PAL & batt. voltage

Here's an observation I had with my PAL Gold. I put an old 9V battery in it, and there was no change in brightness between the high and low settings. The battery had 5.1 volts. Another battery had 6.4 volts and did show the expected difference in brightness between the two settings.

Since 5V is still more than the LED needs to run at full brightness, did the PAL fail to shine brightly due to the losses in the circuitry or is a battery that low unable to provide the amps needed?

Just curious.
RonM

Additional note: Just noticed that the high setting is now DIMMER than the low settng. Battery voltage is at 4.17V.

RonM

my guess it because the Hi, just attaches the battery directly to the LED and the low setting limits the power to the LED. So in hi there is more of a darin on the battery and the LED is dimmer. In low it can only get a limited amount of power so it doesn't run the voltage down as far. Just a theory.

Brock

Brock - IMHO I don't think on high the PAL connects the led directly to the battery. At 9V the led would get smoked.

Regards - RonM

I thought the same thing, but the voltage on the battery was 6.7 v and the LED was pulling 70 mA. Maybe it just has a small resistor, because you right it would probably be higher than that. Who wants to try it? Not me.

Brock

I think that what's happening has to do w/ the drop-out voltage of the linear regulator. I don't remember the series number for the regulator (LM318 or something), but there is an "over-head" in the regulator that essentially subtracts from the battery voltage even before it starts its current limiting. As a bad "fer instance", if you have a LR w/ a 2.5 volt drop-out, you need to have a voltage source that is a minimum of your desired voltage plus 2.5v. At 4 ot 5 volts, you probably aren't getting the linear regulator to do ANYthing. At 9 volts, it scoops its drop-out voltage off the top, and then regulates the remaining voltage.

LRs are NOT very efficient vehicles... Their efficiency rises SLIGHTLY as the source voltage gets closer to the target voltage, but still, the efficiency is around 50% (from hazy memory). With a LDO (low drop-out) regulator, you might have a drop-out voltage on the order of less than 1 v (some, now, down to a few hundred millivolts) which let you exhaust your battery a bit more, but don't to much at all for the efficiency.

Now... step-down SWITCHING regulators at, say, 90% efficiency... THAT's another story. ;-)

(for THAT story, read "9v x 595 mAh = 5355 mWh, x 90% = 4820 mWh / (3.6v x 20 mA) = 67 hours"? I've got to check my math on that... that can't be right. Or maybe it can. Anybody else figure it the same way?
Ah... one thing, that's a the "optimum" current of 20 mA, not the 96 mA that the PAL uses... heh heh heh. Sorry. Been a while since EE school.)

FWIW
There's a new 9.6v rechargeable
Nimh (170mah capacity)
-compared to that old Nicad 7.2v
120ma......
Ordered one for my Pal Gold.
It's from: www.smallpower.com
$9.60