I Don't Understand Transformers/How to Apply Them

[rtareen]

TL;DR Summary

I'm looking to understand transformers as they apply to circuits only. I would prefer minimum background theory, but if necessary please include.

1. We know that a transformer splits a circuit into a primary and secondary circuit. So then, does the secondary coil act as a voltage source for the secondary circuit, or a current source for the secondary circuit? Does it depend on whether the primary circuit uses a voltage/current source as?

2. Following the equation $V_2 = (N_2/N_1)V_1$ , I'll assume the secondary coil acts as voltage source $V_2$? Then hat is $V_1$? Is that the voltage source of the primary circuit? There is no potential difference across the the primary coil (negligible resistance), as it is at a constant potential. Or does it have a potential difference due to the secondary coil? If it does, is this potential difference such that $V_{1} = -V_{battery}$? Or else how does it work?

3. I don't really understand impedance in terms of transformers. The equation is $Z_1 = (N_1/N_2)^2Z_2$. Does $Z_1$ actually exist, or is it just what it would be if we placed the component on the other side? How does that make sense? We learned that impedances depend on frequency only. On the other hand, if $Z_1$ actually exists, what if it is is a capacitor and $Z_2$ is an inductor?

I'm very confused.

 

[Baluncore]

Start here;
https://en.wikipedia.org/wiki/Transformer#Ideal_transformer

 

[rtareen]

Start here;
https://en.wikipedia.org/wiki/Transformer#Ideal_transformer

This doesn't really help. Let me ask one thing at a time. If the amplitude of the voltage source is 1V. Is the amplitude of the voltage drop across the primary coil equal to -1V? In other words would this work?

 

[artis]

Why -1V? Since the only load across the voltage source is the primary winding, you can say that the voltage across the winding is 1V.
Note that it would probably drop due to the load at the secondary side, so it would be somewhat less than 1V

 

[Baluncore]

I'm very confused.

This doesn't really help. Let me ask one thing at a time.

Asking the wrong questions will just confuse you more. All you need to ask the right question is in the wikipedia page on ideal transformers. Read that until you understand the concepts and terminology. Then you can ask the right questions, and the answers will be obvious.
Do you really expect me to cut and paste the relevant text from the wikipedia page?

 

[Borek]

Aren't you trying to treat the winding as a resistor, ignoring the fact is is actually an inductor in an AC circuit?

 

[rtareen]

Why -1V? Since the only load across the voltage source is the primary winding, you can say that the voltage across the winding is 1V.
Note that it would probably drop due to the load at the secondary side, so it would be somewhat less than 1V

So that its 1V + V_{winding} = 0. So that V_{winding} = -1V? I guess what I'm asking is if KVL applies to the primary loop.

 

[rtareen]

Asking the wrong questions will just confuse you more. All you need to ask the right question is in the wikipedia page on ideal transformers. Read that until you understand the concepts and terminology. Then you can ask the right questions, and the answers will be obvious.
Do you really expect me to cut and paste the relevant text from the wikipedia page?

I'm already aware of the ideal transformer equations. I even have them derived. My questions are about the behavior of the transformer in a circuit. I'm aware almost everything is available on Wikipedia. It doesn't help. Thats why I chose to come here.

 

[rtareen]

Aren't you trying to treat the winding as a resistor, ignoring the fact is is actually an inductor in an AC circuit?

For the primary winding I'm trying to see if it can act as something that offsets the voltage source in the primary loop. I'm trying to see if KVL would apply. For the secondary loop I'm trying to have it act like a voltage source that provides the correct current so that the voltage drop across the resistor is equal to it.

 

[artis]

Well @rtareen it's good that you can derive those equations but you also need other knowledge as well.

Think about it this way. You are asking whether when one applies a 1v sine across a primary there is an opposite -1v sine across it. Why would there be ? A transformer is not a generator it doesn't produce power it just transforms it. If you apply a voltage of 1v across the primary then that's it there is just 1V across the primary and that's it. Now sure because a transformer is an inductive load the current will lag the voltage as within all inductors.
If you apply the same 1v across a resistor, or a diode or a capacitor you don't get an equal but opposite voltage across the element , then why would that happen in a transformer, try to explain it in your own thinking and words and if you can't then there is a reason for that...because that's not what happens.

You can ofcourse always perform a practical experiment and measure the voltage across a transformer primary that is connected to an AC source with a multimeter/voltmeter and then compare the measurement with the AC source, but I already told you what you will see.As for the secondary , please go and read the definitions of what is a voltage and what is a current source and then think about your question.

 

[anorlunda]

KVL, and KCL do apply separately for the primary circuit and the secondary circuit. But there is a mutual dependency in the two loops.

Rather than worrying about whether a source is voltage or current, study the math of the ideal transformer as others already recommended.

 

[Charles Link]

See https://www.physicsforums.com/threa...er-core-when-coupling-high-low-power.1001193/
Posts 22 and 43 by @Dale are very good ones to study if you can follow the mathematics. One important feature about the transformer when the power transfer is significant is that $|N_p I_p| \approx |N_s I_s |$ to a very good approximation. That is also discussed in a couple of places in this thread.

 

[DaveE]

Also this post may help:
https://www.physicsforums.com/threa...-coupling-high-low-power.1001193/post-6478419

 

[rtareen]

If you apply the same 1v across a resistor, or a diode or a capacitor you don't get an equal but opposite voltage across the element , then why would that happen in a transformer, try to explain it in your own thinking and words and if you can't then there is a reason for that...because that's not what happens.

Yes you do. That's the point of KVL, that the sum of voltages around a closed loop is zero. If you have a 1V voltage source in a closed loop with 1 ohm resistor, the voltage source will provide the current necessary so that there is a -1V drop across the resistor in the direction of the current.

 

[rtareen]

I'm just going to skip this section. Thanks.

 

[Charles Link]

In understanding the transformer and power transfer, one way of explaining it is by a magnetomotive force (mmf) $N_p I_p$ that gets balanced by the secondary's magnetomotive force $N_s I_s$ in the opposite direction. See https://en.wikipedia.org/wiki/Magnetomotive_force . This IMO is better explained by a modified form of ampere's law for magnetic materials $\oint H \cdot dl=\sum N_i I_i$, but in any case, this power balance is important in learning about the transformer.
(In more detail in the above, when the secondary circuit has a current running through it because of the load, it will create an mmf of $N_s I_s$ that would cause a changing magnetic field that would change the voltages in the primary and secondary considerably, but it doesn't, because the primary responds with an mmf of its own, with $|N_p I_p|=|N_s I_s |$).

Faraday's law comes in handy for computing voltages in the primary and secondary, but if you want to learn how the transformer works with significant current in the secondary, the mmf concepts and/or the modified ampere's law are needed. It is a somewhat complex subject, and the mathematics does get a little complicated.

additional note: Since Faraday's law with the same flux and flux changes in the primary and secondary gives $\frac{V_s}{V_p}=\frac{N_s}{N_p}$, we have with $N_p I_p=N_s I_s$ that $I_p V_p=I_s V_s=P$.

This ignores the zero load current in the primary $I_{po}$ which ideally is rather small. The zero load current (its derivative) along with Faraday's law is what determines the voltages in the primary and secondary, even when $I_p$ and $I_s$ are large, because the system is self-balancing with $N_p I_p=N_s I_s$.

 

[artis]

You are correct @rtareen if viewed from KVL then you can say so, I thought you were speaking about a real life measurement.

 

[Charles Link]

It is customary in going clockwise around the circuit, that the voltage drop is said to be positive if the component has a voltage that drops as you proceed in the clockwise direction, as it does in your example above. The voltage drop is + 1V.

Meanwhile EMF's are positive if they push the current in the clockwise direction.

 

[Charles Link]

Just an additional comment or two about the transformer=see also post 16 above: If all that is wanted is a couple of basic formulas, the two formulas $\frac{N_s}{N_p} =\frac{V_s}{V_p}$, and $P=I_p V_p=I_s V_s$, with $N_s I_s \approx N_p I_p$ will furnish what is needed.

What I find fascinating though is the way the transformer will balance itself, so that if the load draws more current by having a smaller impedance, the magnetic field that would result from this extra current in the secondary never materializes, but instead the primary current increases precisely in such a manner that the increased magnetic field that would result from the extra current in the secondary is offset by increased current in the primary, so that the magnetic field in the transformer core maintains itself at the zero load level, regardless of how much current flows in the secondary. The late Jim Hardy mentions this at the bottom of post 17 of this "link": https://www.physicsforums.com/threads/magnetic-flux-is-the-same-if-we-apply-the-biot-savart.927681/
The modified ampere's law is useful in analyzing the magnetic fields in the transformer core, and in seeing how this balance is achieved.

 

[anorlunda]

We all loved Jim Hardy's posts. But isn't that magical balance simply what is needed to conserve energy on an instantaneous basis?

I try to remind myself is that the ideal transformer could be implemented some way other than a real transformer and without magnetic fields. I do the same with other ideal components.

 

[Charles Link]

We all loved Jim Hardy's posts. But isn't that magical balance simply what is needed to conserve energy on an instantaneous basis?

I try to remind myself is that the ideal transformer could be implemented some way other than a real transformer and without magnetic fields. I do the same with other ideal components.

From the best I can tell, that balance, which is an energy balance, didn't need to necessarily happen. The fact that it did happen though made for ac power systems everywhere. Other possibilities that could have occurred were that the increased load would result in a much lower voltage in the secondary. There could also have been much core heating and much reduced magnetic field in the core from eddy currents (from reverse currents in the core created from the Faraday EMF), but laminations of the core solved that problem. IMO the transformer is a rather remarkable device, where we are very fortunate that its operation is very close to the ideal case.

 

[alan123hk]

In understanding the transformer and power transfer, one way of explaining it is by a magnetomotive force (mmf) NpIp that gets balanced by the secondary's magnetomotive force NsIs in the opposite direction. See https://en.wikipedia.org/wiki/Magnetomotive_force . This IMO is better explained by a modified form of ampere's law for magnetic materials ∮H⋅dl=∑NiIi, but in any case, this power balance is important in learning about the transformer.

In my opinion, the transformer is indeed a great device. Through the use of Ampere's law and Faraday's law, its basic characteristics can be revealed.

Let
I₁ , I₂ = primary side current and secondary side current
N₁ , N₂ = Number of turns on the primary side and the secondary side
V₁ , V₂ = primary side voltage and the secondary side voltage
R_L = load resistance on the secondary side
θ = magnetic flux of the magnetic core
H_c = Magnetic field strength of the magnetic core
l_c = effective length of the magnetic core
A_c = effective area of the magnetic core
u_c = permeability of the magnetic core

$$ H_c ~l_c = N_1I_1 - N_2I_2 ~~~~~~~,~~~~~~~ I_1 = \frac { H_c ~l_c} {N_1} + \frac {N_2} {N_1} I_2 $$
$$ I_1 = \frac { l_c} {u_c A_c N_1^2} (N_1 \theta) + \frac {N_2} {N_1} I_2 ~~~~~~~,~~~~~~~
\frac {dI_1} {dt}= \frac { l_c} {u_c A_c N_1^2} (N_1 \frac {d\theta} {dt}) + \frac {N_2} {N_1} \frac {dI_2} {dt} $$
$$ \frac {dI_1} {dt}= \frac { V_1} { \left[ \frac {u_c A_c N_1^2} {l_c} \right] } + \frac {N_2} {N_1} \frac {dI_2} {dt} ~~~~~~~,~~~~~~~ I_1 = \frac { \int V_1 \, dt} { \left[ \frac {u_c A_c N_1^2} {l_c} \right] } + \frac {N_2} {N_1} I_2 $$
Please note that $~\left[ \frac {u_c A_c N_1^2} {l_c} \right] ~$ represents the magnetizing inductance (L_m) referred to the primary, and $~\frac { \int V_1 \, dt} { \left[ \frac {u_c A_c N_1^2} {l_c} \right] }~$ represents the magnetization current (I_m).

Rearranging the last equation,
$$ \frac {N_1~I_1 - N_2~I_2} { \left[ \frac {l_c} {u_c A_c}\right] } = \frac {\int V_1 \, dt} {N_1 } = \theta $$
where $~N_1I_1 - N_2I_2~$ is the magnetomotive force, and $~ \frac {l_c} {u_c A_c} ~$ is the magnetic reluctance (R_m) of the magnetic core.

This means that the magnetization current provide magnetic flux to generate primary and secondary voltages.

$$ \frac {N_1~I_1 - N_2~I_2} {R_m} = \frac {N_1 I_m} { R_m} = \theta $$

In addition, since $~I_2 = \frac {N_2} {N_1} \frac {V_1} {R_L} ~$, the input current on the primary side can be expressed as follows.

$$ I_1 = \frac { \int V_1 \, dt} { \left[ \frac {u_c A_c N_1^2} {l_c} \right] } + \frac {V_1} {\left[{\frac {N_1} {N_2} }\right]^2 R_L } $$

 

[Charles Link]

Very good @alan123hk :). One comment in the above, is that for the primary current $I_ 1$, its two components are 90 degrees out of phase with each other. Please correct me if that is not the case. I think I have that correct.

 

[alan123hk]

Very good @alan123hk :). One comment in the above, is that for the primary current I1, its two components are 90 degrees out of phase with each other. Please correct me if that is not the case. I think I have that correct.

Yes, there is a 90 degree difference between the magnetizing current and the load current.

Let $V_1 = V_{10}~ sin (\omega t)$, then the equation becomes as follows :
$$ I_1 = \frac { V_{10}~ sin(\omega t-\frac {\pi} {2})} {\omega~ L_m } + \frac { V_{10}~ sin (\omega t) } {\left[{\frac {N_1} {N_2} }\right]^2 R_L }$$
That is, the magnetization current lags the load current (reflected to the primary side) by 90 degrees, and the load current is in phase with the power source $V_1$.

Another case is that the load is not a resistor but a capacitor (C).
$$ I_1 = \frac { \int V_1 \, dt} { L_m} + \frac {N_2} {N_1} C \frac {dV_2} {dt} $$
$$I_1 = \frac { V_{10}~ sin(\omega t-\frac {\pi} {2})} {\omega~ L_m } + \left( \frac { N_2 } {N_1 } \right)^2 \omega C~ V_{10}~ sin \left(\omega t+\frac {\pi} {2} \right) $$
The phase of magnetization current is opposite to the phase of load current, so in the resonance state, $~ \frac { 1 } { \omega~ L_m } = \left( \frac { N_2 } {N_1 } \right)^2 \omega C ~$ , $~I_1~$ will become zero.

 

[Baluncore]

There could also have been much core heating and much reduced magnetic field in the core from eddy currents (from reverse currents in the core created from the Faraday EMF), but laminations of the core solved that problem.

Yes. The transformer is a remarkably elegant and efficient compromise.

Correctly oriented metal laminations do reduce eddy currents, but that is not actually why they are there, nor what determines the optimum lamination thickness. Frequency of operation and skin effect are much more important than eddy currents in selecting the lamination thickness.

The magnetic field gains access to the magnetic core material by passing rapidly through the surface oxide insulation, between the metal laminations, then the field diffuses very slowly into the magnetic material. That is why laminations that are thinner than twice the skin depth should be used. Computation of the eddy currents does not come into the selection of lamination thickness.

Without sufficiently thin laminations, much of the magnetic core material would be inaccessible, the girth of the core would be greater, the windings would be longer, so would have higher resistance, and more power would be wasted. Both the weight and cost of the transformer would increase. Eddy currents can be ignored, if not forgotten.

 

[Charles Link]

In response to post 25, perhaps it is a matter of definitions that is the stumbling block. The conductive core material along with the Faraday EMF and the currents that are generated in the absence of laminations, and the magnetic fields from these currents which would serve to reduce the original magnetic field, is precisely why the magnetic field wouldn't reach the inner core without laminations to limit these currents. In the literature, they are typically called "eddy" currents, although the name sounds like something that would refer to a local swirling. (I didn't originally like the word "eddy" to describe the currents that are created in the absence of laminations, but that's what they call them.) It is these currents that, if I'm not mistaken, limit the propagation of the magnetic field into the material. The effect is similar to the skin effect, but I don't think I have a complete misunderstanding of the principles involved.

 

[Baluncore]

It is these currents that, if I'm not mistaken, limit the propagation of the magnetic field into the material. The effect is similar to the skin effect, but I don't think I have a complete misunderstanding of the principles involved.

I do not criticize your understanding, just the emphasis.

The eddy current is a closed loop current that flows only around the surface of the core, or around the surface of every lamination or insulated magnetic particle. That is why it is called an “eddy” current.

That eddy current is an essential part and must be present in order for the magnetic field to penetrate the surface of the iron. More lamination surface area implies more available magnetic material, so more eddy current.

The argument that laminations prevent eddy currents is somewhat hollow.

I like to consider the core as a magnetic mirror, that reflects and cancels the difference current in the windings.

 

[Charles Link]

Without any currents generated in the core, the Faraday ($\mathcal{E}=-\frac{d \Phi}{dt}$) induced electric field lines are concentric with the circular cross section of the toroidal-like core. The "eddy" currents will travel along these circles. It is apparent that significant currents could still be present with thick laminations, but the laminations will be effective in blocking them if they are sufficiently thin.

These eddy currents are not to be confused with the magnetic surface current, (magnetic surface current density per unit length $\vec{K}_m=\vec{M} \times \hat{n}/ \mu_o$), that is mathematically responsible for the magnetic field in the core. These magnetic surface currents are unaffected (not blocked) by laminations because there is no actual charge transport from these magnetic surface currents. Meanwhile the addition of more surface area from the laminations in principle does not mean more magnetic surface current to consider, (using Biot-Savart) because any contributions will cancel from opposite faces of the laminations. For the purposes of computation of the magnetic field using a magnetic surface current approach, the laminations have insignificant effect on the magnetic surface currents, and the laminations can be ignored. The neat feature is that the "eddy" currents are blocked by the laminations, and thereby the magnetic field that occurs is simply from the magnetic surface currents=the ideal scenario.

For computational purposes, the magnetic pole method with $B=\mu H$ is simpler, (and it does get the exact same answer as magnetic surface current calculations), and with eddy currents almost entirely blocked by laminations, the $B$ field is very nearly that of the ideal case of zero eddy currents to consider.

 

[Vitina]

I would like to state that a transformer is an AC device that basically operates as
1. An AC voltage converter
2. A phase inverter
3. An isolation device separating high and low voltage controls
4. A component in resonant systems. AC voltage regulators, radio frequency amps, Tesla coils ect.

Eddy currents in an solid iron core are not just a surface effect. Lamination thickness generally is a practical accommodation, higher frequencies require thinner laminations, or engineered materials like ferrite, powdered iron, or even rare Earth metals to limit eddy current and hysteresis losses, air cores at high frequencies.

The size of the core generally determines power capacity at 60hz. Higher frequencies, smaller cores.

There also is capacitance both between windings and core, and can be substantial and dangerous in power systems.

 

[Charles Link]

I would like to add that the power transfer capabilities that the transformer has are really a very remarkable property. If all it did was to supply a voltage (by Faraday's law) at the secondary in an open circuit condition that gets transferred from the primary, it would probably see limited use, if this voltage came with a large output impedance. The current balancing properties, where $N_s I_s =N_p I_p$, (basically a power conservation formula because $\frac{V_s}{V_p}=\frac{N_s}{N_p}$), are essential and why it has seen such widespread use.

The current balance can be viewed as offsetting mmf's, (mmf=magnetomotive force=$NI$), so that the magnetic field in the transformer core is basically independent of how much extra current is present in the primary and secondary, when a load is present. Extra current in the secondary would serve to reduce the magnetic field in the core, but the primary current increases in a very precise manner to offset this effect.

 

[Charles Link]

I gave the electrical transformer some additional thought, and there are a couple of things, besides just the formulas, that make it so remarkable. The primary coil with electrical current running through it creates a magnetic field that is greatly enhanced by the iron core. With an alternating current, this creates an alternating magnetic field, and amazingly enough, the secondary coil is able to pick up on this alternating magnetic field, and have a voltage (by Faraday's law) and current generated in the secondary coil, with nearly one hundred per cent power transfer. Having a toroidal geometry with primary and secondary coils around the torus makes for almost complete magnetic flux coupling. The one problem that occurs, with eddy currents in the iron from the Faraday EMF in the iron, is readily solved by having laminations in the iron. The transformer is truly a remarkable invention. It is also truly amazing that nature allows it to be so simple in design.

 

[alan123hk]

The one problem that occurs, with eddy currents in the iron from the Faraday EMF in the iron, is readily solved by having laminations in the iron

This is another outstanding invention related to iron core transformers. Of course, the experiment can prove that the lamination in the iron core can reduce the eddy current loss, but I think the reasoning method described in the link below is also very convincing.

https://www.physicsforums.com/threa...-commercial-transformer.1002491/#post-6489247

 

[Baluncore]

Notice in that link, how between adjacent laminae, the eddy currents counter flow, so cancel internally. But locally they are needed to get the field into the conductive magnetic material. The result after cancellation of internal sheet currents is the peripheral current that flowed without laminations.

Skin effect dictates the thickness of laminations. If you did not have sufficient gaps in the core material the magnetic field could not get in and out again before the field reversed.

It is the orientation of the laminations that reduces eddy current losses.

 

[davenn]

looks like we long since lost the OP

 

[Charles Link]

I think I see the eddy current problem slightly different than that of the link of post 32. Iron is a conductor and the Faraday EMF in the iron will cause significant current flow of the same symmetry (concentric with the core) as the Faraday EMF's that are generated by the coil windings (compare to a solenoid with an iron core). Laminations block this current flow in a capacitive manner=it only takes a little bit of charge to accumulate on the lamination to create a voltage opposing the Faraday EMF. The diagram of the link seems to mix the eddy current concept with the bound surface currents that are not blocked by the laminations. Basically the magnetic surface current per unit length $K_m=M \times \hat{n}/\mu_o$, with or without laminations. Perhaps I am missing something here, but that is my interpretation. (Note: Unlike the eddy currents, the magnetic surface current does not involve any charge transport).

and note, the paths the eddy currents take will be circular, (concentric with the center of the core), and laminations in a horizontal direction will be effective in blocking these circular paths. The neat thing is that the laminations don't block the magnetic surface currents, and the magnetic material (iron) behaves like it needs to=with laminations it is simply minus the eddy currents.