EE331HW7solutions - EE-331 Devices and Circuits 1 Prof. R....

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WLratio 1 R on K n V GS V TN () := WLratio 4.706 = (b) V GS 3.3 volt := WLratio 1 R on K n V GS V TN := WLratio 7.843 = Problem 2: Problem 4.23 (a,b,c,d) of Jaeger and Blalock. K n 400 µ A volt 2 := V TN 0.7 volt := (a) V GS 3.3 volt := V DS 3.3 volt := V GS > V TN , so the MOSFET is ON V DS DSsat = V GS - V TN = 2.4 V, so the MOSFET is SATURATED. (b) V GS 0.0 volt := V DS 3.3 volt := V GS < V TN , so the MOSFET is OFF V DS does not matter. (c) V GS 2.0 volt := V DS 2.0 volt := V GS TN , so the MOSFET is ON V DS DSsat GS TN = 1.3 V, so the MOSFET is SATURATED. (d) V GS 1.5 volt := V DS 0.5 volt := V GS TN , so the MOSFET is ON V DS DSsat GS TN = 0.8 V, so the MOSFET is NON-SATURATED. Side note: parts (e) and (f) are not sensible questions, because the V DS for an n-channel MOSFET can never be negative. The drain is always defined as the more positive terminal. EE-331 Devices and Circuits 1 Prof. R. B. Darling Homework # 7 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 4.16 (a,b) of Jaeger and Blalock. (a) R on 500 := V GS 5.0 volt := K n 100 µ A volt 2 := V To 0.75 volt := V SB 0.0 volt := so that V TN V To := EE-331 HW#7 Solutions Page 1 R. B. Darling
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I D 107.9 µ A = (b) Add in the effect of channel length modulation, noting that V DS = V GS : V GS root 1 2 K n V GS V TN () 2 1 λ V GS + V DD V GS R V GS , := V GS 1.208 volt = The drain current is then: I D 1 2 K n V GS V TN 2 1 λ V GS + := I D 107.9 µ A = (c) Change the W/L ratio to 25/1: WLratio 25 1 := K n k n WLratio := V GS root 1 2 K n V GS V TN 2 V DD V GS R V GS , := V GS 1.046 volt = The drain current is then: I D 1 2 K n V GS V TN 2 := I D 109.5 µ A = Problem 4: Problem 4.39 of Jaeger and Blalock. The drain saturation voltage is V DSsat GS - V TN , and since the transistor is a D-mode, V TN < 0. Thus, V DSsat will be greater than V GS . With V GS DS , we thus have that V DS < V DSsat , so the MOSFET must be operating in the NONSATURATED mode. (It will already be conducting because V GS > V TN ).
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EE331HW7solutions - EE-331 Devices and Circuits 1 Prof. R....

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