Bias current of 78XX series regulators. Good as a battery indicator?

[brett]

Hi.
I've been fooling with regulating a supply to 6 to 7 volts and having a battery indicator (LED) that tells me when  regulation is failing (ie the output voltage is falling).  The regulator is a 78L05 5 volt regulator (it's the low poer version of the 7805) which I've bumped up to 6.3 V by putting 2 x 1N4148s on the "com" leg.  The battery indicator I was working on was based on the usual Zener diode.  (6.3 V - 3.9V Zener - 2V LED diode = a small voltage drop acxross the resistor and tight control of LED brightness).

But wait!  The datasheet for the 78L05 shows that the bias voltage is 4mA.  This is great (I think).  I can stick an LED between the "com" pin and ground.  This should raise the regulated voltage to 7V.

My question is:  What happens to the brightness of the LED with falling battery voltage?  Does the bias current always flow or is it a function of Vin?  (I'm hopeful that once Vout falls, Ibias will also fall and the LED will go out.  Maybe?  Maybe not?)

Thanks for any suggestions.

[Sir H C]

Inside the 78xx regulator is what amounts to an op-amp whose output is driving a large NPN transistor to keep the voltage steady at the output.  This amplifier requires current to operate.  That is the bias current you see out the ground pin.  This current is necessary to know so that you know the efficiency of the regulator, as this is in effect wasted current.  It should not change much with the input voltage.

[Processaurus]

theres an easy, low tech way to do that with a just a zener diode that I've adopted, you put a reverse biased zener in series with your LED and current limiting resistor.  The resistor needs to be lower to get the LED bright, but interestingly it doesn't take any more current to get it as bright as it was before without the zener.
http://www.diystompboxes.com/smfforum/index.php?topic=44109.msg320367#msg320367
See the schematic a couple posts up, and Zack's response a couple down