I have an external LED on my alarm system (outside at my garage) that indicates the status of my alarm system. The LED is solid green when the alarm is off and all zones are closed (no movement). The LED flashes in green when there is an open zone (some movement). When the alarm is activated the LED flashes in red.

The LED is connected to 3 wires. The one wire is +3V, the other -3V and the last is ground.

My requirement is to add a buzzer to my alarm at the location of the LED using the power coming to the LED. The buzzer should beep once when I activate it, and beep twice when deactivated. I plan on removing the LED to accommodate the buzzer. If required to drive the buzzer, there is a 220V mains close to the LED.

So my question is how can I design a circuit to use the +-3V as a control signal to activate a buzzer as described above. Any help would be appreciated.

Table of voltage states:

enter image description here

  • \$\begingroup\$ You said one of the wires is +3v, and the other is -3v. 1) Under what condition did you measure this? When the system was deactivated? (LED solid green?) What is the voltage on these wires when the either of the LEDs is blinking? 2) Do you want any buzzer beeps when the there is an open zone? \$\endgroup\$
    – tcrosley
    Commented May 8, 2014 at 18:15
  • \$\begingroup\$ The voltage levels were measured with the LED disconnected from the wiring. The buzzer is only required to indicate to me when the alarm is activated and deactivated. I am planning on removing the LED if the buzzer works. \$\endgroup\$
    – Rynardt
    Commented May 8, 2014 at 19:30
  • \$\begingroup\$ The thing is, I don't know the function of the two wires is, and what their voltages are when the system is activated. Just saying one is +3v and other is -3v doesn't provide any useful information other than their voltages in the deactivated state. I don't know how to detect when the system becomes activated (and returns to the non-activated state). If you could activate the system and then measure the two leads again it would be very useful. \$\endgroup\$
    – tcrosley
    Commented May 8, 2014 at 20:57
  • \$\begingroup\$ The voltages were measured when the system was active and not active without the LED connected. The voltage was from the one wire to ground and the other to ground. \$\endgroup\$
    – Rynardt
    Commented May 9, 2014 at 6:44
  • \$\begingroup\$ I'm sorry to harp on this, but I still don't have the information I need. You said +3v on one wire and -3v on the other. That's only one state (active or inactive, I don't know). What are the voltages for the other state? Add a little table to your question if you can. \$\endgroup\$
    – tcrosley
    Commented May 9, 2014 at 7:26

1 Answer 1


As is true for a lot of projects on this site that use discrete logic, this could be done with a microcontroller instead with fewer parts, but that would also require the person building it to know how to program (which they may not), and to have the necessary development environment set up, including the hardware to program the microcontroller.

Looking at the table that was added to the OP, one can see that there are times when the system is either activated or not activated and both Wire 1 (GREEN_LED) and Wire 2 (RED_LED) have no voltage on them because they are flashing due to an open zone. So if we want to run the circuit off of just these two leads, this has to be accounted for.

Looking at the circuit below, I am ORing the two input leads using a pair of Schottky diodes, and feeding them into a Supercap to hold the voltage across the gap when no voltage is present. I chose a 0.4 F (Farad) capacitor, after using this Supercap calculator. The voltage following the diodes (Vcc) will be around 2.75v,. which is compatible with the HC (but not HCT) logic family. With no input on either Wire 1 or 2, and with the buzzer on consuming 30 mA, a 0.4F SuperCap will keep Vcc from dropping below 2.5v for 0.73 seconds (which provides a safety margin over the minimum 0.5 seconds needed).

enter image description here

(Right click and select View Image to see a larger version of this schematic.)

Here's how it works. There are four 74HC123 one-shot monostable multivibrators (IC1 & IC3) used for timing. All are set for 1/2 second delay using a 75K resistor and 10 µF cap. Typically one would trigger timer IC3A on just Wire 1 (GREEN_LED) going high, but since the lead may be flashing, this would cause it to trigger multiple times. So I am using a 74HC74 D-type flip-flop (IC6/1) to remember the state. When the flip-flop is set, the output is fed into the B input of the 74HC123. timer. When the B input goes high, it starts the timer. The output of the timer high feeds into one of the inputs of the 3-input NOR gate (IC5), making the output low, turning on the P-channel FET and sounding the buzzer.

After half a second, the timer expires and the buzzer turns off. Even if the Wire 1 (GREEN LED) goes to 0 and back again because it is flashing, the flip-flop won't change state and the timer will not trigger again.

Similar logic occurs when the state goes from activated to deactivated (Wire 2, RED LED goes high). Again a flip-flip is used to keep state. The difference is when the top right timer IC1A expires, the middle timer (IC1B) is triggered. It simply waits a half second with the buzzer off, and then the bottom timer (IC3B) is triggered, sounding the buzzer another half a second, altogether twice per the spec.

The logic could be simplified a little, but I presented it this way so it would be easier to understand. Instead of using the two inverters IC2A/IC2B, the two unused 3-input NOR's could have all three inputs tied together and function as an inverter. In that case the circuit could be realized using just four IC's.

  • \$\begingroup\$ Thanks for the elaborate answer. When I have the time I will further investigate your solution and get back to you with my results. Where will the VCC connections be made to? What do you mean with the comment on the drawing regarding the power and ground and bypass caps? \$\endgroup\$
    – Rynardt
    Commented May 12, 2014 at 13:11
  • \$\begingroup\$ Another questions: You mention "74HC123 astable multivibrators " can it be a monostable multivibrator? \$\endgroup\$
    – Rynardt
    Commented May 12, 2014 at 13:41
  • 1
    \$\begingroup\$ @Rynardt VCC is simply the 3v supply. I probably should change the label. Each of the IC's has a power and ground pin, typically 14 and 7 (except the 74HC123), which you can find in the datasheet. You need to connect these to VCC and ground, and add a 0.1 µF cap between these pins. Yes, I meant a one-shot monostable. I'll correct my answer. Also, I've made some calculations re the Supercap, plus I need to change the diodes to Schottky to reduce the voltage drop. I'll update the schematic tonight. \$\endgroup\$
    – tcrosley
    Commented May 12, 2014 at 14:57
  • \$\begingroup\$ @Rynardt I added the calculation I did to determine the size of the Supercap (now 0.4F), plus included a part number for the Supercap (FT0H474ZF) and the buzzer (PB-12N23P-03Q). I also added the pins for the poweer to all the chips, and the bypass capacitors. \$\endgroup\$
    – tcrosley
    Commented May 13, 2014 at 6:38

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