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Astable multivibrator question

  1. May 17, 2006 #1
    Hello all...

    I am trying to understand astable multivibrators and I keep on reading this phrase in connection to one transistor going into saturation and the other going into cutoff...

    "This change in voltage is coupled through C2 to the base of Q1, forcing Q1 to cutoff."

    This statement is referring to the fact that when the second transistor conducts, the end of the capacitor connected to the collector rapidly drops from (approximately) the voltage of the source to (approximately) ground.

    Im not seeing how this change in voltage is "coupled" thorugh the capacitor to the base of the other transistor.

    Any help would be appreciated.


    Thanks.
     
  2. jcsd
  3. May 17, 2006 #2

    chroot

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    High frequency signals "pass" through capacitors. The signal modulates the voltage on one end of the capacitor, which causes oscillations in the electric field between the capacitor's plates, which causes oscillations in the potential on the other side of the capacitor.

    In fact, the higher the frequency of stimulus you apply to a capacitor, the more like a short circuit it will behave.

    - Warren
     
  4. May 17, 2006 #3
    Thanks for the reply...

    First question......could you explain the connection between the oscillations in the electric field and the change in potential on the other side of the capacitor?

    If I am asking too much, then could you direct me to a good source where I could study this in more depth?

    The way I have always thought about a capacitor acting like a "short" at AC was to picture charge being put on the plates and then taken off very quickly...so quickly that there never really is an accumulation of charge or volatge (kind of like a tug of war with charge)...

    But with this astable multivibrator I don't have a nice clear path (at least I cant picture it) connecting the plates of the capacitors because of the transistors, so my simple picture doesn't seem to work.

    But you are going much deeper into the physics than I was and I would like to try and understand what you are saying, if you will bear with me...

    Here is what I am picturing.....before the transistor 'B' is turned on we have one end of the capacitor at the source voltage (say 6 volts) and the other end is at the base voltage of transistor 'A' (which is approximately 0.7 volts since transistor 'A' is on)....

    So there is about 6 volts accross the capacitor and....since the volatge across a capacitor can't change instantly...after transistor 'B' turns on we have one end of the capacitor at zero volts and the other end at -6 volts, thus maintaining the rule that the voltage can't change instantaneously.

    Is this correct?





    Russ
     
    Last edited: May 17, 2006
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