The following circuit is what I hope to be a functional battery charge state indicator. The Green LED indicating the battery voltage is good and the Red LED indicating that the battery should be charged soon.

Did I botch this up?




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You're trying for nearly 15mA current through the LEDs.

Did you know that the LM339 doesn't like to sink that much current?
I don't like to go much above 5mA; preferring to stay in the range of 3mA-4.5mA.

 

Sink current is 6mA. I'd suggest a bit of hysteresis feedback so it doesn't flicker when voltage is exactly at the set value (instead of two comparators).

With feedback, you could use only one comparator, and a transistor to illuminate the red LED when the green goes out.

 

Why don't you run both leds off one comparator - ie. the 2nd led to ground. You could use the other half comparator for a high voltage level crossover - maybe even think of a tricky way to put LEDs between comparator outputs.

By the way - the indicator is only of battery voltage - not charge (as that requires a whole other level of measurement).

Ciao, Tim

 

OK, running both LED's off of one comparator and using a transistor to drive the LED's
I'm not sure how to draw that schematic.

@timrobbins,
The LED is not to indicate that the battery is charging, but that it needs to be charged, so that I do not runn the battery down too low.

iONic

 

Sink current is 6mA. I'd suggest a bit of hysteresis feedback so it doesn't flicker when voltage is exactly at the set value (instead of two comparators).

Where did you get the 6mA figure from? Calculation shows about 14.7mA in the above circuit.

LM339 Datasheet indicates 16mA typical with ~1.5V Vo but I agree with Sgt that it would be far better to stay below 5mA so that the Vo is kept low.

 

An LM339 is a quad comparator in a 14 pins case. An LM393 is a dual comparator (the same comparators as in the LM339) but it is in an 8 pins case.
Their minimum output current is only 6mA with a max voltage drop of +1.5V.

 

Perhaps it is not necessary to have the "OK" LED (Green) since the device is operating. Thus a single comparator with one LED for low voltage indication is all I need.

[URL=http://img340.imageshack.us/i/goodchargeindicator.png/][/URL]

 

Your NPN driver will not work, because of the resistor and capacitor(R6 & C1) connected at the emitter.

What we are telling you is that the LED current is too high at about 15mA in your original circuit.

The simplest solution is to increase the series resistor value and use a high intensity LED instead.

These LEDs take only a few mA and give out very bright light.

 

Your transistor will never turn on!
1) The output of the LM339 is an open collector that can go low but never goes high. it needs the 2.2k resistor R8 changed to 7.5k ohms and to connect from the output of the LM339 to the positive 12V supply, and the base of the transistor to connect to the output of the LM339.
2) The resistor and capacitor at the emitter of the transistor should be replaced with a 1N4148 diode with its cathode at ground and its anode at the emitter of the transistor.

The LED will be turn on most of the time. It will turn off when the battery voltage drops below 11.7V. I think that is backwards.

 

The cap/Resistor combination is supposed to make the LED Blink at about 2Hz.
The Red LED need to stay off until the battery voltage gets below 11.7V

 

The cap/Resistor combination is supposed to make the LED Blink at about 2Hz.

no. The capacitor turns the LED off after it blinks only one time. Buy a blinking LED and get rid of the capacitor.

The Red LED need to stay off until the battery voltage gets below 11.7V

Of course. Then the transistor should be a PNP type and the circuit needs to be redesigned for it.

 

Perhaps this:

[URL=http://img151.imageshack.us/i/goodchargeindicator.png/][/URL]

@AudioGuru
...Not sure the function of the diode in the circuit you posted.

 

You really don't want to discharge a lead-acid battery down that far.

Look in the Tips & Tricks thread; 4th page, reply 38 I think. There's an image of an Excel spreadsheet. If you discharge your battery below 50%, it's life span will be considerably shorter than if you kept it higher than that.

The more shallow your discharges, the longer the battery will last.

 

Your new circuit uses the transistor as an emitter-follower that is slow and has a high voltage loss. The LED and its current-limiting resistor should be from the collector to ground so the transistor can switch quickly on and off. Then the PNP transistor needs an additional series 7.5k base resistor.

 

@Sgt, Then 12V should be OK?

@AG,



Is R7 then necessary?

 

R7 turns off the transistor and pulls up R5.