elfvenlord
New Member
Hi All
Im trying to build and understand the attached circuit but im a bit lost on how it works.Basically all it is is a low voltage alarm that sounds the buzzer when the battery voltage is below the set voltage alarm level,which i am guessing is set with the voltage divider from the two resistors.when the voltage is below the set alarm voltage the comparator realises that,and switches to the +V.The problem is how can the voltage divider stay constant if the input voltage from the battery is dropping,unless im not looking at it the right way.
This circuit is supposed to be for my rc plane battery wich is 5V and is supposed to trigger the alarm when the battery voltage drops past about 4.0V.
Can anyone just briefly explain how it really works,I would most gratefull.
Thanks :wink:
Im trying to build and understand the attached circuit but im a bit lost on how it works.Basically all it is is a low voltage alarm that sounds the buzzer when the battery voltage is below the set voltage alarm level,which i am guessing is set with the voltage divider from the two resistors.when the voltage is below the set alarm voltage the comparator realises that,and switches to the +V.The problem is how can the voltage divider stay constant if the input voltage from the battery is dropping,unless im not looking at it the right way.
This circuit is supposed to be for my rc plane battery wich is 5V and is supposed to trigger the alarm when the battery voltage drops past about 4.0V.
Can anyone just briefly explain how it really works,I would most gratefull.
Thanks :wink: