Electronic – Battery cutoff circuit with NE555

555low-battery

For a project, I need a simple battery cutoff circuit for protecting a 9V battery from over-discharging.
(by simple, I mean with only a few off the shelf components, and no micro-controllers)

Looking for inspiration on the web, I found this article, which at first sight looked exactly like what I wanted/needed.

It uses the voltage comparators of the NE555 to operate a relay, based on the resistors value.

Everything works as expected on the breadboard, but there's something I don't understand.

The NE555 chip, as well as the voltage dividers and filter cap will obviously consume some current, even when the relay is off, and the main load disconnected.

In such conditions (relay off), my test circuit drains 0.006 amps from the supply.
So in fine, the battery may still be damaged from over-discharge.

So what's happening here? Is such a circuit OK, and 0.006 an acceptable value, am I missing something, or is the article flawed?

For info, here's the (poor) schematic given on the website I mentioned:

enter image description here

Best Answer

I can see a few problems with this circuit.

The first one is about the 7805 voltage regulator.In order to get a stable +5V output,the input voltage has to be at least 7,5V.As a battery is discharged,its outputted voltage drops: enter image d

enter image description here You stated that the circuit is to protect against over-discharge,which means to disconnect when the battery is empty. At some point,the \$V_{in}\$ will become too small for the 7805 to work properly.In other words,the circuit may cut off the power earlier and in an unwanted way,the battery still being able to supply power.You can't really know how is a regulator going to behave if you don't run it within its operating conditions(which can be found in the datasheet).From this point of view,the design is unreliable.

Let's suppose the 7805 supplies the 555 with what's needed as intended and exactly when needed,the relay is turned off.You stated that the timer IC will still continue to draw current(this current is called quiescent current):a minimum of 3mA and a maximum of 6mA.The battery will over-discharge like this.

I believe you should choose another circuit.

Related Topic