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Old 13th December 2004, 05:47 PM   (permalink)
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im sorry, i forgot to change the transistor's orientation. in swcad iii, which is what i used to make the schematic, for some reason, when u place a pnp transistor, it puts it like that. last time i fixed it in paint, but this time i forgot.

Quote:
You could conveniently connect its collector to Q4 emitter but I'm not sure that will give the polarity of the switching you want.
im not sure what you mean by this. could u explain a littl more?

When the alarm goes high resistance (beam is broken) the alarm goes off.
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Old 13th December 2004, 07:37 PM   (permalink)
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I can try explain your problem with the aid of a few circuits. I am substituting the NPN with a switch for simplicity.

Buzz_on is your circuit working just after the NPN switched on, charging the cap. Note the current to the buzzer and the voltage over the buzzer. So far looks OK.

Buzz_Bad is your circuit after the NPN already turned off (for 9 sec) creating an almost open circuit. I have put 50M Ohm over the switch to simulate the transistor turned off. Notice that there is still some current going to the buzzer and some low voltage over the buzzer.

Buzz_good is the modified circuit with the 1M resistor added to the base of the PNP. It shows the conditions again after about 9 seconds from the NPN turning off, BUT see how in this instance the current is zero, voltage over buzzer is zero and the voltage over the cap is under 0.6v.

Now your circuit failed because in your circuit the cap could not fully discharge (or low enough) through the base-emitter junction of the PNP and the leakage resistance 50M from the NPN still kept a very small current flowing, helping the PNP to stay on (slightly) and also work against the discharge of the cap to some degree.

In the working circuit the 1Meg resistor helps to discharge the cap once the b-e junction stopped conducting. Also the value of this resistor is such that the cap is discharged faster than the leakage current from the NPN can charge it up again.

Hope you followed all that
Attached Images
File Type: gif buzz_on.gif (8.3 KB, 346 views)
File Type: gif buzz_bad.gif (8.2 KB, 342 views)
File Type: gif buzz_good.gif (8.5 KB, 342 views)
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Old 13th December 2004, 11:16 PM   (permalink)
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thanx theone, that worked quite nicely. thanx for taking the time to make me the pictures and everything, really appreciate it

i was thinking of putting a resistor there to drain the cap but i wasnt sure wut value. i shouldve just aked that :lol:
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Old 13th December 2004, 11:31 PM   (permalink)
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I thought it would aid in showing not only you, but other learners how a simple circuit can be very interesting once you start analyzing it.

Happy learning! :wink:
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Old 18th December 2004, 06:01 AM   (permalink)
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just wen i think im done . . . well, terrible's law has been proven true once again (down in my signature). i just build a PCB for the circuit today and of course it doesn't work.

the buzzer stays on all the time no matter wut. the very very strange thing is that in the part of the circuit wher the base of Q1, the LDR and R1 meet, i am getting around 10v!! the 10v because i am using a walwart supply and it puts out 10. but on a simulation that i ran, it should only be about 1.6v @ the highest!! i replaced Q1, didnt do nothin, n i know foresure the LDR is okay. wut could be causing this?
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Old 18th December 2004, 06:49 AM   (permalink)
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:idea: Make sure you don't have a PNP in place of Q1 with the emitter tied to the 9v (10v). This would also cause the buzzer to stay on. Check and double check your circuit. Also check that Q1, if not a PNP by mistake is not faulty. I often use a simple trick (after voltage measurements to identify the culprit and before desoldering ) to check transistor switching in low power circuits by briefly shorting the b-e junction of a transistor to force it to turn off, when on.

Make sure that you have not created a "mirror image" of the circuit on the pcb. I have seen this a lot with 1st pcb attempts by students! If this is the case the components needs to be soldered to the track side of the pcb to correct the error. (which gets real messy when IC's are involved, as the legs have to be folded back in the opposite direction so the device is actually upside down)
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Old 18th December 2004, 06:08 PM   (permalink)
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i cannot stand it wen it works on the breadboard n not in the final project!!

okay, ive replaced Q1, still dont work, and yes it definetly is an NPN. i also shorted the be junction of Q1 but the buzzer still stays on.i am beginning to think that the problem is in Q2 Q3 OR Q4. i will short the be junction of each of those transistors to find the problem.

now that ive got that figured out, what is reeeealy driving me insanse is that i am getting 10v between the LDR, Q1's base and R1. i am measuring relative to ground. WHY!?!? is this not very strange??
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Old 18th December 2004, 09:38 PM   (permalink)
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figured it out: Q2 was bad. simple as that :lol:

WHY WHY WHY did it make that junction (base of Q1, R1 and LDR) be 10v!?!?!
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Old 18th December 2004, 09:46 PM   (permalink)
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Remember what I said in a previous post

Quote:
If Q1 ever manage to switch on hard enough (accidents happen), Q2 will be fried because of high base current into Q2. Good practice will be to limit base current to Q2 in the collector of Q1 with a resistor. Maybe move R3 to Q1 collector
Just to prove accidents do happen.....
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Old 18th December 2004, 10:01 PM   (permalink)
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naw, thats not the reason. hate to disprove ur proving :lol:
i was trying to use up some old crap transistors that had been used in other things (bak wen i had less of a clue wut i was doing) and well, it mustve been fried. besides, it never fried on the breadboard (i was using different transistors) and it works now. if i moved the resistor to the collector of Q1 then Q2 wouldnt be able to turn on tho right?

and somebody plz satisfy my curiousity: why did it make that junction (base of Q1, R1 and LDR) be 10v!?!?!
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Old 18th December 2004, 10:10 PM   (permalink)
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Quote:
Originally Posted by zachtheterrible
figured it out: Q2 was bad. simple as that :lol:

WHY WHY WHY did it make that junction (base of Q1, R1 and LDR) be 10v!?!?!
Although it is impossible to know what happened, here is one possible sequence of events:

R1 or Q1 b-c was shorted accidently. This caused a high current into the base of Q1, causing an even higher current through Q1 into the base of Q2. Q1 fails with a short b-c (explains the 10v on the base), open b-e and Q2 shorted c-e (causing the buzzer to stay on)
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Old 18th December 2004, 10:26 PM   (permalink)
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oh, i c wut ur sayin now, i thot u wer talkin about not shorting or anything, just the circuit turning on really hard for some reason or another. im not worried about shorting nething because even in a perfect circuit, if something gets shorted accidently, things can get ruined. its very much mor likely tho that i ruined the transistor previously. in fact i no it for a fact because i cheked the circuit over very good b4 turning it on, to make sure i hadnt shorted anything :lol:

thanx theone, and everyone who helped me, you've all been very kind to me :lol:
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Old 26th August 2008, 11:13 AM   (permalink)
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dude can you send me on my e mail a copy of you invention so i can try it by myself i am a computer Engineer trying to be a electronics someday can you send the copy of you projrct also the schematic diagram and the values of the electronics parts thankssss by the way here is my e mail adress Fermae14@yahoo.com please send it to me i will be waiting on it thank you veery much that there are a person like you thanks.... LOL....
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Old 25th September 2008, 09:26 AM   (permalink)
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can i ask what is the meaning of 4k7 of R2 for the given circuit.
could it really work perfectly,because i'm having problem working out with our school project... we are running out of time to finish our laser burglar alarm project...anyone or someone who has experienced this and had been given help please send me your circuit diagram... thanks a lot & God bless...
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Old 25th September 2008, 09:30 AM   (permalink)
Default please help me huhuhuhu

can i ask what is the meaning of 4k7 of R2 for the given circuit.
could it really work perfectly,because i'm having problem working out with our school project... we are running out of time to finish our laser burglar alarm project...anyone or someone who has experienced this and had been given help please send me your circuit diagram... thanks a lot & God bless...
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