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Banha University

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10x2.<br />

25x10<br />

<br />

10<br />

25x450<br />

<br />

5x10<br />

<br />

<br />

3<br />

19<br />

Is = 1. 6x10<br />

3<br />

3<br />

4<br />

16<br />

2<br />

x10<br />

3.<br />

96x10<br />

A/<br />

cm<br />

The diode current I = Is(e V/Vt – 1)<br />

= 3.96x10 -16 (e 0.65/0.026 – 1) = 2.85x10 -5 A<br />

___________________________________________________________________<br />

b) Describe the operation, the charge flow and the current components in npn<br />

bipolar junction transistor in the forward active mode. What are the operating<br />

modes of the transistor?<br />

Solution<br />

The bipolar transistor consists of three separately doped regions and two pn<br />

junctions. There are two types of transistor structures an npn bipolar transistor and<br />

a pnp bipolar transistor. The three terminal connections are called the emitter,<br />

base, and collector. The width of the base region is small compared to the minority<br />

carrier diffusion length.<br />

The operation of an npn transistor in the forward active mode and the<br />

charge flow<br />

The base-emitter (B-E) pn junction is forward-biased, and the base-collector<br />

(B-C) pn junction is reverse-biased in the forward-active operating mode. Since the<br />

B-E junction is forward-biased so electrons from the emitter are injected across the<br />

B-E junction into the base. Also holes from the emitter are injected across the B-E<br />

junction into the emitter.<br />

The injected electrons create an excess concentration of minority carriers in the<br />

base. The B-C junction is reverse biased, so the minority carrier electron<br />

concentration at the edge of the B-C junction is ideally zero. The large gradient in<br />

the electron concentration means that electrons injected from the emitter will<br />

diffuse across the base region into the B-C space charge region where the electric<br />

field will sweep the electrons into the collector. Some of the injected electrons<br />

recombine with the majority holes in the base. We want as many electrons as

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