Simple LED circuit.. HELP!

One way might be to include an op-amp and a few resistors to your circuit.

If you google for circuit op amp hystresis you can find some examples. The idea is that using a few resistors you can make the op amp output switch from low to high at a fairly specific voltage. You would probably use two resistors to create a voltage divider to make your 6.5 volts. Then you would use two more resistors to create the feedback needed to get the op amp to turn on and off at that voltage. Then you would use the output of your op amp to drive your LED and resistor.

For that level of accuracy you need something like a voltage comparator.

off

Are you talking about the 9V battery that powers the circuit itself discharging to 6.5V and the LED indicating the charge level of that battery, or are you talking about a separate voltage input? Either way, I'd probably use a comparator (see LM393), but it's a little more complicated to measure your own supply voltage reliably.

View this with a fixed pitch font such as Courier

+9V | +--------------------------------+ | | | ------ | \ / LED | \/ | ----- | | | | | \ | / 620 | \ | / | | | 10K |\ | +--/\/\/\/\-----+-----| \ | | | - | \ ______| \ | | / 1.5 / +-------------| / LM339 or similar comparator K \ | | + |/ / | | | | \ +-------+ / 27K | \ -----/ / / ^ | / \ 4.7V | --- 400 mw | | | | | +---------------+ | ----- --- -

Resistors are 1/4W 5% If you want to adjust to exactly 6.5 V for switching, replace the 10K resistor with a 7.5K fixed resistor in series with a 5K trim pot. Caveat... this circuit does not provide hysteresis, so if the source is a bit noisy or fluctuates around the switching point, the LED may flicker on/off. But this certainly meets your original spec.

--
Dave M
MasonDG44 at comcast dot net  (Just substitute the appropriate characters in 
the address)

Never take a laxative and a sleeping pill at the same time!!

Google for "window comparator."

If you want a sharp on/off for the diode, I think a comparator i needed (or just a cheap op amp wired as one) I'd use a resistor and a 3.0v zener to give 3.0v, and put a 35K an

30K resistor in series from + to ground, to give 3.0v when th battery is at 6.5v Then run the 3v reference from the zener & the voltage from th divider to the appropriate inputs of he op amp. And then the outpu of the op amp drives the led, (via a suitable resistor) if you go the inputs the right way about. And, as a bonus.. put another diod to the + rail, to tell when the battery is flat. (maybe a red/gree diode, if the pinout is right)

Alright, thanks for all the info guys! I will look into all your solutions this morning. For the record, I am measuring my own supply voltage.