1
\$\begingroup\$

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

\$\endgroup\$
  • \$\begingroup\$ It looks like the relay would be on all the time.Well,the 7805 would provide the power.Also,I don't understand why the output is connected to the relay and diode. \$\endgroup\$ – Daniel Tork Aug 4 '16 at 17:06
  • 1
    \$\begingroup\$ @DanielTork Pin 3 is low when Vin > 2/3 Vcc \$\endgroup\$ – Doodle Aug 4 '16 at 17:07
  • \$\begingroup\$ The problem I can see here is that it is possible to disconnect the battery... if there was a seperate supply for the 555, any solution I can think of still will pull at least 3mA due to the 555 IC current consumption \$\endgroup\$ – Doodle Aug 4 '16 at 17:13
  • 1
    \$\begingroup\$ @DanielTork No, the relay wouldn't switch on. The diode is a back-emf diode, one of the basics. If the voltage on pin 2 is >2/3 Vcc, then pin 3 will be high (+5V). Therefore there is 0V across the relay, it won't actuate. Once the voltage on pin 2 is < 1/3 Vcc then pin 3 will go low (0V) there is now 5V across the relay and it actuates. The 555 can sink around 200mA so a 5V relay isn't going to kill it. \$\endgroup\$ – Doodle Aug 4 '16 at 17:35
  • 1
    \$\begingroup\$ This whole proposal has many problems - look up the quiescent current of a 7805, and consider how much power the relay coil will waste. There are circuits designed for this job. \$\endgroup\$ – Chris Stratton Aug 5 '16 at 17:34
2
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ If it were an Alkaline (primary, ie, "throwaway") battery it is not all that clear what preventing overdischarge will accomplish. Rechargeable cells typically should not be drained below 1.1v per cell, but 6.6v is marginal for a 7805 (there apparently now also exist 7 cell versions). Also there is the concern of behavior once the regulator goes into saturation - it may not make the circuit stay "off". \$\endgroup\$ – Chris Stratton Aug 5 '16 at 17:36
  • \$\begingroup\$ @Chris Stratton Why would the 7805 enter saturation as you mention? \$\endgroup\$ – Daniel Tork Aug 5 '16 at 17:48
  • \$\begingroup\$ I probably should have said dropout - but entering saturation is typically the cause of the behavior known as dropout. \$\endgroup\$ – Chris Stratton Aug 5 '16 at 17:52
  • \$\begingroup\$ I don't really understand these terms \$\endgroup\$ – Daniel Tork Aug 5 '16 at 17:55
  • \$\begingroup\$ These are the terms and mechanisms behind your awareness that the 7805's input voltage must be substantially higher than its output for regulation to properly occur. There was a question here recently that went into some depth of the details about dropout behavior. \$\endgroup\$ – Chris Stratton Aug 5 '16 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.