maximum distance for digital input signal

i am having a pull cord switch far from the PLC by X meters, i want to check if the signal will reach the PLC input module
I am using 24 VDC

From allen bradley input module, i got the below information:

off-state current max = 1.5mA

on state current min = 2mA at 10VDC
on state current max = 10mA at 31.2 VDC

so my understanding is that since i am using 24 VDC, so that the maximum distance will be calculated as below:
for 1.5mm2 cross section cable, i cable resitance/distance = 12.8 ohm/km

cable resitance = x(km) * 12.8 ohm/km.

so the maximum impedance that i can reach is 5Kohm (10 VDC/2mA)...

however at this, cable distance will be very far so it is not logical, i guess noise should be taken into consideration in this case.

what is the best practice for getting the maximum distance that can be reached between a pull cord switch and a PLC input module.

thanks!

The short answer is in a book I just read. The US president in Washington DC opened the Omaha Nebraska World's Exposition in 1898 by flipping a switch. Needless to say no radio telemetry was involved. They used hard wired telegraphs signals, undoubtedly with repeaters or relays at some points.

If the voltage drop is right, i.e. the wire is big enough, you can go any distance. Noise is a concern, but only if other power wires share the conduit. Noise is not usually as much of an issue on discrete as it is on analog signals, though.

Don't forget the wire is twice the physical distance - you have to get out and back!

dear Tom,

thank you.

so if i am assume that i need a voltage drop of 5%, and in order to reach 10mA (which i believe is necessary to get a clean contact) so the maximum distance it can be reached is 10 km (back and forth) which is 5km maximum of cable distance.

What do u think?

I think you could get more if you increased your power supply to 32 VDC.

your calculation is working because the result remain around a working range but it is wrong.
First you are using 24v but your calculation is for 10???why?
Then you forget to look for the voltage required to trigger on the plc input...In your voltage drop calcul, you need to keep a minimum remaining at the plc terminal...
So your calculation should be
(Power supply voltage - plc minimum input on voltage) / worst case Ampsin the circuit(0.010)
For a Siemens that switch on at 5v:
24v-5v =19V/0.010 = 1900ohms
But i would keep a safety factor and use 20v
20-5 = 15v /0.010 = 1500ohms

then your cable is 12.8ohm/km so 1500/12.8 = 148km

In fact you could reduce your cable size a little and remain ok as your long distance is probably not longer than few km ?

24vdc and long run is a problem when you have to switch relay coil or contactor with an inrush current to magnetize. This inrush could be near of 1 amp for small relay so in this case you would be many more limited in lenght as the relay need that current and an higher voltage to break in the inertia.
to keep 20v at 1amp at the relay terminals:
24-20 = 4v/1 = 4 ohms in this case 4ohms/12.8= 312meters without any safety factor...


for noise...i'm not sure but usually 24v signal cable are run far from power cable and a a 10ma noise above 5v would be an heavy one?
On the other hand a safety pull cord is probably fail safe and remaining on in normal operation meaning the noise would probably increase the voltage and would be of any effect on the ''1'' reading of the plc.
If you are still afraid of it, you could filter the input at the plc: increase load of the signal by installing a parrallele resistor to drain any possible noise comming in(Sometimes a small capacitor also that react in time)
So if you redo all the maths, a 250ohm would bring your cable lenght to 15km

in real, you will have to take into account other resistance point like multiple connection in junction box, terminals and input card terminals + the pull cord closed contact resistance itself. all of them are usually below 1 ohm but this is why you need a safety factor.

since you said AB input module i just take a look to a compact 1769-iq16 to compare vs my calculation
With this module, you need at least 10v to trigger on this reduce your lenght and current needed is 115mA that also reduce lenght but increase noise immunity...
get back to numbers:
24v - safety factor ov 4v = 20v
20v-10=10v
10v/0.115 = 87ohms
87ohms minus some connection resistance = 70ohms
70ohms/12.8ohms = still 5468 meters (2700m long for back and forth)

Increasing powersupply voltage to 30v (To remain ok for the 30v limit of the card)
26-10 = 16v / 0.115 = 139 ohms
128 ohms/12.8= 10km (5+5)

Use N.C. contact on pullswitch

and let the PLC input card tell you if the signal is robust.

I am reading between the lines and surmising that you are adding a pull-switch to the far end of a large system.

This pull-switch is important to you and the application, otherwise you would not post this query.

I hope you are not "programming" and e-stop style event. If so, you need to asses dual-channel hard wired installation, and not simple PLC input.

If the signal in not safety in nature, you are still adding it for a purpose that you can trust.

Simple. Make the installation using the N.C. on the switch.
If there is signal loss, you program will respond.

$0.02

Dear guys,

thank you for your valuable input, special thanks to Jeff23spl

Dear,

i was looking at the attached datasheet,
it says switching voltage of 5V, and switching current 5mA. I dont believe we can get the contact resistance from this (1Kohm)... it is not logical.

what do you think?

Charbel said:

Dear,

i was looking at the attached datasheet,
it says MINIMUM switching voltage of 5V, and MINIMUM switching current 5mA. I dont ...

what do you think?

Click to expand...

/sigh

Dear rdrast,

thanks for your answer

however, my question was not clear, i will clarify it below:

as far as i know the contact resistance should be lower than 1ohm.

what are the minimum switching current and minimum switching voltage used for?

thank you!

What are the minimum switching current and minimum switching voltage used for?

Click to expand...

Those values are used to determine if there is enough power at the end of a wire to cause the device to switch from off to on. If the voltage and current are below the minimum for switching, then there will be no switching.

Lancie1 said:

Those values are used to determine if there is enough power at the end of a wire to cause the device to switch from off to on. If the voltage and current are below the minimum for switching, then there will be no switching.

Click to expand...

dear,

note that this switch will be connected to a input module so my understanding is that to know if the switch is on 5mA needs to be measured at the input module and to know if the switch is OFF, 5V needs to be measured as the input module so the calculation should follow this to know if the control system will work. thanks for confirming.

Jeff23spl said:

since you said AB input module i just take a look to a compact 1769-iq16 to compare vs my calculation
With this module, you need at least 10v to trigger on this reduce your lenght and current needed is 115mA that also reduce lenght but increase noise immunity...
get back to numbers:
24v - safety factor ov 4v = 20v
20v-10=10v
10v/0.115 = 87ohms
87ohms minus some connection resistance = 70ohms
70ohms/12.8ohms = still 5468 meters (2700m long for back and forth)

Increasing powersupply voltage to 30v (To remain ok for the 30v limit of the card)
26-10 = 16v / 0.115 = 139 ohms
128 ohms/12.8= 10km (5+5)

Click to expand...

Dear Jeff23spl,

if i have a minimum current of 10mA (see attached), is it not possible to make the calculation like
20-10 / 0.01 = 1000 Ohms.
1000 Ohms / 12.8 Ohms = 78 Km = 39 km of cables length.

please advice.
thanks.
charbel

Regardless of the cable length I see a problem. The MKLS switch has a minimum current of 10mA. The input module has a maximum current of 10mA.

Allen Bradley has a KB on maximum cable length.

You want to look at the voltage drop which is maximum current * resistance of the wire. (V = I * R * 2L)

So, if we have 5km of 12.8ohm/km wire and the maximum current of the input module is 10mA (V = 0.010A * 12.8ohms/km * 10km)= 1.28V.

So, in a 24 volt circuit the loss would be roughly 5%, as you calculated above. I would consider 5% negligible. Sure, the input module may well sense it over a much greater distance, because the input module only needs 10 volts, but would you trust a signal that has greater than 50% loss?