• This topic has 1 reply, 2 voices, and was last updated 4 days ago by mwpeters75.
Viewing 1 reply thread
  • Author
    Posts
    • #9172
      LectronFan
      Moderator

      Hi all,

      In previous experiment, we saw how the Schmitt trigger reacts when a certain charging level of the capacitor is reached.

      Now, we will test how much current is needed to turn the Schmitt trigger permanently in the ON state.
      After building the circuit, set the potentiometer so that the Schmitt trigger keeps on.

      Some explanation :
      The 1st transistor in the Schmitt trigger needs a certain current to turn on.
      When the voltage across the 100µF capacitor is large enough, current will flow through the B-E junction of this transistor, turning it on.
      Now, a current will also flow through the 47Ω resistor inside the Schmitt trigger module. When this current is larger than the charging current through the 100µF capacitor, it will start to discharge the 100µF capacitor.

      This causes a loop where the Schmitt trigger will be switching from ON to OFF, making the lamp blink.

      We will make use of this phenomenon to create a pulse generator in the next experiment.

      Many greetings

      Schmitt trigger MV2

    • #9175
      mwpeters75
      Keymaster

      Always the leader in Lectron System circuits!  Thank you Frank!

Viewing 1 reply thread
  • You must be logged in to reply to this topic.