Relaxation Oscillator Schmitt Trigger

Discussion in 'Homework Help' started by Confusedcat, Feb 23, 2014.

  1. Confusedcat

    Thread Starter New Member

    Feb 23, 2014
    Hey everyone, was hoping to have some help with two questions I have for the same circuit! I have searched the forums and not yet found something similar to my questions.
    Any help would be appreciated.
    here's the circuit in question
    Say that a 5V top rail is connected to both the trigger inverter and the diode (where there is a switch before the diode connected to 0V, meaning the diode can receive 0V when the trigger receives 5V). I understand that when there is 0V at the anode of the diode and 5V given to the trigger, there will be 0.7V at the input of the trigger.. I know this is something to do with the diode, but why at the input?

    Also my other question is that supposedly the upper trip point of a Schmitt trigger when given 5V is 2.8V. However I also thought this was meant to be 66% of the voltage given to the trigger from the rail... and that 2.8 isn't 66% of 5? or am i completely wrong? so I would've thought that say 10V was given to the trigger, the upper trip point would be 10x0.66 = 6.6 (66%). please help! sorry if this makes no sense at all haha, that's the trouble when you don't really know what you're talking about :p
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    This diode is add to the circuit to control this oscillator.
    Low state at diode cathode will turn-off the oscillator. High states will turn-on the oscillator.

    As for the gate threshold voltage look at datasheet

    Positive-Going Threshold Voltage for VDD = 5V, --->V+ = 3.6V

    Negative-going threshold voltage for Vdd = 5V -->V- = 1.4V
  3. AnalogKid

    AAC Fanatic!

    Aug 1, 2013
    First, the trigger volltage. Depending on the logic family of the inverter (TTL, LSTTL, old CMOS, A series CMOS, AT series CMOS, etc.), the upper and lower threshold voltages of a schmidt input can vary a lot. However, they generallly are not as far apart as 1/3 Vcc and 2/3 Vcc. Those are the thresholds of a 555.

    The diode is acting as an enable input for the oscillator. When the cathode is at VCC, the diode is reverse biased by the output voltage of the inverter in both the low and high states, does not conduct, and the oscilator runs as if it isn't there. When the diode cathode is low (~0 V), it acts as a clamp on the inverter input. The input can never get above 0.6 V, the forward voltage of the diode (0.2 for Schottkey). This is significantly lower than the lower threshold voltage of the inverter, no matter what logic family it is, so the input is camped low and the output sits high.