EE331HW8solutions - EE-331 Devices and Circuits 1 Prof. R....

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V Tn V TO := V Tn 1.00 volt = k n 100 µ A volt 2 := WLratio 6 1 := K n k n WLratio := K n 600 µ A volt 2 = The voltage on the gate is set by the voltage divider of R 1 and R 2 : V G V DD R 1 R 1 R 2 + := V G 3.125 volt = Since this is greater than V Tn , the MOSFET is ON. Start by first guessing the MOSFET to be SATURATED, since this is easier to analyze. V GS 1.0 volt := (Start with a guess for the root finder.) V GS root V G V GS 1 2 K n R 3 V GS V Tn () 2 V GS , := V GS 1.478 volt = I D 1 2 K n V GS V Tn 2 := I D 68.6 µ A = V BS I D R 3 := V BS 1.647 volt = V DS V DD I D R 3 R 4 + := V DS 7.53 volt = V DSsat V GS V Tn := V DSsat 0.478 volt = Since V DS > V DSsat , the guess for saturation was correct. Also, since γ = 0, no body bias corrections to the threshold voltage are needed, even though this circuit does produce a body bias of V SB = 2.151 Volts. EE-331 Devices and Circuits 1 Prof. R. B. Darling Homework # 8 Solutions First, define some useful units and constants: µ m1 0 6 m := eV 1.602 10 19 joule := ms 10 3 sec := µ s1 0 6 sec := q 1.602 10 19 coul := k B 8.62 10 5 eV K 1 := nm 10 9 m := ε rSiO2 3.9 := ε o 8.854 10 14 farad cm 1 := n i 10 10 cm 3 := V T 0.025 volt := ε rSi 11.7 := ohm := k 10 3 ohm := M 10 6 ohm := Problem 1: Problem 4.93 (a) of Jaeger and Blalock. V DD 10.0 volt := R 1 100 k := R 2 220 k := R 3 24 k := R 4 12 k := V TO 1.0 volt := EE-331 HW#8 Solutions Page 1 R. B. Darling
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V Tp 0.75 volt := k n 100 µ A volt 2 := k p 40 µ A volt 2 := Equate the current through both transistors (which are both SATURATED) to give: 1 2 k n W n L n V o V Tn () 2 1 2 k p W p L p V DD V o V Tp + 2 := Simplifying gives: V DD V o V Tp + V o V Tn k n k p := since the W/L ratios are equal and cancel out.
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EE331HW8solutions - EE-331 Devices and Circuits 1 Prof. R....

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