5-7 meter wires from Arduino to LCD 16*2

Hello,

I will need to place the LCD 16*2 about 5-7 meters away from the control (Arduino Uno).

What is the optimal wire size (AWG or mm^2) for such thing - not too small and not too big :slight_smile:

Tnx!

Is the 1602 LCD 4-bit parallel interface or i2c interface (i.e. had an i2c backpack)?

I suspect the choice of AWG is not going to make this work if it does not work. The AWG only affects the resistance, and the capacitance and inductance of the wiring is also important.

Get two Arduinos.

The remote Arduino has the LCD.

The two Arduinos communicated with serial.

The other Arduino does the main work.

I am using 4 bits parallel interface.

What is remote Arduino? I have never heard of it.

Tnx!

It would be an Arduino at the LCD location.

It would look after sending the needed signals to the LCD.

This Arduino connects to the main Arduino via your 7 meter serial cable.

You send serial LCD data from the main Arduino to the LCD Arduino.

Is there any power available to the LCD location ?

larryd:
It would be an Arduino at the LCD location.

It would look after sending the needed signals to the LCD.

This Arduino connects to the main Arduino via your 7 meter serial cable.

You send serial LCD data from the main Arduino to the LCD Arduino.

I think a second Arduino will be too much here, but your idea made me re-think my design - I will have the Arduino at one end of the cables, and then I need 7m wires to activate relay shield (4 relays) - this is probably easier.

But here too - I will need wires...?

Your re-think would work.

Make sure relays have kickback diodes across the relay coil.

Connect the diodes at the relay location.

Time to draw up a schematic.

UKHeliBob:
Is there any power available to the LCD location ?

I am re-thinking my design -

I will have the Arduino at one end of the cables, and then I need 7m wires to activate relay shield (4 relays). What wires do you think I should use here?

larryd:
Your re-think would work.

Make sure relays have kickback diodes across the relay coil.

Connect the diodes at the relay location.

Time to draw up a schematic.

What is the purpose of these diodes?
(I think the relay shield has them on already).

What size of wires would you use (AWG / mm^2) to push the relays - VDD / VSS / command 1-4

Tnx!

trsuthbu888:
What is the purpose of these diodes?
(I think the relay shield has them on already).

They protect the circuit from the reverse voltages which are induced when the relay coils are de-energised. You are correct, a shield, or relay modules will have these installed. If you were connecting bare relays, you would need to add them.

trsuthbu888:
What size of wires would you use (AWG / mm^2) to push the relays - VDD / VSS / command 1-4

These signals are low voltage, low current and low data speed (unlike the LCD signals, which are low voltage, low current, but a higher data speed). So almost any wire will do. 6-core security system cable or 8-core Ethernet cable for example.

Donโ€™t forget about relay coil power, if more than one relay is energized at the same time, you may need a separate 5V supply for the relay module.

PaulRB:
They protect the circuit from the reverse voltages which are induced when the relay coils are de-energised. You are correct, a shield, or relay modules will have these installed. If you were connecting bare relays, you would need to add them.
These signals are low voltage, low current and low data speed (unlike the LCD signals, which are low voltage, low current, but a higher data speed). So almost any wire will do. 6-core security system cable or 8-core Ethernet cable for example.

Thank you!
I will try the Ethernet cable.

JCA34F:
Don't forget about relay coil power, if more than one relay is energized at the same time, you may need a separate 5V supply for the relay module.

I have only 1 relay active at a time.

A general question about this -
Can I get the DC power (+ / -) to the relays directly from my power source, or I should get it only from the Arduino 5V/GND ports?

Also, just in case for the future - when suggesting to have a separated wire to each relay - should all of them connect to the same 5V port on the Arduino?

Can I get the DC power (+ / -) to the relays directly from my power source, or I should get it only from the Arduino 5V/GND ports?

That depends on your relay modules, and the power that is being switched. Some relay modules are described S "5V relays" and some as "12V relays". This refers to the voltage needed to operate their coils. If you are switching 12V DC power, it makes sense to use 12V relay modules, so that the Arduino's 5V regulator is not burdened with providing current for the relay coil. The Arduino's 5V regulator can only provide current for one or maybe two 5V relay modules at once, if that, depending how the Arduino itself is powered. For example, if the Arduino is powered from 12V through its barrel jack, it will struggle to provide current for even one 5V relay module.

Another option, if the power being switched cannot be used to power the relay coils, is to provide a second power supply from the power to be switched. In this case, you would need to have a common ground between the extra power supply and the Arduino. Unless, that is, the relay modules contain opto-isolators.

Tnx for the detailed reply!

The relay module I am using is 3.3-5V for the coils.

I am powering the Arduino via the barrel jack with a 12V-5A power supply (I am transforming it from 24V I have near by).

I am about to switch either 230V AC or 24V DC. Not sure yet. But the switched power will come from a separated source. Switched current will connect to COM / NO of the relay channels.

So my issue is to get enough current to push the relay coils.

The relay module has its own VDD/GND plugs - I guess this is for the LEDs on the module board and for the active-high/active-low options. It also has an "IN" plug for each channel - 4 inputs, that actually activate the coil.

Any idea?
Or maybe I should just get the Ethernet cable and test it...?

trsuthbu888:
I am powering the Arduino via the barrel jack with a 12V-5A power supply (I am transforming it from 24V I have near by).

Ah yes, common but serious mistake unfortunately encouraged by the misleading tutorials on the site (and elsewhere). :astonished:

The "barrel jack" or "Vin" with the internal regulator is for powering the Arduino board itself only for playing about. Not for any "serious" application.

You need a 5 V regulated supply to operate your (5 V) devices such as relays and that same power supply can power the Arduino via the "5V" pin.

I cannot imagine why you would have 3.3 V relays and you really need to cite what these modules are (Web links) as there are tricks in using "opto-isolated" modules. In fact, there are a number of "tricks" or ways of wiring things to avoid problems with relays in general.

My "full on" situation, when all possible devices are working is as follows, with the current draw of each:

  1. LCD 16*2 - max 150 mA
  2. 1 LED (regular 5mm) 20-30 mA (I am using 1k resistor)
  3. 1 relay (75mA)

So total current being drawn from the 5v pin is ~250mA

I am reading more and more tutorials saying that when I am using the barrel jack - I can easily get up to 750mA on the 5v output pin - so your idea to get an external power source is not really understood :slight_smile: ?

The relays are actually 5v - I found the spec here

(I don't know why the web-seller is saying it can be switched with 3.3).

Since the relay module has 4 relays on, BUT, I have only 1 relay active at a time, and never more - I think my current consumption calculation is ok.

What do you think :confused: ?

I am reading more and more tutorials saying that when I am using the barrel jack - I can easily get up to 750mA on the 5v output pin - so your idea to get an external power source is not really understood :slight_smile: ?

If you were to supply the barrel jack with 12 volts and took .75A form the 5V pin you would burn up your Arduino voltage regulator.

12V powering:
Power dissipated by the Arduino voltage regulator = (12V-5V) * .75 = 5.25 watts

7V powering:
Power dissipated by the Arduino voltage regulator = (7V-5V) * .75 = 1.5 watts

There is probably no heatsinking on the Arduino regulator so it will get HOT!

When you power the 5 volt pin from an external 5 volt power supply, the regulator is not used, hence no heat problem.

(I donโ€™t know why the web-seller is saying it can be switched with 3.3).

Relays have a must pick voltage.

Your 5V rated coil more than likely has a must pick voltage of 3.3V.

  1. LCD 16*2 - max 150 mA

No way! I would be surprised if it's more than 30~40mA. Measure it!

  1. 1 LED (regular 5mm) 20-30 mA (I am using 1k resistor)

Colour? If red, probably has forward voltage around 1.8V, so current will be (5.0-1.8 )/1K = 3.2mA

The relays are actually 5v - I found the spec here

No use, sorry. That's the data sheet for the relay itself, which an Arduino cannot operate directly. We need the data sheet for the relay module.

You would be better off using 12V relay modules, given you are powering the Arduino with 12V. If the Arduino only has to power itself, the LCD and one led, then it's regulator should be ok. The relay modules would be powered directly from the 12V supply.

Another option would be to get a 12V to 5V DC-DC convertor module, 1A or higher. This can power the Arduino (via 5V pin), LCD & 5V relay modules without overheating, because they are much more efficient than linear regulators.