We are having a thesis and we are using 20x4 lcd whenever we run the code of the program the lcd will be off and be turned on again.we are assuming that the arduino mega board cannot supply the sensors,6 relays and the lcd by itself so we had an external source .but the lcd will just display squares and it will not work.What shouild we do with it?
We are planning to connect the relays to an external source but when we had it,the relays won't work.We assume it would be best to have an external source rather than the arduino itself,is this a good idea?
An Arduino - UNO or Nano or Pro Mini etc. - does not "support" any current beyond a very few LEDs at 20 mA each or driving the data line of WS28xx devices. The 5 V pin is a reference output only which may be used to supply a few sensors given that they only require a few milliamps. It must never be considered as a "power supply".
The inbuilt regulator has no heatsink, so it is limited to 100 mA or so, and the higher the voltage on the "barrel jack" or "Vin", the more power it must dissipate and the less current it can supply before overheating and (hopefully) safely shutting down.
If you are using LED strips or relay modules, or other such devices requiring a 5 V power supply, you must have an adequately regulated 5 V power supply of some sort. Once you have that, you can use it to power the Arduino via its 5 V pin as a power input given that the requirements for the LEDs/ relays etc. and the microcontroller are exactly the same.
And the rule so frequently explained here: Get one part - such as your LCD - working first, then another such as the relays, then both together. Step by step.
pammy,
At this point we have only a vague idea what components you are using and how you have wired things up.
You should provide a wiring diagram of what you have wired up in your circuit so we can understand how you have things wired up.
A few photos of everything would be helpful.
In other words to understand your environment, we would need enough information to be able replicate it.
Note that if you are simply yanking the power to the LCD and then re-applying it again, that will require the LCD to be re-initialized before it can be used again.
--- bill