IR transmitter and receiver.

[eptheta]

Actually i had the other side filled up with wires, transistors, a resistor and a space for the 3 button cell batteries. Before i knew it, there was no space for the receiver, so i just put it at the back......

EDIT: wait what ? my remote ? up-side down ? I don't understand.....??

 

[(*steve*)]

No, the transmitter/remote control in the acrylic box. All the components are on the copper side.

It looks kinda cool, but it's not quite conventional

 

[eptheta]

Yeah ! I wasn't going to drill holes and stuff on the PCB and i didn't have any surface mount chips with me so i just used normal chips, bended the pins and soldered it onto the PCB as if they were surface mount anyway. It does look cool !
The other side just has my two buttons so it looks kinda sad.......

 

[eptheta]

Sorry for reviving a very old thread but I was writing a report on my remote control and IR communication and had a question relevant to this topic.

Physically(or mathematically), how/why should duty-cycle of the 38khz modulated IR wave affect its strength.
It seems(taking the word of several references) that if the duty cycle is kept at 10%(i forgot where i read this so it may be wrong, but it was definitely not 50%) then the remote control would work at a larger range...

Is it an inherent characteristic of the GaAs diode's band gap/crystal structure etc, or is it because of how the hardware is built into that TSOP1738 sensor chip?

I have not verified the this observation yet, but I plan to vary the duty cycle and see how it affects range soon enough.

I was thinking that maybe it has to do with response time of the sensor and if it has capacitive elements(I don't know?) then a 10% duty cycle would give it more time to return to its original state?

Unfortunately, as much as i know about the functioning of the transmitter--the simple part, I know absolutely nothing about what is inside the TSOP 1738....

If anyone has any idea, then please do help. Thank you.

 

[Resqueline]

I'd expect that to be due to that you can drive the LED's with higher pulse currents at 10% duty than compared to 50%, thus keeping the LED power dissipation constant.
The higher peak flux from the LED then simply reaches farther. Cell phones uses this technique to improve efficiency afaik.
You won't get that effect by varying duty only.

 

[eptheta]

So it's just about a 'cool-down' time of the IR LED itself ?
More current=lots more IR=more heating
More current+less time=less heating + more IR

Just a blend of timing and high currents ?

EDIT:
Also, do you have any idea what is inside that TSOP detector other than a photo-diode ?
How does it check for 38khz square waves ? A comparator and another oscillator running at the same frequency ?
A pulse detector(An anti-missing pulse detector--using a 555 or equivalent) ?
Little trained leprechauns ?

Thank you.

 

[Resqueline]

You want to spend say 0.1W in the LED and so a shorter on-time means higher current meaning higher flux that has the ability to punch through the receiver noise floor.
You could "in theory" have almost 5 times the current (& the flux) at 10% duty, but since the voltage drop rises (& the efficiency decreases) it won't be that good in practice.
You're also expected to modulate the 38kHz, and if you put that on with a 10% duty cycle you could push even more current.
A pulse train like that is in fact also needed for most receivers to work, since their circuits seems to be AC-coupled even after the 38kHz detector.

More current + less time = same heating + more peak IR (but less average IR)

I don't really know what's inside the detector, but I'd expect some kind of "ordinary" bandpass filter right after the photo-diode, before amplification, detection & squaring.

I'm sure I have misbehaving leprechauns in my computer, sometimes harassing me with iexplorer hangups, disconnecting/connecting USB-ports and fan-speed detection..

 

[eptheta]

That cleared up a lot of my doubts, thanks!
I'll post if I have any other questions....

 

[eptheta]

Actually, I am still a little confused. In order to exploit this maximizing-current-minimizing-duty-cycle, does it mean increasing the source voltage of the LED ?

Take this scenario:
I run the LED at 50% duty cycle, and have found experimentally that if I put it at anything more than 3V, the LED will blow up.
Now I change my duty cycle to 10%. Does this mean:
a) for the same voltage (3V), a larger current is used by the LED, thus increasing its range.
b) for the same LED, I can now use a larger voltage (say 5V) without blowing it up, thus automatically increasing current at no extra risk.

It is (b) right, because it's not like resistance is changing or anything...?

Thank you.

 

[Resqueline]

Huh, 3V to an IR led... They require something like 1.2V... And you don't drive LED's with a voltage, you make sure to drive them with a current regardless of voltage drop..

 

[eptheta]

Huh, 3V to an IR led... They require something like 1.2V...

Yeah, I know. I was just making up a hypothetical scenario to clarify my doubts.....

And you don't drive LED's with a voltage, you make sure to drive them with a current regardless of voltage drop..

That's exactly what I was wondering. I know that LEDs are current driven , which got me confused when you said "a shorter on-time means higher current meaning higher flux".

Is energy in both cases constant? Because then the equations make some sense..
Is it more like "If I have a given amount of energy, 'X' which I am willing to use over any time duration, I can get more current out of it in a shorter on-time."
So E1=E2, i1^2Rt1=i2^2Rt2 (where t2<t1)
i2^2/i1^2=t1/t2>1
so i2>i1 ?

 

[Resqueline]

The higher current you can drive through a LED with shorter on-times is nothing that happens automatically, you'll have to calculate and design towards it.
The current could be fixed, or with a current-measuring series resistor and op-amps you could make a circuit that'll adjust the current by itself as the duty-cycle changes.
It's also neccessary to keep in mind the adjustment needed to compensate for the higher power loss caused by the higher forward voltage drop during higher currents.

 

[eptheta]

So changing only the duty-cycle would do absolutely nothing unless I manually increase current through the LED... Makes sense....
Well, thanks.