NMOSNonsaturation region (ups < ups (sat))iD = K[2(UGS - VTN)UDS - VDS]Saturation region (Ups > Ups(sat))ip = K₁(VGS - VTN)²Transition pointVDs (sat) = UGS - VTNEnhancement modeVTN >0Depletion modeVTN<0K.=+H.C. (H)=+*: (7)K₂kT =1-1-²-2) -di D3) L = [2K (Vasq - VIN)²]*¹ = [2]¹ON DS1-poini Dar = K₂ (VGs - VIN) ² (1+2VDS)V₂N =V₂NO+Y(√20, +V'S = √20₁)= 2K (Vas -VTN) = 2√ √K₂1DQic = IseVBE/VT¡E = ¹ = ³/VTiB ==/VrFor both transistorsig=ic + igig = (1+B)iBα = 1Bipolar Junction Transistor (BJT)Summary of the bipolar current-voltage relationships in the active regionNPNPNPic Is exp, exp (* * *)(1+² =)V₁BlaPMOSNonsaturation region (USD < Usp (sat))iD = Kp[2(USG + VTP) USD - USD]Saturation region (VSD > USD (sat))ip = K₂(USG + VTP)²Transition pointVSD (sat) = USG + VTPEnhancement modeVTP < 0Depletion modeVTP > 0K‚= ‡μ‚C« (H) = + *; (H)laic=¡E = ¹ =iB = 1 =Iseva/VT¹/V₂²¹/V₂ic = BiBic = αig = (+)ieß = 12ro10 In the circuit in Figure B1 with circuit and transistor parameters Rp = 20kn, K₂ =0.1 mA/V2, VTN = 0.8 V, and λ = 0.2. V₁=5 V, V₂=03. V₁ V2=5 V- Egn₁ = BT ³/22kgTJ₁ = eD₂alDrift and Diffusion CurrentsJ=oE=(1/p)EVbi =Determine the currents IR, ID1, D2, and voltage Vo in a digital logic gate for the following inputconditions:1. V₁ V₂ = 0 VPhysical Constants & Semiconductor Propertiesq=1.602 x 10-19 C= 8.854 x 10-¹2 F/mkB = 1.380 x 10-23 J/Kme = 9.109 x 10³1 kg1 μm = 1 x 10 m&-Si=11.7V₁_N₁-=N₂dndxVsp-n Junction and Diode Circuits- KT (NAND)eM₁10 = 1 [exp (1987)-1]IDVDD=5 VIDIEg-Si= 1.1 eVEg-GaAs = 1.4 eVEg-Ge=0.66 eVnsi =nsi = 1.45 x 10¹01.45 x1 nm 1 x 10⁹ mphotocurrent:MOS Capacitor and MOSFETSC=244dV₂Figure B1Formula Sheet2 fRCRp&x= (3.9) × (8.854 x 10-¹2) F/mn² = n. p10¹⁰ cm-³cm3V=V₁ Insource regulation and load regulation:M₂o = enfle + epμhdpJp =-eDpdxwheref=Bsi = 5.23 x 10¹5 cm-³K-3/2BGaAs = 2.10 x 10¹4 cm³ K-3/2BGe 1.66 x 10¹5 cm³ K-3/2μe-si = 1400 cm² V-¹ s-¹== V₁ In (NAND) G21 pm = 1 x 10-¹2 m10212TpC = Eax AtaxVoAVL X100%AVPSIph = neDAD Dp KT=H₂Hp 9C₂0IDD = 1 [EXP ( 1 ) - ₁]1+VRμh-St=450 cm² V-1 S-11 fm = 1 x 10-15 mThermal Voltage V₁ = 0.026 VCut-in Voltage V₂ = 0.7 VVL, no load VL, full loodVL, full load-x 100%tax

Given: RD=20 kΩ, Kn=0.1 mA/V2, VTN=0.8 V.

Answered: Determine the currents IR, ID1, D2, and…

Determine the currents IR, ID1, D2, and voltage Voin a digital logic gate for the following input conditions: 1. V₁=V/₂=0 V 2. V₁=5 V, V₂=0 3. Vi=V₂=5 V

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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It's for Digital logic Design.