I'm trying to rig a bike horn to an alarm clock

[Alexander8046]

A few days ago I saw this article on Instructables.com (http://www.instructables.com/id/Build-The-Worlds-Loudest-Alarmclock/) that shows you how to rig a car horn to a cheap alarm clock. I wanted to do this so I bought everything and got to work.

Parts:
Motorbike Horn (12V 1.5A DC)
Wall power adaptor (12V 2A DC)
IRF540N HEXFET Power MOSFET (https://m.ebay.co.uk/itm/IRF540N-HE...043399?varId=490205044826&txnId=1424930910009)
2N3906 PNP TO-92 Transistor (https://m.ebay.co.uk/itm/2N3906-PNP...650234?varId=590358508478&txnId=1477261816019)
Alarm clock with 1.5V DC trigger signal
10K Resistor
1K Resistor

I disassembled the alarm clock and found that the beeping sound was produced by a tiny 1.5V buzzer. I de-soldered it and connected a female 3.5mm jack to the buzzer’s wires so that I could connect an aux cable to the alarm clock when needed.
I then made the driver circuit using the diagram attached but it did not work. The horn would make a constant long sound as soon as I connected the 12V DC source and wasn’t affected by the voltage (or absence of it) from the alarm.

So that’s what I need. A circuit that can take a 1.5V DC signal and use it as a trigger to allow a 12V 1.5A DC current to flow through a horn.

Any ideas are welcome.
Thanks
(Please bear in mind that I only have a limited understanding of how electrical components interact so a diagram would be very helpful!)

 

[duke37]

This would never have worked.
You will need a npn transistor (2N3904, 2N2222, BC548, 2N706, BC108 etc) as the input feeding the pnp driver.

Keep the 1k input resistor, connect to base, connect emitter to ground, connect collector to 2N3906 base via a 10k resistor.

You may need a capacitor gate to ground to hold the fet on between pulses.

 

[AnalogKid]

First, add a resistor between the base and emitter of the 3906. This guarantees that the transistor turns off when the input goes open circuit. Something between 10 K and 100K should work.

Speaking of the input... This circuit assumes that the signal from the alarm clock goes to a very high impedance when the alarm is not beeping. As with almost all instructables circuits, this is a very shakey assumption. If the beeper drive goes to GND (or anything less than 11 V), then the horn will be on all the time. Can you measure the beeper output signal in the on and off states?

If the output signal does go to GND in the off state and goes to something above 4 V in the on state, then you might be able to eliminate the 3906 and drive the MOSFET directly.

ak

 

[CDRIVE]

In addition to the advice already given ...
If your motor bike horn is of the electromagnetic variety you may also want to install a protection diode across it.

Chris

 

[Audioguru]

You copied the circuit from an "Instructable" that was designed by a kid who knows NOTHING about electronics.
So of course the alarm blasts when it should be quiet and it is quiet when it should blast.

 

[CDRIVE]

Yeah, "Distructables" gives just enough bad information to help a nube electrocute themselves or burn their house down!

Chris

 

[Alexander8046]

This would never have worked.
You will need a npn transistor (2N3904, 2N2222, BC548, 2N706, BC108 etc) as the input feeding the pnp driver.

Keep the 1k input resistor, connect to base, connect emitter to ground, connect collector to 2N3906 base via a 10k resistor.

You may need a capacitor gate to ground to hold the fet on between pulses.

Hi, thanks for the suggestion. I bought a 2N3904 and did this but it still doesn't work. When I connect the 12V source the horn sounds and the voltage across the horn is 5V. Then, when the the signal from the alarm clock is emitted (0-1.5V) I can hear the horn sound change and the voltage across the horn is 7V at intervals (although it may be more, but my multimeter can't keep up with the speed at which the voltage goes from 0V change).

 

[Alexander8046]

First, add a resistor between the base and emitter of the 3906. This guarantees that the transistor turns off when the input goes open circuit. Something between 10 K and 100K should work.

Speaking of the input... This circuit assumes that the signal from the alarm clock goes to a very high impedance when the alarm is not beeping. As with almost all instructables circuits, this is a very shakey assumption. If the beeper drive goes to GND (or anything less than 11 V), then the horn will be on all the time. Can you measure the beeper output signal in the on and off states?

If the output signal does go to GND in the off state and goes to something above 4 V in the on state, then you might be able to eliminate the 3906 and drive the MOSFET directly.

ak

Do you mean putting the resistor like this:

Also the beeper signal is 0V when off and 1.5V when on

 

[Alexander8046]

In addition to the advice already given ...
If your motor bike horn is of the electromagnetic variety you may also want to install a protection diode across it.

Chris

Ok I'll do that
Thanks for the advice

 

[Audioguru]

The 2N3906 PNP transistor turns on when the input goes to ground. Then the Mosfet also turns on which turns on the horn. It is backwards.
Instead you need an NPN common-emitter transistor to turn on a P-channel Mosfet that turns on the horn when the input is about +1.5V.

 

[AnalogKid]

Also the beeper signal is 0V when off and 1.5V when on

That is a critical detail, and renders moot most of the responses so far. 1.5 V is not high enough to turn on a MOSFET directly. One way to go is to turn you schematic "upside down". That is, change the PNP transistor to NPN (2N4401, 3904, 2222, etc.) (emitter tied to GND) and change the MOSFET to a p-channel type (source tied to +12 V). Now one end of the horn is tied to GND and the other end is pulled up to the +12 V by the MOSFET.

ak

I started this response then had to go take care of something, and AG beat me to it.

 

[Alec_t]

Another way is to keep the PNP and NFET that you have, but use the beeper signal to switch on an NPN which, in turn, switches on the PNP and hence the NFET.

 

[duke37]

Another way is to keep the PNP and NFET that you have, but use the beeper signal to switch on an NPN which, in turn, switches on the PNP and hence the NFET.

That is what I suggested in #2.

It may be possible to just use a power npn transistor if the clock can drive it, otherwise a Darlington may suffice.

 

[Alec_t]

That is what I suggested in #2.

Sorry Duke, I missed that.

 

[CDRIVE]

Here's a schematic of Duke's and Alec's suggestion.

Chris

 

[duke37]

Thanks CDRIVE, it may be advantageous to put a capacitor across R4 to keep Q3 on between input pulses. Try 100nF.

 

[CDRIVE]

Done.

Cheers
Chris

 

[Alexander8046]

Done.

Cheers
Chris
View attachment 36421

Thanks everyone
I will try this on Saturday and let you know if it works

 

[CDRIVE]

Alexander, just a quick note. Though I specified Q1 as a 2N3904 it's not a critical component value. A 2N2222 will also work just fine. In fact just about any small signal NPN will do the job in this undemanding application.

Chris