automatic light detection curcuit

So I have made this circuit for my project. (Instead of that variable resistor, I have used a single 1k ohm resistor)

I know that a transistor can work as a switch and amplifier, but in this circuit, what role is it playing? Please explain in detail? Thanks guys.

  • \$\begingroup\$ What do you think? Does this circuit need an amplifier or a switch? \$\endgroup\$
    – Eugene Sh.
    Feb 6 '17 at 20:58
  • 1
    \$\begingroup\$ This smells a lot like a homework problems. Could you show some effort before we just give you the answer. \$\endgroup\$
    – vini_i
    Feb 6 '17 at 20:59
  • \$\begingroup\$ It appears LDR is detecting something to activate the LED. As there's more of that something, the BJT enters active mode and then it will eventually hit saturation mode. Does that sound like a switch or an amplifier? \$\endgroup\$
    – user103380
    Feb 6 '17 at 21:05
  • \$\begingroup\$ Think of a transistor of more like a current controlled resistor \$\endgroup\$
    – Voltage Spike
    Feb 7 '17 at 1:24

Depending on the resistance of the LDR, the transitor may be viewed as either an amplifier or as a switch - and there is not a sharp transition between the two modes.

When the LDR is high resistance, the base current will be low, and the collector current will be base current times the transistor's gain (beta). There will be significant voltage drop between the transistor's emitter and collector.

As more light hits the LDR, its resistance will drop, allowing more base current, and thus collector current. At some point the collector current will become limited by the collector load (resistor and LED), so the transistor will be saturated, with only 0.2 volts or so between emitter and collector.

When we use a transistor as an amplifier, we keep the base current low enough that the transistor will not be saturated.

When we use a transistor as a switch, we want to have the transistor either cut off (no collector current), or saturated (current limited by load resistance).


The transistor won't do much to turn on the LED if it isn't bright enough for the LDR conduct enough current to effectively drop in resistance below 25kΩ, which may be bright daylight.

Any Common Emitter with output on collector is an inverting base current amplifier. With no Re value the gain is high above the Vbe threshold of 0.6V.

If you expected the reverse with the LED coming on at dark, then some inversions and bias changes are needed.

If you want to change the sensitivity to sunrise then a larger value for the 1K shunt to ground is needed.


In this circuit probably its a switch. The Ldr and the variable resistor will form a voltage divider. Usually the ldr's resistance is low when the light is falling on it. Hence the applied voltage will appear across the variable resistor. This voltage is fed to the transistor base with respect to its emitter. This causes the transistor to go to saturation mode. At this mode transistor allows the current to pass from its collector to emitter which in turn lights the led. At this stage transistor is a closed switch which connects the led to the ground. In low light or darkness the ldr's resistance is very high and all the applied voltage appears across it therefore there will be no voltage at the variable resistor and across the base of the transistor . So at this state transistor acts as a open switch also known as cut off mode. There is no flow of current through the transistor and hence led will be turned off. Correct me if i am wrong.


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