Transformer

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A transformer is a device for stepping down, stepping up or isolating AC voltages or signals.

Contents

Transformer FAQ

  1. Can I operate a transformer at higher than rated current?
  2. Can I operate a transformer at higher than rated voltage?
  3. Can I operate a transformer backwards? (eg a 120v->12v step down with 120v input to the secondary to give 1200v out on the primary)

Answers

  1. Generally yes. More current -> faster heating -> a shorter run before being allowed to cool.
  2. Generally no. The maximum voltage is determined by the core cross section, core material saturation, number of turns, and operating frequency. Unless you change one of those, no.
  3. Re-read FAQ number 2.

Schematic Symbols

Transformer symbols
Isolation Transformer Step down transformer Step up transformer
Vpri=Vsec Vpri>Vsec Vpri<Vsec
Transformer.GIF Step-down-transformer.GIF Step-up-transformer.GIF
Transformer type symbols
European Transformer symbol Iron Core Transformer Ferrite Core Transformer Tuned Slug ferrite core Transformer
European-transformer-symbol.GIF Transformer.GIF Ferrite-core-transformer.GIF Tuned-slug-ferrite-core-transformer.GIF

Transformer Theory

A transformer usually consists of 1 primary coil and 1 or more secondary coils wound around a common metalic core (iron for low frequency, and ferrite for high frequency). Some very high frequency transformers are air cored only.

A transformer works by inducing a voltage from the primary coil to the secondary coil. (See Faraday's Law) When an alternating voltage is applied to the primary coil, an alternating magnetic field is created around this coil since it acts as an electromagnet. Since the secondary coil is in the alternating magnetic field, a voltage is induced into it in much the same way it does when moving a magnet (alternating magnetic field) in and out of a coil.

The amount of voltage on the secondary coil depends on several factors: the ratio of primary to secondary turns (often just called the turns ratio), the core material, the driving frequency and coupling.

The most important utility of transformers is to convert voltages. With AC (which is supplied by the electricity grid), the voltage is converted several times between the large electrical generators and your house. At this point is is usually around 120 V AC or 240 V AC depending on where you live. When you plug in an electrical appliance, it may require a different voltage or voltages to operate, these appliances will usually use a transformer to convert the voltage which can later be rectified to DC later if required. (note: transformers do not operate on DC, which is much more complex to convert)

To obtain several voltages, transformers can either have several secondaries with different winding ratios or a single tapped secondary (output wires are connected to several places along the secondary coil, allowing the number of turns to be selected)

Ideal vs. Non-ideal Transformers

It is convenient to split transformers into two distinct categories.

  1. “ideal” transformers
  2. everything else

Properties of an ideal transformer

  1. magnetising current is as low as economically possible through high inductance
  2. leakage inductance is as low as economically possible through near 100% coupling
  3. operation does not take the core into saturation

An ideal transformer application may be recognised if further improving magnetising current or leakage inductance results in better operation. A specification for these would be “0 to some maximum permitted”.

Examples of ideal transformers

Properties of non-ideal transformers

Non-ideal transformers make use of one or more of finite primary inductance, finite leakage inductance, and operation into saturation, and specify and control them. This often saves the use of external components, whose function is then built-in to the transformer for little or no extra charge. A specification for magnetising current or leakage inductance would be a target value, +/- a tolerance.

The permeability of the core may be reduced by an air-gap, or by using less magnetic powder and more filler in the core. In power circuits this allows for significant energy storage in the core, in signal circuits it gives more repeatable inductance and generally higher Q. The leakage inductance may be increased slightly by separating the primary and secondary, or much more by the use of a magnetic shunt. Core or shunt saturation may be used to further affect the operation. However, intentionally increasing the losses in either winding or core is almost unheard of.

Examples of non-ideal transformers

  • Flyback transformer (TV EHT or auto ignition coil) - low permeability or air-gap for energy storage
  • Microwave Oven Transformer - small magnetic shunts provide enough leakage inductance to resonate the doubler capacitor
  • Neon Sign Transformer - large magnetic shunts provide enough leakage inductance to limit the short-circuit current
  • Rogowski coil - air-cored, low coupling geometry for field measurement
  • Tesla coil - air cored, low coupling geometry for insulation
  • Ferro-Resonant transformers - include saturable shunts to control the output voltage
  • Magnetic amplifier - can be saturated from a control winding
  • SFCL - Superconducting Fault Current Limiter - saturates a superconducting shorted secondary turn at high current

Magnetic shunt

Shunts can be used to limit current or regulate voltage. This regulation is accomplished by inserting a ferrous magnetic shunt into the transformer core, such that the magnetic flux from the primary winding has an alternate (although high impedance) path around the secondary winding. As the current draw on the transformer secondary winding increases, more primary magnetic flux diverts through the magnetic shunt. Selecting different materials with suitable saturation characteristics it is possible to make transformers with a range of regulation functions.

Types of Transformer

Transformer types
multitap isolated secondaries
Tapped-transformer.GIF Isolated-secondaries-transformer.GIF


Autotransformer

Another type of transformer is known as the autotransformer.
Autotransformer.GIF
It consists of a single tapped primary where the center tap is common to both primary and secondary (not isolated). some of the turns on the coil are used for the primary and some are used for the secondary. an example of a variable autotransformer is known as a variac. A variac is a single coil with a sweeping arm for the center common, this allows the ratio of primary turns:secondary turns to be altered easily.


Current Transformer

Yet another type of transformer is the current transformer.
Central to all of the AC power transducers is the measurement of current. This is accomplished using a current transformer (CT), a "donut" (toroidal) shaped core through which is threaded the wire whose current is to be measured.

Current transformers are designed to produce either an alternating current or alternating voltage proportional to the current being measured.

Ferroresonant transformer

These transformers use a tank circuit composed of a high-voltage resonant winding and a capacitor to produce a nearly constant average output with a varying input. The ferroresonant approach is attractive due to its lack of active components, relying on the square loop saturation characteristics of the tank circuit to absorb variations in average input voltage. The ferroresonant action is a flux limiter rather than a voltage regulator, but with a fixed supply frequency it can maintain an almost constant average output voltage even as the input voltage varies widely.

Ferroresonant transformers output either non-sinusoidal (CVN type transformers) or sinusoidal waveshapes (CVS type transformers).


Potential Transformer

CPt.jpg

Potential transformers are used by the electrical industry for smaller power applications and metering. They are a step-down transformer and range in wattage. This is a common Westinghouse 1500VA 12,000V transformer.





Properties Of A Transformer

Transformers have many properties due to construction. The Ideal transformer is lossless, but the losses in a transformer will be elaborated later as well.

Here we have a simplistic model of our idealized transformer. N1 and N2 are the number of turns for each winding, e1 and e2 are the voltages in the windings, and Φ is the flux.

Transformerex.jpg

With a load attached to the secondary we now see the currents i1 and i2.

Transformerex2.jpg


Primary Induced E.M.F {e_1}={N_1}\frac{d\Phi}{dt}, And Secondary Induced E.M.F {e_2}={N_2}\frac{d\Phi}{dt}

Proportionality of the Transformer \frac{v_1}{v_2}=\frac{e_1}{e_2}=\frac{N_1}{N_2}=\frac{i_2}{i_1}=a, where \textbf{a} is the Transformation Ratio.


Voltage and current relationship {\textbf{v}_1}{i_1}={\textbf{v}_2}{i_2}


Changing Flux relationship \Phi = \Phi_m\sin\omega\textbf{t}, where \omega= \textbf{2}\pi{f} which is the angular frequency in radians/second, \textbf{f} is the source frequency, \textbf{t} is the time at wich the measurment takes place, and \Phi_{\textbf{m}} is the magnetizing Flux density.


Load Impedances {Z_2}=\frac{V_2}{I_2}=\frac{1}{a^2}\frac{V_1}{I_1}=\frac{1}{a^2}{Z_1}


Induced Primary E.M.F. {\textbf{e}_1}= {N_1}\omega{\Phi_m}\cos{\omega}t,


And the R.M.S. value {E_1}= \frac{1}{\sqrt{2}}{N_1}\omega{\Phi_m}\angle{0}^\circ= 4.44f{N_1}{\Phi_m}\angle{0}^\circ


EXAMPLE

A 5kVA Stepdown transformer has a nominal primary voltage V1 = 480, nominal secondary voltage V2 = 120, and a flux Φ = 0.15tWb. What are the nominal currents I1 and I2, as well as the induced emf in the 200 turn primary winding at 1 second?


{I_1}=\frac{5000 VA}{480 V}=10.4A


{I_2}=\frac{5000 VA}{120 V}=41.7A


a=\frac{480}{120}=4


\Phi=0.15\textbf{t}Wb, so taking the derivative w.r.t. time, \frac{d\Phi}{dt}=0.15 Wb


The induced E.M.F. between terminals 1 and 2 of our transformer model above is:

{e_{12}}={N_1}\frac{d\Phi}{dt}=200*0.15=30 V


The secondary turns {N_2}=\frac{N_1}{a}=\frac{200}{4}=50


And from the Right Hand rule we see that the secondary polarity is in opposition to the flux so its E.M.F. is:

{e_{34}}=-{N_2}\frac{d\Phi}{dt}=-50*0.15=-7.5V

We can also find the maximum flux in the core. (assuming f=60Hz)

{\Phi_m}=\frac{V_1}{4.44f{N_1}}=\frac{480}{4.44*60*200}=9.0 mWb

Realistic Models

So now that the basics are taken care of, we must consider the losses in the transformer as well as its reactive components.

Transformer3.jpg

We will consider the Source-side of the circuit the Low-Side, and the Load-side as the High-Side. Primary and Secondary could also be used as references, but there is some confusion when dealing with these designations because each is not very specific.

The losses in the system will vary greatly as you will see in the worked example. This is why a Wattmeter must be used to determine the actual working parameters of the transformer.

The Low side current {I_L}=\frac{P}{V_L}

The High side current {I_H}=\frac{P}{V_H}


Excitation current {\tilde{{I}_\Phi}}=\tilde{I_c}+\tilde{I_m}, which is the current in the transformer under no-load condition, and is the sum of the core loss current \tilde{I_c} and the magnetizing current \tilde{I_m}.

Open Circuit Parameters

The additional subscripts o.c. designate the 'open circuit' parameters of the transformer under test. This will become more clear in the example later.

Core loss resistance {R_{c1}}=\frac{\tilde{E_1}}{\tilde{I_c}}=\frac{\tilde{V_{1o.c.}}^2}{P_{o.c.}}=\frac{{V_{Lo.c.}}^2}{P_{o.c.}}

Power Factor P.F.= \textbf{cos}\angle\theta, And \theta=\textbf{cos}^{-1}\frac{P}{{\tilde{V_1}}{I_L}}=\textbf{cos}^{-1}\frac{P_{o.c.}}{{V_{Lo.c.}}{I_{Lo.c.}}}

The core current loss {I_\Phi}={I_{Lo.c.}}\textbf{sin}\angle\theta

Magnetizing reactance {X_{m1}}=\frac{\tilde{E_1}}{j\tilde{I_m}}=\frac{V_{Lo.c.}}{I_{\Phi}}

Short Circuit Parameters

The additional subscripts s.c. designate the 'short circuit' parameters of the transformer under test. This will become more clear in the solved example.

Equavlient winding resistance {R_1}=\frac{{P_{s.c.}}}{\tilde{I_{1s.c.}}^2}=\frac{{P_{s.c.}}}{{I_{Ls.c.}}^2}

Leakage Reactance j{X_1}=\sqrt{ \left(\frac{{P_{s.c.}}}{{I_{s.c.}}}\right)^2-{R_1}}


Equavilent secondary winding resistance {R_2}=\textbf{a}^2{R_1}

Equavilent secondary Leakage Reactance \textbf{j}{X_2}=a^2j{X_1}

Referring Loads and Sources

To fully understand the source, load, and loss, we must introduce a reference to which our values are 'referred' to. This enables the engineer to analyze the load as one equavilent transformed circuit.

Treferred.jpg

Referred To Primary Parameters

{{\hat{Z}}_1}'=\frac{N_1}{N_2}{{\hat{Z}}_1}, {R_2}'=\left(\frac{N_1}{N_2}\right)^2{R_2}, j{X_2}'=\left(\frac{N_1}{N_2}\right)^2j{X_2}

{{\tilde{V}}_1}={{\tilde{V}}_2}'=\frac{N_1}{N_2}{{\tilde{V}}_2}, {{\tilde{E}}_2}'=\frac{N_1}{N_2}{{\tilde{E}}_2}, {I_1}={{\tilde{I}}_2}'=\frac{N_1}{N_2}{{\tilde{I}}_2}

{{\textbf{R}}_{equivalent}}={R_1}+{R_2}', {{\textbf{X}}_{equivalent}}={X_1}+{X_2}'

Referred To Secondary Parameters

{{\hat{Z}}_1}''=\frac{N_2}{N_1}{{\hat{Z}}_1}, {R_1}''=\left(\frac{N_2}{N_1}\right)^2{R_1}, j{X_1}''=\left(\frac{N_2}{N_1}\right)^2j{X_1}

{{\tilde{V}}_2}={{\tilde{V}}_1}''=\frac{N_2}{N_1}{{\tilde{V}}_1}, {{\tilde{E}}_1}''=\frac{N_2}{N_1}{{\tilde{E}}_1}, {{\tilde{I}}_2}={{\tilde{I}}_1}''=\frac{N_1}{N_2}{{\tilde{I}}_1}

{{\textbf{R}}_{equivalent}}={R_2}+{R_1}'', {{\textbf{X}}_{equivalent}}={X_2}+{X_1}''


EXAMPLE

An Engineer needs to know the parameters of a 46KVA Transformer which has a 2300V/230V winding. His results are:

Open Circuit Test: 230V 11.2A 1150W

Short Circuit Test: 160V 20.0A 1150W

We must first determine wich side Low or High the test was performed on.


For the Open Circuit test we compare the tested voltage to the rated voltage of the transformer. In this example we see that the Open Circuit Test voltage is the same as the Low side rated operating voltage, thus we know the test was performed on the low side and the high side was left open.

Next we need to determine which side the Short Circuit Test was performed on, so we compare the Current this time.

{I_L}=\frac{S}{{V_L}}=\frac{46000W}{230V}=200A

{I_H}=\frac{S}{{V_H}}=\frac{46000W}{2300V}=20A


So now we know where the tests were performed.

Open Circuit Test: 230V 11.2A 1150W -H.V. Left open, and tested on Low side

Short Circuit Test: 160V 20.0A 1150W -L.V. Shorted, and tested on High side


So we know that the Open Circuit parameters are Referred to Primary. So we will use the Referred to Primary parameters for the Open Circuit test.

What we know:

\textbf{V}=230,
{I_o}=11.2A,
S=1150W

{R_c}=\frac{V^2}{S}=\frac{230^2}{1150}=46\Omega

\theta=cos^{-1}\left[\frac{S}{VI}\right]=cos^{-1}\left[\frac{1150}{230*{11.2}}\right]=63.5^\circ


{I_{\Phi}}={I_o}sin\theta=11.6sin63.5^\circ=10A

{X_{\Phi}}=\frac{V}{{I_{\Phi}}}=\frac{230}{10}=23\Omega


Now remember how the Short Circuit Test proved out that the L.V. side was shorted? This means our equivalent calculations are Referred to the H.V. side. So in effect what we are calculating are our X2 and R2 values. If we want our X1 and R1 values, we must divide by our transformation ratio.

{R_2}=\frac{S}{{I_{s.c.}}}=\frac{1150}{20}=2.875\Omega

{X_2}=\sqrt{ \left({\frac{{V_{s.c.}}}{{I_{s.c.}}}}\right)^2 -{R_2}}= \sqrt{ \left({\frac{160}{20}}\right)^2 -2.875\Omega}=7.818j\Omega


To complete our Primary Referred Circuit we must find our R1 and X1 values. And given the fact that the transformer is stepping up voltage from 230 to 2300, we can see that it is a 1:10 ratio, or N1 = 1,N2 = 10.

{R_1}=\left(\frac{{N_1}}{{N_2}}\right)^2{R_2}=\left(\frac{{1}}{{10}}\right)^2{2.875}\Omega=0.02875\Omega

{X_1}=\left(\frac{{N_1}}{{N_2}}\right)^2{j{X_2}}=\left(\frac{{1}}{{10}}\right)^2{7.818j}\Omega=0.07818j\Omega

Now of course one might think this is the end, but that is far from the truth. At this time we know enough about the circuit to go further if we desire. From here we can obtain the terminal voltages for simulation, a very important factor in designing a system!

\tilde{{E_1}}={I_1}\left({R_1}+{X_1}\right)={I_2}\frac{{N_2}}{{N_1}}\left({R_1}+{X_1}\right)=200A(0.02875 + j0.07818)=200A\angle{0}^\circ{R_{eq}}\angle\gamma^\circ

Now here we must convert our resistance into a singular vector for multiplication.

We will momentarily ignore our j operator to get the magnitude of the vector.

{R_{eq}}=\sqrt{{0.02875}^2 + {0.07818}^2}=0.0833\Omega

\angle\gamma^\circ=tan^{-1}\frac{{X_1}}{{R_1}}=tan^{-1}\frac{0.07818}{0.02875}=\angle{69.8}^\circ


Now we can find \tilde{E_1}.

\tilde{E_1}=200\angle{0}^\circ{0.0833}\angle{69.8}^\circ=200*.0833\angle{(0+69.8)}^\circ=16.66\angle{69.8}^\circ V

\tilde{E_2}=\frac{N_2}{N_1}{E_1}= 166.6\angle{69.8}^\circ

So you can see from the determination that 6.6 volts are due to internal losses in the transformer itself. Knowing this magnitude, an Engineer can design a simulation with a source to the exact specifications and losses.

Links

http://sound.westhost.com/xfmr2.htm#s1122

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