What is the Input Current and Output Current of this IR Transmitter Module?

Other Hardware General Electronics
1

Hi

I am using Arduino Nano, which is based on ATmega328.
The ATmega328 can give upto 40mA of current, from an Output leg.

There is this small IR Transmitter Module:

It contains an IR LED, and a Transistor.

I am curious what is the Input Current (which goes to the Base of the Transistor),
and Output Current going out of the Transistor to the IR LED.

The SMD Transistor is marked as "J3",
which according to Google is an S9013 NPN Transistor.

Also there are 2 Resistors:
R1 "103" meaning 10KΩ
R2 471Ω

If I understand the circuit correctly, from this image of the PCB,
then the 10KΩ Resistor is the Base Resistor,
and the 471Ω Resistor is the ICE Resistor..

If so, that means that the Output Current from the Transistor is (5v-1.4v)/471Ω = 7mA?

If yes, for that we do not need a Transistor..
We could simply get that 7mA directly from the MC..

Or maybe I did not understand the PCB correctly?

Thank you for your help

2

470 ohms, not 471, just like 103 means 10000, 471 means 470 and 470 means 47

While 10mA is fine for an ATmega microcontroller pin, many logic chips have much more limited
output current capabilities, for instance the Due pins are rated at various values, some as low as
3mA IIRC. Many standard CMOS logic families would struggle with 10mA.

3

MarkT:
470 ohms, not 471, just like 103 means 10000, 471 means 470 and 470 means 47

Oops, you're totally correct, my mistake..

Since the mistake had little effect on the result this time, then we're still with less than a 10mA output..

MarkT:
While 10mA is fine for an ATmega microcontroller pin, many logic chips have much more limited
output current capabilities, for instance the Due pins are rated at various values, some as low as
3mA IIRC.
Many standard CMOS logic families would struggle with 10mA.

I see,
interesting to know, didn't think some MCs can be so limited in their output legs.

In my case, since I am using ATmega328, 10mA is less than what I could get via directly operating the IR LED from the MC...

So in that case this module is more limiting for me than amplifying..

Let's try another options that look nice..

718×615 103 KB

This module has 2 IR LEDs, and a Transistor.

The Transistor is marked as "J3Y" this time,
and according to google this is S8050: NPN 40v 0.5A.

Since the module is black, it's a bit hard to see the copper traces on the PCB.

So I will try to base my calculation on logics, regarding how the circuit is.

First of all, there is a Power On SMD LED (not IR LED) on the board, and near it a - "102" (1000Ω) Resistor,
so we can put the aside.

Then, we can see that near each IR LED there's a "101" (100Ω) Resistor,
so logically (and hopefully) these are the LEDs resistors..

Then, there's a "102" (1KΩ) Resistor, and the Transistor,
so I assume this 1KΩ resistor is the Base resistor..

Now, assuming the LEDs are connected to the Transistor's output in Parallel,
it means the current going thru each IR LED is (5v-1.4v)/100Ω = 36mA

That's better than the 7mA we got on the Red module,
yet still this does not justify getting this module, since I can get those same 36mA directly from my MC...

Is my calculation correct this time too?

4

Probably - short of tracing all the traces and redrawing the circuit I strongly suspect that's how its
wired.

Some IR transmitter LEDs can take much higher pulse currents than this, upto 1A is not unknown,
which gives extra range. Of course a low duty-cycle is needed to avoid thermal damage at these
high currents, and the power supply has to handle the load.

5

MarkT:
Probably - short of tracing all the traces and redrawing the circuit I strongly suspect that's how its
wired.

I see.
Thank you.
In that case both these modules are useless for what I needed,
and will not give the IR LED any more Current than what I can already achieve using the MC alone.

MarkT:
Some IR transmitter LEDs can take much higher pulse currents than this, upto 1A is not unknown,
which gives extra range. Of course a low duty-cycle is needed to avoid thermal damage at these
high currents, and the power supply has to handle the load.

Right.

I actually have IR LEDs at home.
What I have are from Tayda Electronics:

Vishay - TSAL7400

The important part from the datasheet is:

Forward Current = 100mA
Peak Forward Current = 200mA
Surge Forward Current =1.5A (for upto 100μs)

I also have various Transistors at home.. (also from Tayda)

The only reason I planned to buy a module, is because with a ready-made module, all 4 components (IR LED, Transistor, Rbase, Rce) are already on a (very small) PCB.

It's nice and compact.
But it seems that those modules' parameters are not what I needed..
So I will build myself on a small protoboard..

6

IR has current of up to 100mA and forward voltage of 1-1.8v. I would change the resistors on the red board and adjust the current higher. Maybe 0805 will fit on there. Can't tell from picture, usually all these boards look like ants when I get them in the mail.

If you changed the 470 to 50 you should get close to 70mA.

7

wolframore:
I would change the resistors on the red board and adjust the current higher.

SMD soldering seems like a nigthmare for me..
I would rather use a protoboard and use thru-hole components..

8

I only suggested it since you like the package. It's not impossible. Search youtube. Just need to desolder and wick off. Then lay fresh solder on one side... tack in then go finish.

9

I see.

Thanks.

But I will choose the simpler option of creating a small module myself, using a protoboard.

Thank you all.

10

I think you haven't told us yet what you want to do with this IR module.
Camera night illumination, line follower, remote controller, beam break.
Continuous mode, modulated, pulsed.
Leo..

11

Hi Wawa

The answer is remote controller, and data transfer between 2 Arduinos..

Which in terms of your question are the same: it's pulsed..