[SOLVED] Infrared sensor circuit flow

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i,

I thing it is not IR SENSOR CIRCUIT. It is Light Sensor circuit which is used to indicate the LED .

the Circuit is common emitter. if the base voltage is given, it triggers and the current flows form collector to emitter. so the IR LED emit the signal. Vc=6V .IC =2omA. beta=Ic/Ib ; normally beta for BC547 is 100 .
Ib= Ic/Bta
ib=20mA/100=.2mA. required theoretically.

In general, LDR resistance is minimum when it receives maximum amount of light and goes to maximum when there is no light falling on it.

so need to apply voltage divider and point out the voltage at base. if you use LDR and 1k resistor,
Vb= 1k/(Rldr+1k)*6V ( Rldr at sunlight 200k, night is 1k = Approx)

Vb(sunlight)= 1k/201k* 6v = .03V. (<=0.7 not sense in Vb) so the transistor is not ON.

if iVb reach the 1V , the LED Blinks .

normally this circuit is used to on the LED as based upon the light. At sunlight condition the LED OFF. At dark condition the LED ON.

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WhyWhy said:
Hi all,

I'm confusing the flow of this infrared circuit,can try explain to me ? Thanks.
https://www.buildcircuit.com/wp-content/uploads/2011/12/single-transistor-light-sensor-809x1024.jpg

Regards,
YY
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yes..correct . sorry i reversed it in my reply. this is the easiest way to explain the Flow of circuit. I used sunlight condition 200K and dark condition is 1K. that one is wrong

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shegzzyy said:
when light is falling on the ldr the resistance is low,the voltage drop across r3 will be high thus turning on the xsistor.when the light is obstructed,its resistance become high,voltage drop across r3 become low thus turning off the xsistor
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