i would like to know how this circuit work.

 

Your circuit appears to be an ambient light detector with hysteresis.

When the circuit is in the dark, the LED will be on and when the circuit is exposed to light the LED will turn off. The hysteresis is there to prevent the output from dithering when the light level is at the detection threshold as set by the potentiometer.

There is no power supply indicated so I would guess that the circuit was designed to operate at around 5 volts DC.

hgmjr

 

can you explain about how the current flow and also about the voltage...

 

In the absense of light, the LDR (light dependent resistor) exhibits a very high resistance. The high resistance insures that transistor Q1 is turned off. With Q1 off, Q2 is turned on. This lights the LED. With Q1 turned on, the diode and series resistor present additional loading on the LDR that must be overcome when the LDR is exposed to light. This is the hysteresis that I mentioned.

In the presense of sufficient light, the LDR exhibits a low resistance which turns on Q1. With Q1 on, Q2 is turned off and so the LED is turned off.

Sorry for the abridged description. I don't know how deep you transistor knowledge is.

hgmjr

 

what is the purpose of putting r3, r4 and r5?

 

what is the purpose of putting r3, r4 and r5?

Your question leads me to believe you are unfamiliar with Bipolar Junction Transistors (BJT). You should be able to benefit from reading the material in the AAC ebook that introduces the basics of BJTs at this link. By reading this material you will gain the familiarity you need to better understand these very interesting electronic devices.

Enjoy,
hgmjr

 

why is it when Q1 is on, Q2 is off?
and why is when Q2 is on, Q1 is off ?

 

why is it when Q1 is on, Q2 is off?
and why is when Q2 is on, Q1 is off ?

When Q1 is saturated, it pulls the base of Q2 below 0.7 volts. With Q2's base below 0.7 volts, it turns off.

When Q2 is saturated, the ON/OFF state of Q1 is conditional on the resistance of the LDR. If the LDR is not being exposed to light then its resistance is high. This high resistance allows Q2 to turn Q1 off through the 4.7K resistor and the forward biased diode. If the LDR is exposed to sufficient light, then its resistance will be low. If its resistance is low enough, it will prevent Q2 from turning Q1 off.

What you have here is a circuit with hysteresis. When the LDR is starved of light, Q1 turns off. This permits Q2 to turn on and light the LED. Now with the LED on, if you expose the LDR to bright light Q1 will turn on and Q2 will turn off and the LED will go off.

This circuit appears to be designed to detect sunrise and sunset. It lights the LED at sunset and turns it off a sunrise.

hgmjr