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I want to build a battery level indicator circuit for a 12 V sealed lead acid battery.

The battery indicator circuit will have two functions:

  1. It will power 4 LEDs to show the level of the battery
  2. At the lowest level it should disconnect the load from the battery to prevent deep discharging. To achieve this, I'm considering using a MOSFET (IRLML6244) as a switch. (comments on this approach welcome)

So far I've seen predominately two approaches to achieve this. The first method uses zener diodes and transistors to switch the LEDs as shown here

The other method uses op-amps as comparators to switch the LEDs as used here

I can't really see a clear advantage of the one over the other. I think the components costs will be roughly similar, but maybe a bit more soldering when using the zener approach. Keeping in mind that I also want to switch the MOSFET, which of these methods (if any) is the recommended method to use in my battery indicator circuit?

Not sure if it makes a difference, but my load won't draw more than 1A.

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12V battery level indicator circuit with LED bar /dot display LM3914. On the Internet. ttp://www.circuitstoday.com/12v-battery-level-indicator-circuit-led-bargraph

In the circuit diagram LEDs D1 toD10 displays the level of the battery in either dot or bargraph mode. Resistor R4 connected between pins 6,7 and ground controls the brightness of the LEDs. Resistors R1 and POT R2 forms a voltage divider network and the POT R2 can be used for calibration.

The circuit shown here is designed in order to monitor between 10.5V to 15V DC. The calibration of the circuit can be done as follows. After setting up the circuit connect a 12V DC source to the input. Now adjust the 10K POT to get the LED10 glow (in dot mode) or LEDs up to 10 glow (in bar mode). Now decrease the voltage in steps and at 10.5 volts only LED1 will glow. Switch S1 can be used to select between dot mode and bar graph mode. When S1 is closed, pin9 of the IC gets connected to the positive supply and bar graph mode gets enabled. When switch S1 is open pin9 of the IC gets disconnected to the positive supply and the display goes to the dot mode.

With little modification the circuit can be used to monitor other voltage ranges. For this just remove the resistor R3 and connect the upper level voltage to the input. Now adjust the POT R2 until LED 10 glows (in dot mode). Remove the upper voltage level and connect the lower level to the input. Now connect a high value POT (say 500K) in the place of R3 and adjust it until LED1 alone glows. Now remove the POT, measure the current resistance across it and connect a resistor of the same value in the place of R3. The level monitor is ready.

If you only want 4 LED's Use Only D10, D9, D8, D7 anything below D7 is dead anyway.

Price: $1.82 parts.arrow.com

enter image description here

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  • \$\begingroup\$ Thanks, I saw that circuit but thought using the op amps or zeners will be slightly cheaper. However, looking at this circuit again it looks a lot simpler to build. More importantly, do you think instead of driving a LED, I can connect the last LED to the gate of a MOSFET (in graph mode)? That way when the battery is too low it disconnects the load from the battery? \$\endgroup\$ – Johan Apr 16 '15 at 19:24
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Absolutely you can use any LED output pin to drive a shut off circuit, or function as needed to control your project. The IC will probably require some kind of pre-driver since it appears to only sink current not supply it. Have not checked specs. --*///*--

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