brodin said:
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?.