piezo buzzer using 555 ?

[Thieaux]

High guys im a n00b at this, but i've been practicing a lot , i need some assistance on a project , when i use my keyless entry on my car it doesnt emit any sound or visual alert so i dont know if its open or not .

i want to use the light from the security led to trigger my 555 so it can emit 3 beeps in one second to let me know that i've locked my car (that would be 3 hz?) 3 cycles per second right?...well anyway trigger it from the led , 3 beeps thats it .

But can the 555 handle a very loud beeper im talking in hearing it a couple of feet away , i think in the 90 db range .

but also the trigger could be the momentary 12 v pulse from the door locks  , i would preffer the door locks , but since the pulse its very fast , could tha capacitor have a chance of chargin and  powering the beeper ?

IF anyone can help me with the schematics would really appreciate,
    THanks , sorry for the long post ...

[audioguru]

A piezo beeper might be loud enough when it is powered from only 12V.
A 555 by itself will make a single long output (monostable) , or continuous outputs (astable)without stopping. A counter IC like a CD4017 and some diodes can make 3 pulsed outputs, then drive a transistor that drives the beeper.

[Thieaux]

Thanks man, correct me if im wrong , ill be using the led to trigger the counter , it would start the sequence, drive the beeper trough a transistor and stop, does it mater that the led would be pulsating as long as the car is locked or would go into a loop aqnd reset itself everytime the led flashes.?

[audioguru]

I was thinking of using a CD4017 counter. It needs to have an oscillator to drive it. The oscillator is triggered by the LED signal the the CD4017 would make the 3 beeps and its 4th output would turn off the oscillator for it to be ready to be triggered the next time. If the LED keeps flashing then the circuit will continuously beep.
You need to have another circuit that detects the LED flashing to start the oscillator and the CD4017 stops the oscillator. Then the other circuit is reset when the LED stops flashing.

[Thieaux]

How about using the signal from the keyless ecu, that sends a momentary 12v to the locks

I could use that to trigger it , the problem is that im so new at this , and  i have a lot of questions....

use the signal from the ecu with a diode , to trigger the oscillator, which in terms causes the counter to "count" and on the 4 output to reset of the oscillator .

so ill be using a 555 and a 4017?

Cause remember when i press my kkeyless remote i want 3 beeps confirmating that the car was licked, so ill be using the signal from the ecu to the locks.

I/ve been told to use a self oscillating piezzo buzzer, but it defeats the porpouse of learning electronics.

[Thieaux]

any help with the schematics?

do i need to have a constant supply of voltage or could short pulse do the job with a cap?

[audioguru]

The circuit needs to have a constant supply voltage.
A piezo beeper has a tuned resonant housing and the piezo has an extra feedback terminal so its built-in oscillator oscillates at the tuned frequency for the loudest sound. If you make your own oscillator then it must be manually adjusted to the loudest frequency.
A 555 oscillator to step the CD4017 needs to have a flip-flop circuit to start it and to stop it.

[Thieaux]

Went and got a pulsating buzzer from radio shack , i could use the momentary voltage to the lock solenoid to somehow activate a flip- flop that is power independently from a 12 v source , so when the voltage is applied for a second or so to the flip-flop it could activate it cuont the time and shut down. What do you think?

[audioguru]

Instead of a flip-flop, you need to have a monostable timer circuit that is triggered by the voltage pulse to the locks then it times-out and turns off the buzzer.
A 555 or a Cmos 555 could be used as a monostable timer but their trigger pulse is negative, a transistor inverter is needed.

[Thieaux]

U DA MAN, i was reading on monostables i figured that was what i need it, but u beat me to the punch , so here it  goes :

Pulse from the locks , to trigger the timer , 555 using an constant 12 v from the accesory or near voltage supply on the car .

1- Timing calculator for the buzzer? .
2- can the 555 power the buzzer or do i need a relay (radio shack 273-066) i think is rated at 125 milliamps?
3 does the inverter only works one way or it doenst matter what polarity comes in, he changes it to the opposite ?

Regarding the third question, i asked cause the solenoid from the locks only use 2 cables that change from + to - depending on the function pressed (lock or unlock)

Thanks

[audioguru]

1- Timing calculator for the buzzer?

The datasheet for a 555 or a Cmos %%% has the circuit for a monostable and a graph showing parts values vs time.

2- can the 555 power the buzzer or do i need a relay (radio shack 273-066) i think is rated at 125 milliamps?

How much current does your buzzer use? A 555 can supply up to 200mA and a Cmos 555 can supply up to 10mA.

3 does the inverter only works one way or it doenst matter what polarity comes in, he changes it to the opposite?

An NPN transistor has its emitter grounded and its collector connected to the trigger pin 2 of the 555. A 10k resistor connects pin 2 to the 12V supply. The base of the transistor has a 10k resistor in series and it connects to the locks. When the locks are not powered then their voltage is 0V so the transistor is off and pin 2 of the 555 is at the positive supply voltage so it is not triggered. When the locks get power then the base of the transistor has current so it turns on and grounds pin 2 of the 555 which starts its timing.

Regarding the third question, i asked cause the solenoid from the locks only use 2 cables that change from + to - depending on the function pressed (lock or unlock)

The transistor won't work both ways and might not work properly when the doors lock. Maybe two diodes can be used to detect that one of the door lock wires goes positive then feed the 10k resistor to the transistor's base, and the transistor will need a 100k resistor from its base to its emitter to turn it off when it should be off.

[Thieaux]

DONE!!!!!
monostable done , with a 2 second "ON" time.

All i need now is to make the transistor inverter , correct me if im wrong , but wouldnt that be 12v (resistor ) to base , and grond to the collector , so when the base glows, allows a path to ground?

[audioguru]

Depending on how your door locks are wired, this inverter might work:

[Thieaux]

Everythings is ready and tested , works great , dont know how to thank you !!!

Keep up the good work u sure are THE GURU!!!!!

But i think ill have to tap the signal from the ecu directly to see if i can get a cleaner signal , cause on the locks it oscillates betwen 0.2 and 2.12 volts , making the system unstable.

;D

[audioguru]

Hi Thieaux,
Good it works, but it can be made more stable.
Reduce the value of the resistors like this:

[Thieaux]

remember im new at this, would these changes in resistors make the transistor less vulnerable to voltage fluctuations, cause as i understand the base needs to be low at all times, so i need a way to stop the fluctuation at the base , how would i go about it, im sorry  to be a nagg but i really want to understand this.

So we need to find a way to make the fluctuations betwen 1.5 to 4 volts at the base come to "0" .
iF the solution is by chaging the values of the resistors, could you explain why?

[audioguru]

Hi Thieaux,
Since your locks must lock and unlock by swapping the voltage at their ends, I assume that they are not connected to anything when they are not operating. I also assume that the fluctuating voltage you measure is just interference from other electrical motors that is picked up by the wires acting like an antenna. The lower resistor values will short any interference pickup to ground.

The transistor needs at least 0.6V at its base to turn on. The diodes each have a voltage drop of 0.6V. The resistors are a voltage divider.
I changed the ratio of the resistor values to 3.3:1 so that a voltage of 3.2V or less at the locks should not turn on the transistor. Change the 100 ohm resistor to 39 ohms and the transistior should not turn on if the voltage on the locks is less than about 6.3V.

[Thieaux]

Thanks for your patience , ill let you know tomorrow...

[Thieaux]

Ok. Help me out here , i read on the forum the explanation on a voltage divider .
Since the value of the resistence on the lower one is 100 , it takes less voltage to make the electrons flow , so since the less voltage is required it means that less voltage is flowing trough Vout, what i still cant understand is the ratilo u mentioned on the previous post .