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ece340Fall03HW13sol

# ece340Fall03HW13sol - ECE 340 Homework XIII Fall 2003 Due...

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ECE 340 Homework XIII Fall 2003 Due: Wednesday, November 05, 2003 (solutions, 40 points total) 1. Redraw Fig. 7-3 for an n + -p-n transistor, and explain the various components of carrier flow and current directions. 2. Sketch the energy band diagram for an n-p-n transistor in equilibrium (all terminals grounded) and also under normal active bias (emitter junction forward biased, collector junction reverse biased). With the emitter terminal grounded, determine the signs (positive or negative) of the collector voltage V CE and base voltage V BE , relative to the emitter, that correspond to normal bias.

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3. A symmetrical p + -n-p + Si bipolar transistor has the following properties: Emitter Base A = 10 -4 cm 2 N a = 2x10 17 /cm 3 N d = 10 15 /cm 3 W b =1 μm τ n = 0.1 μs τ p = 10 μs μ p = 180 cm 2 /V-s μ n = 1300 cm 2 /V-s μ n = 600 cm 2 /V-s μ p = 450 cm 2 /V-s (a) Determine if the straight-line approximation can be applied to evaluate the excess carriers in the base region. (b) With V EB = 0.3 V and V CB = - 4 V, calculate the base current I B , assuming perfect emitter injection efficiency. (c) Calculate the emitter injection efficiency γ and the amplification factor β , assuming the emitter region is long compared to L n . Solutions: (a), 2 0.0259 450 11.655 / p p kT D cm vs q µ = = × = 6 11.655 10 10 0.0108 108 p p p L D cm m τ µ = = × × = = therefore, p b L W ± , straight line approximation can be applied. (b), Since Ep Ep En i i i γ = + , 1 0 En i γ = = , Using straight-line approximation, , _ 2 p E b B perfect p p Q qA p W I γ τ τ = = , / / 2 20 0.3/0.0259 10 3 15 2.25 10 2.413 10 10 EB kT EB kT qV qV i E n d n p p e e e cm N × = = = = × , Therefore, 19 4 10 4 12 5 1.6 10 10 2.413 10 10 1.93 10 2 2 10 E b B p qA p W I A τ × × × × × = = = × × , for this, we assume perfect injection efficiency, which is purely due to recombination in base region, we called it .
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ece340Fall03HW13sol - ECE 340 Homework XIII Fall 2003 Due...

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