Introduction to DC Machine Modeling

Learning Objectives

By the end of this lecture, you will be able to:

Understand the need for mathematical modeling of DC machines
Recall the construction and operating principle of DC machines
Derive the fundamental torque and EMF equations
Identify machine parameters and their physical significance
Express equations in compact forms suitable for analysis

Why Model DC Machines?

Importance of Mathematical Modeling:

Predict machine behavior under different operating conditions
Design control systems for speed and torque regulation
Analyze steady-state and dynamic performance
Optimize drive system efficiency
Simulate before physical implementation

Model Applications

Motor starting and braking studies
Speed control design
Energy consumption analysis
Fault diagnosis and condition monitoring

DC Machines in Modern Applications

Traditional Applications:

Rolling mills
Paper mills
Traction systems
Machine tools
Hoists and cranes

Modern Applications:

Electric vehicles
Robotics and automation
CNC machines
Aerospace actuators
Precision servo systems

Key Advantages

Simple and precise speed control
High starting torque capability
Wide speed range operation
Independent control of flux and torque

Review of DC Machine Construction

DC Machine: Major Components

1. Stator (Field System)

Provides mechanical support and magnetic return path
Houses field poles with field windings
Produces the main magnetic field
Can use electromagnets or permanent magnets

2. Rotor (Armature)

Cylindrical laminated core mounted on shaft
Contains armature windings in slots
Carries armature current during operation
Rotates within the magnetic field

3. Commutator and Brush Assembly

Mechanical rectifier for DC operation
Maintains unidirectional torque
Ensures current reversal at appropriate instants

Simplified DC Machine Structure

Operating Principle

Fundamental Principle

When a current-carrying conductor is placed in a magnetic field, it experiences a mechanical force (Lorentz force).

Fleming's Left-Hand Rule (Motor Action):

First finger: Direction of magnetic field (N to S)
Second finger: Direction of current
Thumb: Direction of force/motion

Key Concept

Maximum torque occurs when the magnetic field and armature current interact with conductors positioned perpendicular to the field

Role of Commutator

The Problem Without Commutator

As the rotor rotates, conductors move from one pole to another. Without commutation, the force direction would reverse, causing oscillation instead of continuous rotation.

The Solution: Commutator Action

The commutator reverses the current direction in the armature conductors at the right instant (when crossing the magnetic neutral axis), ensuring:

Unidirectional electromagnetic torque
Continuous rotation in one direction
Automatic switching synchronized with rotor position

Commutator Construction:

Copper segments insulated from each other
Mounted on rotor shaft
Carbon/graphite brushes make sliding contact

EMF Equation Derivation

Induced EMF: Physical Basis

Faraday's Law of Electromagnetic Induction:

\[e = -\frac{d\phi}{dt}\]

For a conductor moving in a magnetic field:

\[e = Blv\]

where:

\(B\) = magnetic flux density (T)
\(l\) = active length of conductor (m)
\(v\) = velocity of conductor (m/s)

In DC Machine

Multiple conductors on the rotor cut the magnetic flux as the machine rotates, inducing EMF in each conductor. Total EMF depends on machine construction and rotational speed.

EMF Equation: Step-by-Step Derivation

Machine Parameters:

\(P\) = Number of poles
\(\phi\) = Flux per pole (Wb)
\(Z\) = Total armature conductors
\(n\) = Speed in rpm

\(a\) = Number of parallel paths
\(\omega_m\) = Angular speed (rad/s)
\(E\) = Induced EMF (V)

Derivation Steps

Step 1: Flux Cut by One Conductor per Revolution

Each conductor cuts flux from \(P\) poles in one revolution:

\[\text{Flux cut per revolution} = P\phi \text{ (Wb)}\]

Step 2: Time for One Revolution

If speed is \(n\) rpm:

\[\text{Time per revolution} = \frac{60}{n} \text{ seconds}\]

Step 3: EMF Induced in One Conductor

Using Faraday's law:

\[e_{\text{conductor}} = \frac{P\phi}{60/n} = \frac{P\phi n}{60} \text{ volts}\]

Step 4: Total EMF with All Conductors

With \(Z\) conductors in \(a\) parallel paths:

Conductors in series per path = \(Z/a\)
EMFs of conductors in series add up

\[\boxed{E = \frac{P\phi Zn}{60a} \text{ volts}}\]

Compact Form of EMF Equation

Expressing in terms of angular velocity:

Angular velocity in rad/s: \(\omega_m = \frac{2\pi n}{60}\)

Therefore: \(n = \frac{60\omega_m}{2\pi}\)

Substituting in EMF equation:

\[E = \frac{P\phi Z}{60a} \times \frac{60\omega_m}{2\pi} = \frac{PZ}{2\pi a}\phi\omega_m\]

Standard Compact Form

\[\boxed{E = K_a\phi\omega_m}\]

where the machine constant is:

\[K_a = \frac{PZ}{2\pi a}\]

For constant flux operation:

If flux \(\phi\) is maintained constant:

\[\boxed{E = K_e\omega_m}\]

where \(K_e = K_a\phi\) is the EMF constant (V·s/rad)

Winding Types and Parallel Paths

Lap Winding:

Number of parallel paths: \(a = P\)
Higher current rating
Lower voltage rating
Used in high-current applications

Wave Winding:

Number of parallel paths: \(a = 2\)
Lower current rating
Higher voltage rating
Used in high-voltage applications

Torque Equation Derivation

Electromagnetic Torque: Physical Basis

Force on a Current-Carrying Conductor:

When a conductor carrying current \(i\) is placed in a magnetic field of flux density \(B\):

\[F = Bil \text{ (Newtons)}\]

where \(l\) is the active length of the conductor.

Torque from One Conductor:

If the conductor is at radius \(r\) from the shaft center:

\[\tau_{\text{conductor}} = F \times r = Bilr\]

Total Electromagnetic Torque

Assumptions:

Uniform flux distribution under pole faces
All conductors carry equal current
Negligible flux fringing

Total Force on All Conductors:

Average flux density: \(B_{\text{av}} = \frac{P\phi}{\pi Dl}\)

where \(D\) is armature diameter and \(l\) is axial length.

Current per conductor: \(i_c = \frac{I_a}{a}\)

where \(I_a\) is total armature current and \(a\) is parallel paths.

Total force on \(Z\) conductors:

\[F_{\text{total}} = Z \times B_{\text{av}} \times l \times i_c = Z \times \frac{P\phi}{\pi Dl} \times l \times \frac{I_a}{a}\]

Electromagnetic Torque:

\[T_e = F_{\text{total}} \times r = F_{\text{total}} \times \frac{D}{2}\]

\[T_e = Z \times \frac{P\phi}{\pi Dl} \times l \times \frac{I_a}{a} \times \frac{D}{2} = \frac{PZ}{2\pi a}\phi I_a\]

Standard Torque Equation

\[\boxed{T_e = K_a\phi I_a}\]

where \(K_a = \frac{PZ}{2\pi a}\) (same constant as in EMF equation)

Compact Forms of Torque Equation

For constant flux operation:

\[\boxed{T_e = K_t I_a}\]

where \(K_t = K_a\phi\) is the torque constant (N·m/A)

Relationship Between EMF and Torque Constants

Recall:

\[\begin{aligned} E &= K_e\omega_m \quad \text{(EMF equation)}\\ T_e &= K_t I_a \quad \text{(Torque equation)} \end{aligned}\]

Power balance (neglecting losses):

\[P_{\text{electrical}} = P_{\text{mechanical}}\]

\[E \times I_a = T_e \times \omega_m\]

Substituting:

\[(K_e\omega_m) \times I_a = (K_t I_a) \times \omega_m\]

Important Result

\[\boxed{K_e = K_t = K}\]

In SI units with consistent definitions (V·s/rad and N·m/A), the EMF constant equals the torque constant numerically!

Alternative Torque Expression

Field flux in terms of field current:

For a linear magnetic circuit (unsaturated region):

\[\phi = \frac{N_f I_f}{\mathcal{R}_m}\]

where:

\(N_f\) = number of field turns per pole
\(I_f\) = field current
\(\mathcal{R}_m\) = magnetic reluctance

Substituting in torque equation:

\[T_e = K_a\phi I_a = K_a \times \frac{N_f I_f}{\mathcal{R}_m} \times I_a\]

Two-Current Torque Expression

\[\boxed{T_e = K_f I_f I_a}\]

where \(K_f = \frac{K_a N_f}{\mathcal{R}_m}\) is a proportionality constant

Summary and Key Equations

Summary of Fundamental Equations

EMF Equations

\[\begin{aligned} E &= \frac{P\phi Zn}{60a} \text{ (V, general form)}\\ E &= K_a\phi\omega_m \text{ (compact form)}\\ E &= K_e\omega_m \text{ (constant flux)} \end{aligned}\]

Torque Equations

\[\begin{aligned} T_e &= K_a\phi I_a \text{ (N·m)}\\ T_e &= K_t I_a \text{ (constant flux)}\\ T_e &= K_f I_f I_a \text{ (two-current form)} \end{aligned}\]

Machine Constants

\[K_a = \frac{PZ}{2\pi a}, \quad K_e = K_t = K \text{ (for constant flux)}\]

Physical Significance of Parameters

Parameter	Symbol	Significance
Number of poles	\(P\)	Flux distribution
Total conductors	\(Z\)	Current-carrying capacity
Parallel paths	\(a\)	Voltage/current rating
Flux per pole	\(\phi\)	Field strength
Machine constant	\(K_a\)	Construction dependent
EMF constant	\(K_e\)	Speed-voltage relation
Torque constant	\(K_t\)	Current-torque relation

Key Insight

For a given DC machine, \(P\), \(Z\), and \(a\) are fixed by construction. Therefore, \(K_a\) is constant. If field flux \(\phi\) is maintained constant, then \(K_e\) and \(K_t\) are also constant.

Typical Parameter Values

Example: Small DC Motor (1 kW, 220 V, 1500 rpm)

Parameter	Typical Value
Number of poles, \(P\)	4
Total conductors, \(Z\)	400–600
Winding type	Lap (usually)
Parallel paths, \(a\)	4 (for lap)
Flux per pole, \(\phi\)	20–30 mWb
EMF constant, \(K_e\)	1.0–1.5 V·s/rad
Rated speed	157 rad/s (1500 rpm)

Note: Actual values vary based on specific machine design and rating.

Practice Problems

Problem 1: A 4-pole DC machine has 600 armature conductors with lap winding. If the flux per pole is 25 mWb and the machine rotates at 1200 rpm, calculate:

The induced EMF
The machine constant \(K_a\)
The EMF constant \(K_e\)

Problem 2: A separately excited DC motor has \(K_e = 1.2\) V·s/rad. When the motor runs at 1000 rpm, what is the induced EMF? If the armature current is 8 A, what is the electromagnetic torque?

Problem 3: Show that \(K_e = K_t\) numerically in SI units by considering power balance in a DC motor.