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automatic switching of street lights by using ldr and relay

 hey me and my friends were ton a project of automatic switching of a street light depending on the brightness in the surroundings.
so we decided to use an ldr(light dependent resistor). here in this circuit ldr has low resistance in brightness(only few ohms) and very high resistance in complete  darkness(nearly 1M ohm).so the transistor Q1 becomes on in day times and Q2 becomes off and hence the relay remains in NC(normally closed) state and the lamp will not glow. at night the transistor Q1 becomes OFF and Q2 becomes ON and hence the relay goes to NO(normally open) state and the lamp glows.and the sensitivity of the device can be varied by adjusting the potentiometer 
now i just want to know are there any defects in this circuit so can u please help me in knowing the drawbacks and correcting them
and i want to know whether i can connect a relay in place of a fuse as shown at bottom


Picture of automatic switching of street lights by using ldr and relay
Circuit 1.jpg
circuit.bmp
Riju952 years ago
The circuit is nice. It works efficiently but I the battery is getting hot. I think somewhere the electricity is being lost as short circuit. Can anyone suggest a method to make it more efficient? Instead of relay can we put any load of limited voltage requirement????
I have made circuit shown above. but its working in reverse.
i have connected bulb instead of relay. The bulb glows brighter when light falls on it and it dims it i cover the ldr. Please help me fixing the issue.
Thanks in advance,
Sandeep.
tirthankar4 years ago
This is the best circuit ever made.......just use a 6V relay..........
I'd probably just wired the transistors as a darlington pair, to get much higher gain, so the relay can snap on when the light gets brighter.

YOur circuit also probably has no hysteresis, so when the trigger lighting level is marginal, the thing will blink on and off.


The final circuit won't work, because nothing ever completes the circuit.

As mentioned - theres no hysterisis - it needs some feedback so that when q2 is conducting, it biases q1 to not come on until its brighter than the level required to turn q1 off - otherwise it may flicker as steve says.
This could be accomplished (I think) by adding a resistor and capacitor between Q2's Base and ground, so that the signal is delayed - some experimentation to figure out the right values might be necessary.


The second circuit needs a smoothing capacitor after the bridge rectifier or the bouncy rectified AC will wreak havoc on the transistors.
You also want a feedback diode on the relay coil to stop the relay from kicking current back into the emitter of Q2 when it turns off, potentially destroying it.

Best of luck with the circuit!  prototype it, and if it works then post an instructable!
Bum. Missed the missing flywheel diode ! +1
spartans (author)  steveastrouk4 years ago
Sir , i cant get you .will you please explain in detail
You must wire a diode across the relay, with the cathode to the +5V supply, and its anode on Q2 collector.
As asteve says a darlington pair would provide higher gain, though if the relay's sensitive it'll not need a pair of transistors.

Though you'll need to finish the rest of the circuit to make it work.

A simpler one would have the LDR above the transistor and then the potentiometer below, with the base of the transistor between, the relay above the collector and the output hitting the ground rail...
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