Question

The back emf constant (Ke) and the rotor resistance (Rm) of a PMDC motor are 30 nV/rad/s and 10 ohm, respectively. The motor is driven by an H-bridge power convereter with 12V power supply. When the H-bridge is controlled using 80% duty cycle the motor runs at 2000 RPM. Due to a change in load torque, the speed is increased to 2400 RPM. What should be the new duty cycle to bring the motor speed back to 2000 RPM?

a). 70%
b). 67%
c). 83%
d). 75%

Answer 1

To determine the new duty cycle required to bring the motor speed back to 2000 RPM, we can use the relationship between motor speed and duty cycle. However, in this case, it is not possible to bring the motor speed back to 2000 RPM with the given scenario.

Step-by-step explanation:

To determine the new duty cycle required to bring the motor speed back to 2000 RPM, we need to use the relationship between motor speed and duty cycle. The speed of a PMDC motor is proportional to the average voltage supplied to the motor, which is determined by the duty cycle of the H-bridge. Since the speed of the motor is directly proportional to the voltage across the motor, we can use the equation:

RPM2/RPM1 = V2/V1

Substituting the given values, we have:

RPM2/2400 = V2/12

Similarly, for the new duty cycle:

RPM2/2000 = V2'/12

By cross-multiplying and solving for V2', we can find the new voltage:

V2' = (RPM2/2000) * 12

Finally, we can find the new duty cycle by dividing V2' by 12 and multiplying by 100:

New duty cycle = (V2'/12) * 100

This can be calculated as:

New duty cycle = ((2400/2000) * 12 / 12) * 100 = 120% * 100 = 12000% = 120%

Therefore, the new duty cycle required to bring the motor speed back to 2000 RPM is 120% which is not within the given options. Since there is no suitable option, we can conclude that it is not possible to bring the motor speed back to 2000 RPM with the given scenario.