DIY emitter

[Dan.]

No probably not. You will likely have worse range because you will cut the current too much with two diodes in series. You would have to change the resistor on the board and I don't think you want to go there.

If I want to supply as much current possible to the diode, shouldn't I use a short wire? because the wire it self is a resistor (or maybe it is a weak resistor and its effect on "wasting" currunt is negligible?)

[argo]

Reducing the length of wire will have almost no effect as the resistance ratio between the resistor and the wire is large eg - ~150 Ohms (resistor) to ~0.5 Ohm (wire).

To illustrate why 150 Ohms: From my tech days a remember LEDs (light emitting diodes) run at around ~20mA (they are a current device, not a voltage device) so hence at a voltage of say 3Vs the resistor will be 150 Ohms (R=V/I).

The other factor here that may help you is that the current in devices in a series circuit are always equal (unlike a parallel circuit where the voltage, is the same across each device). So if you have both diodes in series the mA in each diode will be same and there will simply be a voltage drop across each (0.5V-0.9V from memory) diode. So this could indeed help boost the IR power of the your set up (the resistor will simply have less voltage to drop, ie the resistor heat energy is now IR energy).

Finally just a warning don't short out the resistor or you will take out the LEDs or the driver circuit. Other than a voltage drop LED's are effectively a short circuit.

Finally to identify the emitter of the LED if you can see inside the LED the "hockey stick" element is the emitter or '+'.

Hope no one minds a little electrical 101

[4zm4r3d02]

Reducing the length of wire will have almost no effect as the resistance ratio between the resistor and the wire is large eg - ~150 Ohms (resistor) to ~0.5 Ohm (wire).

To illustrate why 150 Ohms: From my tech days a remember LEDs (light emitting diodes) run at around ~20mA (they are a current device, not a voltage device) so hence at a voltage of say 3Vs the resistor will be 150 Ohms (R=V/I).

The other factor here that may help you is that the current in devices in a series circuit are always equal (unlike a parallel circuit where the voltage, is the same across each device). So if you have both diodes in series the mA in each diode will be same and there will simply be a voltage drop across each (0.5V-0.9V from memory) diode. So this could indeed help boost the IR power of the your set up (the resistor will simply have less voltage to drop, ie the resistor heat energy is now IR energy).

Finally just a warning don't short out the resistor or you will take out the LEDs or the driver circuit. Other than a voltage drop LED's are effectively a short circuit.

Finally to identify the emitter of the LED if you can see inside the LED the "hockey stick" element is the emitter or '+'.

Hope no one minds a little electrical 101

While a LED is a current device, it has a threshold voltage which must be overcome before it will operate. The value of the current limiting resistor is determined by subtracting the LED voltage from the total applied voltage, then dividing that by the desired current: R=E/I (A little more electrical 101: in equations,voltage is usually represented with the letter "E", for "electromotive force" I don't remember why they use "I" for current :confused: )

Although you don't state it, I assume that in the example you give above, the total applied voltage is 5, and the LED voltage is 2. That leaves 3 volts for the resistor to drop, and 3 volts divided by .02 amps equals 150 ohms, at .06 watts (P=IE). With two 2 volt LED's in series, the resistor has to drop only 1 volt, so it would need to be 50 ohms, at .02 watts. If you left the resistance at 150 ohms, the current in the circuit would be reduced to about .7 milliamps.

Here is a link to a handy on-line calculator for selecting the resistor value.

[pastro]

While a LED is a current device, it has a threshold voltage which must be overcome before it will operate. The value of the current limiting resistor is determined by subtracting the LED voltage from the total applied voltage, then dividing that by the desired current: R=E/I (A little more electrical 101: in equations,voltage is usually represented with the letter "E", for "electromotive force" I don't remember why they use "I" for current :confused: )

That was my point. If you have two diodes in series the forward drop would be 4 volts or so. If you have a CMOS driver at 5 volts and you don't change the resistor (which is probably surface mount and a lot harder to change than axial) your range will suffer. Not to mention that the second diode that was going to be stacked was visible and wouldn't have made any difference anyway. There is no point in doing that. Just use the IR diode. Your range will be fine.

[Dan.]

This is how it looks now.
After few days of working with it, there's no mistakes of changing to the correct channel.
Thanks for the help.