ECE 211 · Lecture 18: Ideal Transformers

§1 · The ideal transformer

Two windings, one flux

A transformer transfers energy between two circuits that share no conducting path, and the coupling is carried entirely by a magnetic flux. Two windings are placed on a common core; a changing current in one winding produces a changing flux, and the flux induces a voltage in the other. Under the assumption of ideal coupling, every line of flux produced by either winding passes through every turn of both windings, the windings have no resistance, and the core neither stores nor dissipates energy.

What Test 3 covers

The board notes open with the scope of the third examination: sinusoidal steady state (phasors), single-phase power, three-phase power, and mutual inductance. The ideal transformer developed here is the limiting case of mutual inductance in which the coupling is perfect.

The device

A primary winding of $N_1$ turns and a secondary winding of $N_2$ turns are wound on a common core. The reference directions are the ones used throughout this lecture: $i_1$ enters the primary at its positive mark, and $i_2$ leaves the secondary at its positive mark, so that the primary absorbs what the secondary delivers.

One flux links both windings

Ideal coupling means that a single flux $\phi$ threads every turn of both windings. Faraday's law then applies to each winding with the same $\phi$ :

v = N\,\frac{d\phi}{dt}

The number of turns is the only thing that distinguishes one winding from the other.

Faraday's law at each winding

\tm{tf.v1}{v_1} = \pm\, \tm{tf.l1}{N_1}\,\frac{d\phi}{dt} \qquad\qquad \tm{tf.v2}{v_2} = \pm\, \tm{tf.l2}{N_2}\,\frac{d\phi}{dt}

The sign of each expression is set by the direction in which the winding is wound relative to the assumed voltage polarity, which is information the schematic must supply separately. That information is supplied by the dot convention of Section 2.

The flux cancels: the turns ratio

Dividing the two expressions removes $d\phi/dt$ and leaves a pure ratio of turns:

\frac{\tm{tf.v2}{v_2}}{\tm{tf.v1}{v_1}} = \pm\,\frac{\tm{tf.l2}{N_2}}{\tm{tf.l1}{N_1}} = \pm\, \tm{tf.nn}{n} \;\Longrightarrow\; \boxed{\;v_2 = \pm\, n\, v_1\;}

The quantity $n = N_2 / N_1$ is called the turns ratio, and a transformer drawn as $1 : n$ carries $n$ on the secondary side. The relation holds instant by instant, so it holds equally for phasors.

No power is lost or gained

An ideal transformer stores no energy, so the power entering the primary appears at the secondary at every instant:

\tm{tf.v1 tf.i1}{p_1(t) = v_1 i_1} = \tm{tf.v2 tf.i2}{p_2(t) = v_2 i_2}

Substituting $v_2 = n v_1$ leaves $v_1 i_1 = n v_1 i_2$ , and the voltage divides out:

\boxed{\;\tm{tf.i1}{i_1} = \pm\, n\, \tm{tf.i2}{i_2}\;}

The same sign is used in both boxed relations. Voltage is multiplied by $n$ and current is divided by it, which is exactly what a lossless device must do.

The reference directions are part of the relations

Both boxed results assume the orientation drawn above: the positive marks of $v_1$ and $v_2$ at the top, $i_1$ entering the primary at its positive mark, and $i_2$ leaving the secondary at its positive mark. Reversing any one of those choices flips the sign of the corresponding relation, which is a bookkeeping error that is easy to make and hard to find.

§2 · The dot convention

Which way are the windings wound?

The turns ratio fixes the magnitudes, and the dot convention fixes the sign. Consider first the magnitudes alone. A transformer marked $4 : 3$ has $N_1 = 4$ and $N_2 = 3$ parts of turns, hence a turns ratio of $n = N_2 / N_1 = 3/4$ :

A transformer of turns ratio 4 : 3. The secondary voltage $v_2$ is three quarters of $v_1$ , while the secondary current $i_2$ is four thirds of $i_1$ .

n = N₂ / N₁ = 3/4 v₂ = ± (3/4) v₁ i₂ = ± (4/3) i₁ voltage goes with the ratio, current goes opposite

The remaining question is the sign. A schematic cannot show the direction of winding, so a dot is placed at one terminal of each winding: currents entering the dotted terminals produce fluxes that add. The rule that follows from that definition, stated in the form used on the board, reads:

Both [dots]{ds.d1 ds.d2} on top (or both on the bottom): use [+n]{ds.sign}.

[Dots]{dp.d1 dp.d2} on opposite sides: use [−n]{dp.sign}.

The rule in one line

If both dots sit on the same side of the core, then $v_2 = +n v_1$ and $i_1 = +n i_2$ . If the dots sit on opposite sides, then every $n$ becomes $-n$ . The sign is carried by $n$ itself, which is why the model of Section 3 needs no further sign bookkeeping.

The rule presumes the standard orientation

The dot rule as stated applies to the drawing convention of Section 1, with both positive marks at the top, $i_1$ entering the primary, and $i_2$ leaving the secondary. When a problem is drawn differently, the safe procedure is to redraw it in the standard orientation first, absorbing any reversed reference into a sign, and only then to read the dots.

§3 · The dependent-source model

Two dependent sources replace the coupling

Nodal and mesh analysis have no rule for a magnetic coupling, so the transformer is replaced by an equivalent that contains only familiar elements. The two defining relations are read directly as source definitions: the primary current is set by the secondary current, and the secondary voltage is set by the primary voltage.

The model of a 1 : n transformer. The primary becomes a current-controlled current source $n\,i_2$ , and the secondary becomes a voltage-controlled voltage source $n\,v_1$ . Once the two sources are in place, the circuit is an ordinary resistive (or phasor) network.

The two-step recipe from the board notes

Draw the model: replace the transformer by the dependent current source on the primary side and the dependent voltage source on the secondary side.
Fix the sign and the ratio: read $n = N_2 / N_1$ from the turns labels, and read the sign from the dots. Both sources then use the same signed $n$ .

Worked example: a 3 : 10 transformer

The transformer as given: the turns ratio is 3 : 10, and the dots sit on opposite sides of the core.

The magnitude of the turns ratio is $N_2 / N_1 = 10/3$ , and the opposed dots make it negative, so $n = -10/3$ . The model follows immediately, with the same signed $n$ in both sources:

The completed model. The primary carries $-\tfrac{10}{3} i_2$ and the secondary carries $-\tfrac{10}{3} v_1$ . A transformer stepping the voltage up by a factor of $10/3$ steps the current down by the same factor.

§4 · Worked example: solving for $V_x$

A transformer inside a phasor network

The circuit below is driven at a single frequency, so the whole problem is solved in the phasor domain. The transformer is replaced by its model, and the result is a network that nodal analysis handles without any new machinery.

Notation

The board notes mark phasors with a tilde, as in $\tilde{V}_1$ . On this page the tilde is dropped and the case of the symbol carries the meaning: lowercase $v_1$ and $i_1$ denote time functions, uppercase $V_1$ and $I_2$ denote the corresponding phasors.

Find $v_x$ , the voltage across the bridging 8 Ω resistor. Given: $v_s = 60\cos 100t$ V, a 4 Ω series resistor, a 1 : 2 transformer whose dots sit on opposite sides, a 2 Ω resistor in the secondary loop, and an 8 Ω load.

Two features of the drawing deserve notice before any equation is written. The bridging 8 Ω resistor ties the primary side to the secondary side, so the two loops are not independent, and the bottom rail is continuous, so a single reference node serves both sides.

Replace the transformer by its model

The turns ratio is $N_2 / N_1 = 2$ , and the opposed dots make it negative, so $n = -2$ . The primary becomes the dependent current source $-2 I_2$ directed downward, and the secondary becomes the dependent voltage source $-2 V_1$ with its positive mark at the top. The source is written as the phasor 60 V, since $60\cos 100t$ has the phasor $60\angle 0°$ .

Two unknowns cover the whole network

Both dependent sources are referenced to ground, so no supernode is needed. The primary node voltage is named $V_1$ , and the secondary current delivered through the 2 Ω resistor is named $I_2$ . Every remaining node then follows: the source node sits at $-2V_1$ , and the far node of the 2 Ω resistor sits at $-2V_1 - 2I_2$ .

KCL at the primary node

Currents are summed out of the $V_1$ node. The dependent source draws $-2I_2$ downward, and the bridging 8 Ω resistor leads to the node at $-2V_1 - 2I_2$ :

\tm{vx.r4 vx.vs}{\frac{V_1 - 60}{4}} - \tm{vx.cs}{2 I_2} + \tm{vx.r8x vx.nc}{\frac{V_1 - (-2V_1 - 2I_2)}{8}} = 0

Multiplying by 8 gives $2V_1 - 120 - 16 I_2 + 3V_1 + 2I_2 = 0$ , that is $5 V_1 - 14 I_2 = 120$ .

KCL at the far node, then solve

At the node held at $-2V_1 - 2I_2$ , the current $I_2$ arrives through the 2 Ω resistor while the bridge and the load drain:

-\tm{vx.r2 vx.i2}{I_2} + \tm{vx.r8x}{\frac{(-2V_1 - 2I_2) - V_1}{8}} + \tm{vx.r8l}{\frac{-2V_1 - 2I_2}{8}} = 0 \;\Longrightarrow\; 5V_1 + 12 I_2 = 0

The second equation gives $V_1 = -\tfrac{12}{5} I_2$ . Substituting into the first yields $-26 I_2 = 120$ , hence

\tm{vx.i2}{I_2 = -\tfrac{60}{13}\ \text{A}} \qquad \tm{vx.v1}{V_1 = \tfrac{144}{13}\ \text{V}}

Read the answer off the bridge

$V_x$ is the drop from the primary node to the far node:

V_x = V_1 - (-2V_1 - 2I_2) = 3V_1 + 2I_2 = \frac{432 - 120}{13} \;\Longrightarrow\; \boxed{V_x = 24\angle 0°\ \text{V}}

Returning to the time domain, $v_x(t) = 24\cos 100t$ V.

Check: both node equations balance

With $V_1 = 144/13$ V and $I_2 = -60/13$ A, the far node sits at $-2V_1 - 2I_2 = -168/13$ V. At the primary node: $(144/13 - 60)/4 + 120/13 + (144/13 + 168/13)/8 = -159/13 + 120/13 + 39/13 = 0$ ✓. At the far node: $60/13 + (-168/13 - 144/13)/8 + (-168/13)/8 = 60/13 - 39/13 - 21/13 = 0$ ✓. The bridge therefore carries $24/8 = 3$ A, which is the current the two equations were built to account for.

§5 · The reflected resistance

What the source actually sees

A transformer changes the resistance presented at its primary. This is the property that makes the device useful far beyond the distribution of power: a load of fixed value can be made to look like any other value, simply by choosing the turns ratio.

A Thevenin source drives a 1 : n transformer loaded by $R_L$ . Behind the dashed surface only one number is visible to the source, namely the equivalent resistance at the primary terminals.

The resistance is found exactly as in Lecture 5: the port is driven by a test source, and the ratio of port voltage to port current is formed. A 1 A source is convenient here because the primary relation then reads directly.

Drive the port with a test source

A 1 A test source is connected at the primary terminals, and the resulting port voltage $V_1$ is sought, because $R_{eq} = V_1 / 1\ \text{A}$ . The dependent sources of the model stay active, as dependent sources always do.

The primary relation gives the secondary current

The entire test current flows through the dependent source, so $\tm{im.i1}{I_1} = 1\ \text{A}$ and

\tm{im.i1}{1} = \tm{im.cs}{n\, I_2} \;\Longrightarrow\; \boxed{I_2 = \tfrac{1}{n}}

KVL around the secondary

The secondary source and the load form a single loop:

-\tm{im.ds}{n V_1} + \tm{im.rl}{R_L}\left[\tm{im.i2}{\tfrac{1}{n}}\right] = 0 \;\Longrightarrow\; V_1 = \frac{R_L}{n^2}

The load resistance appears at the primary divided by the square of the turns ratio, because the transformer scales the voltage by $n$ and the current by $1/n$ , and resistance is their quotient.

The reflected resistance

R_{eq} = \frac{V_1}{1\ \text{A}} \;\Longrightarrow\; \boxed{\;R_{eq} = \frac{R_L}{n^2}\;}

A step-down transformer ( $n < 1$ ) therefore makes a small load look large, and a step-up transformer makes it look small.

The matching condition

Lecture 5 established that a source of Thevenin resistance $R_T$ delivers its greatest power to a load equal to $R_T$ . When the load is fixed at some other value, a transformer can still satisfy the condition, because it is the reflected resistance that the source sees. Setting $R_{eq} = R_T$ gives

\frac{R_L}{n^2} = R_T \;\Longrightarrow\; \boxed{\;R_L = n^2 R_T\;}

Orientation matters

The formula assumes a transformer written as $1 : n$ with $n$ on the load side. If the schematic is labelled the other way, the ratio must be rewritten in that form before the square is taken; otherwise $n$ and $1/n$ are exchanged and the answer is wrong by a factor of $n^4$ .

§6 · Matching for maximum power

The same load, four and a half times the power

The value of the reflected resistance is best seen on a numerical case from the board notes. A 120 V rms source with an internal resistance of 800 Ω is to deliver power to a 50 Ω load, which is a severe mismatch.

Direct connection: the [50 Ω load]{nx.rl} receives [0.997 W]{nx.pl}.

Through a [4 : 1 transformer]{wx.ratio}: the same load receives [4.5 W]{wx.pl}.

The required ratio follows from the matching condition, with $R_L = 50$ Ω and $R_T = 800$ Ω:

50 = n^2 \cdot 800 \;\Longrightarrow\; n^2 = \tfrac{1}{16} \;\Longrightarrow\; \boxed{n = \tfrac{1}{4}}

A turns ratio of $1 : \tfrac14$ , that is a $4 : 1$ transformer, is therefore required. The two computations are compared below.

One loop, and no transformer:

-120 + 850\,I = 0 \;\Longrightarrow\; I = \frac{120}{850} = 0.1412\ \text{A}_{\text{rms}}

P_L = I^2 R_L = \left[\frac{120}{850}\right]^2 \cdot 50 \;\Longrightarrow\; \boxed{P_L = 0.997\ \text{W}}

Almost all of the available power is spent inside the source, because 800 Ω of the 850 Ω total lies there.

With the model in place, one KVL equation is written on each side, using $I_1 = \tfrac14 I_2$ and $V_2 = \tfrac14 V_1$ :

\text{left:}\quad -120 + 800\left[\tfrac{1}{4} I_2\right] + V_1 = 0 \qquad \text{right:}\quad -\tfrac{1}{4} V_1 + 50\, I_2 = 0

The right equation gives $V_1 = 200 I_2$ , and substitution into the left gives $-120 + 200 I_2 + 200 I_2 = 0$ , hence

I_2 = 0.3\ \text{A}_{\text{rms}} \qquad P_L = (0.3)^2 \cdot 50 \;\Longrightarrow\; \boxed{P_L = 4.5\ \text{W}}

The figure is exactly the ceiling of Lecture 5, $V_T^2 / 4R_T = 120^2 / 3200 = 4.5$ W, which confirms that the match is perfect. The load power has been raised by a factor of 4.5, and no energy was created: the transformer merely stopped the source resistance from consuming almost everything.

Sweeping the turns ratio

The slider varies $n$ with the load and the source held fixed. The reflected resistance is $50 / n^2$ Ω, and the load power peaks where that value crosses 800 Ω.

Load power against the turns ratio, with the marker at the current setting.

n = N₂ / N₁ 0.25

Turns ratio N₁ : N₂ = 4.00 : 1. Both extremes waste the source: a very small n reflects an enormous resistance and starves the loop, while n near 1 leaves the original mismatch in place.

R_eq = 50 Ωn² = 800 Ω I₂ = 0.300 A_rms P_L = 4.500 W (100.0 % of the 4.5 W ceiling)

Efficiency is still 50 %

At the matched setting the 800 Ω internal resistance dissipates 4.5 W while the load receives 4.5 W. Matching maximises the power delivered, not the fraction of the generated power that arrives, and the distinction matters whenever the source energy is expensive rather than scarce.

§7 · Worked example: transformers in cascade

Two transformers and a capacitor

Nothing in the model restricts a circuit to one transformer. The closing example of the board notes chains two of them and terminates the chain with a capacitor, which makes the problem complex-valued but not harder.

Find $v_o$ . Given: $v_s = 12\cos 10t$ V, a 10 Ω resistor, a 1 : 3 transformer with dots on the same side, an 18 Ω resistor, a 6 : 1 transformer with dots on opposite sides, and a 1/20 F capacitor.

The operating frequency is $\omega = 10$ rad/s, so the capacitor has the impedance

Z_C = \frac{1}{j\omega C} = \frac{-j}{\omega C} = \frac{-j}{10 \cdot \tfrac{1}{20}} \;\Longrightarrow\; Z_C = -j2\ \Omega

Draw both models

The first transformer has $N_2/N_1 = 3$ with the dots on the same side, so $n_1 = +3$ . The second has $N_2/N_1 = 1/6$ with the dots opposed, so $n_2 = -\tfrac16$ . Each transformer contributes a dependent current source on its primary side and a dependent voltage source on its secondary side, leaving three loops joined only through those sources.

The transformer constraints

The current relations tie the three loop currents together, so only one of them is independent:

\tm{cs.i1}{I_1} = 3\, \tm{cs.i2}{I_2} \qquad\qquad \tm{cs.i2}{I_2} = -\tfrac{1}{6}\, \tm{cs.i4}{I_4}

The middle loop current $I_2$ is retained as the unknown, and $I_4 = -6I_2$ follows.

One KVL equation per loop

\text{left:}\quad -12 + \tm{cs.r10}{10}\left[3 I_2\right] + \tm{cs.v1}{V_1} = 0

\text{middle:}\quad -3\,\tm{cs.v1}{V_1} + \tm{cs.r18}{18}\,I_2 + \tm{cs.v3}{V_3} = 0

\text{right:}\quad \tfrac{1}{6}\,\tm{cs.v3}{V_3} - \tm{cs.c}{j2}\, \tm{cs.i4}{I_4} = 0

The three equations, together with the two current constraints, close the system.

Eliminate and solve

The right equation with $I_4 = -6I_2$ gives $V_3 = -j72\,I_2$ , and the left equation gives $V_1 = 12 - 30 I_2$ . Substituting both into the middle equation:

-3\left(12 - 30 I_2\right) + 18 I_2 - j72\, I_2 = 0 \;\Longrightarrow\; \left(108 - j72\right) I_2 = 36

\tm{cs.i2}{I_2 = \frac{36}{108 - j72} = \frac{1}{3 - j2} = \frac{3 + j2}{13}\ \text{A}}

Back-substitution then gives $V_1 = \tfrac{66 - j60}{13}$ V and $V_3 = \tfrac{144 - j216}{13}$ V.

The output voltage

\tm{cs.i4}{I_4} = -6 I_2 = \frac{-18 - j12}{13}\ \text{A} \qquad \tm{cs.vo}{V_o} = -j2\, I_4 = \frac{-24 + j36}{13}\ \text{V}

\boxed{V_o = 3.328\angle 123.69°\ \text{V}} \qquad v_o(t) = 3.328\cos\left(10t + 123.69°\right)\ \text{V}

Check: the same answer by reflecting impedances

Each transformer divides the impedance behind it by $n^2$ , so the chain can be collapsed from the right. The capacitor reflects through $n_2 = -1/6$ as $-j2 \big/ \tfrac{1}{36} = -j72$ Ω; adding the 18 Ω gives $18 - j72$ Ω, which reflects through $n_1 = 3$ as $(18 - j72)/9 = 2 - j8$ Ω. The source loop therefore sees $10 + 2 - j8 = 12 - j8$ Ω, so

I_1 = \frac{12}{12 - j8} = \frac{9 + j6}{13}\ \text{A} \qquad I_2 = \frac{I_1}{3} = \frac{3 + j2}{13}\ \text{A}\ \checkmark

which reproduces the stepper result, and with it $V_o$ . The power balance confirms it independently: the source supplies $\tfrac12 \Re\{12 I_1^{*}\} = 54/13 = 4.154$ W, while the two resistors dissipate $\tfrac12 |I_1|^2 (10) + \tfrac12 |I_2|^2 (18) = 3.462 + 0.692 = 4.154$ W, the capacitor absorbing no average power ✓.

§8 · Your turn

Practice 1: reading a transformer

The turns ratio is 2 : 5 and the dots sit on opposite sides. Given $v_1 = 8\cos 200t$ V and $i_2 = 0.4\cos 200t$ A, find $n$ , $v_2$ , and $i_1$ , then verify that the instantaneous powers agree.

Hint

The magnitude of the turns ratio is $N_2 / N_1 = 5/2$ . Opposed dots make it negative. Both relations of Section 1 then use that one signed number.

Solution

n = -\frac{N_2}{N_1} = -\frac{5}{2}

v_2 = n v_1 = -\tfrac52 \left(8\cos 200t\right) \;\Longrightarrow\; \boxed{v_2 = -20\cos 200t\ \text{V}}

i_1 = n i_2 = -\tfrac52 \left(0.4\cos 200t\right) \;\Longrightarrow\; \boxed{i_1 = -\cos 200t\ \text{A}}

The power check: $p_1 = v_1 i_1 = -8\cos^2 200t$ W and $p_2 = v_2 i_2 = -8\cos^2 200t$ W, so $p_1 = p_2$ at every instant ✓. Both are negative, which simply records that in this problem the energy travels from the secondary side to the primary side; the transformer itself is indifferent to the direction.

Practice 2: designing the match

A 60 V rms source behind 45 Ω is to drive a 5 Ω load through a 1 : n transformer with the dots on the same side. Find (a) the load power with the transformer removed and the load connected directly, (b) the turns ratio that maximises the load power, and (c) that maximum power.

Hint 1: part (a) needs no transformer theory

With the load connected directly, the circuit is one loop of $45 + 5 = 50$ Ω across 60 V rms.

Hint 2: parts (b) and (c)

The source sees $R_{eq} = R_L / n^2$ . Matching requires $R_{eq} = R_T$ , that is $R_L = n^2 R_T$ . Once matched, the ceiling of Lecture 5 applies directly and no further circuit analysis is needed.

Solution

(a) Directly connected:

I = \frac{60}{50} = 1.2\ \text{A}_{\text{rms}} \qquad P_L = I^2 R_L = 1.2^2 \cdot 5 = \boxed{7.2\ \text{W}}

(b) Matching:

5 = n^2 \cdot 45 \;\Longrightarrow\; n^2 = \tfrac19 \;\Longrightarrow\; \boxed{n = \tfrac13}

which is a $3 : 1$ transformer.

(c) With the match in place,

P_{L,\max} = \frac{V_T^2}{4 R_T} = \frac{60^2}{4 \cdot 45} = \boxed{20\ \text{W}}

The transformer therefore recovers a factor of $20 / 7.2 = 2.78$ . The answer may be confirmed on the model: $R_{eq} = 5 \cdot 9 = 45$ Ω, so $I_1 = 60/90 = \tfrac23$ A, $V_1 = 30$ V, $I_2 = I_1 / n = 2$ A, and $V_2 = n V_1 = 10$ V. The load then satisfies $10 = 2 \cdot 5$ ✓ and receives $2^2 \cdot 5 = 20$ W ✓.

Note finally that the dots do not affect any of these numbers. Power depends on $n^2$ , so opposed dots would reverse the sign of $v_2$ and $i_2$ while leaving every power in the problem unchanged.