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For any condition of light or dark the photoresistor will have a value of some number of ohms. That value in relation to R2 and R3 will determine if the transistor is on or off or somewhere in between.
If the transistor is on then current will flow through the coil and contact will close. It the transistor is off then no current will flow through the coil. As the transistor makes its transition from on to off or vice versa the coil will draw current until it either pulls in the contacts or releases them.
This might happen if the circuit was "observing" a sunrise or sunset.
Right now, i bulit this circuit which switches on a fan powered by a 9V battery. So my LDR sensor circuit is powered by 6V battery. Is there any way to not use the Relay and only use one battery supply ?
The circuit will work when it is powered by a 9V battery. Then simply replace the relay coil with the fan, leaving the diode where it is. If the transistor is strong enough to drive the fan then it will work.
I connected the fan parallel to the diode.. it doesnt work. Looks like that part of the circuit cant supply enough voltage to power a fan. It can however power an LED.
Is there any way to add a battery or increase the voltage ?
It isn't voltage that's the problem. Your motor will run striaght from the battery won't it? So how can it be a problem with the voltage?
It's current that's the problem, the small transistor can't supply enough current to drive your motor. You're lucky the transistor still works and you havent destroyed it.
I don't like these circuits. The light level at which the relay turns on depends on the charicteristics of the transistor, the relay and the power supply voltage. There isn't a propper switching action either, if you connect an LED it won't cleanly switch on or off but gradually get brighter or dimmer.
Now you could fix this circuit so it will drive the motor. You could use the relay or replace the transistor with a darlington pair (two transistors connected together, Google for it if you don't understand).
A better solution would be to use a comparator, like this: **broken link removed**
Note you can connect your motor in place of the realy, note you'll still need to use a darlington pair or you could use a MOSFET in which case you should remove R5 and short circuit R4 (the 741's output when low is about 2v, this is enough to tun a ransistor on but not enough for a MOSFET; this is why they are needed for the former and not for the latter.
Meanwhile here's come reading for you if you don't know about comparators.
Hint:
You can make the circuit I posted above even more reliable by adding a 1M resistor from the output of the uA741 to it's + input as this will introduce some hysteresis.
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