Choice of barrery?

[brodin]

I am building a mobile IR-transmitter. It is going to be used outdoors with a rather high current through the LED. In my first prototype i used a 9V battery to a 5V vreg. Then there is a PIC12F675 controlling the LED. I am using a transistor connected to the PIC with a 20 ohm resistance between the LED and +5v. That would be 250 mA through the LED.

Can a 9V battery deliver that much current? It is only for a short while (2 us) the LED is on. Then it is of much much longer (24 us). I thought it might be better to use AAA batterys? To use 3 of them insted of one 9V battery. Can a AAA battery deliver more current than a 9V?

What do you think?

 

[Nigel Goodwin]

I am building a mobile IR-transmitter. It is going to be used outdoors with a rather high current through the LED. In my first prototype i used a 9V battery to a 5V vreg. Then there is a PIC12F675 controlling the LED. I am using a transistor connected to the PIC with a 20 ohm resistance between the LED and +5v. That would be 250 mA through the LED.

Can a 9V battery deliver that much current? It is only for a short while (2 us) the LED is on. Then it is of much much longer (24 us). I thought it might be better to use AAA batterys? To use 3 of them insted of one 9V battery. Can a AAA battery deliver more current than a 9V?

What do you think?

It's standard practice in IR remotes to have a large electrolytic on the supply to the LED (to provide the high currents for the short pulses), I explain this in my IR PIC tutorial. So the high current comes from the stored charge in the capacitor, not the battery - it's also usually higher than 250mA, 1A pulses are common.

This is why with low batteries a remote usually still works, but with longer and longer periods between button presses - as the low battery gradually recovers from the last button press the capacitor slowly charges up, this usually gives you one or two button presses off the charge in the capcitor.

Using a 9V battery and 5V regulator isn't a very good idea, you are wasting huge amounts of your power in the regulator. Feeding off three or four AAA batteries is a better idea, also you could add a second LED in series with the first one, reducing the value of the series resistor - this reduces the power wasted in the resistor, and increases the IR output, without increasing battery consumption at all (again, it's common practice in remote controls).

PIC's seem quite happy off 6V, but if you don't like it, feed the PIC through a series diode to drop 0.7V or so, but feed the LED from the full 6V supply (the large capacitor must obviously be on the battery side of the diode).

Hope this helps?.

 

[brodin]

Yes thanks that helped, i think.
I have made a scematic, can you take a look at it and see if i have understood you right. The cap must be connected before the resitance, or else there will be nothing that controls the LED current, right?

My project isn't going to be used as a remote control. This device will be turned on all times when using it. It will make a light barrier over a road and when i passes with my motorcycle, a sensor on the bike will give a signal to a laptimer onboard the bike. I have tried it and it works rather good with my scematic, but it is a bit to weak power trough the LED so the distance is getting to short.

The signal the PIC is transmitting is a HIGH for 1 ms and LOW for 25 ms and when signal is HIGH it gives a 38KHz output for the led with 20/255 on/off duty.

Will your schematic work then. I mean with the cap, will there be time enough to rechagre. What value of the cap should i use?

How do i find a good value for the resistance?
If i want 1A pulses.

4.5 - 3(2x1.5 for led) = 1.5
r = u/i
r = 1.5 / 1 = 1.5 ohm?


or do i count like this:
r = u/i
r = 4.5/1 = 4.5 ohm?

Or am i totally wrong with both?

Where can i read your IR guide (if it is public)?

 

[Nigel Goodwin]

Yes thanks that helped, i think.
I have made a scematic, can you take a look at it and see if i have understood you right. The cap must be connected before the resitance, or else there will be nothing that controls the LED current, right?

That looks OK, I think I used a 470uF in my tutorials.

My project isn't going to be used as a remote control. This device will be turned on all times when using it. It will make a light barrier over a road and when i passes with my motorcycle, a sensor on the bike will give a signal to a laptimer onboard the bike. I have tried it and it works rather good with my scematic, but it is a bit to weak power trough the LED so the distance is getting to short.

The signal the PIC is transmitting is a HIGH for 1 ms and LOW for 25 ms and when signal is HIGH it gives a 38KHz output for the led with 20/255 on/off duty.

You need to make the output pin go high and low much faster than that to generate 38KHz (13uS on, 13uS off). You do that for a number of cycles, then leave the output low for the 25mS.

Will your schematic work then. I mean with the cap, will there be time enough to rechagre. What value of the cap should i use?

How do i find a good value for the resistance?
If i want 1A pulses.

4.5 - 3(2x1.5 for led) = 1.5
r = u/i
r = 1.5 / 1 = 1.5 ohm?


or do i count like this:
r = u/i
r = 4.5/1 = 4.5 ohm?

Or am i totally wrong with both?

Where can i read your IR guide (if it is public)?

Have a look at http://www.winpicprog.co.uk for my PIC tutorials, including the hardware for them - my circuit shows the current limiting resistor between the two LED's - it makes no difference, I put it there because I fitted it like that on the board.

 

[brodin]

You need to make the output pin go high and low much faster than that to generate 38KHz (13uS on, 13uS off). You do that for a number of cycles, then leave the output low for the 25mS.

I think you understood me wrong.
Sy signal that i am sending is ON 1 ms and OFF 25 ms. But when it mean ON, i mean 38KHz with 20/255 duty which means 2us HIGH and 24 us LOW. I don't use 50% duty when 38KHz ON beacuse of saving battery.
Look at my simple drawing.

Then the sensor on the bike meassures the OFF time (25ms). and If it is 25ms it knows that it is the correct transmitter.


But the question is how do i figure out the size of the cap. I mean if i use the T = R * C
Then R = 0 and T = 0
Will that work? Can a cap get charged that fast? (i am really bad at analog electronics).

Should i have a resistor between +4.5V and the cap AND a resistor between the cap and the LED. Or can i just have a resostor between +4.5V and the cap, and connect the cap directly to the LED´s (but then the current wwould be very high in the LED, wouldn't it?)

 

[Nigel Goodwin]

I think you understood me wrong.

Sy signal that i am sending is ON 1 ms and OFF 25 ms. But when it mean ON, i mean 38KHz with 20/255 duty which means 2us HIGH and 24 us LOW. I don't use 50% duty when 38KHz ON beacuse of saving battery.

I wondered if that was what you meant, it wasn't very clear.

Look at my simple drawing.

Then the sensor on the bike meassures the OFF time (25ms). and If it is 25ms it knows that it is the correct transmitter.


But the question is how do i figure out the size of the cap. I mean if i use the T = R * C
Then R = 0 and T = 0
Will that work? Can a cap get charged that fast? (i am really bad at analog electronics).

You don't need to work it out really, just use a decent size one, I used 470uF in my tutorial - but if you are only sending 1mS in 26mS, smaller would be fine (but using a 470uF wouldn't do any harm).

Should i have a resistor between +4.5V and the cap AND a resistor between the cap and the LED. Or can i just have a resostor between +4.5V and the cap, and connect the cap directly to the LED´s (but then the current wwould be very high in the LED, wouldn't it?)

Your circuit is correct as drawn, compare it with my tutorial circuit, you can't have a resistor feeding the cap - it would slow it's charging down.