Astable question.

Discussion in 'Homework Help' started by chaosdestro0, May 1, 2011.

  1. chaosdestro0

    Thread Starter New Member

    Apr 30, 2011
    So this is what I did for the first bit. Ignore that misplaced T1 at the right.

    Then for the next bit

    T_{1}=0.69C(R_{b})<br />
 30*10^{-3}=0.69(3.3*10^8)R_{b}<br />
  30*10^{-3}=2.277*10^{-6}R_{b}<br />
 \frac{30*10^{-3}}{2.277*10^{-6}}=R_{b}<br />
R_{b}\sim 13k

    Right for the last question I am really not sure how the mark scheme got 3R_{b}=R_{a} , I really can't see how they derived that.
    Last edited: May 1, 2011