greatpeople1 said:
well will any body explain a bit of working of that circuit thts posted by negil :?: :idea:
what will hysteris do i know what hysteris is but who will it affect the circuit plz.... ! :idea:
I'll explain the circuit as drawn before introducing hystersis.
The threshold voltage is set by P1. Say, for example, it is set to 6 Volt.
With no light, LDR1 has a high resistance so the voltage at pin 2 will be low. Thus the output of the Op Amp will be high and current will flow through R4 into the base - emitter of Q1. Q1 will be on and so the relay is operated.
If the intensity of the light falling on LDR1 is gradually increased, the voltage at pin 3 will rise in sympathy. As it approaches 6 Volt, the output of the Op Amp will start to fall. Thus the current into the base of Q1 will reduce (as will the collector current) and as the light intensity increases further, the current will reduce to the point where the collector current falls below the relay release current and the relay will release.
When the light level is reduced, the relay will operate again once the voltage at pin 2 falls to about 6 Volt.
The point to note is that the base current (and therefore the collector current) changes slowly as the light level changes near the threshold level. So the threshold is not clearly defined.
This can be solved by introducing hystersis. Connect a 100 k resistor between pins 6 and 3. This provides positive feedback. Now when the voltage on pin 2 approaches the threshold, the feedback speeds up the change so the base current into Q1 changes very rapidly.
You can use an Op Amp such as the LM301 or whatever you can buy. Let us know which one is the cheapest and we will advise you if it is suitable.
The supply voltage will depend upon the voltage that the relay needs. The Op Amp will operate over a wide range of voltage. So if you have a 12 Volt relay, then make the supply 12 Volt.