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 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 = 0al3. V₁= V/₂=5 VM₁VDD=5 VIDIV₂RpŢFigure B1M₂Determine the currents IR, ID1, D2, and voltage V, in a digital logic gate for the following inputconditions:1. V₁ V₂ = 0 V102

Answered: Image /qna-images/answer/776cb3db-eac4-41f5-9d73-a791527848a6.jpg

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 = 0 al 3. V₁= V/₂=5 V M₁ VDD=5 V IDI V₂ Rp Ţ Figure B1 M₂ Determine the currents IR, ID1, D2, and voltage V, in a digital logic gate for the following input conditions: 1. V₁ V₂ = 0 V 102

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 = 0 al 3. V₁= V/₂=5 V M₁ VDD=5 V IDI V₂ Rp Ţ Figure B1 M₂ Determine the currents IR, ID1, D2, and voltage V, in a digital logic gate for the following input conditions: 1. V₁ V₂ = 0 V 102

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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Question

Physical Constants & Semiconductor Properties
q=1.602 x 10-19 C
a = 8.854 x 10-¹2 F/m
kB = 1.380 x 10-23 J/K
me = 9.109 x 10-³1 kg
1 μm = 1 x 10 m
&-Si=11.7
- Eg
n₁ = BT ³/22kT
Drift and Diffusion Currents
J=σ E=(1/p)E
dn
dx
NMOS
p-n Junction and Diode Circuits
Vbi
w = KeTmn (NAND)
e
J₁ = eD,
r₂ =
qVD
10 = 1. [exp (2012-)-1]
=
"N₁
Vs N₂
MOS Capacitor and MOSFETS
&A
d
C=
VDs (sat) = VGS - VTN
Enbomaan
Enhancement mode
VTN >0
Depletion mode
VIN<0
K.=+H.C. (17)=+*: (1)
سال)۔
Eg-Si = 1.1 eV
Eg-GaAs = 1.4 eV
Eg-Ge=0.66 eV
nsi = 1.45 x 10¹0 cm3
1 nm = 1 x 10-⁹ m
source regulation and load regulation:
r₂
Nonsaturation region (ups < ups (sat))
iD = K₂ [2(VGS - VTN)UDS - VDS]
Saturation region (ups > Ups(sat))
iD = K (VGS - VTN)²
Transition point
ic IseBE/Vy
=
photocurrent:
ic= 1, exp
Sox (3.9) x (8.854 x 10-12) F/m
=
n² = n. p
¡E = ¹ = e/Vr
18 = ¹=¹/V₂
For both transistors
ig=ic +iB
ig = (1+B) B
α = 1+B
V, z
_V₁B
ᏙᏰ
la
* EXP (*/ /* )( 1+ ²+ )
Formula Sheet
8₂ =
= 2K (V - VTN) = 2√ √K,IDQ
Lovas -
iDeal = K₂ (VGS - VIN) ² (1+2VDs)
V₂N = V₂NO+Y(√20₁ + VS - √201)
2 fRC
V=VT In
ic
o=enfle + epμh
dp
dx
Jp =-eDp
where f=
læ
8m V₁
In (NAND)
=
= [2K (VGs - V₁N)²] = [¹
Bsi 5.23 x 10¹5 cm-³K-3/2
BGaAs = 2.10 x 10¹4 cm³ K-3/2
BGe 1.66 x 10¹5 cm²³ K-3/2
μe-si= 1400 cm² V-1 s-l
1 pm = 1 x 10-¹2 m
C = Eax A
tax
PMOS
Bipolar Junction Transistor (BJT)
Summary of the bipolar current-voltage relationships in the active region
NPN
PNP
AVEX 100%
AVps
Iph = neDA
1
27,
IsevEB/V
¡E = ¹ = ¹e¹a/V₁
iB = = ¹/V₂
ic = BiB
E
1+8
ic = aig = (i
B = 12a
C₁ =
10 = 1 [expP(+)-1]
D₂ D₂ KT
=
=
H₂ Hp 9
V.
1+K
Nonsaturation region (USD < USD (sat))
iD = Kp[2(USG + VTP) VSD - USD]
Saturation region (VSD > VSD (sat))
ip = K₂(USG + VTP)²
Transition point
USD (sat) = USG + VTP
Enhora
Enhancement mode
VTP < 0
Depletion mode
VTP > 0
W
K₂ = + H₂Ca(²+) = + k, (1/7)
V₁
Ic
Thermal Voltage V = 0.026 V
Cut-in Voltage V₂ = 0.7 V
VL,no load VL, full load
VL, full load
μhs = 450 cm² V-1 S-1
1 fm = 1 x 10-15 m
ON
-x 100%
Eax
t

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 = 0
al
3. V₁= V/₂=5 V
M₁
VDD=5 V
IDI
V₂
Rp
Ţ
Figure B1
M₂
Determine the currents IR, ID1, D2, and voltage V, in a digital logic gate for the following input
conditions:
1. V₁ V₂ = 0 V
102

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