Fancy 12V car battery level indicator

Hi, I have been trying to design a circuit to monitor my 12V leisure car battery, I'm thinking this should be fairly simple, but its been years since I did any real electronics.

So the basic idea: Use the circular bargraph to indicate the level of the battery when a momentary push button is pressed
the circular LED bar graph: https://www.sparkfun.com/products/11492

So the range of the battery would be something like 10.5V to 13.5V, there are 32 LEDs in pairs, so essentially 16.

1. So if we take the first LED pair, this should illuminate when less than 10.5V, so that should be easy, as long as there is some life in the battery it should illuminate

2. The next LED should only illuminate at the next step, 16 LEDs, 3V range 16/3= steps of 0.2V So at the following steps:
10.5, 10.7, 10.9, 11.1, 11.3, 11.5, 11.7, 11.9, 12.1, 12.3, 12.5, 12.7, 12.9, 13.1, 13.3, 13.5.
So for 10.7V I guess I should use resistor before each LED.

3. The LEDs should light up after a momentary push switch is triggered, I would like this to be graceful, so the ring lights up smoothly one LED at a time as the voltage increases, I guess capacitors between the resistors and LEDs?

This is the simplest way I can think of doing it, I have seen circuits which use a LM339 chip, but that can only monitor 4 voltages, perhaps 4 of these in series could do the job? http://www.electroschematics.com/7068/lm339-lm239-lm2901-datasheet/ I think this is the way to go rather than using inline resistor/ capacitor pairs?

So I am a clear novice when it comes to electronics (but I'm enthusiastic), so if anyone can help me with this I would be more than happy to put together a how to so others can copy/develop it, something similar has been done before: http://www.instructables.com/id/Car-battery-tester/
In that example only one bar is lit, I want the other bars to stay lit.
I can use Eagle to make the PCB diagram and get it printed, but I need to figure out what components I need and how to do it!

Thanks for any help.

Gary

 

Hi Thanks for your reply,

Well, this is a battery status indicator, so jumps of 0.2v or thereabouts as I have 16 LED positions, otherwise some LEDs would never be used.

With more searching I have discovered the LM3914. This can cope with 10 LEDs, so I am 6 short, perhaps 2 LM3914 chips?

A few resistors and potentiometers are needed, I have found a similar post here: http://www.electro-tech-online.com/...gn-ideas-reviews/108340-lm3914-voltmeter.html

Thanks again

 

Attached is a partial circuit, I got stuck.

So I've gone with the LM3914 IC set to Bar mode (pin 9)
On the wiring diagram I saw the LEDs were connected cathode to anode from the IC, with this ring LED setup they share 4 common cathodes, so how do I get around this?

I am returning to electronics after 15 years and a very bad teacher, so I'm learning as I go, any obvious problems?
The capacitors are going to be used to create a fade out effect, I guess this is where they should go?
You will notice I have only concentrated on the right side LEDS, I will use a second IC for them once I have this side working.

Thanks for any input

Gary

 

The LED diagram should be right as it is from the retailer, they provide the eagle libraries.
https://www.sparkfun.com/products/11492

so for now if we ignore the fade effect, can this LED setup be controlled per LED? Or do I have to have a major rethink?

thanks

 

Thanks Rleo, that is along the lines of what I want, although I have 16 LEDs so I will have to configure 2 LM3914s accordingly, but for now If I can get half of them designed the other half should be fairly straight forward.

 

Thanks Dave, that makes sense now, I will start playing with transistors!
I guess no other version of the 3914 which works the other way?

 

So, the transistor route...
As the 3914 is a sink I need a PNP Transistor, right?
So if I only use 1 LED for example (see attached) this should work?

Thanks again for all your work, I wish the internet was around when I was being taught (badly) this at school!

 

So if the previous is correct my transistors will look a little like this?
(Note pins 11 and 10 are not needed)

 

Ok, so before anyone points out some clear errors, here is a corrected version with the +V coming into the PNPs and out to the common cathode. I have include the 2nd IC, I'm pretty sure I will need some more resistors between R2 and the power source otherwise the current will go straight to ground! So I will need to do some calculations.

But for now, how is it looking?