Choose Blocks to Model Transformers

The Simscape™ and Simscape Electrical™ libraries include several blocks that can model the same transformer device. For example, the Mutual Inductor, Ideal Transformer, and Nonlinear Transformer blocks all model two electromagnetically-coupled windings. However, these blocks make different modeling assumptions. To choose the right block for your application you need to understand these assumptions and how they impact the block behavior as a function of frequency.

Compare Equations

To understand the differences between blocks that model transformers, it is useful to compare the equations that they use. In this section, you first derive a generalized transfer function for a magnetic domain representation of a two-winding component and some simplified approximations for idealized cases. Then, you compare the magnetic equations to the electrical equations that different transformer blocks use. This analysis focuses on two-winding components but some of the key points extend to three-winding components.

For a brief summary of the differences between transformer models, the blocks that use these models, and guidance on when to use each block, see Choose the Right Block.

Generalized Magnetic Circuit

This figure shows the generalized magnetic equivalent circuit for a two-winding transformer.

Equivalent magnetic circuit for a two-winding transformer. A Fundamental Reluctance block along the closed loop between the primary and secondary windings models the core reluctance. Two additional Fundamental Reluctance blocks, in parallel with the primary and secondary windings, model the leakage reluctances.

The Electromagnetic Converter blocks, labeled Primary electromagnetic converter and Secondary electromagnetic converter, represent the two windings that connect the electrical and magnetic circuits. In this orientation, these blocks follow the winding configuration that all transformer blocks use, in which a positive-going voltage on the primary winding results in a positive-outgoing current on the secondary winding.

These equations define the behavior of the primary and secondary winding.

F_{1} = N_{1} i_{1}

(1)

F_{2} = N_{2} i_{2}

(2)

v_{1} - i_{1} r_{1} = - N_{1} \frac{d ϕ_{1}}{d t}

(3)

v_{2} - i_{2} r_{2} = - N_{2} \frac{d ϕ_{2}}{d t}

(4)

The equations have these block parameter values:

r₁ is the primary ohmic resistance.
r₂ is the secondary ohmic resistance.
N₁ is the number of turns on the primary winding.
N₂ is the number of turns on the secondary winding.

The equations have these variables in the magnetic and electrical domain:

Φ₁ is the flux flowing into the north terminal of the primary electromagnetic converter.
Φ₂ is the flux flowing into the north terminal of the secondary electromagnetic converter.
F₁ is the magnetomotive force (MMF) across the north and south terminals of the primary electromagnetic converter.
F₂ is the MMF across the north and south terminals of the secondary electromagnetic converter.
i₁ is the current flowing into the positive terminal of the primary electromagnetic converter.
i₂ is the current flowing into the positive terminal of the secondary electromagnetic converter.
v₁ is the voltage across the positive and negative terminals of the primary electromagnetic converter.
v₂ is the voltage across the positive and negative terminals of the secondary electromagnetic converter.

The Fundamental Reluctance blocks, labeled R1, R2, and Rc, represent magnetic reluctances which exhibit a drop in MMF when a magnetic flux 𝜙 flows through them. The leakage reluctances, R1 and R2, represent the part of the flux that the electromagnetic converter generates that does not flow around the main magnetic path. These blocks represent a loss of coupling between the two connected electrical circuits. The reluctance Rc is the core reluctance. These parameter values describe the magnetic behavior of the Fundamental Reluctance blocks:

R₁ is the primary leakage reluctance.
R₂ is the secondary leakage reluctance.
R_c is the core reluctance.

A smaller value of R_c means that a smaller MMF creates a given level of flux linkage between the two windings. This smaller MMF means that a smaller current creates the same level of flux linkage, because MMF is equal to the current multiplied by the number of turns on the winding.

Now, consider Kirchoff's equations. Summing the MMFs around the main magnetic path gives

N_{1} i_{1} - R_{c} ϕ_{1} + N_{2} i_{2} = 0.

(5)

Summing the fluxes at the primary electromagnetic converter north terminal gives:

- ϕ = ϕ_{1} + \frac{N_{1} i_{1}}{R_{1}} = 0,

(6)

where Φ is the flux flowing through the core reluctance R_c from the north terminal to the south terminal. Summing the fluxes at the secondary electromagnetic converter north terminal gives

- ϕ = ϕ_{2} + \frac{N_{2} i_{2}}{R_{2}} = 0.

(7)

Now, consider the input-output behavior. Suppose you apply a voltage V to the primary winding and short circuit the secondary winding. Noting that v₁ = V and v₂ = 0, and using equations 3–7, you can derive this expression for the secondary winding current:

i_{2} = \frac{- N_{1} N_{2} R_{1} R_{2} s}{N_{1}^{2} N_{2}^{2} (R_{1} + R_{2} + R_{c}) s^{2} + (N_{1}^{2} R_{1} R_{2} r_{2} + N_{1}^{2} R_{2} R_{c} r_{2} + N_{2}^{2} R_{1} R_{c} r_{1}) s + R_{1} R_{2} R_{c} r_{1} r_{2}} V,

(8)

where s is the Laplace transform variable. The numerator of this equation is zero for s = 0, which implies no transmission of DC inputs, as you would expect for mutual coupling.

Equation 8 is the general case transfer function. You can simplify this expression for idealized cases when:

The core reluctance is very small — There is low opposition to magnetic flux around the main magnetic path.
The leakage reluctance is very large — There is a large opposition to magnetic flux outside of the main magnetic path.

Small Core Reluctance. If the core reluctance R_c is small enough that you can neglect it, the equivalent magnetic circuit simplifies to:

Equivalent magnetic circuit for a two-winding transformer with negligible core reluctance. Two Fundamental Reluctance blocks, in parallel with the primary and secondary windings, model the leakage reluctances.

Equation 8 simplifies to

$i_{2} = \frac{- N_{1} N_{2}}{N_{1}^{2} N_{2}^{2} \frac{(R_{1} + R_{2})}{R_{1} R_{2}} s + N_{1}^{2} r_{2}} V .$

For direct current, s is equal to zero and this equation further simplifies to

$i_{2} r_{2} = \frac{- N_{2}}{N_{1}} V .$

This equation is the ideal transformer equation. This equation does not depend on R₁ and R₂.

If R₁ and R₂ are very large, the $\frac{R_{1} + R_{2}}{R_{1} R_{2}}$ term becomes very small. If you also neglect this term, equation 8 becomes the ideal transformer equation at all frequencies.

Large Leakage Reluctances. Now, consider the case where the core reluctance R_c is not small and the leakage reluctances are very large. The equivalent magnetic circuit simplifies to:

Equivalent magnetic circuit for a two-winding transformer with infinite leakage reluctances. A Fundamental Reluctance block along the closed loop between the primary and secondary windings models the core reluctance.

Divide the numerator and denominator of equation 8 by R₁R₂ to give

$i_{2} = \frac{- N_{1} N_{2} s}{\frac{N_{1}^{2} N_{2}^{2} (R_{1} + R_{2} + R_{c}) s^{2}}{R_{1} R_{2}} + (N_{1}^{2} r_{2} + \frac{N_{1}^{2} R_{c} r_{2}}{R_{1}} + \frac{N_{2}^{2} R_{c} r_{1}}{R_{2}}) s + R_{c} r_{1} r_{2}} V .$

If you neglect the small terms in the denominator, this equation simplifies to

$i_{2} = \frac{- N_{1} N_{2} s}{N_{1}^{2} r_{2} s + R_{c} r_{1} r_{2}} V .$

In this case, DC does not transmit at any frequency due to the s on the numerator. There is also a first-order time constant with a value that only depends on the core reluctance and primary electromagnetic converter parameters,

$τ = \frac{R_{c} r_{1}}{N_{1}^{2}} .$

Summary. An analysis of the equivalent magnetic circuit of a two-winding transformer provides insight into understand the differences between blocks that model two-winding transformers in the electrical domain:

If the core reluctance R_c is very small, power transmits at DC. This transmission can be an issue for some circuits because it effectively adds a ground path that does not exist in the real system.
If the leakage reluctances R₁ and R₂ are very large, power does not transmit at DC. Each winding presents an impedance with an associated time constant to the connected circuit.
If the core reluctance is very small and the leakage reluctances R₁ and R₂ are very large, the transformer behaves like an ideal transformer.

Mutual Inductor Block

These equations define the behavior of the Mutual Inductor block, without perfect coupling:

v_{1} - i_{1} r_{1} = L_{1} \frac{d i_{1}}{d t} + M \frac{d i_{2}}{d t}

(9)

v_{2} - i_{2} r_{2} = L_{2} \frac{d i_{2}}{d t} + M \frac{d i_{1}}{d t},

(10)

where

M = k \sqrt{L_{1} L_{2}},

(11)

and k is the coupling factor. This parameter represents the coupling between the electrical and magnetic circuits, not the magnetic coupling between the primary and the secondary. Hence, k relates to leakage reluctance.

These equations must represent identical behavior to the generalized magnetic circuit (equations 3–7). The magnetic domain has three extra equations for the magnetic domain and you have three more variables to solve for: Φ₁, Φ₂, and Φ. By comparing the electrical and magnetic equations, you can deduce these relationships between the parameters.

$L_{1} = N_{1}^{2} (\frac{1}{R_{c}} + \frac{1}{R_{1}})$

$L_{2} = N_{2}^{2} (\frac{1}{R_{c}} + \frac{1}{R_{2}})$

$k = \frac{1}{\sqrt{(1 + \frac{R_{c}}{R_{1}}) (1 + \frac{R_{c}}{R_{2}})}}$

The defining electrical equations (equations 9, 10, and 11) are valid only for magnitudes of k that are less than one. When the magnitude of the coupling factor is precisely one, there must be just one differential equation, but equations 9 and 10 define two equations. To understand why this is, you need to revisit the magnetic domain equations. Assuming that the leakage reluctances R₁ and R₂ are very large, which corresponds to k = 1, equations 5–7 simplify to:

$N_{1} i_{1} + R_{c} ϕ_{1} + N_{2} i_{2} = 0$

$ϕ_{1} = ϕ_{2} .$

Using these equations along with equations 3 and 4, you get the ideal transformer equation relating input and output voltages:

$\frac{v_{1} - i_{1} r_{1}}{v_{2} - i_{2} r_{2}} = \frac{N_{1}}{N_{2}} .$

Differentiate the first equation to give

$N_{1} \frac{d i_{1}}{d t} + R_{c} \frac{d ϕ_{1}}{d t} + N_{2} \frac{d i_{2}}{d t} = 0.$

Eliminate Φ₁ from this equation using equation 3 to give

$N_{1} \frac{d i_{1}}{d t} + R_{c} \frac{- (v_{1} - i_{1} r_{1})}{N_{1}} + N_{2} \frac{d i_{2}}{d t} = 0.$

Now, rearrange this equation to give

$\frac{v_{1} - i_{1} r_{1}}{v_{2} - i_{2} r_{2}} = \sqrt{\frac{L_{1}}{L_{2}}} .$

Hence, the two electrical domain equations for the perfect coupling case, with k = 1, are:

\frac{v_{1} - i_{1} r_{1}}{v_{2} - i_{2} r_{2}} = \sqrt{\frac{L_{1}}{L_{2}}}

(12)

v_{1} - i_{1} r_{1} = L_{1} \frac{d i_{1}}{d t} + \sqrt{L_{1} L_{2}} \frac{d i_{2}}{d t} .

(13)

Both sets of electrical equations, equations 9–10 and 12–13, are consistent with the magnetic equations.

The Mutual Inductor block:

Isolates the two windings electrically, providing no path for DC.
Presents an inductive impedance to each of the circuits connected to the windings.
Presents an impedance of L₁ at the primary, if the secondary is open circuit.
Presents an impedance of L₂ at the secondary, if the primary is open circuit.

In the case of perfect coupling, there is only one differential equation and associated time constant.

Changing k without changing L₁ and L₂ corresponds to reducing leakage only and does not change the properties of the core magnetic path. To set the coefficient of coupling to one while retaining the correct low-frequency behavior, blocking DC, use the Mutual Inductor block in the Simscape Foundation library and select the Perfect coupling - no leakage parameter.

Standard Transformer Equations and the Ideal Transformer Block

The standard transformer equations make the simplifying assumption that the main magnetic path in the transformer has negligible reluctance. Hence, the core reluctance R_c is zero and the equivalent magnetic circuit simplifies to:

Equivalent magnetic circuit for a standard two-winding transformer model which has negligible core reluctance. Two Fundamental Reluctance blocks, in parallel with the primary and secondary windings, model the leakage reluctances.

Equations 5, 6, and 7 simplify to:

$ϕ_{2} = ϕ_{1} + \frac{N_{1} i_{1}}{R}$

$F_{1} = - F_{2},$

where $R = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$ .

Differentiate the equation for Φ₂ to give

$\frac{d ϕ_{2}}{d t} = \frac{d ϕ_{1}}{d t} + \frac{N_{1}}{R} \frac{d i_{1}}{d t} .$

Eliminate Φ₁ and Φ₂ from this equation using equations 3 and 4 to give

$\frac{- (v_{2} - i_{2} r_{2})}{N_{2}} = \frac{- (v_{1} - i_{1} r_{1})}{N_{1}} + \frac{N_{1}}{R} \frac{d i_{1}}{d t} .$

Rearrange this equation to give:

$v_{2} - i_{2} r_{2} = \frac{N_{2}}{N_{1}} (v_{1} - i_{1} r_{1} - \frac{N_{1}^{2}}{R} \frac{d i_{1}}{d t}) .$

Substituting for F₁ and F₂ in the F₁ = -F₂ equation using equations 1 and 2 gives

$N_{1} i_{1} = - N_{2} i_{2} .$

These last two equations are the standard transformer equations. This figure shows the equivalent electrical circuit:

Equivalent electrical circuit for a standard two-winding transformer model which has negligible core reluctance. An Inductor block connects to the primary winding positive terminal of an Ideal Transformer block.

where $L_{T} = \frac{N_{1}^{2} (R_{1} + R_{2})}{R_{1} R_{2}}$ .

The Ideal Transformer block does not model leakage inductance L_T. This simplification gives just these two equations:

$N_{1} i_{1} = - N_{2} i_{2}$

$\frac{v_{2} - i_{2} r_{2}}{v_{1} - i_{1} r_{1}} = \frac{N_{2}}{N_{1}} .$

The equivalent magnetic circuit simplifies to:

Equivalent magnetic circuit for a two-winding transformer with negligible core reluctance and infinite leakage reluctance. The primary and secondary windings connect in a closed loop with no Fundamental Reluctance blocks.

To summarize:

The standard transformer equations neglect core reluctance. In Simscape, neglecting the core reluctance is equivalent to modeling the transformer by connecting an Inductor block to the primary winding positive terminal of an Ideal Transformer block.
Standard transformer models do not isolate the two connected circuits electrically, providing a path for DC. This path is nonphysical and creates a ground path that does not exist in practice.

The results of standard transformer models differ from those of mutual inductor models when one winding is open-circuit and you perform an impedance test on the other. Standard transformer models present an infinite impedance, whereas mutual inductor models present a finite impedance.

The Ideal Transformer block uses a special case of the standard transformer equations that does not model leakage inductance.

Nonlinear Transformer Block

The Nonlinear Transformer block represents a transformer with a nonideal core. This figure shows an equivalent electrical circuit.

Equivalent electrical circuit for the Nonlinear transformer block. Resistor and Inductor blocks connect in series with the primary and secondary winding positive terminals of an Ideal Transformer block. A Resistor and Nonlinear Inductor block connect in parallel with the primary terminals of the Ideal Transformer block.

In this diagram:

r₁ is the primary winding resistance.
L₂ is the primary leakage inductance.
r₂ is the secondary winding resistance.
L₂ is the secondary leakage inductance.
r_m is the magnetization resistance.
L_m is the magnetization inductance.

The Nonlinear Transformer block models the core reluctance from the magnetizing inductance Lm. The input impedance for an open-circuit secondary is the sum of the primary leakage and magnetizing inductances. The Nonlinear Transformer block does not transmit DC because Lm behaves like a short circuit at DC, shorting the primary winding.

Choose the Right Block

This table summarizes the analysis of the magnetic and electrical equations, grouping transformer blocks into three sets of models: mutual inductors, ideal transformers, and nonlinear transformers. Use this table to decide which model you need.

Model	Block	Library	Models Core Reluctance	Transmits DC	Input Impedance for Open-Circuit Secondary	Applications
Mutual inductors	Mutual Inductor	Simscape Foundation library	Yes	No	L₁	Audio, RF, and switching power converters when the impedance of the mutual inductor plays a part in correct circuit operation.
	Mutual Inductor	Simscape Electrical library
	Three-Winding Mutual Inductor	Simscape Electrical library
Ideal transformers	Ideal Transformer	Simscape Foundation library	No	Yes	Infinite	AC power and power converters. These blocks do not support analysis of when one of the windings disconnects from its circuit.
Ideal transformers	Ideal Transformer with an Inductor connected to the primary winding positive terminal	Simscape Foundation library	No	Yes	Infinite
Nonlinear transformers	Nonlinear Transformer	Simscape Electrical library	Yes	No	L₁ + L_m	AC power and power converters
Nonlinear transformers	Three-Winding Nonlinear Transformer	Simscape Electrical library	Yes	No	L₁ + L_m	AC power and power converters

For more information about the Mutual Inductor and Ideal Transformer blocks in the Simscape Foundation library and their equivalent magnetic circuits, see A Comparison of the Mutual Inductor and Ideal Transformer Library Blocks.

Mutual Inductors Models

This table compares blocks that use mutual inductor models. Use this table to choose a block and option for the Perfect coupling - no leakage parameter. This choice depends on the number of windings your transformer has, whether you need to model winding leakage, and whether you need hardware-in-the-loop (HIL) testing.

Block	Library	Number of Windings	Perfect coupling - no leakage Parameter Value	Models Winding Leakage	Good for HIL (Avoids Fast Time Constants)
Mutual Inductor	Simscape Foundation library	Two	`Off`	Yes	No, if k is close to 1
Mutual Inductor	Simscape Foundation library	Two	`On`	No	Yes
Mutual Inductor	Simscape Electrical library	Two	N/A	Yes	No, if k is close to 1
Three-Winding Mutual Inductor	Simscape Electrical library	Three	`Off`	Yes	No, if k is close to 1
Three-Winding Mutual Inductor	Simscape Electrical library	Three	`On`	No	Yes

The Mutual Inductor block in the Simscape Foundation library and the Three-Winding Mutual Inductor block support perfect coupling. To set the coefficients of coupling to one while retaining the correct low-frequency behavior, blocking DC, select the Perfect coupling - no leakage parameter. If you need to simulate ideal coupling for a two-winding mutual inductor model, use the Mutual Inductor block in the Simscape Foundation library.

The Mutual Inductor block in the Simscape Electrical library models an equivalent series resistance and a parallel leakage path of the primary and secondary winding, which the Mutual Inductor block in the Simscape Foundation library does not. The Simscape Electrical block also has optional models for tolerances, faults, and operating limits.

Ideal and Nonlinear Transformer Models

This table compares blocks that use ideal and nonlinear transformer models. Use this table to choose a block depending on whether you need to model winding leakage and whether you need HIL testing.

Block	Library	Number of Windings	Models Winding Leakage	Good for HIL (Avoids Fast Time Constants)
Ideal Transformer	Simscape Foundation library	Two	No	Yes
Ideal Transformer with an Inductor connected to the primary winding positive terminal	Simscape Foundation library	Two	Yes	No, if the ratio of leakage inductance to the connected circuit impedance is small compared to the sample time.
Nonlinear Transformer	Simscape Electrical library	Two	Yes	No, if the ratio of leakage inductance to the connected circuit impedance is small compared to the sample time.
Three-Winding Nonlinear Transformer	Simscape Electrical library	Three	Yes	No, if the ratio of leakage inductance to the connected circuit impedance is small compared to the sample time.