Singly-Excited and Doubly Excited Systems



Excitation means providing electrical input to an electromechanical energy conversion device such as electric motors. The excitation produces working magnetic field in the electrical machine. Some electrical machines require single electrical input whereas some others require two electrical inputs.

Therefore, depending on the number of electrical inputs to electromechanical energy conversion systems, they can be classified into two types −

  • Singly-Excited System

  • Doubly-Excited System

Singly Excited System

As its name implies, a singly-excited system is one which consists of only one electrically energized coil to produce working magnetic field in the machine or any other electromechanical energy conversion device. Hence, the singly-excited system requires only one electrical input.

A singly excited system consists of coil wound around a magnetic core and is connected to a voltage source so that it produces a magnetic field. Due to this magnetic field, the rotor (or moving part) which is made up of ferromagnetic material experiences a torque which move it towards a region where the magnetic field is stronger, i.e., the torque exerted on the rotor tries to position it such that it shows minimum reluctance in the path of magnetic flux. The reluctance depends upon the rotor angle. This torque is known as reluctance torque or saliency torque because it is caused due to saliency of the rotor.

Analysis of Singly Excited System

We made following assumption to analyze the singly-excited system −

  • For any rotor position, the relationship between flux linkage ($\psi $) and current ($\mathit{i}$) is linear.

  • The coil has negligible leakage flux, which means all the magnetic flux flows through the main magnetic path.

  • Hysteresis loss and eddy-current loss are neglected.

  • All the electric fields are neglected and the magnetic field is predominating.

Singly Excited System

Consider the singly-excited system as shown in Figure-1. If R is the resistance of the coil circuit, the by applying KVL, we can write the voltage equation as,

$$\mathrm{\mathit{v\:=\:iR\:+\:\frac{\mathit{d\psi }}{\mathit{dt}}}\cdot \cdot \cdot (1)}$$

On multiplying equation (1) by current $\mathit{i}$, we have,

$$\mathrm{\mathit{vi\:=\:i^{\mathrm{2}}R\:+\mathit{i}\:\frac{\mathit{d\psi }}{\mathit{dt}}}\cdot \cdot \cdot (2)}$$

We are assuming initial conditions of the system zero and integrating the equation (2) on both side with respect to time, we obtain,

$$\mathrm{\int_{0}^{\mathit{T}}\mathit{vi\:dt}\:=\:\int_{0}^{\mathit{T}}\left ( i^{\mathrm{2}}\mathit{R}\:+\mathit{i}\:\frac{\mathit{d\psi }}{\mathit{dt}} \right )\mathit{dt}}$$

$$\mathrm{\Rightarrow\int_{0}^{\mathit{T}}\mathit{vi\:dt}\:=\:\int_{0}^{\mathit{T}}\mathit{i^{\mathrm{2}}R\:dt}\:+\:\int_{0}^{\psi }\mathit{i\:d\psi }\cdot \cdot \cdot (3)}$$

Equation-3 gives the total electrical energy input the singly-excited system and it is equal to two parts namely,

  • First part is the electrical loss ($\mathit{W_{el}}$).

  • Second part is the useful electrical energy which is the sum of field energy ($\mathit{W_{f}}$) and output mechanical energy ($\mathit{W_{m}}$).

Therefore, symbolically we may express the Equation-3 as,

$$\mathrm{\mathit{W_{in}}\:=\:\mathit{W_{el}}\:=\:\left (\mathit{W_{f}} \:+\:\mathit{W_{m}} \right )}\cdot \cdot \cdot (4)$$

The energy stored in the magnetic field of a singly-excited system is given by,

$$\mathrm{\mathit{W_{f}}\:=\: \int_{0}^{\psi }\mathit{i\:d\psi }\:=\:\int_{0}^{\psi }\frac{\psi }{\mathit{L}}\mathit{d\psi }\:=\:\frac{\psi ^{\mathrm{2}}}{2\mathit{L}}\cdot \cdot \cdot (5)}$$

For a rotor movement, where the rotor angle is $\mathit{\theta _{m}}$, the electromagnetic torque developed in the singly-excited system is given by,

$$\mathrm{\mathit{\tau _{e}}\:=\:\frac{\mathit{i^{\mathrm{2}}}}{\mathrm{2}}\frac{\mathit{\partial L}}{\mathit{\partial \theta _{m}}}\cdot \cdot \cdot (6)}$$

The most common examples of singly-excited system are induction motors, PMMC instruments, etc.

Doubly Excited System

An electromagnetic system is one which has two independent coils to produce magnetic field is known as doubly-excited system. Therefore, a doubly-excited system requires two separate electrical inputs.

Analysis of Doubly Excited System

A doubly-excited system consists of two main parts namely a stator and a rotor as shown in Figure-2. Here, the stator is wound with a coil having a resistance R1 and the rotor is wound with a coil of resistance R2. Therefore, there are two separated windings which are excited by two independent voltage sources.

Doubly Excited System

In order to analyze the double-excited system, the following assumption are made −

  • For any rotor position the relationship between flux-linkage ($\psi$) and current is linear.

  • Hysteresis and eddy current losses are neglected.

  • The coils have negligible leakage flux.

  • The electric fields are neglected and the magnetic fields are predominating.

The magnetic flux linkages to two windings are given by,

$$\mathrm{\psi _{\mathrm{1}}\:=\:\mathit{L_{\mathrm{1}}i_{\mathrm{1}}}\:+\:\mathit{Mi_{\mathrm{2}}}}\cdot \cdot \cdot (7)$$

$$\mathrm{\psi _{\mathrm{2}}\:=\:\mathit{L_{\mathrm{2}}i_{\mathrm{2}}}\:+\:\mathit{Mi_{\mathrm{2}}}}\cdot \cdot \cdot (8)$$

Where, M is the mutual inductance between two windings.

By applying KVL, we can write the equations of instantaneous voltage for two coils as,

$$\mathrm{\mathit{v}_{\mathrm{1}}\:=\:\mathit{i_{\mathrm{1}}R_{\mathrm{1}}}\:+\:\frac{\mathit{d\psi _{\mathrm{1}}}}{\mathit{dt}}}\cdot \cdot \cdot (9)$$

$$\mathrm{\mathit{v}_{\mathrm{2}}\:=\:\mathit{i_{\mathrm{2}}R_{\mathrm{2}}}\:+\:\frac{\mathit{d\psi _{\mathrm{2}}}}{\mathit{dt}}}\cdot \cdot \cdot (10)$$

In case of doubly-excited system, the energy stored in the magnetic field is given by,

$$\mathrm{\mathit{W_{f}}\:=\:\frac{1}{2}\mathit{L_{\mathrm{1}}i_{\mathrm{1}}^{\mathrm{2}}}\:+\:\frac{1}{2}\mathit{L_{\mathrm{2}}i_{\mathrm{2}}^{\mathrm{2}}}\:+\:\mathit{Mi_{\mathrm{1}}i_{\mathrm{2}}}\cdot \cdot \cdot (11)}$$

And, the electromagnetic torque developed in the doubly excited system is given by,

$$\mathrm{\mathit{\tau _{e}}\:=\:\frac{\mathit{i_{\mathrm{1}}^{\mathrm{2}}}}{\mathrm{2}}\frac{\mathit{dL_{\mathrm{1}}}}{\mathit{d\theta _{m}}}\:+\:\frac{\mathit{i_{\mathrm{2}}^{\mathrm{2}}}}{\mathrm{2}}\frac{\mathit{dL_{\mathrm{2}}}}{\mathit{d\theta _{m}}}\:+\:\mathit{i_{\mathrm{1}}i_{\mathrm{2}}}\frac{\mathit{dM}}{\mathit{d\theta _{m}}}\cdot \cdot \cdot (12)}$$

In Equation-12, the first two terms are the reluctance torque and the last term gives the co-alignment torque due to interaction of two fields.

The practical examples of doubly-excited system are synchronous machines, tachometer, separately-excited DC machines, etc.

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