S06hw2s

S06hw2s - EE3220 Homework #2 Solutions February 23, 2006...

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Unformatted text preview: EE3220 Homework #2 Solutions February 23, 2006 7.79 Two n-channel enhancement-mode MOSFETs with K = 0 . 5 mA/V 2 and V T R = 0 . 8 V are connected as below. Find the resulting values of I D 1 , I D 2 , V DS 1 , and V DS 2 . V DD Q 1 Q 2 R A R B 1.8 M 2.2 M 12 V First, find V G 2 : V G 2 = R B R A + R B V DD = 6 . 6 V Now, note that V GS 1 + V GS 2 = V G 2 and V GS 1 = V GS 2 , therefore V GS = V G 2 / 2 = 3 . 3 V. I D 1 = I D 2 = K ( V GS- V T H ) 2 = K ( V G 2 2- V T H ) 2 = 3 . 125 mA V DS 1 = V GS 1 = V GS = 3 . 3 V V DS 2 = V DD- V GS = 8 . 7 V 1 Check for constant-current operation: V GS- V T H = 3 . 3 V- . 8 V = 2 . 5 V 7.89 For a BJT with the v- i characteristics shown in the book, evaluate the parameters F and at the following operating points: Given that: F = I C /I B = i C / i B a) 2.2 mA, 4 V I B = 40 A F = 55 i C = 1 . 6 mA , i B = 20 A = 80 b) 6 mA, 4 V I B = 80 A F = 75 i C = 2 mA , i B = 20 A = 100 c) 7 mA, 16 V I B = 70 A F = 100 i C = 1 . 4 mA , i B = 10 A = 140 d) 10 mA, 8 V I B = 120 A F = 83 i C = 2 . 2 mA , i B = 20 A = 110 e) 11 mA, 4 V I B = 140 A F = 79 i C = 2 mA , i B = 20 A = 100 2 f) 16 mA, 14 V I B = 160 A F = 100 i C = 2 . 2 mA , i B = 20 A = 110 7.90 Derive a suitable small-signal model for a pnp BJT following the procedure of Section 7.4.2. First, i C is given by: i C =- I E ( e- V BE /V T- 1) Then, find g m and r : g m = i C V BE =- I E- V T e- V BE /V T =- I C V T 1 r = i B V BE =- I B V T r =- V T I B Now, note that because the currents are defined in the same was as for the npn BJT, the small-signal model will be the same, but with the above values of g m and r ....
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S06hw2s - EE3220 Homework #2 Solutions February 23, 2006...

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