# Astable question.

#### chaosdestro0

Joined Apr 30, 2011
10
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.

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