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Timers ___________________________________

Applications 01: For the starter control shown below, when the start button S1 is pressed, the line contactor K1 is energized. Contactors K4, K3 and K2 are then energized with a delay of 5 seconds between each to short-circuit their relevant resistor groups. When the last contactor K2, has energized, the slip rings of the armature are short circuited and the motor runs at its rated values. By pressing the stop button S0 or when the protection fuse is tripped, the control is stopped.

Ladder Diagram STL

A "Start" S M 0.0 A( O "Stop" O "Fuse" ) R M 0.0 NOP 0

A M 0.0 AN "K2" AN "K3" AN "K4" S "K1" AN M 0.0 R "K1" NOP 0

A T 4 S "K4" A( O "K3" ON M 0.0 ) R "K4" NOP 0

A T 3 S "K3" A( O "K2" ON M 0.0 ) R "K3" NOP 0

A T 2 S "K2" AN M 0.0 R "K2" NOP 0

A M 0.0 L S5T#5S SS T 4 A( ON M 0.0 O "K3" ) R T 4 NOP 0 NOP 0 NOP 0

A M 0.0 L S5T#10S SS T 3 A( ON M 0.0 O "K2" ) R T 3 NOP 0 NOP 0 NOP 0

A M 0.0 L S5T#15S SS T 2 AN M 0.0 R T 2 NOP 0 NOP 0 NOP 0

FBD

Applications 02: Design a ladder program to solve the following control problem: For the shown liquid level control, when the start button is pressed, the inlet valve is

opened until receiving a high signal from the level switch. After filling the tank the

motor runs for 10 seconds. Then, the outlet valve is opened until receiving a low signal

from the level switch. The filling process is restarted automatically. The stop button

ends the process.

Ladder STL

A I 0.0 S M 0.0 A I 0.1 R M 0.0 NOP 0

A M 0.0 AN I 0.2 S Q 4.0 A( ON M 0.0 O Q 5.0 O I 0.2 ) R Q 4.0 NOP 0

A M 0.0 A I 0.2 L S5T#10S SE T 1 AN M 0.0 R T 1 NOP 0 NOP 0 A T 1 = Q 4.1

A M 0.0 A I 0.2 AN T 1 S Q 5.0 A( O I 0.3 ON M 0.0 ) R Q 5.0 NOP 0