Does this burglar alarm circuit design function correctly?

[swty todd]

We have to make a miniproject as part of my engineering course. I chose the above burglar alarm. I thought that there were some mistakes and I have made some changes (therefore some things are written in black). I did be grateful if you guys could confirm whether the circuit will work theoretically and practically. Let me know if I have made some mistakes. Here IC2 uA 741 is wired as a sensitive comparator ,whose set point is set by R6 & R7.The voltage divide by LDR and R9 is given at non inverting pin of IC2.At standby mode these two voltages are set equal by adjusting R9.Now the out put (pin6) of comparator will be low.Transistor Q1 will be off.The voltage at trigger pin of IC1 will be positive and there will be no alarm.When there is an intruder near the LDR the shadow causes its resistance to increase.Now the voltages at the inputs of comparator will be different and the out put of IC2 will be high.This makes Q1 on.This makes a negative going pulse to trigger the IC1 which is wired as a monostable multivibrator.The out put of IC1 will be amplified by Q2 (SL 100) to produce alarm.

 

[dlgoff]

When there is an intruder near the LDR the shadow causes its resistance to increase.

How does the occur? What is R9; a photoresistor?

 

[Adjuster]

Here are a few comments which I hope may help:

• I cannot see what you mean by

  I have made some changes (therefore some things are written in black).

  What exactly have you changed?
  
• To define the standby mode you should adjust for a definite "off" bias. Making the voltages at the comparator input exactly equal would give an undefined state, subject to the input voltage offset of IC2.
  
• An alarm working directly on brightness would be very sensitive to changes in ambient lighting e.g. due to varying daylight or changes in mains voltage affecting electric light. (That's not your problem if your instructor has told you to use this circuit.)
  

 

[Adjuster]

You have me confused here, you seem to be saying that the output of IC2 is going from low to...low?

At standby mode these two voltages are set equal by adjusting R9.Now the out put (pin6) of comparator will be low.Transistor Q1 will be off.The voltage at trigger pin of IC1 will be positive and there will be no alarm.When there is an intruder near the LDR the shadow causes its resistance to increase.Now the voltages at the inputs of comparator will be different and the out put of IC2 will be low.

 

[swty todd]

R8 is a 47k pot and R9 is an LDR. I made those changes in the above figure(wrote them in black) but seems like they weren't saved.Extremely sorry about that, I have posted the correct figure. Also output goes from low to high

 

[swty todd]

Here are a few comments which I hope may help:

• I cannot see what you mean by What exactly have you changed?
  
• To define the standby mode you should adjust for a definite "off" bias. Making the voltages at the comparator input exactly equal would give an undefined state, subject to the input voltage offset of IC2.
  
• An alarm working directly on brightness would be very sensitive to changes in ambient lighting e.g. due to varying daylight or changes in mains voltage affecting electric light. (That's not your problem if your instructor has told you to use this circuit.)
  

To avoid situations you have stated in your second and third comments, any idea how I should modify the circuit.One way I can think of would be to adjust the 47K pot so that its resistance is little more than the LDR. By increasing the value of the pot , I can also make the alarm less sensitive (to avoid situation no.3 ), though am not sure how much a particular change will affect its sensitivity.

Also is it okay if I use SL 100 for Q1?
And thanks a lot!