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light dark sensor

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trymyx

New Member
need help in explaining the circuit.
 

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birdman0_o

Active Member
1) The potentiometer and the LDR form a voltage divider, which, when it reaches Above 2* 0.65V at the base of the first transistor, it slowly turns on the darlington pair of transistors which will turn on what looks like a relay. The diode is to protect the back EMF from the relay when it turns off.

To summarize, when there is enough light / or not enough, the relay will turn on.
 
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trymyx

New Member
1) The potentiometer and the LDR form a voltage divider, which, when it reaches Above 2* 0.65V at the base of the first transistor, it slowly turns on the darlington pair of transistors which will turn on what looks like a relay. The diode is to protect the back EMF from the relay when it turns off.

To summarize, when there is enough light / or not enough, the relay will turn on.




can u give me detailed explanation.
 

mneary

New Member
can u give me detailed explanation.
That *is* a detailed explanation. :( Perhaps you can tell us what part you do not understand.

Feel free to research key terms first, such as LDR, voltage divider, transistor, relay, and darliington.
 

trymyx

New Member
thanks for the help from both of u.

there is some more questions i would like to ask.
when i did the circuit in bread board there was a chattering sound from the relay when the light is not stable to the ldr. to eliminate the chattering sound i have put a capicitor valued 10uf to the base of the tr1 and the emiter of tr2(ground). and the chattering sound is solved.
but i would like to know how to calculate the correct value for the capacitor? ( i tried with a high valued capacitor(22pf) and low valued(1.4uf) capacitor which i had. but only the 10uf didnt make any noise and stable in switching. )

"thanks for the help in advance"
 

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trymyx

New Member
help

when i did the circuit in bread board there was a chattering sound from the relay when the light is not stable to the ldr. to eliminate the chattering sound i have put a capicitor valued 10uf to the base of the tr1 and the emiter of tr2(ground). and the chattering sound is solved.

How to calculate the correct value for the capacitor?

( i tried with a high valued capacitor(22pf) and low valued(1.4uf) capacitor which i had. but only the 10uf didnt make any noise and stable in switching. )

:):confused::)
 

sheldonstv

New Member
can u give me detailed explanation.
be careful- txt speak is not allowed on this forum- you will have your post jumped on by audioguru and nigel-im the only txt speak poster here according to audioguru.....
 

Mr RB

Well-Known Member
It's a terrible circuit. The darlington will have a really soft turn-on and light changes very slowly. The relay will be pulled in/out unreliably with chattering etc.

For 2 NPN transistors and similar circuit complexity (maybe adding 3 resistors) you can make a schmidt type version of the circuit with some hysteresis that will switch very cleanly.
 

trymyx

New Member
It's a terrible circuit. The darlington will have a really soft turn-on and light changes very slowly. The relay will be pulled in/out unreliably with chattering etc.

For 2 NPN transistors and similar circuit complexity (maybe adding 3 resistors) you can make a schmidt type version of the circuit with some hysteresis that will switch very cleanly.



where do ineed to connect the resistors and which values???
 

trymyx

New Member
circuit with all values

this is the circuit with all values to the components
 

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Mr RB

Well-Known Member
Ok, I added 2 more resistors and 1 small cap.

The relay transistor is the detector, the other transistor is inverse operation and provides hysteresis via the 1meg resistor.

To reduce the hysteresis use a higher value than 1meg, but I think the 1meg will work ok. It's been a few years since i built a light operated switch like this but those parts values look ok. It should switch clean and fast.

The value of the small cap can be anything, probably 0.1uF or less should be ok.

Note too if you are using a 9v battery and 6v relay you might want to put a resistor (R ?) in series with the relay to keep it's operating voltage at about 6v on the relay coil, this will use less battery power.
 

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trymyx

New Member
hi

Ok, I added 2 more resistors and 1 small cap.

The relay transistor is the detector, the other transistor is inverse operation and provides hysteresis via the 1meg resistor.

To reduce the hysteresis use a higher value than 1meg, but I think the 1meg will work ok. It's been a few years since i built a light operated switch like this but those parts values look ok. It should switch clean and fast.

The value of the small cap can be anything, probably 0.1uF or less should be ok.

Note too if you are using a 9v battery and 6v relay you might want to put a resistor (R ?) in series with the relay to keep it's operating voltage at about 6v on the relay coil, this will use less battery power.




hi. is there any way that i can keep the the darlington pair and put the resistors to make the circuit stronger.


"in my earlier circuit i put the one without the capacitor. now there is the circuit with the capacitor to eliminate the chattering."

thanks for the help you are giving :)
 

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  • light darksensor circuit with values.jpg
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