Schmitt Trigger with zener diode as a feedback

Discussion in 'General Electronics Chat' started by Xufyan, Nov 21, 2011.

  1. Xufyan

    Thread Starter Member

    Aug 3, 2010
    114
    0
    Hello everyone,

    please look at these two circuits (attachments) , both has been solved by me but the value of Vutp and Vltp in the 2nd circuit is incorrect :/

    the book says its 2.54volts and -2.54volts while my output is 1.72v and -1.72v,
    what is the problem here ?

    please explain
     
  2. praondevou

    AAC Fanatic!

    Jul 9, 2011
    2,936
    488
    The OPAMP will try to adjust zero voltage difference between it's inputs. The difference between output and inverting input is clamped to 5.4V. The max voltage drop on R1 will therefore also be 5.4V.

    You can then calculate the current through R1 and R2. The current through R2 will give you the voltage drop on R2 which is also the turning point.

    The opamp output voltage is therefore not +-5.4V but the sum of the voltage drops of R1 and R2.
     
    Last edited: Nov 22, 2011
    Xufyan likes this.
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If the +ve feedback factor is K, then the output voltage excursions will be given by

    Vout=±(Vz+Vf)/(1-K)

    Since Vout-K*Vout=(Vz+Vf)

    So if K=47k/147k=0.32, Vz=4.7V & Vf=0.7V then

    Vout=±5.4/0.68=±7.94V

    The Vutp and Vltp values will then be ±K*7.94 or ±0.32*7.94 = ±2.54V
     
    Last edited: Nov 22, 2011
  4. T.Jackson

    New Member

    Nov 22, 2011
    328
    14

    That's an OP AMP in a nutshell.

    Bit of an overkill for a Schmitt trigger though. Assuming that we even know what one is, or even need one for that matter
    .
     
  5. Xufyan

    Thread Starter Member

    Aug 3, 2010
    114
    0
    thankyou :)
     
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