Need a simple alarm (buzzer) that is triggered when the voltage drops.

Thread Starter

Shimra01

Joined Nov 13, 2017
35
Who can help: I need to create a simple alarm (buzzer) that is triggered when the voltage drops.
The alarm should be powerd by a seperate 12V supply.
The circuit should be as simple as possible.
The voltage we are measuring and treiggering the alarm on is on an external ciruit with its own 24V DC power supply. Our alarm circuit should not have an effect on the exisiting circuit. The trigger must operate when the standard voltage drops from 21V (officially this is 24V DC) to below 5V (it seems to drop to about 1.8V and not to 0V, not sure why). I have 8 of these points and the alarm should be able to be triggered on any of these. Once the buzzer goes we should be able to disable to buzzer for 5 minutes by pressing a button.
Is there anybody that can help me with this design? I can manage most of the circuit but the trigger voltage drop from 21V to 1.8V is causing me some issues. I thought of using an Optocoupler or comparator but most designs are based on the Alarm circuit and the external citcuit having the same ground and this is not the case here as both systems will have their own power supply.
 

Thread Starter

Shimra01

Joined Nov 13, 2017
35
Can you post a schematic ?

Regards, Dana.
Hi sorry dont have schematic. but let me try to explain. the existing circuit is from an existing alarm system that does not have a buzzer. The Alarm system cannot be altered or tampered with so not much room there. This is a 40 year old circuit. The alarm currently has 8 outputs in the form of 24V lamps. The whole system runs on 24V (we are not allowed to use the existing power supply). when a button is being pressed a few hundred meter away a small lamp lights up. There are 8 buttoms at 8 different locations, making 8 lamps to light up. After measuring for long I found that for each of these 8 lamps has a line coming in that usually has 21V on it but when the button is pressed this voltage drops to about 1.8V. I do believe that the 21V is in fact the system's operating voltage of 24V but being so old components may have deteriorated such that the voltage has become less. When a button on a panel is pressed the situation returns back to before, the voltage goes up to 18V. And when a 2nd button on a panel near the lights is pressed the voltage goes back to 21V and the lamps goes off.

So what I had in mind was to use the light control line as my trigger. I want to connect a wire to this point that goes down from 21V to about 1,8V and use as trigger for my buzzer circuit. Somehow I will need to put the 8 signals through a logic circuit or so such that we only need one buzzer in case more than one light would go on. My thought was to use an optocoupler, but not sure how. Or maybe better connect these points to a 24V low signal relay (the coil side) such that when the 21V drops to 1.8V the relais stops being triggered and this closes it, this can trigger a simple buzzer circuit. But I am afraid that this relais may have an impact on the existing circuits.

I want to run the buzzer circuit on its own 12VDC (all voltages mentoned are DC). I am a bit stuck in how to pickup this voltage drop as my trigger point. I have been at this for almost a whole day and with my limited electronics knwoledge have been thinlking about using, relais, optocouplers, SCR's Transistors, comparators etc. have been googling but not found what I need. I hope that someone here can assist me with some guidance in how to go about it.
 

Thread Starter

Shimra01

Joined Nov 13, 2017
35
This is a natural for ATTINY (45 or 85) UP running Basic interpreter. It
has A/D, onchip reference, can easily handle timeouts and buttons.

Just a thought.

https://www.mcselec.com/

Regards, Dana.
Sorry I am not good at all in Microcontrollers and their programming so have not looked at this. And at the same time it gives me still the problem of how to pick-up the signal.
 

ElectricSpidey

Joined Dec 2, 2017
1,333
So you want to detect the voltage drop at 8 points, and trigger a single buzzer...correct?

Are all of the 8 points positive relative to a common ground point?

How much current do you think you can "borrow" from the 8 points?

I have an idea that "might" work, but it will need a little current from the detect points, it would need 8 Zener diodes, 8 optocouplers a single 8 channel Darlington driver chip and some assorted components. (like the buzzer…resistors…PC board…etc) And of course the 12 volt supply.

I would have no way to test the circuit, so you would be on your own to breadboard and test, and then tweak it if it doesn't work, but I doubt anyone here can suggest something guaranteed to work.

Also I will say that there are people here that could suggest a circuit that would require very little current from the device using OP amps or such, but I'm not that great with OP amps, or counting on their reliability in the field.

And there are also plenty of people here who use a sim to test circuits, I do not.

So let me know if you want me to post a block diagram, of my idea, unless someone posts something in the meantime.
 

Thread Starter

Shimra01

Joined Nov 13, 2017
35
So you want to detect the voltage drop at 8 points, and trigger a single buzzer...correct? - YES

Are all of the 8 points positive relative to a common ground point? - YES

How much current do you think you can "borrow" from the 8 points? - Unknown but if any it must be as little as possible

I have an idea that "might" work, but it will need a little current from the detect points, it would need 8 Zener diodes, 8 optocouplers a single 8 channel Darlington driver chip and some assorted components. (like the buzzer…resistors…PC board…etc) And of course the 12 volt supply.

I would have no way to test the circuit, so you would be on your own to breadboard and test, and then tweak it if it doesn't work, but I doubt anyone here can suggest something guaranteed to work.

Also I will say that there are people here that could suggest a circuit that would require very little current from the device using OP amps or such, but I'm not that great with OP amps, or counting on their reliability in the field.

And there are also plenty of people here who use a sim to test circuits, I do not.

So let me know if you want me to post a block diagram, of my idea, unless someone posts something in the meantime.
I have replied in your post.
And yes please do go ahead and post block diagram, ideas etc.
I will make a test circuit and test this on the live system tomorrow.
I will feedback the results.
I think this is a great place to cooperate and bring brilliant minds together to solve problems. Hope others can also contribute and comment.
 

ElectricSpidey

Joined Dec 2, 2017
1,333
Ok, so understand this is only a working concept, not a working circuit, and I have given no component values, because they will probably have to be determined with testing.

So the Zener develops a voltage that drives the input of the opto, the opto in turn keeps the output Darlington in the “off” state, the opto should be a MOSFET type but a bjt would probably work.

Then when the voltage drops, the input to the opto drops, turning off its output transistor, and allowing the Darlington to conduct…turning on the buzzer.

All of the Darlington outputs are in parallel so you get the “logic” you need to turn on the buzzer regardless of which line voltage drops.

I would probably start with 12 volt Zeners, that way other values can be determined accordingly.

8_Volt_Drop_Detect.jpg
 
Last edited:

ElectricSpidey

Joined Dec 2, 2017
1,333
I think that Zener is in the wrong place....sorry

I will update.

EDIT:

Ok so I updated the image...I was only a little confused...:oops:
 
Last edited:

Thread Starter

Shimra01

Joined Nov 13, 2017
35
Thanks this is a very good start and worth trying. I do have a few suggestion/question. If we use a normal Opto instead and a PNP transistor, would that work? and would that make the design easier? (cheaper as you can use very commonly available parts)?
Opto's I have lots in stock are the EL817.
 

Thread Starter

Shimra01

Joined Nov 13, 2017
35
I also would need to add a simple timer such that I can kill the buzzer for like 5min.
Question about your design. Do you have any model or part suggestion for the OPTO?
 

AnalogKid

Joined Aug 1, 2013
8,688
Let's clarify some of the control conditions.
Is a simple comparator set to 5 V sufficient?
OR, do you want the system to sound the alarm when the input drops to below 5 V, but not turn off the alarm until it returns to at least 21 V?
If you press the button for 5 minutes of silence, what happens if the failed voltage returns to its nominal value -
Keep going through the 5 minute period?
Immediately clear the error and reset the 5 minute timer for the next fault?
What happens if during the 5 minute silence period, the first input that failed restored itself but another input fails?

My thinking for the inputs is that a simple CMOS comparator powered by 12 V has a transition level of approx. 6 V.

I think I've got the entire circuit down to one CD4060 oscillator/divider for the 5 minute delay, and one CD4001 quad NOR for the input comparator and control logic. Because the entire system has a common ground, no opto's needed.

ak
 

Picbuster

Joined Dec 2, 2013
1,022
Who can help: I need to create a simple alarm (buzzer) that is triggered when the voltage drops.
The alarm should be powerd by a seperate 12V supply.
The circuit should be as simple as possible.
The voltage we are measuring and treiggering the alarm on is on an external ciruit with its own 24V DC power supply. Our alarm circuit should not have an effect on the exisiting circuit. The trigger must operate when the standard voltage drops from 21V (officially this is 24V DC) to below 5V (it seems to drop to about 1.8V and not to 0V, not sure why). I have 8 of these points and the alarm should be able to be triggered on any of these. Once the buzzer goes we should be able to disable to buzzer for 5 minutes by pressing a button.
Is there anybody that can help me with this design? I can manage most of the circuit but the trigger voltage drop from 21V to 1.8V is causing me some issues. I thought of using an Optocoupler or comparator but most designs are based on the Alarm circuit and the external citcuit having the same ground and this is not the case here as both systems will have their own power supply.
important to know:
Drop and recover time ( should a spike trigger system?)
Accuracy?
Do you accept a common negative and ground? if not use a galvanic separator.

The most simple way is to use a microcontroller like pic 12F or 16F
used MIC2920A-3.3WS few resistors and capacitors and hardware is ready.
outputs via one or more pins to relays, led or other things.
Program size in "C" : one A4 maximum.

Using a PIC16F690 you have serial output allowing to connect PC or something else.
This pic will carry flash so thresholds and timing could be changed and read back on the fly.
This will take a line driver and a few lines of C programming.

Picbuster
 

Thread Starter

Shimra01

Joined Nov 13, 2017
35
Let's clarify some of the control conditions.
Is a simple comparator set to 5 V sufficient? - Dont think so as they usually need the same ground. But I was contemplating using a comparator.
OR, do you want the system to sound the alarm when the input drops to below 5 V, - Yes
but not turn off the alarm until it returns to at least 21 V? - Yes but any voltage above 15V should be ok
If you press the button for 5 minutes of silence, what happens if the failed voltage returns to its nominal value -
Keep going through the 5 minute period? - If the failed voltage return and the timer is still going, nothing happens and it stays silent, if the timer is ended and the failed voltage is still failed then the alarm should sound again. Alarm should go off when failed voltage goes up above 15V.. Another press of the button would silence the alarm again.. The way I see it the timer would open/close a relais in the buzzer circuit. ant it gets reset by the failed voltage going above 15V.
Immediately clear the error and reset the 5 minute timer for the next fault?
What happens if during the 5 minute silence period, the first input that failed restored itself but another input fails?

My thinking for the inputs is that a simple CMOS comparator powered by 12 V has a transition level of approx. 6 V. - Why 6V? I dont understand this.

I think I've got the entire circuit down to one CD4060 oscillator/divider for the 5 minute delay, and one CD4001 quad NOR for the input comparator and control logic. Because the entire system has a common ground, no opto's needed. - No the circuit has its own ground and the circuit we are measuring and activating upon has its own ground. We cannot mix these. Or am I missing something here?

ak
Thanks for your reply, see my answers to your questions in italic and underlined.
 

Thread Starter

Shimra01

Joined Nov 13, 2017
35
important to know:
Drop and recover time ( should a spike trigger system?) - No and i dont expect spikes, but no spikes should not trigger system.
Accuracy? - Should work always when the voltage drops. The 21V is already fluctuating a little so that is why I think we can use 15V as the high and 5V as the low. thus if voltage above 15V alarm is not activated and if voltage below 5V alarm gets activated., Voltages in between should not be considered. Does this make sense?
Do you accept a common negative and ground? if not use a galvanic separator. - No cannot use common ground. Any suggestions how I can use Galvanic separation with a DC signal?

The most simple way is to use a microcontroller like pic 12F or 16F
used MIC2920A-3.3WS few resistors and capacitors and hardware is ready.
outputs via one or more pins to relays, led or other things.
Program size in "C" : one A4 maximum.

Using a PIC16F690 you have serial output allowing to connect PC or something else.
This pic will carry flash so thresholds and timing could be changed and read back on the fly.
This will take a line driver and a few lines of C programming.

- I am open to this solution but as I dont have microcontroller experience I would need someone to help me all the way with this. However I still see an issue in how to pickup the voltage drop.

Picbuster
Thanks for your help. See my answers to your questions marked in italic and underlined.
 

AnalogKid

Joined Aug 1, 2013
8,688
For "normal" CMOS gates, the transition level of the input is approx. 50% of the power supply voltage Vdd. So if Vdd = 12 V, then the output changes state when the input crosses 6 V. (It is different for Schmitt Trigger input gates.) If you diode-AND all of the inputs together, then whenever any of then goes below 5.4 V (6 V - one diode voltage drop), the output changes state.

That's all great, but it means that the output changes back when the low input returns above 5.4 V. If you define the ok state as being above 15 V, then you need either a dual comparator circuit called a window comparator, or a single comparator circuit with 10 V of hysteresis. This is more complex than a single CMOS gate, but since you don't need to know which circuit is low, only that one of them is, then you still can the diode-AND the inputs.

I thought there was something in an earlier post about everything having a common ground. Please confirm that the 8 alarm circuits have a common ground. Also, what is the max current you think a sensor circuit can steal from each alarm circuit? For example, could 4 mA for the input of an optocoupler come directly from an alarm circuit?

If the 12 V supply for the buzzer circuit is fully floating, then there should be no problem connecting the alarm and buzzer circuit grounds together and eliminating the optocoupler.

ak
 

Thread Starter

Shimra01

Joined Nov 13, 2017
35
For "normal" CMOS gates, the transition level of the input is approx. 50% of the power supply voltage Vdd. So if Vdd = 12 V, then the output changes state when the input crosses 6 V. (It is different for Schmitt Trigger input gates.) If you diode-AND all of the inputs together, then whenever any of then goes below 5.4 V (6 V - one diode voltage drop), the output changes state.

That's all great, but it means that the output changes back when the low input returns above 5.4 V. If you define the ok state as being above 15 V, then you need either a dual comparator circuit called a window comparator, or a single comparator circuit with 10 V of hysteresis. This is more complex than a single CMOS gate, but since you don't need to know which circuit is low, only that one of them is, then you still can the diode-AND the inputs.

I thought there was something in an earlier post about everything having a common ground. Please confirm that the 8 alarm circuits have a common ground. Also, what is the max current you think a sensor circuit can steal from each alarm circuit? For example, could 4 mA for the input of an optocoupler come directly from an alarm circuit?

If the 12 V supply for the buzzer circuit is fully floating, then there should be no problem connecting the alarm and buzzer circuit grounds together and eliminating the optocoupler.

ak
Aha ok got it. The alarm circuit has its own power supply and as such its own ground. the circuit we are acting on has av 24V DC supply and all the 8 inputs (for our alarm circuit and where the power drops) have the same ground (linked to their own 24DC supply). So we somehow will need to use an optocoupler I think or will your diode -AND input do?

My 15V was just a save voltage but I think your circuit may work as well: "the output changes back when the low input returns above 5.4 V." This should be ok enough for my design and I am willing to try this out.
 
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