4) Construct the circuit in figure 1, By using the values you obtained in step-3

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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given that
3) Based on Bpc obtained in step2, calculate the values of R., B₂, Bc
and RE using your derived equations step1
Vec
R₁
Ne
SR2
Ica
VCE
vet RE
A
Tos
Bac obtained in step is Fac=0.53392
• VCE = V₁ - V₁ •Vc = V₁E + V₁ = 6 + 1-2 • V₁= 7.2V
but V₁ = √₂ x R₂ • Rc = √₂/1₂ = 7-2/2x10-3 = 3.6 kv
Re=3.6 Kv
from the IBO = ICB/Bac = 2X10²3/6.53392
IBQ = 3-7454xlo-³ A
IEQ = 100+ Ica: S. 7459x10-³ A
No.
Vicc - 12V, VGA=0.65V, V₁E = 1/₂ x 12 = 6V ₁ VG = 0.1 XV₁6 = 0.1X12
Ve = 1.2V, Ice = 2mA, IR₂ = 10 IBO
VE TE X RE
VBE VB-VE
Date
●
1. RE: VE/TE = 208.8453 V • TR₂ = 10 x IBQ = 37-459 x10³A
VB = VBE + V₂ = 6.65 +1.2 UG = 1.85V IR₂ = VB / Rz
R₂ = 49-38232
VB = Vcc X R₂ / R₂ + R₁
-49.3873 +R₁ = 12x 49.3873/1.85
R₁= 320.3501-49.3873
=
R₂ = UB/IR 1.85/37.459x16-3
By using voltage division rule
1-85= 12 x 49.5875/49-3823 +R₁
49,3873+R₁ = 320, 350ll
R₁ - 270.9628 22 R₂ = 49.3873-12 R₁₂3.8K2 R$ = 268.843352
for Bac 0-53392
4) Construct the circuit in figure 1, By using the values you obtained in
step-3
R₁=220.96282
Transcribed Image Text:given that 3) Based on Bpc obtained in step2, calculate the values of R., B₂, Bc and RE using your derived equations step1 Vec R₁ Ne SR2 Ica VCE vet RE A Tos Bac obtained in step is Fac=0.53392 • VCE = V₁ - V₁ •Vc = V₁E + V₁ = 6 + 1-2 • V₁= 7.2V but V₁ = √₂ x R₂ • Rc = √₂/1₂ = 7-2/2x10-3 = 3.6 kv Re=3.6 Kv from the IBO = ICB/Bac = 2X10²3/6.53392 IBQ = 3-7454xlo-³ A IEQ = 100+ Ica: S. 7459x10-³ A No. Vicc - 12V, VGA=0.65V, V₁E = 1/₂ x 12 = 6V ₁ VG = 0.1 XV₁6 = 0.1X12 Ve = 1.2V, Ice = 2mA, IR₂ = 10 IBO VE TE X RE VBE VB-VE Date ● 1. RE: VE/TE = 208.8453 V • TR₂ = 10 x IBQ = 37-459 x10³A VB = VBE + V₂ = 6.65 +1.2 UG = 1.85V IR₂ = VB / Rz R₂ = 49-38232 VB = Vcc X R₂ / R₂ + R₁ -49.3873 +R₁ = 12x 49.3873/1.85 R₁= 320.3501-49.3873 = R₂ = UB/IR 1.85/37.459x16-3 By using voltage division rule 1-85= 12 x 49.5875/49-3823 +R₁ 49,3873+R₁ = 320, 350ll R₁ - 270.9628 22 R₂ = 49.3873-12 R₁₂3.8K2 R$ = 268.843352 for Bac 0-53392 4) Construct the circuit in figure 1, By using the values you obtained in step-3 R₁=220.96282
Q VEL
Ve →
IR₂
R₂
Tea
VBE
DIMR
HA
Re
Figure-1
1) Figure 1 shows a voltage divider circuit. By referring to figure - 1,
Produce desgin equation for each resister (R₁, R₂, Rc and Re) based on
the following information:
VCE = 1/2 Vec, 1R2 = 10 18a, Ica = 2mA, Vcc = 12 V, VBFCON) = 0.65 V, V₁ = 0.1 Vcc
• V₁ = V₁B + V₂ = 6+1·2= 7.2 V • Rc= V₂c-Vc/Ica = 12-7-2/2 = 2.4 ks2
take ß= loo, 1BQ = 1cQ/B = 2/100
0.02mA
LEQ = IBA + Ico= 0·02+2=2.0 m A
R₁ = V₁ / 1₁ = 1.2/2.02 = 0.59KR
• VB = VBE + VE = 0.65 + 1.2 = 1.85 V •IR₂ = 10x 0.02 20.2m A
• R₂ = VB/1R₂ = 1.85/0·2=9.25k
By voltage division Rule, V₁ = Vcc R₂/R₁+R₂
R₁ + 9.25= 60
R. So.75 KQ
lok s
+
VCE
RE
Figure-2
LKJ
(A)
RE=0.59K R₁ = 2.4K 22
2) Construct the circuit shown in figure 2, determine the value of
Boc at the given Q-point.
Vcc= 12V
2N3904
IMD
lokn
(2
FFL
VCE=(V
B
IB +
V66=0.6515
2
No.
●
KUL in loop (2)
|-6-10-³ x 1₂ x 10³ +12=0 I₁ = 6 A
Boc Tc/10
BAC = 6/11-2376
DE
Bpc = 0.533392
DC voltage, Vec - 12V collector to emitter voltage, VcR = YZVCc
Base to emitter to voltage, VBE = 0.65
9 V₁z12V
160
12x1.25/
So, R.-So.75k R₂=9.25 k
Date
/RH 9.25 = 1.85
KVL in Loop (1)
- 12+ 1x 10² x 10-61B + lo x10³ x 106 La to, 85=0
1.0113= 11.35 18 211.35/8-01
IB= 11. 2376 A
Que
IR
Risa
RL
↓
VEEL RE
IBA
• Vcc= 12 V, Ves = 1/2X12=6V, V BE = 0.65 V
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Transcribed Image Text:Q VEL Ve → IR₂ R₂ Tea VBE DIMR HA Re Figure-1 1) Figure 1 shows a voltage divider circuit. By referring to figure - 1, Produce desgin equation for each resister (R₁, R₂, Rc and Re) based on the following information: VCE = 1/2 Vec, 1R2 = 10 18a, Ica = 2mA, Vcc = 12 V, VBFCON) = 0.65 V, V₁ = 0.1 Vcc • V₁ = V₁B + V₂ = 6+1·2= 7.2 V • Rc= V₂c-Vc/Ica = 12-7-2/2 = 2.4 ks2 take ß= loo, 1BQ = 1cQ/B = 2/100 0.02mA LEQ = IBA + Ico= 0·02+2=2.0 m A R₁ = V₁ / 1₁ = 1.2/2.02 = 0.59KR • VB = VBE + VE = 0.65 + 1.2 = 1.85 V •IR₂ = 10x 0.02 20.2m A • R₂ = VB/1R₂ = 1.85/0·2=9.25k By voltage division Rule, V₁ = Vcc R₂/R₁+R₂ R₁ + 9.25= 60 R. So.75 KQ lok s + VCE RE Figure-2 LKJ (A) RE=0.59K R₁ = 2.4K 22 2) Construct the circuit shown in figure 2, determine the value of Boc at the given Q-point. Vcc= 12V 2N3904 IMD lokn (2 FFL VCE=(V B IB + V66=0.6515 2 No. ● KUL in loop (2) |-6-10-³ x 1₂ x 10³ +12=0 I₁ = 6 A Boc Tc/10 BAC = 6/11-2376 DE Bpc = 0.533392 DC voltage, Vec - 12V collector to emitter voltage, VcR = YZVCc Base to emitter to voltage, VBE = 0.65 9 V₁z12V 160 12x1.25/ So, R.-So.75k R₂=9.25 k Date /RH 9.25 = 1.85 KVL in Loop (1) - 12+ 1x 10² x 10-61B + lo x10³ x 106 La to, 85=0 1.0113= 11.35 18 211.35/8-01 IB= 11. 2376 A Que IR Risa RL ↓ VEEL RE IBA • Vcc= 12 V, Ves = 1/2X12=6V, V BE = 0.65 V http://www.tanenghong.com
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