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Chapter 4. Section 4.3.supplement-Solved Problems on Design of BJT.
Problem 1. A BJT has I C = 1mA and I B = 10μA. What are α F , β F and I E ?
Solution:
You can check for consistency that :
Problem 2. Heavy Emitter Doping effect on Short Circuit CE BJT Current Gain. In a BJT D nB (diffusion coefficient of electron in Base)= 3 D pE (diffusion coefficient of holes in Emitter); Emitter Bulk Width = W E = 3W B ; Base Doping is uniform = N B = 10 18 /cc; n iB 2 =n i 2 ;
Determine β F for (a) N E =10 19 /cc; and (b) N E = 10 20 /cc ? [include band gap narrowing due to heavy emitter doping]
Solution: Rewriting Equation 4.3.17 we get:
Case (a) At N E =10 19 /cc , Band-gap narrowing = 50meV
Therefore:
Case (b) At N E =10 20 /cc ,, Band-gap narrowing= 95meV
Therefore β F = 900/38.6 = 23.3.
Band-gap narrowing has a very adverse effect. If it was not considered then S.C.Current Gain would have been 90 and 900 but now it has been reduced to 13 and 23.
Problem 3. Determine the forward Base Transit Time if W B = 70nm and D nB = 10cm 2 /s ? Also determine the improvement in forward Transit Time if W B = 10nm and D nB = 10cm 2 /s ?
This corresponds to Transit angular Frequency or Unity Current Gain angular BW:
This corresponds to :
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