This is not an instructable but a question. See the images for circuit here the 9v battery represents a 12 DC V from an Adapter and the Capacitor represents the Chargeable Battery. Diodes are placed before Chargeable Battery and LED in the direction of current to prevent current in opposite direction. Suppose I have a Chargable Battery of 9v, A 12 Volt adapter (12 volt to charge 9v battery), and a segment LED Strip (3 LEDs in Series.) Now I connect discharged 9V Battery to LED, and adding the 12V DC Input in Parallel as well. I want 12V DC source to charge Battery and Light up LED. I will add protection Diodes as well in positive terminal of led and battery so the current wont flow from the battery. Will the LED light up at 12V at its full intensity? I found somewhere that while a battery is charging, adding something in parallel will draw less current at first and then the Amperage will increase with increase in battery level and then after a lot of time, it will draw as much as current as it would in 12V directly? Is it so, how can I avoid it, i.e, I need the Load (here LED) to get 12 V while that same 12V source should charge the Battery. Sorry for any grammatical mistakes because I am in a hurry. Please answer me. Thanks.
Topic by The Engineer Pro - Mohammed Ayaz Quadri 2 years ago | last reply 2 years ago