hw13_solutions

# hw13_solutions - ECE 440 Homework XIII Solutions Spring...

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ECE 440 Homework XIII Solutions Spring 2006 1. 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. Drawing a transistor band diagram is essentially like drawing a band diagram for two p-n junctions back to back. First draw the equilibrium band diagram. When the transistor is in active bias or “normal” mode operation, the emitter base junction is forward biased while the base collector junction is reversed biased. After biasing the device, the band diagram now becomes: It is important to realize that the right bias polarity is required to operate the transistor in normal mode. Let’s take a look at the n-p-n transistor in normal mode. If V EB < 0, this would indicate that V BE > 0 and similarly V CB > 0. In addition, we were asked to find the sign of the potential between the collector and emitter, V CE . We know from a basic circuits course that V CE = (V C - V B ) + (V B - V E ) = V C + V BE . From our above analysis, we can see that V CE > 0.

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2. A symmetrical p + -n-p + Si bipolar transistor has the following properties: Emitter Base A = 10 -4 cm 2 N a = 10 18 /cm 3 N d = 10 16 /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. In the narrow-base p + n handout we learned that the straight-line approximation can only be applied when the width of the neutral region of the narrow base is much less than the diffusion length, L p . For more information see homework #12.
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