gnupic: Re: [gnupic] sanity check, pic12f675 first project
Subject:
Re: [gnupic] sanity check, pic12f675 first project
From:
Jason ####@####.####
Date:
22 Jul 2009 16:13:24 -0000
Message-Id: <4A672C8D.3060706@lakedaemon.net>
Mark,
Mark Rages wrote:
> On Tue, Jul 21, 2009 at 9:04 PM, Peter ####@####.#### wrote:
>> Jason wrote:
>>> Nice. I found a circuit a while back [1] for increasing the LED
>>> brightness by pulsing it. I thought I'd replace the 555 timer with
>>> a more flexible pic...
>
> I don't think you'll find a significant brightness increase with modern LEDs:
>
> http://members.misty.com/don/ledp.html
>
Good read. Thanks for the pointer.
>> Aha. Watch out with the voltage though - that circuit will make those
>> R2-R3 resistors sweat a bit, and the Q1 transistor will also be a bit
>> stressed.
>
> R2 and R3 have the same low duty cycle as the LEDs.
>
> If the LED current is really 500 mA each, that exceeds the design
> rating of the BC547B by about 10X (Abs max Ic is 100 mA for that
> part). Better to use a logic-level FET for this application I think.
>
Is there a "typical" FET I could try? Regardless, your link, above,
spells out the pointlessness of the circuit in this case.
I may still use the transistor if the recommended current can't be
sourced/sinked by the 12F675... I'll just adjust the R values to get a
steady current at the recommended amperage for the LED.
>> Better match the supply voltage to the forward voltage of your LEDs,
>> then there's no extra voltage that needs to be dropped by other
>> components.
>
> The resistor is doing more than dropping voltage. It is acting as a
> crude current source, to maintain the LED current as the forward
> voltage shifts over temperature, age, manufacturing tolerances etc.
> So no matter how carefully you select the supply voltage, you can't
> remove the resistor without some other means of regulating current.
Ok, my EE degree is getting old, maybe I'm missing something. If I have
a LED (say, 2V fwd voltage) and a resistor in series with a 3V battery,
the voltage drop across the resistor is going to be 1V, which would give
it a value of 50ohm for 20mA.
Assuming that's all correct, if the forward voltage falters over time,
say to 1.5V, the drop across the resistor changes also, to 1.5V. This
would change the current to 30mA...
Also, as temperature changes, I don't think the resistor would
compensate in any way for changes in the forward voltage across the LED.
If memory serves, an LED drops voltage, but has no (or little) internal
resistance. So when hooked directly to a battery, the current surges
until the LED blows or the battery is drained. I think it would be more
accurate to say the resistors are "current limiting resistors", but they
don't help with temperature fluctuations, or aging (except premature
aging ;-) ).
So, if a PIC can sink 20mA on it's own, then an LED and a resistor
should be all I need. The transistor is only needed if I'm sinking
multiple LEDs controlled by one pin. Neat.
Now on to the programming bit. :-)
thx,
Jason.
