Amplifier for IR signal

Other Hardware General Electronics
1

Hello electronic specialists,

this "problem" has been addressed already two times here in the forum one time in nilskilz thread: Amplifying IR LED with Transistor
and then in Cessna172's thread:
IR LED Amplifacation

However It doesn't really seem that any of both came to a working solution at the end.
So, sorry for coming up with it again.

I am already through attempts with a MOSFET IRLZ34 and a transistor amplifier with a TIP120.

Using the MOSFET example circuit @SteveMann provided, signal from an ESP32 S2 Mini, powered with 5V,2A power bank I am getting these peak values:

busV,shuntV,loadV,mA,mW
0.87,0.04,0.87,6.50,6.00

the results with a TIP120 / 1000uF capacitor circuit are:
busV,shuntV,loadV,mA,mW
0.87,0.97,0.87,9.90,10.00

circuit
circuit1000×698 115 KB

photo
photo1000×549 222 KB

You may notice that I didn't add a resistor to the LED in the circuit, however that IR LED can handle 3A for 10 microseconds and I replaced it with a cheap yellow one for visibility reasons.

Well yes, I may burn the LED, however that at least means I finally passed high current to it. :grimacing:
I will adjust the resistor then of course but I guess that should be around 0.5Ohm.

So why I am only getting ~ 10mA instead of >= 1A?

The IRRemote sequence is 29ms and the switching time of the 274-3 diode is 0.5 microseconds.

2

How are you measuring the current and are you providing a modulated carrier wave (say 38kHz, 50% duty cycle) to the base of the transistor.
I'm not so sure about the 2 diodes. Is this an attempt to make a Baker clamp?

3

Yes 38kHz, measuring with an ammeter based on a INA219, so probably not the most accurate way but it should at least show something closer to 1A I guess.

I already thought it might be too slow but shouldn't the ammeter catch a peak once a while in a cycle of ~100 signals? Plus wouldn't the LED be burned?

The circuit is based on Constant current infrared LED emitter circuit - AnalysIR Blog

4

The two diodes do nothing, unless you have a resistor in the emitter lead, as shown in the linked article (reproduced below).

What you posted is not a constant current source, and you may have burned out the IR LED.

What is the maximum current that can be drawn from the "VBUS" pin? If you exceeded that, you may have burned out something on the MCU board.

Constant current source:

Capture
Capture361×624 33 KB

5

You will need a good fast MOSFET and a fast gate driver to be able to modulate an IR LED at 38kHz and 1A

6

According to the description at analysir the value of R1 would be 0.5 Ohm for 1A, with the less it should be even higher than 1A but nothing like that happened. How low could the resistor be to make any sense at all? 0,5 Ohm is already something which is hard to get.
Not the IR LED nor the cheap yellow one were blown.

Do I draw from the VBUS pin? Duh, I thought I'd be bypassing the MCU. The photo is a bit more accurate than the schematics, I am not pulling the VCC from the MCU pin. Well at least that is what I thought, when connecting to the power bank.

As mentioned in my initial post, this here comes to even weaker results:

IR_Driver3

Yes I used a IRLZ34 instead but that should make a major difference in my plan?
Correct me if I am wrong.

7

You will need a gate driver to be able to switch it on/off at 38kHz.
Not possible with a IRLZ34 with a 3.3V drive signal

8

That led LD 274-3 can tolerate peak pulses for only a very limited time (a few uS) then needs a long recovery period. A pulse at 38kHz is about 26uS. The next may come 26uS later.
The simplest way to get a better IR range is to choose a led with a narrow viewing angle. These can go down to 2 or 3 degrees but then more directional accuracy is required when pointing it at the target.
That mcu module is not very suitable for a breadboard with its double pin headers.

9

Okay, what would be your suggestion? The 38kHz signal works with both of these circuits, though both are weak. It is not exactly that I need 1A, I just want the most powerful signal I can get and 10mA is far away from it. Turn on rise time (max) for the IRLZ34 is 0,11microseconds, if that is the bottleneck.

10

The IRLZ34 and the TC1413N gate driver will work. You need to supply the gate driver with 5V to 12V.
You can only operate that LED a 1A for less than 100us and probably need to be off for at least 500us

11

Yes I know it is 10 microseconds for 3A or 100 microseconds for 1A. The 38kHz pulse is 29ms in total but not constant. Not sure how the LED would handle that but I am not even close to 100mA. A burning of one LED would at least mean I am getting closer.

Yes that MCU is not made for the breadboard, that is why I put it on single row sockets.

The next diodes with more narrow angle I found are actually laser diodes and I tried to avoid these for obvious reasons. I could walk that path but then I could also take one of the IR laser modules with <1mW I already have.

12

Okay thanks, I guess that is what I'll try next then.

13

The commercial way to get that is to use multiple LEDs and heat sink them.

14

Not useful for lasertag, once you put these through a lens you get multiple spots, if you cut them down with an aperture later you win actually nothing. Been there already, it would require more hardware/diffusors/reflectors to make use of multiple LEDs.

15

Ok, probably true. Have you tried any IR LEDs that do NOT have diffuser shaped package shapes? Like the yellow LED in the picture. Have you tried IR LEDs that are not black?

16

For the circuit posted by AnalysIR, also in reply #4, the current is given by the following simple formula (one diode voltage drop across the resistor), unless it is further restricted by the transistor characteristics.

0.7V/R = 0.7V/(2.5 Ohms) = 0.28A.

For the TIP120 Darlington transistor, which is two transistors in series, you need three diodes in series to get the same current limiting behavior.

17

Yes according to AnalysIR the resistor HAS to be there. I tried with it and with 2 and 4.5 Ohm, It makes a difference with resistor, the signal gets even weaker :wink:

That formula actually irritates me, say it is
0.5/0.5 Ohm I get 1A.
0.5/0.25Ohm I get 2A
0.5/0.00001 Ohm I get what? That isn't possible at all.

18

Of course not, with any commercially available transistor or power supply. You do have to use some common sense when applying these formulas.

As mentioned, with the TIP120 transistor, 3 diodes are required to overcome the two internal emitter-base voltage drops.

That formula actually irritates me

The correct formula is 0.7V/R, and it works pretty well for practical applications.

19

What diffusor shaped package? I need a fast one with power and a narrow angle, best choice I found is the 274-3. All with more narrow angles I found are considered laser diodes and need special safety treatment. And because I put a lens in front of it the angle also needs to match the focal length of the lens or at least getting close. One with 7° half angle would be nice but couldn't find one.

20

Ok, well that is easy for a test, got plenty of them.

YES, THAT is much better already:

|18:35:23.263 -> 13715158|0.86|6.10|0.87|60.10|52.00|
|18:35:23.310 -> 13734335|0.87|6.07|0.87|56.40|42.00|
|18:35:23.310 -> 13753568|0.87|-0.01|0.87|45.30|0.00|
|18:35:23.310 -> 13772676|0.86|0.58|0.86|18.20|0.00|