Maybe we have already known what is transformers. But what about linear transformers?
Here we introduce the transformer as a new circuit element. A transformer is a magnetic device that takes advantage of the phenomenon of mutual inductance.
The linear transformer will have some important explanation such as:
- Mutual inductance and dot convention
- What is ideal transformer
- Three phase transformer
- Transformer as isolation device
- Transformer as matching device
Linear Transformers
A linear transformer may also be regarded as one whose flux is proportional to the currents in its windings.
A transformer is generally a four-terminal device comprising two (or more) magnetically coupled coils.
As shown in Figure.(1), the coil that is directly connected to the voltage source is called the primary winding.
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| Figure 1. A linear transformer |
The coil connected to the load is called the secondary winding. The resistances R1 and R2 are included to account for the losses (power dissipation) in the coils.
The transformer is said to be linear if the coils are wound on a magnetically linear material—a material for which the magnetic permeability is constant.
Such materials include air, plastic, Bakelite, and wood. In fact, most materials are magnetically linear.
Linear transformers are sometimes called air-core transformers, although not all of them are necessarily air-core.
They are used in radio and TV sets. Figure.(2) portrays different types of transformers.
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| Figure 2. Different types of transformers: (a) copper wound dry power transformer, (b) audio transformers. (Courtesy of: (a) Electric Service Co., (b) Jensen Transformers. |
We would like to obtain the input impedance Zin as seen from the source because Zin governs the behaviour of the primary circuit.
Applying KVL to the two meshes in Figure.(1) gives
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| (1a) |
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| (1b) |
In Equation.(1b), we express I2 in terms of I1 and substitute it into (1a). We get the input impedance as
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| (2) |
Notice that the input impedance comprises two terms.
The first term, (R1 + jωL1 ), is the primary impedance.
The second term is due to the coupling between the primary and secondary windings.
It is as though this impedance is reflected to the primary. Thus, it is known as the reflected impedance ZR, and
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| (3) |
It should be noted that the result in Equations.(2) or (3) is not affected by the location of the dots on the transformer, because the same result is produced when M is replaced by −M.
For this reason, it is sometimes convenient to replace a magnetically coupled circuit by an equivalent circuit with no magnetic coupling.
We want to replace the linear transformer in Figure.(1) by an equivalent T or π circuit, a circuit that would have no mutual inductance.
Ignore the resistances of the coils and assume that the coils have a common ground as shown in Figure.(3).
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| Figure 3. Determining the equivalent circuit of a linear transformer. |
The assumption of a common ground for the two coils is a major restriction of the equivalent circuits.
Common ground is imposed on the linear transformer in Figure.(3) in view of the necessity of having a common ground in the equivalent T or π circuit; see Figures.(4) and (5).
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| Figure 4. An equivalent T circuit |
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| Figure 5. An equivalent π circuit |
The voltage-current relationships for the primary and secondary coils give the matrix equation
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| (4) |
By matrix inversion, this can be written as
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| (5) |
Our goal is to match Equations.(4) and (5) with the corresponding equations for the T and π networks.
For the T (or Y) network of Figure.(4), mesh analysis provides the terminal equations as
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| (6) |
If the circuits in Figures.(3) and (4) are equivalents, Equations.(4) and (6) must be identical. Equating terms in the impedance matrices of Equations.(4) and (6) leads to
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| (7) |
For the π (or Δ) network in Figure.(5), the nodal analysis gives the terminal equations as
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| (8) |
Equating terms in admittance matrices of Equations.(5) and (8), we obtain
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| (9) |
Note that in Figures.(5) and (6), the inductors are not magnetically coupled. Also, note that changing the locations of the dots in Figure.(3) can cause M to become −M.
Read also : truth table of JK flip flop
Linear Transformers Examples
For better understanding let us review the examples below:
1. In the circuit of Figure.(6), calculate the input impedance and current I1. Take Z1 = 60 − j100 Ω, Z2 = 30 + j40 Ω, and ZL = 80 + j60 Ω.
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| Figure 6 |
Solution:
From Equation.(2),
Thus,
2. Determine the T-equivalent circuit of the linear transformer in Figure.(7a).
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| Figure 7. (a) a linear transformer, (b) its T-equivalent circuit. |
Solution:
Given that L1 = 10, L2 = 4, and M = 2, the T equivalent network has the following parameters:
The T-equivalent circuit is shown in Figure.(7b). We have assumed that reference directions for currents and voltage polarities in the primary and secondary windings conform to those in Figure.(3).
Otherwise, we may need to replace M with −M.
3. Solve for I1, I2, and Vo in Figure.(8) using the T-equivalent circuit for the linear transformer.
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| Figure 8 |
Solution:
Notice that the circuit in Figure.(8), the reference direction for current I2 has been reversed, just to make the reference directions for the currents for the magnetically coupled coils conform with those in Figure.(3).
We need to replace the magnetically coupled coils with the T equivalent circuit. The relevant portion of the circuit in Figure.(8) is shown in Figure.(9a).
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| Figure 9. (a) circuit for coupled coils of Figure.(8), (b) T-equivalent circuit |
Comparing Figure.(9a) with Figure.(3) shows that there are two differences.
First, due to the current reference directions and voltage polarities, we need to replace M by −M to make Figure.(9a) conform with Figure.(3).
Second, the circuit in Figure.(3) is in the time-domain, whereas the circuit in Figure.(9a) is in the frequency domain.
The difference is the factor jω; that is, L in Figure.(3) has been replaced with jωL and M with jωM.
Since ω is not specified, we can assume ω = 1 or any other value; it really does not matter.
With these two differences in mind,
Thus, the T-equivalent circuit for the coupled coils is as shown in Figure.(9b).
Inserting the T-equivalent circuit in Figure.(9b) to replace the two coils in Figure.(8) gives the equivalent circuit in Figure.(10), which can be solved using nodal or mesh analysis.
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| Figure 10 |
Applying mesh analysis, we obtain
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| (3.1) |
and
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| (3.2) |
From Equation.(3.2)
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| (3.3) |
Substituting Equations.(3.3) into (3.1) gives
Since 100 is very large compared to 1, the imaginary part of (100 − j) can be ignored so that 100 − j ≃ 100. Hence,
From Equation.(3.3),
and
The advantage of using the T-equivalent model for the magnetically coupled coils is that in Figure.(10) we do not need to bother with the dot on the coupled coils.
