Schmitt Trigger Asymmetric Vth

Discussion in 'Homework Help' started by ECE101, Dec 11, 2011.

  1. ECE101

    Thread Starter New Member

    Nov 25, 2011
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    Hi,

    Im just wondering how I use the following formula to compensate for an asymmetric Vth?

    [​IMG]

    Vth = +/- Vsat(R1/(R1+R2)) + Vref(R2/R1+R2)

    and also is T = 2.2CR the correct formula to use for modifying for an astable multivibrator?



    Any help would be greatly appreciated!
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    See attachment for example with different switching values to your requirement but a similar concept on which you could base your working. I designed for +4V and -6V switching points. The op-amp saturation output is 2V from the supply rail for either output polarity - rather than 0.5V. I assumed forward diode drops of 0.65V.
     
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  3. ECE101

    Thread Starter New Member

    Nov 25, 2011
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    Thanks t_n_k thats great,

    How did you calculate those resistor values though?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    My expectation was that you might figure that out for yourself.

    Here's a start.

    In my posted circuit R1 is 10kΩ.

    If the proposed positive input required to switch the output from -ve to +ve state is +4V, then the current into the 10kΩ (at the transition condition) will be 400uA - the positive and negative op-amp inputs are assumed to be at ground potential (at the transition point) when the positive terminal voltage 'matches' that of the negative terminal which is tied to ground.

    The same 400uA in the 10kΩ must also flow through R2 and diode D2 to the potential at the op-amp output terminal. In my case the output can only swing to ±10V. Immediately prior to output transition with a positive going input, if the output is at -10V then the 400uA is flowing through R2 and D2 to a potential of -10V. Assuming the diode has a forward drop of 0.65V what will be the voltage drop across R2 [V_R2] that will satisfy the circuit conditions? If you know V_R2 and the current in R2 you can then find R2.
     
    Last edited: Dec 12, 2011
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  5. ECE101

    Thread Starter New Member

    Nov 25, 2011
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    Perfect thanks for the explanation :)

    Just one more question,

    After I've modified the circuit as an astable multivibrator, how do I achieve an asymmetric period? For different charge and discharge periods do I need 2 different capacitors?
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    BTW the asymmetric period is 1.33RC not 2.2RC as you originally stated.

    To achieve a symmetric astable operation you would simply replace the dual (Rx + diode) branches with a single feedback resistance.
     
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