0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

I am working on a battery charger with indicator. I am using a LM358 to compare the voltage of my battery with a reference voltage of 8V from 7808. My input supply is 12V. The battery is connected to pin 2 (inverting input) and 7808 output is connected to pin 3 (non inverting input) of LM358. The Comparator is connected to 12V VCC through a 4.7k Resistor. Indicator LED is connected on pin 1 of LM358.

The expected output is that once the battery crosses 8V, the output of the comparator must swing low and LED will turn OFF. Initially, the output is high and LED glows as expected. But when the inverting pin crosses 8V, the output doesn't swing (High to low). However, when I touch any metal part to any component (lead of the component), the output swings low (from High) and LED goes OFF. But this doesn't happen automatically, every time I touch a metal I see the swing.I want this to happen automatically.

And if the battery is above 8V, and when I connect it for charging, the LED does not glow; which is correct. It is just the transition from High to Low I need. Can anyone help/suggest me what am I missing here?

Note: LM358 is used in open loop configuration and no feedback is given.

P.S.: I have inserted the schematic. Sorry if it is not clear, I am new to the interface. Thanks in advance.

\$\endgroup\$
20
  • 4
    \$\begingroup\$ You're really going to have to at least show us your schematic if you seriously expect any help. \$\endgroup\$
    – brhans
    Commented Jun 7, 2018 at 13:28
  • 3
    \$\begingroup\$ The battery is connected to pin 2 (inverting input)... Welcome to this site! We really like schematics here so we have a tool to draw them. Edit your question and add a schematic, you can also include a picture of as long as it is clear enough. No schematic, no answer I'm afraid. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 7, 2018 at 13:30
  • 3
    \$\begingroup\$ your circuit simply doesn't make sense and hence doesn't work – this is not a noise problem. You've basically build a car where the motor sits on the front seat, isn't connected to the gas tank at all, and are complaining that your car only moves if parked downhill, and whether that's a problem with the tire pressure. \$\endgroup\$
    – Marcus Müller
    Commented Jun 7, 2018 at 14:02
  • 6
    \$\begingroup\$ "I will eliminate R3 only if someone can assure me that is what causing the problem" I sense this will not bode well for getting help here. \$\endgroup\$
    – winny
    Commented Jun 7, 2018 at 14:04
  • 3
    \$\begingroup\$ I will eliminate R3 only if someone can assure me that is what causing the problem R3 is a problem. Using a resistor, especially such a high value one, in the supply line is "silly", it serves no purpose. It is not said that that is the only problem this circuit has, there might be more. Example: if D1 is really a 1N4148 then for sure it will break itself as the current through it is more than double what a 1N4148 can handle. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 7, 2018 at 14:08

1 Answer 1

Reset to default
1
\$\begingroup\$

The current driving the LED has to come through R2 (1 kohm). That's fed by the LM358 from its power pin, which means it has to come through R3 (4.7 kohm). That's part of the problem. 5.7 kohm in series with a LED would probably still light it OK, but I suspect that when the LED is on the 358 browns out and stops working until the supply voltage recovers. If you really want R3 in there, try 4.7 ohm instead of 4.7 kohm.

\$\endgroup\$
1
  • \$\begingroup\$ Yes, I eliminated the 4.7K in the circuit and connected my comparator directly to 12V. And Voila! it worked. That was silly of me to use a 4.7K, I was working with some different logic/reason with that Resistor. I just need to check it with my battery completely discharged and see how much time it takes to reach the reference and get the transition.Thanks to all the guys for your valuable information and support. And do I need to update the schematic? \$\endgroup\$
    – New_user
    Commented Jun 8, 2018 at 11:46

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.