Physical Constants & Semiconductor Propertiesq=1.602 x 10-19 Ca = 8.854 x 10-¹2 F/mkB = 1.380 x 10-23 J/Kme = 9.109 x 10-³1 kg1 μm = 1 x 10 m&-Si=11.7- Egn₁ = BT ³/22kTDrift and Diffusion CurrentsJ=σ E=(1/p)EdndxNMOSp-n Junction and Diode CircuitsVbiw = KeTmn (NAND)eJ₁ = eD,r₂ =qVD10 = 1. [exp (2012-)-1]="N₁Vs N₂MOS Capacitor and MOSFETS&AdC=VDs (sat) = VGS - VTNEnbomaanEnhancement modeVTN >0Depletion modeVIN<0K.=+H.C. (17)=+*: (1)سال)۔Eg-Si = 1.1 eVEg-GaAs = 1.4 eVEg-Ge=0.66 eVnsi = 1.45 x 10¹0 cm31 nm = 1 x 10-⁹ msource regulation and load regulation:r₂Nonsaturation region (ups < ups (sat))iD = K₂ [2(VGS - VTN)UDS - VDS]Saturation region (ups > Ups(sat))iD = K (VGS - VTN)²Transition pointic IseBE/Vy=photocurrent:ic= 1, expSox (3.9) x (8.854 x 10-12) F/m=n² = n. p¡E = ¹ = e/Vr18 = ¹=¹/V₂For both transistorsig=ic +iBig = (1+B) Bα = 1+BV, z_V₁BᏙᏰla* EXP (*/ /* )( 1+ ²+ )Formula Sheet8₂ == 2K (V - VTN) = 2√ √K,IDQLovas -iDeal = K₂ (VGS - VIN) ² (1+2VDs)V₂N = V₂NO+Y(√20₁ + VS - √201)2 fRCV=VT Inico=enfle + epμhdpdxJp =-eDpwhere f=læ8m V₁In (NAND)== [2K (VGs - V₁N)²] = [¹Bsi 5.23 x 10¹5 cm-³K-3/2BGaAs = 2.10 x 10¹4 cm³ K-3/2BGe 1.66 x 10¹5 cm²³ K-3/2μe-si= 1400 cm² V-1 s-l1 pm = 1 x 10-¹2 mC = Eax AtaxPMOSBipolar Junction Transistor (BJT)Summary of the bipolar current-voltage relationships in the active regionNPNPNPAVEX 100%AVpsIph = neDA127,IsevEB/V¡E = ¹ = ¹e¹a/V₁iB = = ¹/V₂ic = BiBE1+8ic = aig = (iB = 12aC₁ =10 = 1 [expP(+)-1]D₂ D₂ KT==H₂ Hp 9V.1+KNonsaturation region (USD < USD (sat))iD = Kp[2(USG + VTP) VSD - USD]Saturation region (VSD > VSD (sat))ip = K₂(USG + VTP)²Transition pointUSD (sat) = USG + VTPEnhoraEnhancement modeVTP < 0Depletion modeVTP > 0WK₂ = + H₂Ca(²+) = + k, (1/7)V₁IcThermal Voltage V = 0.026 VCut-in Voltage V₂ = 0.7 VVL,no load VL, full loadVL, full loadμhs = 450 cm² V-1 S-11 fm = 1 x 10-15 mON-x 100%Eaxt The signal source in Figure B3 is v5-5sinot mV. The transistor's current gain B-120 and the Earlyvoltage VA=∞. Neglect the impacts of capacitors. Assume VBE(on) = 0.7 V, V₁ = 0.026 V.V+ = +5 VRς = 2.5 ΚΩwwwwwwRcV=-5 V2. Find small signal parameters 9m and TREFigure B33. Draw the small-signal equivalent circuit.CcHH4. Determine the small-signal voltage gain A-AHIلسل1. Design the circuit's RE and Re such that Icq = 0.25 mA and VCEQ = 3 VOvR₁ = 5 KQ2

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The signal source in Figure B3 is v5-5sinot mV. The transistor's current gain B-120 and the Early voltage V₁=00. Neglect the impacts of capacitors. Assume VBE(on) = 0.7 V, V₁ = 0.026 V. Rg = 2.5 ΚΩ www V+ = +5 V www Rc V=-5 V 2. Find small signal parameters 9m and T RE Figure B3 3. Draw the small-signal equivalent circuit. Cc HH CE Vo 1. Design the circuit's RE and Re such that Icq = 0.25 mA and VCEQ = 3 V R₁ = 5 KQ

The signal source in Figure B3 is v5-5sinot mV. The transistor's current gain B-120 and the Early voltage V₁=00. Neglect the impacts of capacitors. Assume VBE(on) = 0.7 V, V₁ = 0.026 V. Rg = 2.5 ΚΩ www V+ = +5 V www Rc V=-5 V 2. Find small signal parameters 9m and T RE Figure B3 3. Draw the small-signal equivalent circuit. Cc HH CE Vo 1. Design the circuit's RE and Re such that Icq = 0.25 mA and VCEQ = 3 V R₁ = 5 KQ

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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