[Help] Battery Monitor Circuit

[Slasherbaven]

I've created a schematic of a Battery Monitor, where it monitors three voltage ranges:

0~5V
5V~9V
9V~12V

However.. I am unsure of my circuit.
So yeah, well before I build the circuit, can any of you guys double check this for me?
Really sorry for the bother

Thanks in advance. =]


(Attached to this post is the Battery Monitor schematic)

[blocco a spirale]

This won't work. I think it would be best to start again rather than try and fix it.

[SgtWookie]

Yes, it needs work.
1) Use comparators instead of operational amplifiers. Op amps should not be used as comparators; they weren't really designed to be run in open-loop mode indefinitely. Power consumption will be far higher than comparators, as the outputs of the opamp will be in constant saturation. Consider using an LM339 quad comparator, or better yet use an IC that's really designed for this type of thing, an LM3914N. Comparators and 10 LED drivers are built right into this IC.

2) Your LEDs do not have current limiting resistors.

3) The op amps have only negative supply and ground; no positive supply.

4) Although you've provided a 2.5v reference to the resistor ladder for the inverting inputs, the noninverting inputs see full battery voltage. As a result, the outputs will not change states until the battery voltage falls below the minimum input voltage of the TLE2425, at which point the outputs will be unpredictable.

5) Battery voltage remains remarkably constant from when they are fully charged until when they are nearly discharged. An automotive 12.6v battery is considered fully discharged at 11.4v. Dry cells have somewhat different levels, but it's a similar concept.

[Slasherbaven]

I'm a bit confused...

Sergant Wookie, I am required to use an LM324 quad (a low-power quad Op-Amp) for the circuit.


And here, I re-did the whole thing.
Hopefully it's better than the one before.


The Vcc's purpose is to power up the op-amps.

[thingmaker3]

There is an open circuit. There is no current path through the resistors. There will be no voltage across them.

[Slasherbaven]

>_< Alright, here's another one.
It seems quite right.



I'll start fixing things from here.


=EDIT=
The earlier post was deleted by accident.
Sorry about that.

[KMoffett]

U1 (78L12) will not work for this because the battery on the input is only 10V. You were thinking of a 78L05 maybe? And the + terminals on the op amps are hooked to the (-) end of the 10V battery.

Ken

[Slasherbaven]

Ah, my bad on the 10v battery and it's connection.
Here, I made the minor adjustment:


And as for the 78L12, why would using a 78L05 be better?

[KMoffett]

1. For the 78L12 to regulate it needs an input of +14V or more. If your battery is 12v then the regulator has have an output of 10v or less to work. 78L05 is a common IC, but could be a 78L09 or 78L06.

2. The negative end of your battery should be hooked to ground, not the regulator's input. The regulators input should be tied to the battery's (+) terminal.

For the reference voltage, you need to research: three-terminal voltage regulators...how they work.

For level sensing you need to research: voltage dividers...for both of the comparator's inputs.

Ken

[Slasherbaven]

Thank you for being patient, Ken.

Originally Posted by KMoffett

1. For the 78L12 to regulate it needs an input of +14V or more. If your battery is 12v then the regulator has have an output of 10v or less to work. 78L05 is a common IC, but could be a 78L09 or 78L06.

Originally Posted by KMoffett

For the reference voltage, you need to research: three-terminal voltage regulators...how they work.

I just read up on it, and it's pretty interesting.
Seems that the excess voltage dissipates as heat, and the hole on a regulator has to be attached to a heat sink.

Since I'm putting in a 12V, there'll be an excess of up to 5V, so I shall put a heat sink to be safe.

Thank you.

Originally Posted by KMoffett

For level sensing you need to research: voltage dividers...for both of the comparator's inputs.

I'm sorry, but level sensing is..?

Originally Posted by KMoffett

2. The negative end of your battery should be hooked to ground, not the regulator's input. The regulators input should be tied to the battery's (+) terminal.

There we go:


I'm still a bit confused as to why, but will reading up more details on regulators help?