Circuit to trigger on pulse and let current pass for x seconds

Discussion in 'General Electronics Chat' started by wuhtzu, Oct 28, 2011.

  1. wuhtzu

    Thread Starter New Member

    Oct 24, 2011
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    Hi everyone

    Sorry for the very bad title, I don't know the names of the more specific components, circuits and concepts.

    What I want to do is have an wireless doorbell activate an rotating light (http://www.belys.dk/Graphics/Products/1847.jpg).

    So I figured that some pulse must be generated in the doorbell receiver when the transmitter is pressed and that I could let my circuit trigger on this pulse - be it a pulse to activate the audio or the audio signal itself. When I detect this pulse/signal "my" circuit should activate some form of electrical switch which lets current from the batteries pass to the rotating light.

    Could you give me some pointers as to which components / concepts I need to look into?

    Thanks a million
    Wuhtzu
     
  2. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    It depends on your level of experience what would be the best approach to do this.
    The easiest thing is to buy a sound activated switch, like this:
    http://www.quasarelectronics.com/3126-sound-activated-relay-switch.htm The off-delay can be adjusted,

    If the rotating light's current rating is higher than 1A, you would need to put another relay or MOSFET in between.

    If you have the experience you can also directly try to take the signal from the receiver board and build your own circuit on a small PCB. This won't get necessarily cheaper and will take more time.
     
  3. wuhtzu

    Thread Starter New Member

    Oct 24, 2011
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    Thank you so much for your reply.

    I might have exaggerated my inexperience.

    The sound activated switch is not an option, the speaker part of the doorbell has been deactivated. So there is no sound. Actually all i want to do is wirelessly activate my rotation light and I just thought that a wireless doorbell had all the wireless communication already working.

    So whatever small voltage/current (pulse) I can detect in the wireless doorbell receiver when i press the transmitter, I plan on using for triggering a power mosfet.

    I guess one of the things I was looking for was the "power mosfest". That is what I will use to to deliver power to the rotating light from an external power supply.

    So "all I need now" is a circuit / component which when triggered can trigger the power mosfet for x seconds.

    Thanks agian
    Wuhtzu
     
  4. shortbus

    AAC Fanatic!

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  5. Wendy

    Moderator

    Mar 24, 2008
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    I am also very active on this site. As you can see my reward was to be allowed to work harder! :D

    If you a have any specific questions feel free to start a public post, then send me a PM directing me to it.
     
  6. wuhtzu

    Thread Starter New Member

    Oct 24, 2011
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    Thank you for your 555 suggestion - that certainly looks like what I need.

    Would you recommend me to start a new thread with questions related to this 555 monostable circuit?
     
  7. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    You may continue here if you want. In order to trigger a 555 circuit to your doorbell, we need to know what signal you get from there when it is "ringing".
     
  8. wuhtzu

    Thread Starter New Member

    Oct 24, 2011
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    I will carry on here for a bit then.

    After reading about the 555 monostable circuit I am under the impression that the circuit triggers when the voltage applied to pin 2 (the trigger pin) of the 555 timer goes below \frac{1}{3} V_{cc} and V_{cc} is the voltage applied to pin 8 (2-15V).

    So if one were just playing around, trying to learn about the 555 timer, one might have attached a simple mechanical switch between pin2 and the V_{cc} supply. 0V must be below \frac{1}{3} V_{cc} . Am I right?

    So I somehow need something which cuts power to pin2 when it receives the pulse from the doorbell receiver? That was what you sort of referred to when saying "we need to know the output / pulse from the doorbell receiver", right?
     
  9. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    Yes , measure on the doorbell pcb and find any signal that is being activated when the bell rings. It could be anything, positive or negative logic, any voltage level, pulse train anything. Once the signal you want the 555 to trigger with is known, we can help you elaborating a circuit to adapt the doorbell signal to what the 555 needs.
     
  10. wuhtzu

    Thread Starter New Member

    Oct 24, 2011
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    Great, thank you. I'll measure on the doorbell on Tuesday (that's the next time I get by the electronics lab)

    One additional question. To test the 555 timer circuit (which has to be build eventually) could I just use a mechanical switch to trigger pin2? Simply switch pin2 from Vcc to ground to trigger?
     
  11. praondevou

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    Jul 9, 2011
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    For your purpose you could do this, because the time constant you want to adjust is a few seconds long.
    The standard 555 astable multivibrator will be retriggered each time you pull the trigger input to LOW, so for shorter time constants you need to consider the bouncing of a simple switch.
     
  12. wuhtzu

    Thread Starter New Member

    Oct 24, 2011
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    Today I have tried out the 555monostable circuit explained here (http://forum.allaboutcircuits.com/showthread.php?t=14018) and it worked perfectly (surprise... :))

    I also found that I am able to get an 2-5 seconds 1.6-1.8V signal from the doorbell receiver when it is activated. This voltage is supplied to a LED in the doorbell, but I figure it will be perfect for triggering the timer.

    I have successfully taken the output from the 555 timer and used it to trigger a power mosfet which passes current to the rotating light.

    All I need now is to trigger the timing circuit via this 2-5s 1.6-1.8V signal. I tried to have a transistor collect ground and emit it to point A (or B), but I didn't work. Probably because i try to trigger the transistor with power source which powers the circuit which seems to lead to a short circuit (my power supply says so).

    I will draw a circuit of this later, but maybe the text above is enough to get some thoughts going :)

    thanks again
    Wuhtzu
     
  13. sheldons

    Well-Known Member

    Oct 26, 2011
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    Heres a driver schematic with a mosfet you can use, replace the led and resistor and switch whatever load you want(within the limits of the fet)and the second schematic you can use to test your 555 .....for quickness you can trigger your 555 by shorting pin 2 to ground briefly


    [​IMG]



    [​IMG]
     
    Last edited: Nov 2, 2011
  14. wuhtzu

    Thread Starter New Member

    Oct 24, 2011
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    Thank you for your reply sheldons.

    I am not exactly sure what either of your circuits do. But if we just focus on the first one, the driver circuits. Is all that really necessary? Isn't it just hooking up a single mosfet and (maybe) adjust the output of 555 pin 3 (output pin) to a voltage suitable to trigger the mosfet?
     
  15. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    You can trigger the 555 with the circuit attached. I did only draw the trigger input of the 555 and also the MOSFET looks horrible, don't have the time to do it better. It's a low power n-channel MOSFET.

    The MOSFET needs to be one with a gate threshold voltage lower than your doorbell output signal of course. There are many of them on digikey that work with voltages of about 1V. I don't have the time right now to provide a list of them.

    It's important that the MOSFET turns ON fast enough to pull the trigger input down through the capacitor. Otherwise you may need to increase the cap value a bit.
     
  16. Wendy

    Moderator

    Mar 24, 2008
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    A 555 can handle 200ma, no more. A transistor can handle much more.
     
  17. wuhtzu

    Thread Starter New Member

    Oct 24, 2011
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    Thanks alot praondevou.

    I have tried your circuit, but I did not have any spare mosfets and 1 nanofarad capacitors laying about, so I used a BC557B transistor and a 10 nanofarad capacitor - and it did not work.

    Is that to be expected? Although transistors and mosfets perform the same general task, there are differences between them.

    I seem to have a problem with the whole thing just short circuiting through the collector-base connection.

     
  18. sheldons

    Well-Known Member

    Oct 26, 2011
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    the 555 op is active hi-which will turn the first transistor on grounding the base drive resistor connected the the pnp transistor-which then turns on the fet which will switch whatever load is connected to it...as far as the mosfet symbol goes thats what came with the schematic program im using to draw my schematics....of course you could simplify things and use a relay.....

    [​IMG]

    the second circuit can be used to test your 555 and it can again be changed partswise

    [​IMG]
     
  19. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    I don't get this:confused:

    If using a npn transistor you have to make sure that it works as a switch. What base resistor did you use?

    When using this circuit switching speed is important, if too slow the cap is charged before the trigger level of the 555 is reached. You can also try to increase the 47k resistors.

    If using the npn transistor it would be interesting to know what the slope is of the doorbell signal... Are you sure it can drive the 557B?
    You can find out by measuring the doorbell signal while activating your circuit / driving the npn. The voltage level should not change considerably.
     
  20. praondevou

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    Jul 9, 2011
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    Sheldon we are talking about triggering a 555 with a positive 1.5 to 1.8V signal. I don't see how your circuit could do this.
     
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