Alarm circuit

Discussion in 'Homework Help' started by dcar2013, Apr 26, 2013.

  1. dcar2013

    Thread Starter New Member

    Jan 18, 2013
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    I need to design a alarm circuit using 2 NOR gates, DPST switch for ready- standby enable and momentary switch for reset. Also, has 4 SPST switches that when any one of them is activated an LED will light indicating an alarm condition. See attachment.
     
  2. panic mode

    Senior Member

    Oct 10, 2011
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    why don't you explain how is this device supposed to be used? what are the issues you are having?

    why are you using inverter rather than another 7402 gate? you do have two spares...

    where is LED supposed to be? who is supposed to see it? if someone random stops by and presses some button and sees red led after one button press, he will certainly know not to press that button again. the way i see it, few attempts can reveal the "password".
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    You give a specific list of things you are allowed to use and then post a schematic that has many things not on the list. So what are the rules?

    I think the SPST switches simulate the door and window sensors, not some kind of password entry mechanism.

    What is the purpose of R3?

    Would you need U4B if you took the input off the other side of the latch?

    If the output of U2A is LO, how will anything every change it?
     
  4. dcar2013

    Thread Starter New Member

    Jan 18, 2013
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    SPST switches are used for doors, windows, etc. No specific purpose for R3 just trying different things to make the circuit work. LED to show alarm condition when a low is on the output. Also, 7404 not necessarily part of circuit, again just trying different things. U2A and U3B are cross -coupled as given in the original question. "The burglar alarm could be constructed with two cross-constructed NOR gates. Complete the circuit so the data inputs represent switches connected to doors, windows. Enable input is pulled high when system is activated and low for standby. ( ie DPST sw.) Momentary ground on Q output to reset system, using momentary switch. U2A has one input and the only output."
     
  5. WBahn

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    Mar 31, 2012
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    Thanks.
    This is known as "design by happening" -- you try random things and hope that, at some point, something good just happens.
    More design by happening.
    This sheds a lot more light on the subject, but several points are still unclear. When it says that U2A and U3B are cross-coupled as given in the original question, how are they shown in the original question? Which is the Q output?

    What is supposed to happen when the system is "activated" and what is supposed to happen when the system is in "standby"?
     
  6. dcar2013

    Thread Starter New Member

    Jan 18, 2013
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    U2A has the only input and output, ( see attachment Scan 7 ). When system is activated this means that the DPST is in the Ready position and the burglar alarm is armed. Only when one of the SPSTs are open will the LED come on. When in Standby the system is no longer activated and nothing happens. Also, see Scan 6 attachment for original circuit as constructed with NAND gates.
     
  7. WBahn

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    Mar 31, 2012
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    So, if I understand it correctly, Scan7 is a collection of circuit fragments that you are supposed to use in order to solve the problem and, I'm assuming, you are not supposed to use anything other than those fragments. Correct?

    The first thing to do is be sure that you understand how the cross-coupled NORs work. So draw a truth table that has the single input and the outputs of both NORs as your input columns and then determine what the outputs would be for each case. Once that is done, we can worry about identifying which rows in the table can be ignored because they can't happen (unless we add other features, which we will).
     
  8. dcar2013

    Thread Starter New Member

    Jan 18, 2013
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    Let me see if I understand you correctly. If a 0 is applied to the input of the upper NOR gate, that produces a 1 on the output, that in turn produces a 1 on combined inputs of the lower NOR gate, which in turn produces a 0 on the other input of the upper gate then output of upper gate is 1 and LED is off. If a 1 is applied to input on upper gate output becomes 0 applying 0 on inputs of lower gate producing a 1 on output of lower gate and B input of upper gate then output of upper gate is 0 and LED is on. Maybe I just don't see it correctly, tell me if I'm wrong.
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    You can't analyze it simply as a combinatorial circuit because it isn't one. It is a feedback sequential circuit, so you have to consider things under different stored states as well as different input conditions.

    So let's consider when the input (to the upper NOR gate) is LO. There are two possibilities, that the output of that NOR gate is LO or that it is HI. You need to consider both and see if one, the other, or both represent stable conditions. Then do the same thing for when the input is HI. Do that much and then we will proceed to turning that information into a state transition table. But first things first. So complete the following table:

    IN OUT_n INV OUT_n+1
    0 0
    0 1
    1 0
    1 1


    IN is the input signal
    OUT_n is the output signal before IN was applied
    INV is the output of the inverter after IN is applied
    OUT_n+1 is the output signal after IN is applied and things settle
     
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