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Lecture 8_ Jan 31 2011

# Lecture 8_ Jan 31 2011 - Chapter 5 Bipolar Junction...

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Unformatted text preview: Chapter 5 Bipolar Junction Transistors Lecture 8 Cover Chapter: 5.7, 5.8.6-5.8.8 Simplified Cutoff Region Model In cutoff region both junctions are reverse-biased, transistor is said to be in off state v BE < 0, v BC < 0 If we assume that where -4 kT/q = -0.1 V, then the transport model terminal current equations simplify to v B E < - 4 kT q and v B C < - 4 kT q i C = + I S β R i E = - I S β F i B = - I S β F- I S β R Chap 5.7 Simplified Cutoff Region Model (Example) • Problem: Estimate terminal currents using simplified transport model • Given data: I S = 10-16 A, α F = 0.95, α R = 0.25, V BE = 0 V, V BC = -5 V • Assumptions: Simplified transport model assumptions • Analysis: From given voltages, we know that transistor is in cutoff. I C = I S 1 + 1 β R = I S α R = 4 × 10- 16 A I E = I S = 10- 16 A I B = - I S β R = - 3 × 10- 16 A Chap 5.7 For practical purposes, all three currents are essentially zero. Simplified Forward-Active Region Model In forward-active region, emitter-base junction is forward-biased and collector-base junction is reverse-biased. v BE > 0, v BC < 0 If we assume that then the transport model terminal current equations simplify to v B E ≥ 4 kT q = 0.1 V and v B C ≤ - 4 kT q = - 0.1 V i C = I S exp v B E V T + I S β R = I S exp v B E V T i E = I S α F exp v B E V T + I S β F = I S α F exp v B E V T i B = I S β F exp v B E V T - I S β F- I S β R = I S β F exp v B E V T i C = α F...
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Lecture 8_ Jan 31 2011 - Chapter 5 Bipolar Junction...

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