Question

A motor with KT = 1.657 in-oz/A, Ke = 1.23 V/krpm and terminal resistance R = 20Ω is driven at 12 V. Assume VPWM = duty cycle*Vdd. a) How fast will the motor spin under a constant 0.15 in-oz load if it is driven by PWM with a 25% duty cycle and a frequency well above the motor’s time constants? b) What is the speed if the duty cycle is increased to 90%?

Answer 1

Answer:

(a) 9607 rpm

(b) 7308 rpm

Solution:

As per the question:

$K_{T} = 1.657\ in-oz/A$

$K_{E} = 1.23\ V/ krpm$

R = $20\Omega$

$V_{dd} = 12\V$

From the question:

V = $D\times V_{dd}$

$V = 12\times 0.25 = 3 V$

$Constant\ Load, T = 0.15\ in-oz$

Duty cycle, D = 25% = 0.25

(a) Angular speed of the motor, $\omega$ can be calculated as:

EMF of the motor is given by:

E = V - IR

where

Also,

$E = k_{E}\omega$

$\omega = \frac{E}{K_{E}} = \frac{V - IR}{K_{E}}$                (1)

Now,

Torque, T = $K_{T}I$

Thus

$I = \frac{T}{K_{T}}$                 (2)

From eqn (1) and (2)

$\omega = \frac{E}{K_{E}} = \frac{V}{K_{E}} - \frac{TR}{K_{E}K_{T}}$

V = $D\times V_{dd}$

V = $0.25\times 12 = 3\ V$

$\omega = \frac{3}{1.23} - \frac{0.15\times 20}{1.23\times 1.657} = 0.967\ krpm = 967\rpm$

(b) Speed of the motor when D = 90% = 0.9

V = $D\times V_{dd}$

V = $0.9\times 12 = 10.8\ V$

$\omega =  \frac{V}{K_{E}} - \frac{TR}{K_{E}K_{T}}$

$\omega = \frac{10.8}{1.23} - \frac{0.15\times 20}{1.23\times 1.657} = 7.308\ krpm = 7308\ rpm$