How to shorten the length of a trigger signal?

Thread Starter

Central Texan

Joined Jun 8, 2020
10
I’m looking for some knowledge. I have a 2005 Tacoma. It has keyless entry but doesn’t give a short beep of the horn when it is locked like other cars. I installed a relay using the lock signal for the locks as a trigger to activate the horn. The relay functions fine except that the horn honk is too long. The trigger signal is a positive 12 VDC. Is there any cheap and easy fix that I could install to shorten the trigger time length? Thanks for any help.
 

Thread Starter

Central Texan

Joined Jun 8, 2020
10
I’m new to the forum and my electronics knowledge is limited. I was thinking that I would like something that is adjustable if possible. I have done some searching on this site and found reference to something called a “one shot”. Not sure if something like that could be wired up using 12 VDC?

My ideal solution would be something I could buy off the shelf if it is not too pricey. If not then building something would be my other option. I’m just not sure where to start. Again thanks for any help. I am attaching a pic of the current relay switch I am using.

DE897F21-F7EE-4E2B-91F2-75565F20F8BA.pngDE897F21-F7EE-4E2B-91F2-75565F20F8BA.pngDE897F21-F7EE-4E2B-91F2-75565F20F8BA.png
 

drc_567

Joined Dec 29, 2008
855
...Merely speculating, you might experiment by placing an electrolytic capacitor at the 87 terminal, in line with the horn. You will have to check and see what kind of audible modulation this will yield. ... Is it sufficiently loud?
Maybe start with 100 uF capacitor, with the + terminal oriented towards the relay terminal. A bleeder resistor of maybe 100 kohms (100,000) could be placed in parallel with the electrolytic capacitor to allow repetitive operation. Be sure that you get the electrolytic type capacitor, with maybe a voltage rating of 25 volts or so. There should be + and - markings to indicate the different terminals.
There are a multitude of more complicated, and more precise methods, which can be employed.
However, you might test the series capacitor method to see what the result is.
 

Thread Starter

Central Texan

Joined Jun 8, 2020
10
Thanks for the ideas. So do you think it would be better to place the capacitor on the trigger, or the device output? I believe the trigger is in milliamps where as the output is probably around 6-7 amps. Is there a way to calculate how long the discharge time of the capacitor would be? Thanks for all your help.
 

drc_567

Joined Dec 29, 2008
855
... There is an index in capacitor circuits called the RC time constant, whereby the voltage level takes about 5 time constant to go to zero. So, with the capacitor placed on the horn side, most likely all you would get is a blip, since resistance would be nearly zero. If the capacitor is placed on the relay coil input, a coil resistance of 1200 ohms and a capacitor of 100 uF might turn the horn relay on for one of two tenths of a second before it decayed away. ... The one variable that you can change is the size of the capacitor.
 

Thread Starter

Central Texan

Joined Jun 8, 2020
10
Thanks for your help. Just curious why did you recommend electrolytic style? Is it because they tolerate heat better?

Also what would happen if I didn’t put a resistor in parallel?

If I put the capacitor on the horn side do I need to get a special type of capacitor since the horn runs 5-6 amps? Or could I try the same capacitor on the horn and trigger side?

What store do you get your components from?
 

AnalogKid

Joined Aug 1, 2013
8,489
For large capacitance values that don't have to be very precise, electrolytics and the only reasonable choice - lotsa uF for the $.

A simple R-C time constant calculation will not work here, because the relay dropout voltage probably is not 36% of its nominal operating voltage.

IF
This is a 12 V system
The relay drops out at 9 V
The relay coil resistance is 1200 ohms

THEN
You need 290 uF for a 0.1 s pulse, and 725 uF for a 0.25 s pulse.

ak
 

Thread Starter

Central Texan

Joined Jun 8, 2020
10
Also forgot to ask do you think I could use a 16V or 50V capacitor? Not sure how the voltage rating effects everything. I know my voltage is 12VDC, but you suggested 25 V earlier.
 

Thread Starter

Central Texan

Joined Jun 8, 2020
10
For large capacitance values that don't have to be very precise, electrolytics and the only reasonable choice - lotsa uF for the $.

A simple R-C time constant calculation will not work here, because the relay dropout voltage probably is not 36% of its nominal operating voltage.

IF
This is a 12 V system
The relay drops out at 9 V
The relay coil resistance is 1200 ohms

THEN
You need 290 uF for a 0.1 s pulse, and 725 uF for a 0.25 s pulse.

ak
Thanks a lot for the info! Should I put the capacitor on the trigger or horn side? Do I need a resistor in parallel? Would a 16 V capacitor be ok?
 

Hymie

Joined Mar 30, 2018
836
You could achieve this by adding a relay, the contacts of which switch 12V to the horn circuit. The trick is rather than the negative of the relay coil being taken to 0V (chassis) it is connected to 0V via a capacitor of circa 1000µF.

In operation, when 12V is applied to the relay coil, the relay is momentarily energised as the 1,000 µF capacitor is charged (via the relay coil), once the capacitor is charged the relay will de-energise, giving a short beep to the horn.

A bleed resistor of around 10kΩ should be wired in parallel with the capacitor to discharge it, ready for the next beep.

You might need to experiment with the capacitor value to obtain the desired beep length.
 

AnalogKid

Joined Aug 1, 2013
8,489
Would a 16 V capacitor be ok?
As a rule of thumb, run electronic components at 50% or less of their ratings for better long-term reliability. For example, use a 25 V cap in a 12 V system, use a transistor rated for 2 A when driving a 1 A load, etc.

An automotive environment is harsh. Be sure to select an electrolytic with a 105 degree C rating.

ak
 

AnalogKid

Joined Aug 1, 2013
8,489
Yes. The power dissipated in a DC circuit resistor equals the voltage across it squared divided by the resistance value:

P = E^2 / R

Rearranging and plugging in 1/8 watt for margin, any resistance value over 1152 ohms will dissipate less than 1/8 w. Note that the resistor is not dissipating its max power all the time. When the circuit is first energized, most of the current is charging the capacitor; the voltage across the resistor (and hence its power dissipation) increases exponentially over the timing period.

When the circuit is de-energized and the resistor is discharging the capacitor for the next use, a 10 K resistor will take 30 seconds to discharge a 1000 uF capacitor to 5% of full charge.

ak
 

Thread Starter

Central Texan

Joined Jun 8, 2020
10
It took awhile to get my components delivered, but I tried them out today. I wasn’t able to get any type of beep with the ones I tried. Below is a list of the different ones I tried.

100 uF 16V with 20 kohm in parallel
330 uF 25V with 20 kohm in parallel
2200 uF 25V with 20 kohm in parallel

I wired the cap on the device side with the minus hooked up closest to the horn. I didn’t try hooking up to the trigger side because I wasn’t sure if the capacitor would damage anything since I teed into the door lock signal harness of my truck. Does anyone have any ideas? Did I hook something up wrong? Thanks for any help.
 

Thread Starter

Central Texan

Joined Jun 8, 2020
10
If I put the cap on the signal side do I need to worry about the capacitor causing any damage to my trucks electronics since the signal is teed off the unlock motor signal?
 

ElectricSpidey

Joined Dec 2, 2017
1,175
I would be more worried about the added relay doing damage, did you also install a reverse EMF diode with the relay?

Although I would suspect if you placed the relay in parallel with the lock, it should already have a diode.

If you are worried about damage placing the cap on the signal side then just keep in on the horn side, but you will need a larger cap to beep a car horn.
 
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