Calculating Resistors for a Photocell Dark Switch Circuit

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To create a photocell dark switch circuit for an LED, a cadmium sulfide photocell is used to control a transistor that switches the LED on and off based on light levels. The circuit operates at 12V and requires careful calculation of resistors to ensure the correct current flows to activate the transistor. It's important to measure the photocell's resistance under the desired light conditions and ensure that the current exceeds the transistor's base requirements. Using a Darlington transistor or a MOSFET can enhance performance, and a resistor should be placed between the base and emitter to achieve the necessary voltage. Experimentation and reference to the photocell's datasheet are recommended for optimal results.
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Hi I am trying to make an LED 3.5V 0.02A run in a circuit that will turn it off when visible light comes on. I have a cadmium sulfide photocell I took out of a night light. The power input is 12V 250mA. I understand I need to use a transistor as a switch for the LED and I know how to calculate the resistor for the LED. I am having trouble figuring how to calculate and place resistors for the photocell parallel so that the current turns on the transistor to the LED. Any tips on where to start or formulas? I am getting about 4V on after the photocell when lights are on if that helps. Thank you!
 
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A schematic would be helpful.
 
The photocell's resistance changes an awful lot depending on the incoming light. You light find some clues in its datasheet, but it's better to experiment in the real situation. Ah, and don't hope two photocells to be identical.

Once you've measured the resistance at the desired light threshold, compute how much current flows through it from 12V, check it exceeds what the transistor's base needs (take a Darlington if needed, like BC517, or a MOS like BSS170), add a resistor between base and emitter that gives 0.65V at this curernt.
 
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