CHAPTER 8 TROUBLESHOOTING. Mitsubishi Electronics LY20S6, LY40NT5P, LY18R2A, LY40PT5P, LX40C6, LY10R2, LY41NT1P, LX28, LY42PT1P, LX41C4

CHAPTER 8 TROUBLESHOOTING

CHAPTER 8

TROUBLESHOOTING

8.1

Troubleshooting for Input Circuit

Cause

(1) An input signal does not turn off.

(a) Case 1

Case

• The leakage current of an input switch occurred. (e.g. drive by a contactless switch)

AC input

C

Leakage current

Input module

Action

Power supply

• Connect an appropriate resistor so that the voltage across the terminals of the input module is lower than the OFF voltage.

AC input

C

R

Input module

Cause

Action

Power supply

For the CR constant, 0.1 to 0.47µF + 47 to 120

 (1/2W) is recommended.

(b) Case 2

Case

• The leakage current of an input switch occurred. (e.g. drive by a limit switch with neon lamp)

AC input

Leakage current

Input module

Power supply

Take either of following actions.

• Connect an appropriate resistor so that the voltage across the terminals of the input module is lower than the OFF voltage (same action as the case 1).

• Separately configure a display circuit that is independent from the existing circuit.

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71

Cause

(c) Case 3

Case

• A leakage current occurred due to the line capacity of a wiring cable. (The line capacity C of a twisted pair cable is approx. 100pF/m.)

AC input

Leakage current

Input module

Power supply

• Connect an appropriate resistor so that the voltage across the terminals of the input module is lower than the OFF voltage (same action as the case 1). Note that a leakage current does not occur if the power supply is located in the side where an input equipment is connected as shown below.

AC input

Action

Input module

Cause

Action

Power supply

(d) Case 4

Case

A current exceeding the off current of the module leaks even after a switch with LED indicator is turned off.

Connect an appropriate resistor so that a current through the module may become lower than the off current.

LX40C6

2.82mA

Iz=2.0mA

Input impedance

3.8k

R

I

R

=0.82mA

24VDC

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CHAPTER 8 TROUBLESHOOTING

Calculation example

Case

The resistance value of a connected resistor is calculated by the following formula.

Ex.

A switch with LED indicator that generates a current leakage of 2.82mA when 24VDC is supplied is connected to the LX40C6.

Check the following with the specifications of the module.

• Off current: 2.0mA

• Input resistance. 3.8k



I(Leakage current)=I

Z

(Off current of the LX40C6)+I

R

(Current flowing to connected resistor)

I

R

=I-I

Z

=2.82-2.0=0.82[mA]

To hold the current leakage through the LX40C6 equal to or lower than the off current (2.0mA), connect a resistor so that 0.82mA or more current flows to the resistor. Calculate the resistance value (R) of the connected resistor as follows.

I

R

: I

Z

=Z(Input impedance): R

I

Z

R <

I

R

Z(Input impedance) =

2.0

0.82

3.8=9.27[k ]

The resistance value R < 9.27k must be met.

<Checking a connected resistor by calculating the power capacity.>

When the resistor (R) is 8.2k

, for example, the power capacity (W) of the resistor (R) is calculated as follows.

W=

(Input voltage)

2

=

R

28.8

2

8200

=0.101[W]

Since the resistor requires the power capacity of 3 to 5 times as large as the actual power consumption, the resistor connected to the terminal should be 8.2k

 and 1/3 to 1/2 W.

Off voltage when the resistance (R) is connected is calculated as follows.

1

+

1

3.8[k ]

2.82[mA]=7.32[V]

1

8.2[k ]

This meets the condition: less than or equal to the off voltage of the LX40C6, 8V.

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73

Cause

Action

(e) Case 5

Case

By using two power supplies, a sneak path has been configured.

Input module

Lamp

E2 E1

• Use one power supply.

• To prevent the sneak path, connect a diode as shown below.

DC input

Input module

Lamp

E2

E1>E2

E1

Diode

(2) An input signal does not turn on. (AC input module)

Case

Stepwise distortions as shown below appear to the zero cross voltage of an input signal (AC).

Cause

Zero cross voltage

Action Improve the input signal waveform by using online UPS.

Cause

Action

(3) A signal incorrectly inputs data.

Case

Noise has been taken as input data.

Set the input response time longer. (

Page 68, Section 7.1)

Ex.

1ms

 5ms

If this action is not effective, also take the following two measures.

• To prevent excessive noise, avoid installing power cables together with I/O cables.

• Connect surge absorbers to noise-generating devices such as relays and conductors using the same power supply or take other noise reduction measures.

If excessive noise is periodically generated, setting the response time shorter may be effective.

Ex.

70ms

 20ms

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CHAPTER 8 TROUBLESHOOTING

8.2

Troubleshooting for Output Circuit

Cause

(1) When the output is off, excessive voltage is applied to the load (triac output).

Case

• The load is half-wave rectified internally. (In some cases this is true of solenoids.)

TRIAC output module

D1

[1]

Load

[2]

• When the polarity of the power supply is as shown in [1], C is charged. When the polarity is as shown in [2], the voltage charged in C plus the line voltage are applied across D1. Maximum voltage is approx. 2.2E. (If a resistor is used in this way, it does not pose a problem to the output element. But it may cause the diode, which is built into the load, to deteriorate, resulting in a fire, etc.)

• Connect a resistor (several tens to hundreds of k

) across the load.

Resistor

Action

Load

Cause

(2) The load does not turn off (triac output).

(a) Case 1

Case

• A leakage current occurred due to a built-in surge suppressor.

TRIAC output module

Load

Leakage current

8

Action

• Connect a resistor across the load. (Note that a leakage current may occur due to the line capacity when the wiring distance from the output module to the load is long.)

Resistor

Load

75

Cause

(b) Case 2

Case

• The load current is lower than the minimum load current of the output module.

Surge suppressor

TRIAC output module

Phototriac

Control circuit

Load

Action

Triac

• In such a case, the load current flows into a phototriac as shown above because the triac does not operate. If an inductive load is connected with the load current flowing into a phototriac, the load may not turn off because the surge at the time of off is applied to the phototriac.

• Connect a resistor across the load so that the load current is equal to or higher than the minimum load current.

Resistor

Load

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CHAPTER 8 TROUBLESHOOTING

Cause

(3) A load momentarily turns on when powering on the external power supply.

Case

An incorrect output occurs due to floating capacitance(C) between collector and emitter of photocoupler.

When a high sensitivity load (such as solid state relay) is used, this incorrect output may occur.

Output module

Photocoupler

C

Ic

Tr1

Y0

Output Y0

Approx.100 s

Load

Action

Constant-voltage circuit

12/24VDC

COM

SW

24VDC

SW: External

power supply

(24VDC) at ON

10ms or less

When the rise time of voltage of the external power supply is 10ms or less, current (Ic) flows to gate of transistor (Tr1) of next stage due to floating capacitance (C) between collector and emitter of photocoupler. Then, output Y0 turns on for approx. 100µs.

Action 1: Check that the rise time of the external power supply is 10ms or more. And then, install a switch (SW1) for turning on or off external power supply to the primary side of it.

Primary side

SW1

Secondary side

Output module

External power supply

Programmable controller

External power supply terminal

Action 2: When installing the SW1 to the secondary side of it is required, make the rise time to 10ms or more and connect a capacitor and resistor as shown below.

For the following source output modules, take Action 1 on the above due to no effect of Action 2 by the characteristics of the external power supply circuit.

• LY40PT5P

• LY41PT1P

• LY42PT1P

• LH42C4PT1P

Sink type output

Y0

12/24VDC

COM

C1

R1

24VDC

Load

R1: Several tens of ohms

Power capacity ≥ (external power supply current*

1

)

2

× resistance value × (3 to 5)

*2

C1: Several hundreds of microfarads 50V

SW

Example R1 = 40Ω, C1 = 300μF

Time constant = C1 × R1 = 300 × 10

-6

× 40

= 12 × 10

-3

[s]

= 12 [ms]

*1 Check the consumption current of the external power supply for modules used.

*2 Select the power capacity of resistance to be 3 to 5 times lager than the actual power consumption.

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77

Cause

(4) A load momentarily turns on from off when the system is powered off

(transistor output).

Case

When an inductive load is connected, 2) Load may turn on from off due to a diversion of back EMF at 1) Shutoff.

Transistor output module

Sink type output

ON

Y0

Back EMF

Load

3)

Transistor output module

Source type output

Y0

ON

Back EMF

Load

3)

OFF

Y1

2)

Load

OFF

Y1

2)

Load

1) Shut off

12/24VDC

COM

1) Shut off COM

0V

Take one of two actions shown below.

Action 1: To suppress the back EMF, connect a diode to 3) parallel to the load where back EMF is generated.

Sink type output

Back EMF

Load

Source type output

Back EMF

Load

Action

Action 2: Configure another current path by connecting a diode across positive and negative of the external power supply. When taking the action described in "A load momentarily turns on when powering on the external power supply" ( connect a diode parallel to C1 and R1.

Page 77, Section 8.2 (3)) at a time,

For the following source output modules, take Action 1 on the above due to no effect of Action 2 by the characteristics of the external power supply circuit.

• LY40PT5P

• LY41PT1P

• LY42PT1P

• LH42C4PT1P

Sink type output

ON

OFF

Y0

Y1

Back EMF

Load

3)

2)

Load

D1 is in the following status.

Reverse voltage VR (VRM) Approximately 10times higher than

the rated voltage in the specifications

Example 24 VDC Approximately 200V

Forward current IF (IFM) More than twice as much as

the maximum load current (common)

in the specifications

Example 2A/1 common 4A or more

12/24VDC

C1

R1

D1

1) Shut off

COM

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CHAPTER 8 TROUBLESHOOTING

Cause

Action

(5) The load operates due to powering on the external power supply (transistor output).

Case

• The polarity of the external power supply is connected in reverse.

Transistor output module

Y0

Load

Incorrect

External power supply

Correct

COM

Output element protection diode

• When the polarity is connected in reverse, current may flow across an output element protection diode.

Connect the polarity correctly.

Cause

Action

(6) The load operates by incorrect input due to chattering of the external power supply.

Case

The device whose input response speed is too fast is connected to the contact output module.

Use a transistor output module.

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79

Cause

(7) When an output is turned on, a load connected to the other output is also turned on (transistor output (source type)).

Case

If the wire connecting 0V of an external power supply and a common of a load is cut off or disconnected, a current flows to the load that is off due to a parasitic circuit of the output element that is off.

ON

Transistor output module

Source output

Output element

Output control circuit

Y0

Load

OFF

Output element

Output control circuit

Y1 Load

COM

0V

24V

Switch off or disconnection

Action

If a current keeps flowing under the above condition, a failure may occur.

Connect the external power supply and loads correctly.

To prevent the condition described above, connect a diode to each output terminal as shown below.

Source output

Y0

Load

Y1

Load

COM

0V

24V

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