, the current surge would have occurred immediately, at 2.2 cm.

Notice that the current surge is so large that it pulls the grid voltage down when it occurs, just after the peak of the applied grid sine wave.

I then reconnected the secondaries and captured the surge current when the line was connected just at the peak of the grid sine. This should give the maximum surge due to charging the capacitors in the secondary bridge rectifier circuit. The peak surge was about 10 amps:

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4. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The losses in the core are larger when the flux excursions are larger. The flux in the core is proportional to the integral of the applied voltage.

When there is a load on the secondary, the increased current in the primary due to that load causes a drop in the voltage applied to the primary (due to the IR losses in the copper winding of the primary). That reduced voltage applied to the primary results in somewhat less flux excursion in the core, which in turn results in somewhat less core loss.

5. ### MaxHeadRoom Expert

Jul 18, 2013
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The inrush is categorized during the first cycle where the DC resistance of the primary is the only limiting factor to current.
This will occur without a rectified or loaded secondary.
Max.

6. ### WBahn Moderator

Mar 31, 2012
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Uhm... you guys do realize that the OP hasn't shown a lick of effort and that all he is doing is sucking your descriptions onto his paper and turning them in, don't you?

That's this guy's mode of operation: bombard the forum with questions he isn't willing to put forth any effort into hoping that someone will come along and eventually give him answers that he can snag as his own.

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The DC resistance is not the limiting factor if the grid sine wave is applied at the peak (positive or negative) of the sine wave. See the second scope capture above; there is no 180 amp surge in that case. Something other than the DC resistance must have limited the current; what was it?

In the first capture above there was a 180 amp surge. Why didn't the same thing that limited the surge in the second capture also limit it in the first capture?

Well, yes, that's what is shown in the first scope capture above. In fact, I said "I put a current shunt in series with the primary winding (120 VAC here in the U.S.), and disconnected the secondaries from the rest of the circuitry."

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
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If you do a history search on his user name, it appears that he mostly posts these questions in the chat forum. I am used to him asking on the chat forum and I didn't even notice that this thread was in the homework forum.

His questions somehow don't have the flavor of homework questions, and I'm not sure he's actually in school. Maybe he doesn't pay close attention to what forum he's posting in.

9. ### WBahn Moderator

Mar 31, 2012
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He definitely posts similar style posts in both, but you are right that he tends to put them more in the General Chat than in Homework.

10. ### Austin Clark Active Member

Dec 28, 2011
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A few times when I first started hanging around this forum, I was accused of posting homework and just collecting the answers when, in fact, I was just really really curious and in need of a different explanation/perspective.

I usually give people the benefit of the doubt in such matters.

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The simple answer in that situation would be not to post in the homework forum. The assumption is that if it's posted in homework then the homework rules will apply.

If one posts homework in another forum under the guise of not being homework, it doesn't take too long before suspicions are aroused and the OP is occasionally challenged to make an effort.

12. ### JoeJester AAC Fanatic!

Apr 26, 2005
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I guess if someone was seeking clarification vice homework, I would direct them to the appropriate chapter in the eBook with an invitation to return to general chat after reading. I don't consider the questions and worksheets from the Ebook as "homework", but the responsibility of the poster to tell us the source of his information and the question.

13. ### Alan brad New Member

Aug 29, 2013
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0
Wow thanks for this valuable information.i would like to say basic principle of transformer actually two windings are present in the transformer primary and secondary winding..when current flows through primary winding that current is induced in the another winding that means mutual inductance bw two windings.

14. ### daviddeakin Active Member

Aug 6, 2009
207
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Your example shows the opposite of this behaviour. I am curious about this- can you explain why in more detail?

15. ### WBahn Moderator

Mar 31, 2012
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One can choose to give the benefit of the doubt or choose to go the other way. The first approach may be softer on the poster's ego, but it may not help them develop the proper attitude. The latter may be a bit harsher, but the barrier to entry is not that great. You were a case in point, IIRC. I believe I was one of those assuming you were just looking for a free handout, but all we ask that anyone do to destroy that assumption is to show some effort and work. In your case, again, IIRC, the assumption that you were looking for a free handout was driven by the fact that you seemed resistant to show that work - not because you were actually looking for a handout, but because you believed that just beeing shown the answer would be sufficient for you to learn what you needed to learn (and it frequently is). But as soon as you figured out (and it didn't take long) that you had to show some effort, you did and were immediately accepted into the fold, so to say.

So my approach is a bit different than yours -- I believe it's better (otherwise I wouldn't use it), but that doesn't mean I'm right. I will give a new poster the benefit of the doubt and assume that they are just ignorant of the learning process we espouse and so needs to be told -- some times explicitly, sometimes a bit tongue-in-cheek, and sometimes rather sarcastically depending on the tone of the post. It's when a repeated pattern of unwillingness to show effort and work develops that the harshness really increases.

16. ### The Electrician AAC Fanatic!

Oct 9, 2007
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When I just now had a look at post #3, the first image had stopped displaying in the body of the post, even though the thumbnail was still there; very strange. I was able to edit and fix the problem.

I don't see how my examples show the opposite of the part of my text you cited.

If you look at the first image in post #3, you will see that the applied grid voltage (green trace) is just passing through zero on the way to the positive peak of the (somewhat distorted) sine wave. About 5 mS later, there is a 180 peak amp surge (purple trace).

In the second image, the grid voltage (green trace) is applied at the peak of the sine wave and there is no 180 amp surge due to saturation of the core; there is a much smaller 10 amp surge (purple trace) due to charging of the electrolytic following the bridge rectifier.

These two examples are just as I described, are they not?

17. ### daviddeakin Active Member

Aug 6, 2009
207
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Ah, I think I get it. So you're saying that switching at the zero-crossing does indeed causes the largest inrush current pulse, but that it is delayed.
So what delays it?

18. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Faraday's law tells us that the flux is proportional to the integral of the applied voltage.

After having been un-energized for a while, the flux in the core is essentially zero (the iron used in transformers is chosen for low remanent flux).

When voltage is then applied to the primary winding starting at the zero crossing of the sine wave, the flux increases as the integral of the applied voltage. After sufficient volt-seconds (integral of the voltage with respect to time) are applied, the flux in the core reaches the saturation point (this happens at about the peak of the sine wave); further volt-seconds accruing as the voltage passes the peak of the sine push the core further into saturation, and the current is then limited only by the DC resistance of the winding.

For steady state operation, when the applied voltage crosses zero on the way up, the core has been magnetized to maximum flux in the negative direction (so to speak) by the previous negative half cycle (rather than starting at zero flux as when the voltage has been disconnected for a while), so application of the volt-seconds in the positive half cycle doesn't take the core beyond saturation in the positive direction.

But, when applying a full positive half cycle when the core has become unmagnetized by having been turned off for a while, those volt-seconds are sufficient to drive the core well past saturation, leading to a large current surge.

If the applied voltage begins at the positive peak, the integration of the voltage gives only half as many volt-seconds until the voltage reverses, and this many volt-seconds is not enough to push the core into saturation.

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Feb 17, 2009
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