Rising/Falling edge trigger hardware oriented

Discussion in 'The Projects Forum' started by FuzzballJack, Apr 12, 2014.

  1. FuzzballJack

    Thread Starter New Member

    Apr 12, 2014
    13
    3
    Hija,

    Trying to make a hardware oriented edge trigger. I have a lot(!) of op-amps lying about. I thought of making some with them. I have downloaded LTspice and tried some designs I had in mind. Problem is, spice never is the same as IRL. I haven't tested my latest design. But spice is giving me some unexpected results. Are there some spice gurus about? Check this schematic.

    [​IMG]

    Now this is the output of spice with the input visible. I'll comment this.

    [​IMG]

    V(n001) (green) is the sensitivity voltage. Measured at output U1.
    V(n005) (blue) is the positive edge trigger voltage. Measured at positive input U2.
    V(n006) (red) is the output trigger signal. Measured at output U2.
    V(n004) (teal) is the switch signal to be triggered. Measured at output V1.

    Now a graph without v(n004) to clean up the graph for clarity sake. I will address my problem next.

    [​IMG]

    V1 will IRL be a switch that switches 5V to 0V
    V3 will IRL be a pot from positive rail to negative rail to adjust the sensitivity.

    Now. Every signal is as expected except the output signal V(n006) (red).

    Some explanation of how I thought of the circuit. C1 R1 is a high pass filter. So the spike of the transition of the switch will be the triggered signal (step response). The diode D1 is just there to prevent the negative spike. U1 is a buffer to buffer the sensitivity pot 's signal. I wanted to make a comparator out of U2. So in my head the signal was bottom rail when there is no spike. and top rail when there is a spike, according to the sensitivity signal going to the negative input of U2.

    What is wrong? The output of U2 seems to hang around 0.9V bias. Now as a comparator I feel that this is not possible. I will try this IRL but I can't see where I messed up.

    Anybody care to share?

    Peace!
     
  2. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    Without looking at the spec, I suspect U2 can't get to the rail. Unless it's a rail-2-rail device, it's going to get about as far is it's getting.
     
  3. ScottWang

    Moderator

    Aug 23, 2012
    4,856
    767
    Go to check the internal structure of LT1001 on page 12, and to see the low side of output.
     
  4. FuzzballJack

    Thread Starter New Member

    Apr 12, 2014
    13
    3
    Oh boy, I feel kind of silly already!
    Is it the 20 Ohm resistor and Q32 you mean? Does it bring the voltage to 0.9V at minimum?
    Right ... *scratches head*
    Now I wonder. Will this not happen with say a UA741 universal op-amp?
    I was going to try this out first with a 741 and if it worked then switch to a 2022 because I have plenty. Would this happen with those two op-amps?

    I had no idea what I was doing with the LT1001. I didn't find the 741 in the list in LTspice so I gunned it. :(
     
  5. ScottWang

    Moderator

    Aug 23, 2012
    4,856
    767
    If you using 741 could be worse, maybe the output voltage low level will up to 1.4V or 1.8V.
     
    Last edited: Apr 13, 2014
    FuzzballJack likes this.
  6. FuzzballJack

    Thread Starter New Member

    Apr 12, 2014
    13
    3
    Realy? Wow ... I misjudged this! Thanks!

    Mystery solved I guess. Lock it.
     
  7. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,546
    1,252
    When in doubt, use something that's 40 years old. The LM358 dual opamp can swing to within 20 mV of the negative rail (GND in your case).

    Note that your circuit will not de-bounce the switch under all conditions.

    ak
     
  8. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    We don't lock threads once the OP is satisfied. Members are allowed to continue the discussion. You are free to unsubscribe to the thread, if you choose.
     
  9. FuzzballJack

    Thread Starter New Member

    Apr 12, 2014
    13
    3
    Ah, so this will dwell forever and ever? :)
    @AK, there was no need for debouncing. The switch is already debounced with a resistor going to positive rail followed by a capacitor negative rail, switch over capacitor. Real old school deboucing :) . I live for that S :)
     
  10. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    No, we leave it open for discussion unless it goes off-topic, like we're doing now.
     
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