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I would like to add a relay output circuit to this alarm device to allow it to operate my car horn.

The alarm device has 4 different tunes/tones. Each tone plays for about 5 seconds and the next 3 tones follow.

Related: I am confused what the component encircled in red really does. How will the proposed relay behave with this red component? My purpose is to sound the horn 4 times in sequence with about 5 seconds interval. enter image description here

Nar T
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  • Device shown is a capacitor to block DC from the speaker. The circuit around the speaker is wrong. – ATCSVOL Sep 20 '22 at 13:44
  • It's not a capacitor. It has 3 terminals. See the RED CIRCLE. – Nar T Sep 20 '22 at 14:16
  • it can be done ... https://youtu.be/8zEH5GxPNO8 – jsotola Sep 20 '22 at 14:23
  • Most likely, it's a small transistor of some kind. – Simon B Sep 20 '22 at 14:45
  • It can't be a capacitor, if it was the transistor would have no DC on the collector. It also can't be a transistor, one terminal is connected to +9V, one to the collector of the transistor, one to the speaker which has it's other lead connected to +9V. I'm guessing it's an autotransformer, but it would help to see a picture of the other side of the board. – GodJihyo Sep 20 '22 at 16:33
  • Thered circled device appears most likely to be a tapped inductor. It may step up the AC signal from the transistor immediately below it on the diagram. (This is similar to what some orhers have said). The DC level shift on the tansistor MAY be able to be used to drive a relay driver stage. OR the AC signal at the piezo may be able to be rectified and used to drive a driver stage. – Russell McMahon Sep 24 '22 at 11:08
  • Pin 6 on the 8 pin left upper IC has a tone on it at alarm time. This is presumably otherwise EITHER high or low. A transistor - maybe with schottky diode with RC filter, could be triggered on by this and drive a relay driver. – Russell McMahon Sep 25 '22 at 04:25

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I believe that component may be an autotransformer, if so the circuit would look a bit like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The phase dots might be wrong, I'm limited to what the schematic editor has.

As for connecting a relay, hard to say if that would work. You'd be trying to drive the relay with audio tones. Depending on the amount of power from the circuit and the waveform it might be able to close the relay, the relay might just chatter back and forth, or it might do nothing.

You could possibly use a sound activated relay circuit, I would suggest looking into that.

Edit: A bit of searching turned up this question showing piezo speakers being driven by just such a device, so it looks like my guess was correct.

GodJihyo
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Added: I have "rolled back" this answer to my prior version as an edit made by NarT was completely different to what I had said. People are welcome to suggest their own new answers. I note that the same person proposed the edit and then immediately approved it. That is not how the system is meant to work.

As others have noted, the mystery component is likely to be a tapped inductor, to provide higher voltage piezo drive signal.

A method that is likely to be successful is to detect the audio output and use it to provide an alarm drive signal.

In the diagram below:

  • Point A is the initial audio drive. It is probably lower level than at later points but may rest at a DC high or low level when not active. If so this is easily detected.

  • B is similar but capacitively coupled - but clamped somewhat by the be junction with unknown effects from the R222 : R250 divider.

  • C is a transistor collector drive to the inductor - so probably quite a lot higher level.

  • D is the piezo drive and higher again.

Recitfyingsignal from any of these points with a diode will give a DC signal modulated by the AC drive. Smoothing that with an RC will give a DC signal of length controlled by the RC time constant. Adding a forward diode across the R gives fast attack and slower decay. Point C seems the most likely to be useful. Point D may be, but voltage will probably be "rather high".
This can then be used to drive a relay driver transistor.

Whether this DC signal meets your timing needsis to be determined. You may then need to add some extra delay and pulse forming circuitry.

enter image description here

Russell McMahon
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  • its getting more interesting. Can i remove the 3 pins, and directly connect the D and C points on the npn transistor? If thats the case, I think I can used one optocoupler instead of the speaker. please more elaborate on this. Thanks. – Nar T Oct 03 '22 at 08:34
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    The 3 pin device which is almost certainly a tapped inductor MAKES the voltage at D and provides a DC path to C for the transistor. So D is not suitable without the inductor. C would work with a resistor to V+ - say 1k to start. A may have a DC level shift from on to off - which would be useful. B has AC. – Russell McMahon Oct 04 '22 at 02:06
  • There is constant voltage 0.47V at B and only about constant 8V (from 9V supply) at D when the alarm is sounding. Both values are measured with multimeter. So, can a 4 pin optocoupler be used in place of the speaker? and then removed the 3 pin like inductor, then subsequently connect directly D and C. – Nar T Oct 04 '22 at 05:16
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    @NarT Re " ... removed 3 pin ..." --> No!!! As I said above, the 3 pin device MAKES the AC voltage at D. What is your multimeter setting? AC volts or DC volts or ... ? You did not mention the voltages at C with and without alarm signal. An oscilloscope would be extremely useful. – Russell McMahon Oct 04 '22 at 07:24
  • Multimeter is set in DC. I mean there is about constant 8V at C (not D) when the alarm is sounding. And 9V at C when alarm is NOT sounding. – Nar T Oct 04 '22 at 09:29
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    @NarT 3 pin device removed and say 1k resistor C to 9V "should" produce an AC signal at C when alarm sounds. Also try say 10k. See what DC difference there is with 1k and 10k and on and off. It is quite likely that you will have to deal with the AC signal in some way. Try the above first. – Russell McMahon Oct 04 '22 at 09:41
  • oh if its going to deal with AC, then i would not touch anything. – Nar T Oct 04 '22 at 10:34
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    @NarT I do not mean AC mains. I mean that signals (such as the battery or digital input-output pins) are at DC high or low levels AND audio signals are AC - such as the siren output – Russell McMahon Oct 04 '22 at 10:42
  • hello, after thorough studying of the circuit, I came up with another correction. 1. At labelled D in your attached image, there is about 170 Ohms of resistance between the right side pin and the bottom pin. 2. Also, the first pin is directly connected to +9V source. This same pin is again directly connected to the bottom pin. 3. The bottom pin remains the same connected to the one pin of transistor. – Nar T Nov 14 '22 at 12:31