Solution4 - 5.1 Design a resistive-load inverter with R =...

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Unformatted text preview: 5.1 Design a resistive-load inverter with R = 1kg such that VOL = 0.6 V when an enhancement-type nMOS driver transistor has the following parameters: - VDD = 5.0 V - Vm = 1.0 V - 'y: 0.2 V”2 ' l = 0.0 V" ' k' = 22.0 W2 (a). Determine the required aspect ratio, W/L (b). Determine V“ and V” (0). Determine noise margins NM L and NMHr SOLUTION: (a) When Vm = VOL. Vin = VD”: 5.0V, the driver transistor operates in linear region. Using Eq.(5.12) . v —-v ' w [a = 4,91% = ETiZWOI-I ' Vro)V0L ‘ VOLZ] 5—0.6 22x10'6W 2 —=——2s-1 .6—0. 1000 2 L[( )0 6] SolveforWL E=90.1 L (b)When Vin = V1], driver transistor Operates in saturation region. Using Eq.(5.19) Va ‘ V95; =kyfl (th ‘me RL L 2 (iv Taking derivative with respect to Vin and set (Wmr —l v“ uvm I de W l ‘—' =k—(VIL'VT0)=_“ RL dig” mm L Rt Thus. 1 V =V +———=1.0+———-——=L5V "— 7" R JEE 1000-22x10*.90.1 [ 1 L L Vin = VI“, driver transistor operates in linear region. Using Eq.(5.24) repeated here: VDD'”V - k. W 2 TflL_?Z[2(Hn_VT0)me-Vw ] . . . . . . dvour —_1 Dlfierenualmg both Sides wuh respect to Vm and set Wm ' , Vk=Vm 1 GM! , k' W (W (W -——- 0“ =—— 2(Vm-Vm M +2vw-2vm M RL dun Vu‘fiu 2 dun Eur-“H 0 a db?” Vh=fifl é=k¥l<wrwen+2vm1 1 W Ric'— L L Plug in back to Eq.(5.24) repeated ane to solve for Vm “H = Vro +2Vm — VDD_VM=£E 1 RL 2 L VTfl+zvm_m_VT0 Vow—Vow: L V I = 2.122.— “ Vau‘vm 3 W RL'k': s 5 1 =L0+J——-—-—-—-—-——=109V 31000-22x10‘6-90J 1000-22x10'6-90J [ l (C) NML= vIL - V0515 -015 = 0.9 [V] NMH= 11W - v,” = 5.0 - 3.09 =1.91 [V] ...
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Solution4 - 5.1 Design a resistive-load inverter with R =...

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