Mighty IR remote control

[8juancho]

Greetings to all
I am making a project to send a signal via infrared between arduinos but I have noticed that I have to point the transmitter to the receiver to establish communication, the opposite with my tv remote control; the receiver picks up the signal even pointing in the opposite direction, I use the arduino nano as a transmitter by connecting a normal IR diode, what could I do to increase the power of the remote control?

#include <IRLib.h>
#include <IRLibMatch.h>
#include <IRLibTimer.h>

IRsend emitter;
int count;

void setup() {
  Serial.begin(9600);
  count = 0;
}

void loop() {
  count++;
  emitter.send(NEC, 0xFF00926D, 32);
  Serial.print("SENDING DATA: ");
  Serial.println(count);
  delay(80);
}

This is the code in the arduino nano

[jremington]

What are you using for the IR receiver?

Please post a wiring diagram of the sender (hand drawn is fine).

For more IR transmit power, you need a high power IR LED and transistor driver.

[LarryD]

Both the IR receiver and transmitter are involved in this.

Some receivers are better than others, same with the IR TX LEDs.

Your RX and TX need to be the same wavelength.

The TX frequency needs to be correct too, 38kHz is very common.

You can parallel several TX LED circuits to get more IR light.

My favourite receiver is the TSOP4838; you can shine the TX LED to the floor and there is enough gain in the system to work just fine.

[8juancho]

for the IR receiver I'm using the module ky-022

this is the diagram but with nano

[8juancho]

I think that the receiver it's fine because my tv remote can activate it in different angles I think It's better to increase the power signal, how can I make it 38kHz?

[jremington]

This circuit will supply about 100 mA to a 100 mA high power LED. The Arduino has to produce the modulation frequency required by the receiver, usually but not always 38 kHz.

[LarryD]

The IR LEDs I use have a Vf of about 1.7v.

(5v - 1.7v) / 220R = 15mA

My LEDs are rated for 100mA

Suggest using 80mA.

(5v - 1.7v) / .08 = 41 Ω

For this amount of current you will need a BJT or MOSFET.

Note, you can easily place two or more LED/resistor combination in parallel on one transistor.

[8juancho]

Hi, the connection to the arduino is between the digital output? and do yu have some ref of high power IR LED?

[8juancho]

those values can be apply it in the diagram of post #6?

[LarryD]

You need to get a LED that matches the RX wavelength.

DIGIKEY and MOUSER are good suppliers.

My TSOP4838 IS 950:

[LarryD]

@8juancho

those values can be apply it in the diagram of post #6?

Yes

[8juancho]

Ok man I understand you I will tell you how thinks go on

[jremington]

The values in post #6 are approximately correct for a 3.3V supply, or 2XAA batteries.

For my IR LEDs, Vf = 1.4V, so the current is (3.3-1.4)/18 = 110 mA, (ignoring the saturation voltage of the transistor, typically about 0.2V).

Taking that into account, (3.3-1.6)/18 = 94 mA.

[8juancho]

I get it men thank you I'm going to get a 100mA diode, I will keep you informed