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I am building a power failure alarm circuit as a project for College on a breadboard.

The Circuit Diagram, that i am using is as follows

http://pcbheaven.com/scripts/imagepresent.php?filename=%2Fwikipages%2Fimages%2F555circuits_1237476139.jpg

I think there is something wrong with circuit, I have re-wired it a dozen times and it isn't working, the resistors are just preventing the flow of current, maybe the resistance is too high? Could it be?

what should i do? please help...I am new to circuits and this one is just killing me...

@panther1,

Welcome to CL.

Sorry but your circuit has several errors.

i) you have grounded pin 2;

ii) you have grounded pin 6;

iii) if you connect a monitoring voltage (which is by definition a low resistance voltage source) to pin 7 then if pin 7 ever goes low there will be an excessive current flow into pin 7 which may damage the 555 device.

I suggest you read:

http://en.wikipedia.org/wiki/555_timer_IC

and:

http://electronicsclub.info/555timer.htm

then:

https://www.circuitlab.com/docs/

and:

https://www.circuitlab.com/docs/faq/

then:

About the 555 timer model in CL

The high level voltage at the OUTPUT pin of the CL 555 timer model does not track the vcc supply. It is fixed at 5V.

The CL model uses a fixed internal supply of 5V and ignores the voltage on the vcc pin. This means that the internal switching thresholds are fixed at 5/3V and 10/3V.

This is incorrect and can be misleading.

The circuit below correctly models the output swing of the real 555 timer although it does not accurately model the output stage behaviour vs. load currents.

Simulate > Time Domain > Run Time-Domain Simulation

https://www.circuitlab.com/circuit/9fu6wh/555-timer-as-astable-multivibrator-oscillator-model-error/

There’s a bug report about this here:

https://www.circuitlab.com/forums/support/topic/5yc7a6mu/bug-report-high-level-voltage-at-output-pin-of-cl-555-timer/

and a lengthy discussion about some of the other “faults” on the CL 555 timer model here:

https://www.circuitlab.com/forums/support/topic/e4wyttj4/bug-report-555-timer-model-vcc/#comment_3885

then redraw your circuit carefully in CL (using only a 5V supply!) and then run a time domain simulation on it so you can see how it behaves.

You will need to set up a time varying voltage source to simulate dropouts in the the voltage you are monitoring.

:)

More info on the 555 and some applications:

http://www.williamson-labs.com/pdf/555AN.pdf

http://www.electro-tech-online.com/content/440-power-supply-failure-alarm.html

Is this circuit correct?

Until you define exactly what you require your "power failure alarm circuit" to do, your question is difficult to give a meaningful answer to.

You need to think about and answer these questions before you can design a suitable circuit:

i) define "power failure".

Is it a voltage source:

a) going above some reference voltage by more than some defined amount?

b) going below some reference voltage by more than some defined amount?

c) going from a low resistance to a high resistance, i.e. going open circuit (which is not the same as going to 0V)?

ii) What "alarm" action should it take on the defined power failure event?

a) light an LED?

b) flash an LED:?

c) generate a sound?

iii) how long should the alarm persist for?

:)

Power Failure: Voltage goes to 0, means the power supply is cut off.

Generate a sound

Just a beep would Suffice.

In that case,

http://www.electro-tech-online.com/content/440-power-supply-failure-alarm.html

would meet your requirements.

Note that it will not detect the situation where the power supply goes open circuit because then there may be nothing to pull the transistor base to 0V or whatever resistance there is across the supply rails may not be enough to turn the transistor on enough to drive the buzzer.

In that situation you may need to add a second resistor (R2) of something like another 10k or less from the supply rail to 0V so that if the supply goes open circuit the transistor base is pulled down by R1 + R2.

Can you please specify on the diagram, where to insert the second resistor?

There is also another variation of this circuit diagram, but it isnt working either.. http://www.electronicspoint.com/power-supply-failure-alarm-t249322.html

Much Appreciated,

@panther1,

"Can you please specify on the diagram, where to insert the second resistor?"

It's quite explicit:

"from the supply rail to 0V"

I'm sorry but I don't have time to go chasing all over the web looking at random circuits for you.

If you want to ask a question about a circuit then please put some effort into it, make it easier for everyone else and draw it in CL.