Week-7 Challenge: DC Motor Control

1 . Run MATLAB demo ‘Speed control of a DC motor using BJT H-bridge’. Modify the model such that armature current doesn’t shoot up when motor changes direction from forward to reverse.

The block set of the speed control of a DC motor using BJT H-bridge

• The DC motor is of 5HP, 240V, 1750rpm
• 300V is source voltage provied to the DC motor
• The IGBT and diodes are used as the power switching devices
• The IGBT turns on when the collector-emitter voltage is positive and greater than Vf and a positive signal is applied at the gate input
• The input signels are generated with the help of pulse generater is and the and the armature voltage circuit

The pulse generater :

• The pulse generater is of 500Hz
• The armature voltage signal is 0 for the first half of time and 1 for the rest half
• The Q1 and Q4 are operative for the first half the time (0-0.5 s) , Q2 and Q3 are operative for the next half (0.5- 1 s).The green signal is of acclerating and the red signal is of decclerating.
• The duty cycle i,e pulse width is of 75% means the signal is ON for the 75% of time and OFF for the next 25% of time.

Outputs :

PLOT 1:

The scope is connected to the Q3 IGBT

.

for Q3 and Q4 the input will be provided after 0.5 sec so the IGBT signal prenent between 0.5 - 1 sec .Where as in the diode current passes only if the current passes in the flow of circuit ,if the current passes in reverse direction the signal cant enter the diode as shown in 2nd subplot

PLOT 2:

• In the Q1 the motor acclerates till 1000rpm, due to load condition it takes 0.1 sec to acclerate . in Q3 the motor decelerate to -1000rpm.
• The armature current initly increses suddenly due to the load condition and later moves with same IA because of constant speed.
• load on the motor dur to the speed.

To reduce shoot up of armature current when motor changes direction from forward to reverse.

• By reducing the duty cycle to 50% we can reduce the shoot up of the armature current in reverse flow .
• By putting the armature current , final value near to 0 (like 0.4). the armature current dosent shoot up in reverse flow

3. Make a suitable EV model using DC7 block and make result report.

The DC7 four quadrant chopper block set :

The three phase source supply with 380V and 50Hz, The motor torque equation is as given in the week 6 challange i,e Tm=24.7+0.0051w^2

The speed is generated with the help of the signal generater and in 2 steps initally 500 and -1184

Results:

• The four quadrent chopper is simplar to H bridge contains Q1 Q2 Q3 Q4 quadreants , Q1 Q4 are in forwars cycle and Q2 Q3 are working at the reverce cycle .

• The armature current is required to acclerate in the first 2 sec,after that due to decelerating the current consumption is in negative side and at the time near to 6 sec the motor comes to rest the armature current come to rest
• The motor speed is also acclerates and decelerate slowly due to its load conditions as shown in graph.

2 . Refer to help section of ‘The Four-Quadrant Chopper DC Drive (DC7) block’. Compare it with H-bridge model.

• In H-bridge model the duty cycle can be varied where as in DC7 block we cant.
• PI controllers are used in the DC7 bloskset but its not in the H-bridge.
• The H-bridge model is made with the help of IGBT blocks and diodes where as in DC7 'Four-Quadrant Chopper DC Drive(DC7)' simscape block is used .
• The pulse generator is used to generate the signal to operate the IGBT blocks in the H-bridgebut in the DC7 its self operates on the speed limit.
• The load torque inputted is calculated as proportional to the square of the speed in H-bridege but in DC7 load torque is proportional to the speed.the load torque inputted is calculated as proportional to the square of the speed