hw11_solutions - ECE 440 Homework XI Fall 2008 Due: Monday,...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 440 Homework XI Fall 2008 Due: Monday, Nov. 17, 2008 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. 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. Lets 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. 3. A symmetrical p +-n-p + Si bipolar transistor has the following properties: Emitter Base A = 10-4 cm 2 N a = 4x10 17 /cm 3 N d = 6x10 15 /cm 3 W b =1 m n = 0.2 s p = 4 s p = 140 cm 2 /V-s n = 1160 cm 2 /V-s n = 500 cm 2 /V-s p = 420 cm 2 /V-s (a) Determine if the straight-line approximation can be applied to evaluate the excess carriers in the base region....
View Full Document

This note was uploaded on 10/14/2009 for the course ECE 440 taught by Professor Lie during the Spring '09 term at University of Illinois at Urbana–Champaign.

Page1 / 8

hw11_solutions - ECE 440 Homework XI Fall 2008 Due: Monday,...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online