I hope I'm posting this into the right section, if not you can move it to wherever.
I want to know if when constructing a dark activated circuit like the one below:
Does the circuit always draws current from the batteries?
What can I do to prevent it from drawing current from the batteries during the day? if at all possible? or draw as less power as I can and how do I calculate how much power I'm drawing?
The picture isn't mine, I'm going to follow the diagram tho, but I came here for suggestions and to understand how to calculate the power consumption of the circuit during the day.
Thanks in advance,
Hmmm I want it to be automated so getting the batteries out is not good enough.
Is there a switch the can toggle it self through the LDR?
I'm pretty new to electronics so I would also like to understand how do I calculate the amount of power I'm losing during the day with the resistors and transistors and all the rest.
Thank you for the reply.
Current is always flowing through the 100K resistor and the light dependent resistor (plus a small amount of leakage current through the transistor and LED).
The total resistance is the sum of the resistors (ignoring the leakage) and the current is determined by [u]Ohm's Law[/u]:
Current in Amps = 3V / (100K + LDR)
Any circuit is going to draw some current while it's running, even if it's just monitoring the light level waiting for it to get dark (or light).
Much more current flows through the LED when it's on, so if it's on all night and off all day the current during the day isn't significant to the battery life.
Great Thanks so it's not a lot... I'll be able to deal with the power consumption during the day...
And if I put even higher transistor it'll draw even less right?
But I'll need to make sure the transistor will have higher resistance than the transistor.... right?
Thanks again for answer.