Hello!
I am trying to control my on wall AC unit using a MIDEA library I found, and while my programming is good and my circuit works to turn on the unit its VERY weak by comparison to the original remote.
The original remote works in the hallway (it has one IR - model R51M/BGCE) not even within site of the unit and my circuit struggles in the room.
My circuit is based on this:
What can be improved to make it stronger? I have a bunch of TSAL emitters on hand (6100,6200,6400).
Thanks!
--Phil
It may be that the constant current this circuit produces isn't enough to effectively drive the LEDs. It would be interesting to know the voltage drops across the LEDs, the transistor and the resistor when the control line is fixed in the high state.
Theoretically, if we assume the voltage drop across the LEDs is 1.2V each, or 2.4V in total, and the drops across the regular diodes and the BE junction of the transistor are all 0.7V. Then your circuit peggs the Base voltage at 1.4V maximum, which puts the Emitter voltage at 0.7V, which is also the voltage across the resistor. If the resistor is 2.5 ohms and drops 0.7V, then the current must be 280mA. I suspect that's not enough even with two IR LEDs. Then if the supply voltage is 5V, you are dropping 2.4V across the LEDs, 1.9V across the transistor, and .7V across the resistor.
The attached alternative circuit has everything rearranged, and the constant current feature removed. I haven't actually tried it, so you would have to see if it works. Anyway, by placing the LEDs at the bottom and the resistor at the top, I think the transistor will be fully saturated, and there will be no voltage drop across it. In other words, there will be 2.4V dropped across the LEDs and 2.6V dropped across the resistor, the total current will be 1.04A. In addition, this arrangement even runs the base current through the LEDs, so nothing is wasted.
By the way, 1 amp may be too much. You might try a 4.7 ohm resistor at first, and see if that's enough power, or better yet, add a third LED if you have an extra. Either would drop the current to about half an amp.
Edit: You also have to be sure your transistor and whatever's driving the base are up to the task of handling 1 amp.
Edit2: I forgot to deal with the base resistor. If the transitor has gain of 50, you might need a base resistor as low as 100 ohms to drive 1A through the transistor. If supplying enough base current is a problem, you might have to switch the NPN to a darlington.
Circuit from post#0 is fine, but I would reduce the 3k3 resistors to 1k.
That increases base current from ~1mA to ~3.5mA, which a common NPN transistor really needs for that collector current when not reverse-biased yet.
What is actually the transistor. A 2N2222/BC337 would be ok for that ~280mA CC circuit.
0.7volt/2.5ohm= ~280mA is quite a lot for those LEDs.
At that current the LEDs drop about 1.7volt each, which leaves about 0.9volt for the transistor.
It would be better to use the three LEDs in series on a 9volt supply (not PP3).
100mA LED current and narrow beam LEDs can bridge 50 meters.
There could be something else wrong, like transmit frequency.
Leo..
Hello,
I ended up changing R3 to 330 Ohm, and R4 to 0, I doubled the circuit (two of the circuits with a total of 4 IR in parallel off the signal pin).
Works amazingly great and reaches everything in the room (even works from outside the room).
Thanks!
PSteward:
I ended up changing R3 to 330 Ohm, and R4 to 0...
The transistor and the LEDs could be heavily overloaded with those parts values, depending on the output impedance of the power source. There is no constant current source any more with R4 removed, so you could as well remove the diodes. Make sure you have spare transistors and LEDs.
Maybe the power source was the problem all along.
If that's a 9volt smoke alarm battery, then it can't provide those current spikes (protecting LEDs and transistor).
You must support it's Ri by sufficient capacitance (1000uF from 9volt to ground).
Leo..
You might try putting R4 back in and see if it still works ok. You kinda need something to limit the current.
