# Astable question.

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

1. ### chaosdestro0 Thread Starter New Member

Apr 30, 2011
10
0
So this is what I did for the first bit. Ignore that misplaced T1 at the right.

Then for the next bit

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

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