Wiring 2 Red LED's Into IR Amplifier

[GazzaS]

Hi,

I hope someone can help.

I originally took an IR Amplifier circuit that takes a 5v IR signal (from IR extender or USB-UIRT) and amplifies the low power emitter into a high power room blaster. The circuit is here:

**broken link removed**

I modified the circuit to give me 3 IR LED's in parallel so that my entire room is covered (these are high powered LEDs with 1.2v forward voltage and 100ma forward current), reduced the 15 ohm resistor to 10 ohms and am powering it with a 9v battery instead of the 12v source. I am also using L940-36V IR's and an MPSA13 transistor.

This circuit works perfectly, however I now want to modify it to add a Red LED at the blaster area and another at the power source area with the goal of allowing me to see when an IR signal is emitted. I added one forward voltage 2v Red LEDs in parallel to the other three IR LED's with 400k resistor and the other one in parallel to the circuit with another 400k resistor. However they do not seem to light at all. The circuit tests fine and connecting power to the LED's lights them up, but in the circuit they just will not light up

I've very new to electronics (as you can probably tell) so am probably doing something very silly. Any guidance is most welcome.

Thanks,

Gary

 

[ericgibbs]

hi,
RED leds have a forward working voltage of ~2V, so by placing them across the 1.2V IR leds, they will not work.
You need a resistor to drive the RED led and connect it to the supply that drives the IR leds resistors...OK.?

 

[GazzaS]

Thanks for the reply.

Ok, so it sounds like I would need to place the Red LED in series before the IR LED's. Would I need to reduce the 400k resistor down? Would having two resistors and two Red LED's in series on this circuit effect the power of the IR blaster?

Forgive my ignorance

Gary

 

[audioguru]

The typical saturation voltage loss of an MPSA13 darlington is 1.4V at 200mA so if the battery is a brand new 9V alkaline the current in the 3 IR LEDs is a total of (9V - 1.2V - 1.4V)/10 ohms= 640mA which will burn them out if it is continuous. Maybe your circuit pulses the LEDs.

But the 2.2k input resistor produces a base current of (5V - 1.3V)/2.2k = 1.7mA so the darlington will saturate well.

A little 9V battery cannot supply such a high current as 640mA. Adding two more LEDs might kill the little 9V battery very soon.

 

[GazzaS]

Thanks for the advice.

Yes the circuit pulses the LED's as it only sends a short burst when the remote control signal is sent (100ms max). Also it will only be used when the computer controls the AV equipment. If I find the battery drains quickly then I'll swap for a 9v power supply.

So if I add two more LED's where would they go in the circuit and what supply resitance would I need? I presume from the last post that they would need to go in series and not parallel. If so then does that reduce the intensity of the IR LED's?

Thanks,
Gary

 

[audioguru]

Yes the circuit pulses the LED's .... (100ms max).

Then add a 100uF capacitor acros the +9V and 0V to hold up the voltage during each burst.

So if I add two more LED's where would they go in the circuit and what supply resitance would I need? I presume from the last post that they would need to go in series and not parallel.[/quote/]
Of course. Then you must calculate the series current-limiting resistor for them.

If so then does that reduce the intensity of the IR LED's?

With a small 9V battery then its voltage (and current) will fade away quickly.

 

[GazzaS]

Thanks for all the replies.

Now placed just one red LED in the amplifier end in series with the minimum resistor calculated based on the 9v – 1.2v – 1.4v. Now works perfectly and gives me a visual indication when IR signal is sent.

My only query now is that when the device is unplugged from the IR emitter port I see sometimes the RED LED lights very dim when the emitter lead is moved. Any ideas why? It's like it's like the NPN transistor is picking up just enough voltage to let some current through.

Cheers,

Gary

 

[ericgibbs]

hi,
When unplugged, the Base of the transistor is floating, so there maybe sufficient induced current, from your hand, to partly switch on the transistor.

Add a 100K resistor across the transistors Base, Emitter circuit.