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I am new on electronics and i was practicing on transistors. I have a working circuit below. The LDR I use allows current and letting LED smoothly based up to amount of darkness. The Voltage after 2.3 k resistor is 0.7 maximum .. All i want to do is to let led run on the darkest moment. So only if it is fully dark the LED must light up .. Not smoothly relative to darkness. What should i do for this ? FIX: the source voltage is 6 Volts

Circuit Scheme

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  • \$\begingroup\$ Are you willing to consider a 2nd BJT? \$\endgroup\$
    – jonk
    Commented Nov 29, 2017 at 17:49
  • \$\begingroup\$ Maybe you should find the LDR's resistance at the darkest moment. And design R2 accordingly. Possibly would be very high value. \$\endgroup\$
    – Mitu Raj
    Commented Nov 29, 2017 at 18:02
  • \$\begingroup\$ So another 2N2222 will do the job ? \$\endgroup\$
    – user7685914
    Commented Nov 29, 2017 at 18:03
  • \$\begingroup\$ MITU.. If I increase r2 so how can I provide 0.7 volt 2N2222 transistor requires.? \$\endgroup\$
    – user7685914
    Commented Nov 29, 2017 at 18:13
  • \$\begingroup\$ Add a comparator... \$\endgroup\$
    – Trevor_G
    Commented Nov 29, 2017 at 20:32

1 Answer 1

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If you want it to accurately switch you really need something more active like a comparator. This circuit will pull around 15mA through your LED when the voltage at the top of the LDR exceeds the voltage set by VR1.

schematic

simulate this circuit – Schematic created using CircuitLab

There are other ways to do it with discrete parts, but seriously, you just end up with many more parts and a bigger circuit.

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