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Unformatted text preview: February 26, 2008 ECE315 Homework # 4 Solutions (Rev. 0) Spring 2008 1. Textbook Chapter 6, Problem 1. Series Connected MOSFETs. • Since we are treating M 1 and M 2 as resistors (small V DS ) then in general: I D = μ n C ox ( W L )( V GS V T ) V DS . • Since M 1 and M 2 are identical and they are in series then: I D = μ n C ox ( W L )( V GS 1 V T ) V DS 1 = μ n C ox ( W L )( V GS 2 V T ) V DS 2 . If follows that V GS 1 = V GS 2 , V DS 1 = V DS 2 , and V DS 1 + V DS 2 = V DS . • So the equivalent transistor M eq has a drain current given by I D = μ n C ox ( W L ) eq ( V GS V T )( V DS 1 + V DS 2 ) = μ n C ox ( W L ) eq ( V GS V T )(2 V DS 1 ) . • Equating the drain currents of the M eq device to M 1: I D = 2 μ n C ox ( W L ) eq ( V GS V T ) V DS 1 = μ n C ox ( W L )( V GS V T )( V DS 1 ) . this is satisfied if ( W L ) eq = 1 2 ( W L ) L eq→ 2 L . 2. Textbook Chapter 6, Problem 3. The charge per unit area (stored) in the nmos device at zero drain bias is given as: Q = C ox ( V GS V T ) . The total charge is given by: Q T = WLC ox ( V GS V T ) = 5(0 . 1)(10 14 )(1) = 5(10 15 ) [ C ] . 3. Textbook Chapter 6, Problem 5. Plot both V ( x ) and dV ( x ) /dx for L = 1 μ m and V GS V T = 1 volt for the following three cases, with the drain current equal to I sat D , 1 2 I sat D and 1 4 I sat D . Solving for nmos electric field along channel. Equation (6.7) in text states that: I D = WC ox [ V GS V ( x ) V T ] μ n dV ( x ) dx . Rearranging,we obtain: I D Wμ n C ox dx = [ V GS V T V ( x )] dV ( x ) . ECE315 February 26, 2008 Page 2 Integrating with V (0) = 0: I D Wμ n C ox x = ( V GS V T ) V V 2 2 . Solving this using the quadratic formula: V ( x ) = ( V GS V T ) ± s ( V GS V T ) 2 2 I D x Wμ n C ox . Note that we must take the negative sign in the solution as V ( x ) < ( V GS V T ). Differen tiating this to give the field along the channel: dV ( x ) dx = I D Wμ n C ox q ( V GS V T ) 2 2 I D x Wμ n C ox . Note that I sat D = 1 2 μ n C ox ( W/L ). The following graphs are then produced. 4. Textbook Chapter 6, Problem 11. In the ohmic (or triode if you used to work with vacuum tubes) regime: I D = k P 2 2( V GS V T ) V DS V 2 DS ....
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 Spring '07
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 Microelectronics, Volt, Trigraph, VDS, Threshold voltage

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