2
\$\begingroup\$

I have designed a GSM/GPS based tracker which is installed at my bike. But it is draining battery too quickly.

So, I want to add a cutoff circuit when the bike is not running(battery not charging) for a long time and in addition, the cutoff circuit should reconnect power to load (my tracker) whenever the battery voltage rises to a preset leve (For say 12.6v).

I decided to use an op amp comparator based cutoff circuit. But the problem is I am unable to calculate hysteresis when the supply voltage of the op amp is changing (Because I will power the op amp from the battery itself).

Can I use a comparator hysteresis circuit if I have a variable voltage rail? If not what kind of changes to the circuit would I need to make this functional? enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ You may want to crop that image and re-post it, posts are forever and we want them to look good for future reference. \$\endgroup\$
    – Voltage Spike
    Mar 27, 2017 at 6:30

1 Answer 1

-1
\$\begingroup\$

If you want LOW POWER, at limited Temperature Excursion, try this:

schematic

simulate this circuit – Schematic created using CircuitLab

By the way, this schem is edited (added output Zener/diode clamp) to provide more stable Vhysteresis.

\$\endgroup\$
5
  • \$\begingroup\$ How to calculate the value of R5 when the supply voltage of te op-amp itself is changing??That is my concern!! \$\endgroup\$
    – S.Das
    Mar 27, 2017 at 5:56
  • \$\begingroup\$ I am attaching a circuit that i designed so far.. \$\endgroup\$
    – S.Das
    Mar 27, 2017 at 5:59
  • \$\begingroup\$ @S.Das downvote answers that don't answer the question \$\endgroup\$
    – Voltage Spike
    Mar 27, 2017 at 6:29
  • \$\begingroup\$ @s.das You did not ask for constant hysteresis. Nor does the circuit you attached accomplish constant hysteresis. \$\endgroup\$ Mar 27, 2017 at 14:20
  • \$\begingroup\$ Consider added an output clamp, to achieve more stable hysteresis. \$\endgroup\$ Mar 27, 2017 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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