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Help with a very simple rc circuit..

Discussion in 'General Electronics Discussion' started by mathphobe, Sep 28, 2018.

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  1. mathphobe

    mathphobe

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    Sep 3, 2017
    .. for you guys, me.. not so much :(

    I've installed a car alarm, literally made the final connection only to find out its faulty.

    There is a fuel cut off relay that isn't receiving a control voltage from the unit. I could replace the alarm but for many reasons it's a ball ache I could do without tbh.

    So.. looking for work-arounds there's a Vout to the indicators, it could replace the lost source as it's triggered at the same time, if it were a constant voltage and not operating at a frequency of once per second.

    I'm looking to hobby out a solution in my down time this weekend. I like tinkering but for some odd reason I dont like buying alarms and installing them for free o_O.
    I figure a simple rc circuit could work? If so I've no clue of the values the resistor and capacitor need to be, though I have a good variety of both.

    It's a 12V circuit, the current from the unit would be 10mA (if it were present) though I doubt the relay would care much about specific mA as long as it's getting close to 12v.

    Objective; make a 12v 1sec pulse into a relatively stable 12v source the simplest way possible. Can anyone help with this please?
    Thanks, Mike.
     
    Last edited: Sep 28, 2018
  2. Tha fios agaibh

    Tha fios agaibh

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    Aug 11, 2014
    I think you need to clarify.
    "Current from the unit would be 10ma"
    Are you saying the alarm output current is limited to 10ma?
    Are you trying to energize the relay coil for the fuel relay with a 1 second (one shot pulse)?
    What are the ratings of this relay your trying to drive?

    "Make a 12v 1 sec pulse into a relatively stable 12v source." Huh??

    Do you mean create a separate 12v output pulse for 1 sec from a 12v source?
     
  3. mathphobe

    mathphobe

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    Sep 3, 2017
    The output from the alarm to the relay 's coil should be 12v 10mA when the alarm is triggered.
    As that isn't functioning the only other useable outputs are switching on and off in one second periods, I want to spur off that in a way that constantly energises the relay's coil when the alarm is triggered.
    My idea was to make a small board that sees a capacitor discharge fill in the gaps, if that's feasible.
    The 'pulsing' lives are feeding indicator lights at a 5A current, there are two of these, one for each side, there is also one for the sounder siren. Tapping off one of those shouldn't cause an over draw issue. 2× 20w bulbs @12v should draw ~3.33A.
    I assumed the relay should only need 10mA as that is the given parameter from the missing live, the relay is an 'nc jd1915 12v 40a'
    Hope that clarifies?
     
  4. kellys_eye

    kellys_eye

    4,275
    1,146
    Jun 25, 2010
    From your cryptic description I get the impression that the alarm INDICATOR is an LED that takes around 10mA (usually 20ma but...meh...) to activate and you want to use that 'signal' to drive a relay.

    Since the LED is pulsing at 1 second intervals you need a 'pulse extender' which otherwise goes by the name of a re-triggerable monostable and the ubiquitous 555 timer is good for this purpose.

    Use timing components in the circuit to ensure the output stays high for the interval of the LED flashes - the alarm output will remain active whilst the LED is flashing PLUS the time interval of the 555 re-triggerable monostable period (say another 1.5 seconds)

    mono.png
     
    Tha fios agaibh and Terry01 like this.
  5. Alec_t

    Alec_t

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    Jul 7, 2015
    Here's an alternative :-
    PulseExtend.PNG
     
  6. mathphobe

    mathphobe

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    Sep 3, 2017

    You should try talking to me in person, its even worse :D
    I meant the cars directional indicators but the 'pulse extender' answer is still the same so thanks!
     
  7. mathphobe

    mathphobe

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    Sep 3, 2017
    I think making this circuit might be easier than the 555 based one, given I might not need to etch a board. Thanks!
     
  8. airgap

    airgap

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    Sep 18, 2018
    Here is a solution using a comparator. As long as the voltage on the comparator's + pin remains above ~1/3 Vcc, the relay will remain on constantly. Ensure this happens by setting the RC time constant to be LONGER than the time the pulse is off (0V).
     

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  9. airgap

    airgap

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    Sep 18, 2018
    OT: In my adolescence my friends and I used to "steal" car alarms...we would break in the car, unhook everything from the alarm box both inside the car and under the hood, then leave the box, sensors, relays, horns, flashers, wiring harnesses and anything else we could pull relating to the car's alarm system in a huge pile on the driver's seat for the owner to find in the morning hehe.

    Left the radio and any money/valuables we found in the car completely alone, however. We just went after the alarms! Fun days.
     
  10. airgap

    airgap

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    Sep 18, 2018
    Had an idea on how to get it done with just one transistor.

    Edit: The capacitor should be polar type with + facing up.
     

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  11. Alec_t

    Alec_t

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    Jul 7, 2015
    I think you'll find the base current needed to switch the transistor fully on will drain the cap rather too quickly, unless the transistor is a Darlington type..
     
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