# Hw4Soln - ECE 137A Winter 2009 Homework 4 Solutions 4.33(a...

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ECE 137A Winter 2009 Homework 4 Solutions 4.33 (a) Since V DS = V GS and V TN > 0 for both transistors, both devices are saturated. Therefore I D 1 = K n ' 2 W L V GS 1 - V TN ( 29 2 and I D 2 = K n ' 2 W L V GS 2 - V TN ( 29 2 . From the circuit, however, I D2 must equal I D1 since I G = 0 for the MOSFET: I = I D 1 = I D 2 o r K n ' 2 W L V GS 1 - V TN ( 29 2 = K n ' 2 W L V GS 2 - V TN ( 29 2 which requires V GS1 = V GS2 . Using KVL: V DD = V DS 1 V DS 2 = V GS 1 V GS 2 = 2 V GS 2 V GS 1 = V GS 2 = V DD 2 = 5 V I = K n ' 2 W L V GS 1 - V TN ( 29 2 = 100 2 μ A V 2 10 1 5 - 0.75 ( 29 2 V 2 = 9.03 mA (b) The current simply scales by a factor of two (see last equation above), and I D = 18.1 mA. (c) For this case, I D 1 = K n ' 2 W L V GS 1 - V TN ( 29 2 1+ λ V DS1 ( 29 and I D 2 = K n ' 2 W L V GS 2 - V TN ( 29 2 1+ λ V DS2 ( 29 . Since V GS = V DS for both transistors I D 1 = K n ' 2 W L V GS 1 - V TN ( 29 2 1+ λ V GS1 ( 29 and I D 2 = K n ' 2 W L V GS 2 - V TN ( 29 2 1+ λ V GS2 ( 29 and I D1 = I D2 = I K n ' 2 W L V GS 1 - V TN ( 29 2 1+ λ V GS1 ( 29 = K n ' 2 W L V GS 2 - V TN ( 29 2 1+ λ V GS2 ( which again requires V GS1 = V GS2 = V DD /2 = 5V.

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I = K n ' 2 W L V GS 1 - V TN ( 29 2 1 + λ V DS ( 29 = 100 2 μ A V 2 10 1 5 - 0.75 ( 29 2 V 2 1 + .04 ( 29 5 ( 29 = 10.8 mA 4.54 (a) For V IN = 0, the NMOS device is on with V GS = 5 and V SB = 0, and the PMOS transistor is off with V GS = 0, V O = 0, and V SB = 0. R on = 1 100 x 10 - 6 ( 29 10 ( 29 5 - 0.75 ( 29 = 235 (b) For V IN = 5V, the NMOS device is off with V GS = 0, and the PMOS transistor is on with V GS = -5V, V O = 5V, and V SB = 0. R on = 1 40 x 10 - 6 ( 29 25 ( 29 - 5 + 0.75 ( 29 = 235 4.103 a ( The transistor is saturated by connection. V GS = 12 - 10 5 I D and I D = 100 x 10 - 6 2 10 1 A V 2 V GS - 0.75 V ( 29 2 50 V GS 2 - 74 V GS + 16.13 = 0 V GS = 1.214 V , - 0.266 V V GS = 1.214 V since V GS must exceed the threshold voltage. | I D = 100 x 10 - 6 2 10 1 A V 2 1.214 - 0.75 V ( 29 2 I D = 104 μ A | Checking : I D = 12 - 1.21 10 5 = 108 μ A | Q - Point : 108 μ A,1.21 V ( 29 (b) Using KVL, V DS = 10 7 I G +V GS . But, since I G = 0, V GS = V DS . Also V TN = 0.75 V > 0, so the transistor is saturated by connection.
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