Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

Solution beginning at we note that the absence of net

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Solution Beginning at , we note that the absence of net charge yields . At , each positive donor ion contributes to the electric field, i.e., the magnitude of rises as approaches zero. As we pass , the negative acceptor atoms begin to contribute negatively to the field, i.e., falls. At , the negative and positive charge exactly cancel each other and . The direction of the electric field is determined by placing a small positive test charge in the region and watching how it moves: away from positive charge and toward negative charge.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 39 (1) Sec. 2.2 Junction 39 n p + + + + + + + + + + + + + + + + + + + + + + + + + + + + E x 0 a b N D N A + + + x 0 a b E Figure 2.21 Electric field profile in a junction. Exercise Noting that potential voltage is negative integral of electric field with respect to distance, plot the potential as a function of . From our observation regarding the drift and diffusion currents under equilibrium, we may be tempted to write: (2.61) where the subscripts and refer to holes and electrons, respectively, and each current term contains the proper polarity. This condition, however, allows an unrealistic phenomenon: if the number of the electrons flowing from the side to the side is equal to that of the holes going from the side to the side, then each side of this equation is zero while electrons continue to accumulate on the side and holes on the side. We must therefore impose the equilibrium condition on each carrier: (2.62) (2.63) Built-in Potential The existence of an electric field within the depletion region suggests that the junction may exhibit a “built-in potential.” In fact, using (2.62) or (2.63), we can compute this potential. Since the electric field , and since (2.62) can be written as (2.64) we have (2.65) Dividing both sides by and taking the integral, we obtain (2.66)
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 40 (1) 40 Chap. 2 Basic Physics of Semiconductors n p n n n p p p n p x x x 1 2 Figure 2.22 Carrier profiles in a junction. where and are the hole concentrations at and , respectively (Fig. 2.22). Thus, (2.67) The right side represents the voltage difference developed across the depletion region and will be denoted by . Also, from Einstein’s relation, Eq. (2.50), we can replace with : (2.68) Exercise Writing Eq. (2.64) for electron drift and diffusion currents, and carrying out the integration, derive an equation for in terms of and . Finally, using (2.11) and (2.10) for and yields (2.69) Expressing the built-in potential in terms of junction parameters, this equation plays a central role in many semiconductor devices. Example 2.13 A silicon junction employs and . Determine the built-in potential at room temperature ( K).
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