1
\$\begingroup\$

I just need a single LED to light up when the voltage drops below a threshold like say 3.3 volt ( sharpness is not that important ). I know basic electronics like transistors,logic gates etc. On searching a bit, I think a comparator is what I need (all about which I know is that it gives a logic signal based on which input voltage is higher). Can I use just the comparator for the 3.4 volt LED and achieve what I need, from the link given below ? http://www.instructables.com/id/Li-Ion-battery-level-indicator/

\$\endgroup\$
1
\$\begingroup\$

A regulator is a good, simple idea; especially if you already have a regulated voltage (buck converter or similar) present. If that is the case, you can use this as the threshold voltage and simply scale down the battery voltage accordingly.

If not, you need to build a small circuit that produces a voltage independent of the battery voltage: The easiest solution is to utilize a zener diode in reverse bias (and a current limiting resistor.)

schematic

simulate this circuit – Schematic created using CircuitLab

Make sure to calculate the values for your specific design. There are also different types of comparator outputs, perhaps you need to drive the LED differently. You can determine the value of the voltage divider by considering that the voltage at the negative input needs to be the same as the threshold voltage (at the positive input). If I understood you correctly, you want the LED to turn on for battery voltages lower than 3.3V (V_th); that means you calculate $$ \frac{V_{threshold} * R3}{R2 + R3} = V_{@inverting} $$ where you arbitrarily choose R3 to have some resonable resistance (based on how much current it would draw) and your arbitrarily chosen \$ V_{th} \$ of 3.3V.

Consider using two comparators so you can have two thresholds; if you use single package with two comparators this hardly takes up more space.

\$\endgroup\$
  • 1
    \$\begingroup\$ Keep in mind that this circuit will always be draining the battery though R2+R3. This could impact battery life even with very large resistors here. I've actually seen a product that would drain the attached battery within 6 months even if you never turned it on because of a voltage divider used to low battery detect. \$\endgroup\$ – bigjosh Jun 15 '16 at 20:17
  • \$\begingroup\$ The battery has 4.2 Volts when fully charged. I need the threshold to be 3.3 Volts and so I would use a 3.3 volt zener diode. Since the negative input should also get the 3.3 volt, I would use R2 and R3 such that R3/R2=33/9. But as the potential across the battery drops over time (R2 and R3 being unchanged), the potential at the negative input will not be equal to that at the positive input (3.3 v) and hence wouldn't the LED light up for any significant drop in the battery's potential much before it reaches 3.3 volts ? \$\endgroup\$ – Adarsh Pryce Jun 16 '16 at 8:07
  • \$\begingroup\$ @user20962 You have to do the same line of reasoning but with the voltage that you want as threshold for switching the LED on, not the maximum voltage if the battery. The consequence is of course that the LED is off for any larger voltage and on for any smaller voltage. V_th * R3 / (R2+R3) = V_ref \$\endgroup\$ – caconyrn Jun 16 '16 at 13:17
  • \$\begingroup\$ @caconyrn I am not sure I completely understand. Please help me out a bit here. For what voltage of the battery should the voltage at the negative input of the op amp be equal to that at the positive input? \$\endgroup\$ – Adarsh Pryce Jun 16 '16 at 13:48
  • \$\begingroup\$ @user2062 I edited my anwer with an example, I hope that will help. \$\endgroup\$ – caconyrn Jun 16 '16 at 17:02
5
\$\begingroup\$

Simplest “battery low” indicator for a 3.7 volt lithium ion battery?

I don't think you can get simpler than this circuit...

enter image description here

Here is the datasheet for the chip shown, although there are lots of under-voltage sensing chips available from numerous vendors...

http://www.microsemi.com/document-portal/doc_view/11145-sg3546-pdf

Note that the battery must obviously have enough voltage left in it to light the LED to see anything so pick an LED with a low forward drop (red is a good choice).

\$\endgroup\$

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.