AND OPERATION IN PNEUMATIC APPLICATION

Problem Description:

 Operation of the Double Acting Cylinder by using Two Buttons B1 and B2. When both B1 and B2 both are pressed then Cylinder will move forward and get retracts when any one of them releases.

 Pneumatic Diagram

In this Pneumatic Diagram We have used one Double Acting Cylinder, one 5/2 Pilot operated valve and two 3/2 Push Button valve with spring return.

 

Circuit Description

 

Here One Double Acting Cylinder is used with 5/2 pilot operated valve with spring return i.e. when in absence of pressure at pilot point it will automatically return to home position. Input at pilot point comes from the output of one of the 3/2 Push Button valves and input of that Push Button valve is the output of another 3/2 Push Button valve whose input relates to the pressure line. In other words, Both 3/2 Push Button valves are connected in series and output is given to a 5/2 pilot operated valve.

 

Here all valves are of Normally close (NC) type so, it won’t allow pressure to pass through the valve when they are in rest condition, or we can say it as a home position.

Working

 Case 1: PB1 is pressed.

When PB1 is pressed then as it is of NC type it will allow pressure to pass through it but at valve 2 where PB2 isn’t pressed pressure will not be able to pass through it so at pilot valve there is absence of pressure and therefore, Cylinder won’t move from it home position.

Case 2: PB2 is pressed.

When PB2 is pressed then the valve will get open and allow pressure to flow through it but as PB1 hasn’t pressed so there is absence of pressure at input of the second valve as a result there is absence of pressure at the pilot valve of 5/2. So, Cylinder will not move in this case also.

 Case 3: Both are pressed.

When both switches get pressed, our pilot valve is directly connected with the pressure line as both are of NC type so the cylinder will move in forward direction and when any of them get released then the cylinder will get retracted. 

 

This whole operation is similar to AND gate operation in Digital Electronics whose truth table is given below.

Note: - Above application may be different from actual application. This example is only for explanation purpose only. We can use this concept in other examples also. All parameters and graphical representations considered in this example are for explanation purpose only, parameters or representation may be different in actual applications. Also, all interlocks are not considered in the application.

                                                                                     Written by Sneh jain

Double acting pneumatic cylinder operation using 5/2 direction control valve and PLC

 Problem Statement: -

A heavy door is to be opened or closed by a double-acting cylinder. This door will open or close with two push buttons, one push button is located inside and another push button is located outside. Draw a Ladder Diagram for the Giving Condition.

a) The push button should only work when the door is either fully open or fully closed.

Solution: -

Number of inputs = 4

PB1: Inside push button

PB2: Outside push button

L1:  Limit switch located at closing position.

L2:  Limit switch located at open position.

 

Number of outputs = 2

Y1: Left Solenoid of 5/2 control valve

Y2: Right Solenoid of 5/2 control valve

Pneumatic Circuit Diagram:

Answer - (a): -

 Number of inputs = 4

Possibilities = 2⁴ =16

Description: -

Initially, as the door is fully closed the L1 limit switch is pressed (L1=1 & L2=0), the door will only open when any one of the push buttons is pressed (PB1 =1 & PB2=0 OR PB1=0 & PB2=1)

It is not possible that both the Limit switches L1 & L2 are pressed at the same time   (L1=1 & L2=1).

Both the switches are unpressed (L1=0 & L2=0), which means the door is neither fully opened nor fully closed

As the door is fully closed the L1 limit switch is pressed (L1=0 & L2=1), the door will only open when any one of the push buttons is pressed (PB1 =1 & PB2=0 OR PB1=0 & PB2=1).

Hence, Y1= (L1) (L2)’ (PB1)’ (PB2) + (L1) (L2)’ (PB1) (PB2)’

               = (L1) (L2)’ [(PB1)’ (PB2) + (PB1) (PB2)’]

 

             Y2= (L1)’ (L2) (PB1)’ (PB2) + (L1)’ (L2) (PB1) (PB2)’

               = (L1)’ (L2) [(PB1)’ (PB2) + (PB1) (PB2)’]

 

Note: - Above application may be different from actual application. This example is only for explanation purpose only. We can use this concept in other examples also. All parameters and graphical representations considered in this example are for explanation purpose only, parameters or representation may be different in actual applications. Also, all interlocks are not considered in the application.

                                                                                            Written by Subham Prajapati