Forward-Reverse Motor Simulation in TLP LogixPro
Problem:
A box is at the West end of the conveyor
A start switch is pressed
The box moves to the East end.
The box moves back to the West end.
The box moves to the East end.
….
In other words, the box is at the West end. Once the start switch is pressed, the box moves back and
forth between two limiting positions. Assume that the conveyor is activated by a motor. When the
switch is pressed, the motor rotates causing the conveyor to move one way. Then it reverses direction
to bring the box back to its starting position and the cycle continues until the Stop switch is pressed.
Assume that the limiting positions are signaled by two normally open limit switches and the start switch
is of the momentary, normally open type. The stop switch is momentary, normally closed.
Implement the solution using the LogixPro simulator.
Inputs:
I:1/0,
I:1/1,
I:1/4,
I:1/5,
West-end limit switch
East-end limit switch
Start switch
Stop switch
Outputs:
O:2/0, Forward direction output
O:2/1, Reverse direction output
Auxiliary logic inputs and outputs used to counteract the simulator limitations:
O:2/2
O:2/3
O:2/4
Logic equation:
O:2/0 = (I:1/4)(I:1/1not) + (O:2/0)(I:1/1not) + (O:2/1not)(I:1/0)
O:2/1 = I:1/1 + (O:2/1)(I:1/0not)
In words:
O:2/0
is true (motor runs forward) when
(I:1/4)(I:1/1not) +
The start switch is open and the East limit switch is not closed
(remember that the East limit switch is NO. The negated function shows continuity when the switch is
open), or when
(O:2/0)(I:1/1not) +
The motor is running forward and the East limit switch is not closed
(hold function), or when
(O:2/1not)(I:1/0)
The motor is not running in reverse and the West limit switch is closed
(reset function).
The East limit switch had to be included in all the elements to prevent reverse action when moving
forward.
O:2/1
is true (motor runs reverse) when
I:1/1 +
The West limit switch is open (Remember it is NO. We want reverse
operation only when the switch is in its true, open state.), or when
(O:2/1)(I:1/0not)
The motor is running in reverse and the West limit switch is open. (The
negated function provides the needed continuity in the rung).
Rung 1 is the implementation of the first two members of the forward equation.
Rungs 2 and 3, combined are the DeMorganized representation of the third member of the forward
equation, ored to rung 1. Their output, O:2/4 is applied to rung zero.
If we call rung 1 = A and the third element = B then A + B = not(Anot Bnot). Rung 1 is A, Rung 3 is B, rung
2 is (Anot)(Bnot), and O:24 in rung 1 is not(Anot Bnot).
Rung 4 is the implementation of the reverse equation.
Sequence of operations:
1.
2.
3.
4.
5.
6.
I:1/0 is closed
I:1/1 is open
Start
O:2/0 closes (Motor is turning in forward direction)
Open I:1/0 to indicate that the box is departing the West end (The NO switch opens)
Close I:1/1 to indicate box arriving East end
6a. O:2/0 opens
6b. O:2/1 closes (Motor is turning in reverse direction)
7. Open I:1/1 (Box departs East end)
8. Close I:1/0 (Box arrived West end)
8a. O:2/1 opens (Motor stops turning in reverse)
8b. O:2/0 closes (Motor starts turning forward)
9. Go to 5
Note: The stop switch was omitted to simplify the development.