Transcript Ladder Diagram Example - College of Engineering | SIU

et438b-7.pptx
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A manual mixing operation is to be automated using
sequential process control methods. The process
composed of three steps:
a.) filling a tank to a predetermined level
b.) agitating the liquid for 30 minutes
c.) draining the tank for use in another part of
process
Does the ladder logic schematic that follows perform
this function correctly?
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Energized
Press start
open
Tank fills to
limit
Timer goes
To 30 minutes
Solenoid A
energized
tank begins to
fill
open
Timer energized
Motor starts
closed
When tank
drains flow switch
resets. Timer resets
open
Mixer motor off
Solenoid B on
closed
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Let contact state represent a logical value
= Logic 0 Called Form A
Contact
= Logic 1 Called Form B
Contact
Implement AND gate
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grd
potential
output
AB=C
Energized
inputs
Conditions A AND B must be present to energize
output C
Note: all contacts are considered instantaneous
and not held unless modified
With electromechanical relays fan-in and fan-out limited by
number of contacts in relays
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OR
Function
potential
grd
Energized
A + B = C Boolean expression
Either A OR B will cause coil C to be energized
Contacts A, B represent conditions or states in
the sequential process
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NOT Function
Boolean Expression
B=A
Contact of opposite state creates inversion
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Combine the AND function with the NOT function to get a NAND operation.
120 Vac
grd
Energized
open
De-energized
Rung 1 implements the AND function
Rung 2 implements the NOT function
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Any contact
associated with coil
D will change
state like a NAND
TTL gate.
8
Energized
Close
AND
open
NAND
A B C = E and A B C = E
Can add a memory action to the above by including
a feedback from the output coil to the inputs
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Can add a memory action to the above by including
a feedback from the output coil to the inputs
Energized
B and
C are
not
sealed
Close
open
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Energized
The output can not
change unless the
circuit is de-energized.
Close
open
Contact E in rung 2 is a feedback from the output
that makes circuit ignore state changes of A, B and C
after the condition A B C is detected.
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Three-wire control- used for manual and automatic motor starting.
STOP
M1 seals-in the start
PB. Motor stops
when power lost
START
OL1
M1
120 Vac
OL3
OL2
Control wiring
GRD
motor
Thermal overloads
actuate the control
contacts OL1 to
OL3
Motor
Runs
Power
wiring
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OR
A OR B OR C
Energize E
OR
output
NOR
output
Outputs
Notice that Relay logic is similar
to TTL. Can use
Truth tables and Boolean
expressions to do designs
A+B+C = E
A+B+C = E
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Mechanically latched relay - maintains state even when power removed.
Has two coils (operate, reset)
Typical wiring
Operate
Latched
Reset
Latched
Inputs A and B set the output
contacts E and
Close
reset then respectively. This
give toggle action
that “remembers” the last
Typical Applications
input state even when
Reversing Motor starters. Reclose Relay Cut-out power is removed
Close
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E
On
1
2
E1
On
Load1
off
Open
Load 2
off
Press A: continuity
rung 2 Both loads
on
3
4
Close
Energize and re-energize
circuit - Load 2 on
No continuity in rungs 1-4
Continuity in rung 6
Press B: continuity
rung 4 Both loads off
1
5
6
2
Remember all contacts are drawn with the coils
de-energized
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TR-E
On
Open
Close
Load off
Rung 1: when input
A is energized timer
TR-E starts
1
Load on
2
Open
Load off
3
Schematic indicates that this is a on-delay timer. After defined interval
TR-E in rung 2 opens and TR-E in rung 3 closes
Load 1 is deactivated after time delay
Load 2 is activated after time delay
Load 3 is instantaneously deactivated by TR-E
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Loads are toggled between a common point
Typical “Form C” contacts
include both a NO and NC
contact arrangement.
Used in some sensors for
more flexibility
Contact A creates a remote control toggle switch
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Combinational Systems
• Detect patterns of inputs
• Use true tables, Boolean Algebra
• Multiple inputs and/or outputs
• Sum of Products or product of sums Boolean
Implementations
• Reduce to minimum implementation
Sequential Systems
• Follow steps, transition from one step to another.
• Use state transition diagrams or tables with Boolean
Algebra
• State Machine implemented in software or hardware
• Decisions made base on current condition of system and
input information
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Boolean Variables
False =0
True =1
Boolean
Operators
EOR=XOR
Alternate
Implementation
X  A B  AB
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Axioms of Boolean Algebra
Idempotent
AA A
A A  A
Identity
A 0  A
A 1 1
A 0  0
A 1  A
Associative
Distributive
( A  B )  C  A ( B  C ) A ( B  C ) ( A  B )( A  C )
( A  B )  C  A ( B  C )
Complement
A ( B  C ) ( A  B ) ( A  C )
DeMorgan’s
Theorem
Absorption
A A 1
( A  B)  A B
A  A B  A  B
A A  0
( A B)  A  B
A  A B  A
( A) A
Order of Operations
1. NOT
2. AND
3. OR
10
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Example: Simplify the following expression using the axioms of
Boolean Algebra.
X ( A  B  C )  A ( B  C )
X ( A ) ( B  C )  A ( B  C )
Add
Parentheses
Apply DeMorgans’s Theorem to first term
A ( B  C ) ( A ) ( B  C )
X  A ( B  C )  A ( B  C )
X  A ( B  C )  A ( B  C )
X  A B  A  C  A B  A  C
Collect common terms and factor
Apply
DeMorgan’s
Here
( B  C ) ( B  C )
Expand
Expressions
C ( A  A )  A  C  A  C
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Example Continued
X  A  B  A  B  C ( A  A )
A  A 1
X  A  B  A  B  C 1
C 1  C
Use
Complement
Axiom
Use Identity
Axiom
X  A B  A B  C
Simplified Expression
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1.) Obtain description of process
2.) Define control action
3.) Define Inputs and Outputs
4.) Develop Truth Table or Boolean Equation of Process
Process control description
A heating oven with two bays can heat one ingot in each bay.
When the heater is on it provides enough heat for two ingots. If
only one ingot is present, the oven may overheat so a fan is used
to cool the oven when it exceeds a set temperature.
Control Action
When only one ingot is in the oven and the temperature exceeds
the setpoint, turn on the fan
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Define I/O variables
Create Truth Table
T
B2 B1
Inputs: B1 = bay1 ingot present
B2 = bay2 ingot present
T = temperature sensor
Output: F= fan start
F
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1
If there is no over temperature
don’t start the fan
Over temperature in empty oven: safety fan start
Start fan in lightly load ovens with over temp.
Over temperature in full oven: safety fan start
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Select elements from truth table in SOP (sum-ofproducts) form then simplify.
T
B2 B1
F
F  T  B1 B2  T  B1 B2  T  B1 B2  T  B1 B2
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
1
1
0
1
1
F  T  (B2  (B1 B1)  B2  (B1 B1))
1
1
0
1
1
1
1
1
F  T  (B2  B2)
F  T  (B1 B2  B1 B2  B1 B2  B1 B2)
FT
Requires only Temp
control
Ignore unloaded and full load cases and try again
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F  T  B1 B2  T  B1 B2
Revised Truth Table
T
B2 B1
F
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
0
1
0
1
1
1
1
0
1
1
1
1
F  T ( B1 B2  B1 B2 )
Ladder Logic Representation
B1
B2
T
0
B1
et438b-7.pptx
F
B2
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Avoid multiple complemented variables in
ladder logic (No NAND, NOR)
NOR
NAND
X  A B  A  B
X  A  B  A B
NAND/NOR can not be implemented effectively using software.
(Programmable Logic Controllers)
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Definitions
State - current operational mode of system
Examples: On/Off, Idle, Tank filling, dispensing product.
Conditions (inputs) - inputs required for leaving the
current state and moving to another state
Examples: Coins inserted, button pressed, OL activated
Actions (outputs) - actions performed by system when
the transition from one state to another take place
Examples: Start motor, turn on light, sound alarm.
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When a set of inputs (conditions) become valid for
leaving a state, the system is directed to the
destination state
Current
State
State exit
input
conditions
To other
states
Next
State
State
outputs
State
entry input
conditions
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State transition diagrams allow designers to examine the
interaction between desired conditions and find their logical
relationships and sequence. Use in digital computer design
Else
Else
B
Else
State
1
A
State
2
If C true go to State 1
Else State 3
C
If Condition A true go to State 2
Else stay in State 1
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State
3
If B true go to State 3
Else State 2
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Informal:
State
n
.
.
State
j
.
.
.
State
2
State
1
State X =(State X +Arrival from another state) and has not left
for another state
setn
seti
Re-seti,= logical condition to
reset state variable i and leave
State
m
resetm
State
i
set2
.
.
State
k
reseti
.
.
.
reset2
set1
State
2
reset1
Seti =logical condition to
set state variable i and enter state
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State
1
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Set
Conditions
Functions of
state and inputs
Formal Definition:
n

 m
statei   statei   set(statej , I)i    reset(statei , I) k 
j1

 k 1
1
outi  h i (state1, state2 ,...stateN )
Where:
statei = a variable that reflects state i is on
statei+1 = next value of state variable
outi = desired outputs of state i
hi( ) = output function of state variables
n = number of transitions into state i
m = number of transitions out of state i
N = total number of system states
seti= logical condition to set state variable i
reseti = logical condition to reset state variable i
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Reset
Conditions
Functions of
state and
inputs
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Write the state equation for a motor starting control described in the state
diagram below with the following input and outputs
S1
(Start)
X1=1
I1
S0
(Stop)
X1=0
I0=pressed stop button (PB1)
I1= pressed start button (PB2)
I2 = motor overload condition (OL)
O1 = start motor (M)
I0  I 2
Only 1 state variable required
for two conditions X1=0 or X1=1
reset1  I0  I2
set1  I1

X11  X1 set1   reset1


X1  X1  I1  I0  I2
1

Output equation
M  O1  X1
X11  X1  I1  I0  I2
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Boolean Equation to ladder logic diagram
X11  X1  I1  I0  I2
M (M  PB2)  PB1 OL
Construct
Ladder
Substitute variable names
X1  M
I0  PB1 Stop
I1  PB2 Start
I2  OL Overload
START
PB2
STOP
PB1
M
OL
M
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A work piece must travel back and forth on a conveyor. The location
of the work piece is determined by two limit switches. When the
location is detected control signal are sent to a reversing motor
contactor. The machine is started and stopped from a local set of
push button switches. Develop a ladder logic diagram to implement
this control.
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Determine the inputs, outputs and states of system
Inputs:
I0: press start
I1: press stop
I2: Table at reverse limit (1LS)
I3: Table at forward limit (2LS)
Outputs:
I3
I2
O0
I0
I1
O1
et438b-7.pptx
O0: Start motor forward (2CR)
O1: Start motor reverse (3CR)
States:
S0: off
S1: on-forward,
S2: on reverse,
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Assume machine starts at reverse limit. (1LS changes state)
T1, 2  I3
T0,1  I0
S1
(onforward)
T1,0  I1
T2,1  I2
S2
(onreverse)
S0
(Stop)
I0: press start
I1: press stop
I2: Table at reverse limit (1LS)
I3: Table at forward limit (2LS)
O0 start forward
action
O1 start reverse
action
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Define set and reset conditions
set X1  I0  I2  X2
Define 2 state variables X1 and X2
resetX1  I1  I3
X2
X1
Condition
0
0
Off (S0)
0
1
On-Forward (S1)
1
0
On-Reverse (S2)
1
1
Not allowed
T0,1  I0
X11  (X1  set x1 )(resetx1 )
T1, 2  I3
S1
(onforward)
T2,1  I2
T1,0  I1
S0
(Stop)
set X 2  I3  X1
resetX 2  I2
X1=1
X2=0
X11  (X1  (I0  I2  X 2))(I1  I3)
X11  (X1  (I0  I2  X 2))(I1 I3)
X 2 1  (X 2  set x 2 )(resetx 2 )
S2
(onreverse)
X 2 1  (X 2  I3  X1)(I2)
O0
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X1=0
X2=1
O1
Outputs X1  O0, X2  O1
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Convert state equations into ladder diagram
2CR 1  (2CR  Start  1LS 3CR)(Stop 2LS)
3CR 1  (3CR  2LS 2CR)(1LS)
1LS
2CR = O0
3CR =O1
I0=start
I1=stop
I2=1LS
I3=2LS
3CR
STOP
2LS
START
2CR
2CR
1LS
2LS
2CR
3CR
3CR
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Systems with multiple entries and exits from a state require blocking of
Alternatives.
D
B
S0
S2
Two Choices
IF A THEN block C
IF C THEN block A
A
S1
C A given priority to C
S01  (S0  A  S1)  (B  S0)
S11  (S1  (B  S0  D  S2))  (A  S1  C  S1)
A or C can occur independently to exit S1.
Must give one transition priority over other.
Block setting of conflicting state
S21  (S2  C  S1 A)  (D  S2)
C over A
S0 1  (S0  A  S1 C)  (B  S0)
S2 1  (S2  C  S1)  (D  S2)
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FS
A(C  D)
S1
S0
FE
Inputs
A
B
C
D
E
F
FS
Outputs
P
Q
R
BA
E(C  D  F)
Write state equations
for this system. Give state
S2 priority over S0
S2
T01  (A(C  D))  S0
T10  (F  E)  S1
Output Map
State P
Q
R
T1  FS
S0
0
1
1
T12  B  A  S1
S1
1
0
1
T21  (E(C  D  F))  S2
S2
1
1
0
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Write state equations using transitions
S01  (S0  T10  T12 )  T21
S11  (S1 T01  T21  T1 )  (T10  T12 )
S0 blocked if
S2 is active
S11  (S1 T01  T21  T1 )  (T10  T12 ) Simplify using
S21  (S2  T12 )  T21
DeMorgam’s Theorem
Output Equations
Output Map
State P
Q
R
S0
0
1
1
S1
1
0
1
S2
1
1
0
P  S1  S2
Q  S0  S2
R  S0  S1
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