Need help with a Power Failure Alarm Circuit

[jdp]

I need help with my project. I am trying to make an alarm (buzzer) go off when the circuit loses its AC power signal, and have a reset button to shut off the alarm so you don't have to keep listening to it.

I am using a falling edge detector (using nor gate delay), a nor latch, and a PNP transistor. A 9 volt battery powers the nor gates and the buzzer. Here is my current circuit:

[url]http://i.imgur.com/R2nZ0.jpg?1[/url]

I have modeled this circuit using multisim, and it performs perfectly in simulation. When I breadboard it, however, it doesn't work. Here is my parts list:

CD4001BE - quad 2 input cmos nor gate IC
NTE12 - PNP BJT
1N4007 - diode
1N4744A_Q - zener diode
254-EMB105-RO - buzzer
The rest of the components are just resistors and a capacitor. The switch is just a push button switch, initially open circuit.

What happens is that it will not trigger off the falling edge and the circuit thinks it still has power after the AC power cable has been unplugged. I am also having problems with the buzzer, either it is always on, on at a different sound level, or something else. It is just unreliable. Please help with any comments and suggestions you can give me.

[pebe]

Hi,
Could you clear up a few anomalies in your circuit?
1. Your parts list shows a 1N4744A zener which is a 15V one, but the circuit shows a 1N4459 which is a 5A rectifier.
2. What is the supply voltage to the 4001?
3. What voltage and current is specified for the buzzer?

[jdp]

[quote="pebe"]Hi,
Could you clear up a few anomalies in your circuit?
1. Your parts list shows a 1N4744A zener which is a 15V one, but the circuit shows a 1N4459 which is a 5A rectifier.
2. What is the supply voltage to the 4001?
3. What voltage and current is specified for the buzzer?[/quote]

Of course, thanks for checking out my post.

1. The one on the parts list, the 1N4744A, is the one I am using in the breadboard and the one I meant to model. The image has the wrong one - I just now changed it to a 1N4744A, and the circuit still operates correctly in simulation.
2. The 4001 ICs are being supplied with the 9 volt battery shown in the image. These chips are rated up to 20 volts.
3. The buzzer has a 3-7 voltage range, rated voltage is 5 volts. Mean current consumption is 25mA, peak is 75mA.

[pebe]

Hi,
Well there are quite a few things wrong with it. Here is what I found.

1. You have shown the negative side of the supply grounded. This should not be done because you are supplying from 120V mains. In operation that negative rail will be swinging at up to 170V positive relative to ground and that would be dangerous.

2. The voltage of 18.7V across C1 is limited to 15V by the 1N4744A zener. That voltage is being applied to an input of the gate. However, you have a 9V supply to the 4001s. That means that the protection diode of the gate is holding the voltage down to 9.6V so the zener is not being used. It’s not a good practice to rely on protection diodes being used that way.

3. It looks like you are using the propagation delay in U3B/C/D to make U3A act like an XOR gate to get a pulse from its output. That would work OK with fast acting waveforms, but it is being fed with a very slowly falling pulse because of C1. So your gating needs to be redesigned.

4. Coming now to the buzzer. When the output of U4C goes high it should cut off Q1. But D1 is always in conduction so the base of Q1 will always be forward biased and Q1 will always be conducting, to some extent, and the buzzer will stay on.

The circuit could be simplified using only one 4001, and I could draw up a circuit if you want.

[jdp]

[quote]The circuit could be simplified using only one 4001, and I could draw up a circuit if you want.[/quote]

I would really appreciate that.

[quote]1. You have shown the negative side of the supply grounded. This should not be done because you are supplying from 120V mains. In operation that negative rail will be swinging at up to 170V positive relative to ground and that would be dangerous.

2. The voltage of 18.7V across C1 is limited to 15V by the 1N4744A zener. That voltage is being applied to an input of the gate. However, you have a 9V supply to the 4001s. That means that the protection diode of the gate is holding the voltage down to 9.6V so the zener is not being used. It’s not a good practice to rely on protection diodes being used that way.

3. It looks like you are using the propagation delay in U3B/C/D to make U3A act like an XOR gate to get a pulse from its output. That would work OK with fast acting waveforms, but it is being fed with a very slowly falling pulse because of C1. So your gating needs to be redesigned.

4. Coming now to the buzzer. When the output of U4C goes high it should cut off Q1. But D1 is always in conduction so the base of Q1 will always be forward biased and Q1 will always be conducting, to some extent, and the buzzer will stay on.[/quote]

Thanks for the criticism.

1. What should be done with the negative side of the supply? I want this to be a single circuit, capable of operating in parallel with any device that uses 120v mains power, in order to detect when power is no longer reaching the device.

2. Could this be fixed by simply stepping down the rectified AC signal more by changing the resistor divider in R1 and R3?

3. This is a good point, something I overlooked. I am starting to look at new logic designs, but your circuit would help me a lot if you have time.

4. It is fine if its always conducting as long as it is a small enough current to not drain the 9v battery too much, or turn the buzzer on loud enough to hear.

[pebe]

1. For safety reasons you should feed the unit from an isolated supply. You can get a low voltage (7V to 9V) wall adapter very cheaply.
2. You won't need a potential divider or the big cap. Take the + supply lead via a 100K resistor to the gates input.
3. OK. I'll do you a circuit.
4. You don't need the diode there at all.

[I_Daniel]

http://pcbheaven.com/wikipages/555_Circuits/

The foregoing link indicates a simple IC555 circuit. The input monitor voltage should be a few milli-amps mains transformer with a 6.3volt output.

[pebe]

I_Daniel,

I'd like to point out an error in that circuit. As it stands, the circuit is unsafe. With pin 6 high the internal transistor whose collector is connected to pin 7 is conducting. With a high on the input voltage being tested, the 1N4148 into pin7 would blow. That diode should be removed.

[I_Daniel]

I dug out my old 110 IC Timer Projects by Jules H Gilbert, printed in 1979, and there is a circuit in it which confirms what you have said.

It show only one 1N4001. It goes to the junction of the resistor and capacitor which is connected to pins 6 and 2.
ie R1 from 9v+ to pin 7. R2 from pin 7 to pin 6 and 2 and C1 from pin 6+2 to the 9V negative line. The 1N4001 is also connected to Pin6+2. He says the the timing capacitor has a full charging voltage on it all the time which prevents the astable from oscillating.

More plainly his is a "standard" astable multivibrator with a voltage on pin 6 which prevents the timing capacitor from discharging. When the mains is interrupted/fails then this voltage falls away and the charge/discharge cycle commences.

[pebe]

Hi jdp,

Here is the circuit as promised. It is derived from your circuit but only uses one CD4001.

A and B form a Schmitt trigger that gives a sharp pulse into D as the supply voltage falls. C and D form a bistable that is triggered to turn the buzzer on and off. I have fed it from 6V because of the 7V top limit on the buzzer.

Normally, when the mains is on, the battery drain is negligible.

[pebe]

Following up I_Daniel's idea of using a 555 timer, this is probably about the simplest possible circuit .