Assigment #4 Solution DC Biasing- BJTs

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Ain Shams University Faculty of Engineering Course: MANF 360 Instructor: Prof. Ismail Mohammed Hafez

Assignment #4

DC Biasing of BJTs

Question #4.1 (Txt 4.1 & 4.4 & 4.11)) For the fixed-bias configuration of Fig. 4.1, if VCC =16 V, RC=2.7 kΩ, RB=470 kΩ, and β =90, analyze this circuit analytically and then determine: (a) IBQ

(b) ICQ (c) VCEQ (d) VC

(e) VB (f) VE (g) ICsat Fig 4.1 Fixed-bias configuration (h) Calculate the percentage change in IC and VCE when β change to 135. Solution (a) IBQ = 16 V 0.7 V 15.3 V

470 k 470 kQ

CC BEB

B

V VIR− −

= = =Ω Ω

=32.55 μA

(b) ICQ = β IBQ = (90)(32.55 μA) = 2.93 mA

(c) VCEQ = VCC - ICQ RC = 16 V − (2.93 mA)(2.7 kΩ) = 8.09 V (d) VC = VCEQ = 8.09 V (e) VB = VBE = 0.7 V (f) VE = 0 V (g) ICsat = 16 V

2.7 kCC

C

VR

=Ω

= 5.93 mA

For β = 135

(h) IBQ = 32.55 μA (the same)

ICQ = β IBQ = (135)(32.55 μA) = 4.39 mA VCEQ = VCC - ICQ RC = 16 V − (4.39 mA)(2.7 kΩ) = 4.15 V

%ΔIC = 4.39 mA 2.93 mA2.93 mA

− × 100% = 49.83%

%ΔVCE = 4.15 V 8.09 V8.09 V

− × 100% = 48.70%

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Less than 50% due to level of accuracy carried through calculations. Question #4.2 (Txt 4.2) For the fixed-bias configuration Fig. 4.2, if VCC=12 V, VC=6 V, IBQ=40 μA and β =80, analyze this circuit analytically and then determine: (a) IC

(b) RC (c) RB (d) VCE Solution (a) IC = βIB = 80(40 μA) = 3.2 mA Fig 4.2 Fixed-bias configuration

(b) RC = 12 V 6 V 6 V3.2 mA 3.2 mA

CR CC C

C C

V V VI I

− −= = = = 1.875 kΩ

(c) RB = 12 V 0.7 V 11.3 V40 A 40 A

BR

B

VI μ μ

−= = = 282.5 kΩ

(d) VCE = VC = 6 V Question #4.3 (Txt 4.3) For the fixed-bias configuration Fig. 4.3, if VCE=7.2 V, IB=20 μA and IE=4 mA, analyze this circuit analytically and then determine: (a) IC

(b) VCC (c) β (d) RB Solution Fig 4.3 Fixed-bias configuration (a) IC = IE − IB = 4 mA − 20 μA = 3.98 mA ≅ 4 mA

(b) VCC = VCE + ICRC = 7.2 V + (3.98 mA)(2.2 kΩ) = 15.96 V ≅ 16 V

(c) β = 3.98 mA20 A

C

B

II μ

= = 199 ≅ 200

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(d) RB = 15.96 V 0.7 V20 A

BR CC BE

B B

V V VI I μ

− −= = = 763 kΩ

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Question #4.4 (Txt 4.5) Given the BJT transistor characteristics of Fig. 4.4, if the circuit parameters are: VCC=21 V and RC=3 kΩ. For this circuit do the following: (a) Draw the load line. (b) Chose the operating point midway between cutoff and saturation. Determine the

value of RB to establish the resulting operating point. (c) Determine ICQ and VCEQ. (d) Calculate β. (e) Calculate α. (f) Calculate ICsat. (g) Calculate the dc power dissipated by the device. (h) Power dissipated by VCC. (i) Pdis by the resistors.

Fig. 4.4 (a) A transistor network Fig. 4 (b) The transistor characteristics

Solution

(a) Load line intersects vertical axis at IC = 21 V3 kΩ

= 7 mA and horizontal axis at

VCE = 21 V. (b) IB = 25 μA: RB = 21 V 0.7 V

25 ACC BE

B

V VI μ− −

= = 812 kΩ

(c) ICQ ≅ 3.4 mA and VCEQ ≅ 10.75 V

(d) β = 3.4 mA25 A

C

B

II μ

= = 136

(e) α = 136 1361 136 1 137

ββ

= =+ +

= 0.992

(f) ICsat = 21 V3 k

CC

C

VR

=Ω

= 7 mA

(g) PD = VCEQ ICQ = (10.75 V)(3.4 mA) = 36.55 mW

(h) Ps = VCC(IC + IB) = 21 V(3.4 mA + 25 μA) = 71.92 mW

(i) PR = Ps − PD = 71.92 mW − 36.55 mW = 35.37 mW

•

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Question #4.5 (Txt 4.6 & 4.9 & 4.11) For the emitter-stabilized bias configuration of Fig. 4.5, if VCC=20 V, RC=2.4 kΩ, RB=510 kΩ and β =100, analyze this circuit analytically and then determine: (a) IBQ

(b) ICQ (c) VCEQ (d) VC

(e) VB (f) VE (g) ICsat (h) Calculate the percentage change in IC and VCE when β change to 150.

Fig 4.5 emitter-stabilized bias configuration Solution

(a) IBQ = 20 V 0.7 V( 1) 510 k (101)1.5 kQ

CC BEB

B E

V VIR Rβ

− −= =

+ + Ω + Ω = 19.3 V

661.5 kΩ= 29.18 μA

(b) ICQ = β IBQ = (100)(29.18μA) = 2.92 mA

(c) VCEQ = VCC − IC(RC + RE) = 20 V − (2.92 mA)(2.4 kΩ + 1.5 kΩ) = 20 V − 11.388 V = 8.61 V

(d) VC = VCC − ICRC = 20 V − (2.92 mA)(2.4 kΩ) = 20 V − 7.008 V = 13 V

(e) VB = VCC − IBRB = 20 V − (29.18 μA)(510 kΩ) = 20 V − 14.882 V = 5.12 V

(f) VE = VC − VCE = 13 V − 8.61 V = 4.39 V

(g) ICsat= sot

20 V 20 V2.4 k 1.5 k 3.9 k

CCC

C E

VIR R

= = =+ Ω + Ω Ω

= 5.13 mA

For β = 150

(h) IBQ = 20 V 0.7 V( 1) 510 k (150 1)(1.5 k )Q

CC BEB

B E

V VIR Rβ

− −= =

+ + Ω + + Ω = 26.21 μA

ICQ = β IBQ = (150)(26.21 μA) = 3.93 mA

VCEQ = VCC − IC(RC + RE) = 20 V − (3.93 mA)(2.4 kΩ + 1.5 kΩ) = 4.67 V

%ΔIC = 3.93 mA 2.92 mA2.92 mA

− × 100% = 34.59%

%ΔVCE = 4.67 V 8.61 V8.61 V

− × 100% = 46.76%

(g) For both IC and VCE the % change is less for the emitter-stabilized.

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Question #4.6 (Txt 4.7) For the emitter- stabilized bias configuration Fig. 4.6, if VCC=12 V, VC=7.6 V, VE=2.4 V, IC=2 mA and β = 80, analyze this circuit analytically and then determine:

(a) RC (b) RE

(c) RB (d) VCE (e) VB

Fig 4.6 Emitter-stabilized bias configuration

Solution (a) RC = 12 V 7.6 V 4.4 V

2 mA 2 mACC C

C

V VI− −

= = = 2.2 kΩ

(b) IE ≅ IC: RE = 2.4 V2 mA

E

E

VI

= = 1.2 kΩ

(c) RB = 12 V 0.7 V 2.4 V 8.9 V2 mA/80 25 A

BR CC BE E

B B

V V V VI I μ

− − − −= = = = 356 kΩ

(d) VCE = VC − VE = 7.6 V − 2.4 V = 5.2 V

(e) VB = VBE + VE = 0.7 V + 2.4 V = 3.1 V Question #4.7 (Txt 4.8) For the emitter-stabilized bias configuration Fig. 4.7, if VCE=7.3 V, VE=2.1 V, RC=2.7 kΩ, RE=0.68 kΩ and IB=20 μA, analyze this circuit analytically and then determine:

(a) β (b) VCC

(c) RB

Solution Fig 4.7 Emitter- stabilized bias configuration

(a) IC ≅ IE = 2.1 V0.68 k

E

E

VR

=Ω

= 3.09 mA and β = 3.09 mA20 A

C

B

II μ

= = 154.5

(b) VCC = CRV +VCE +VE = (3.09 mA)(2.7 kΩ)+ 7.3 V+2.1 V= 8.34 V+7.3 V+2.1 V

= 17.74 V

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(c) RB = 17.74 V 0.7 V 2.1 V20 A

BR CC BE E

B B

V V V VI I μ

− − − −= = = 14.94 V

20 Aμ = 747 kΩ

Question #4.8 (Txt 4.10) Given the BJT transistor characteristics of Fig. 4.8, if the circuit parameters are: VCC=24 V, ICQ=4 mA, VCEQ=10 V and RE=1.2 kΩ. For this circuit do the following: (a) Draw the load line. (b) Calculate RC and RB (c) Calculate β. (d) Pdis by the transistor. (e) Power dissipated by RC

Fig. 4.8 Emitter-bias network and the device characteristics

Solution (a) The load passes through the points (VCE = 24 V, ICQ=0 mA) and (VCE = 10 V,

IC=4 mA). From the load line IBQ=30 μA and , IC= 6.8 mA at VCE = 0 V

(b) satCI = 6.8 mA = 24 V

1.2 kCC

C E C

VR R R

=+ + Ω

RC = 1.2 kΩ = 24 V6.8 mA

= 3.529 kΩ RC = 2.33 kΩ

RB = 24 V 0.7 V (4 mA)(1.2 k )30 A

BR CC BE E

B B

V V V VI I μ

− − − − Ω= = = 18.5 V

30 Aμ = 616.67 kΩ

(c) β = 4 mA30 A

C

B

II μ

= = 133.33

(d) PD = VCQ ICQ = (10 V)(4 mA) = 40 mW

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(e) P = 2C CI R = (4 mA)2(2.33 kΩ) = 37.28 mW

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Question #4.9 (Txt 4.12 & 4.15& 4.20) For the voltage-divider bias configuration shown in Fig. 4.9, if VCC=16 V, R1=62 kΩ, R2=9.1 kΩ, RC=3.9 kΩ, RE=0.68 kΩ and β = 80 determine:

(a) IBQ (b) ICQ

(c) VCEQ (d) VC

(e) VE (f) VB (g) ICsat (h) Calculate the percentage change in

IC and VCE when β change to 120. Fig. 4.9 Voltage-divider bias network Solution (a) RTh = R1 || R2 = 62 kΩ || 9.1 kΩ = 7.94 kΩ ETh = 2

2 1

(9.1 k )(16 V)9.1 k 62 k

CCR VR R

Ω=

+ Ω + Ω = 2.05 V

IBQ = 2.05 V 0.7 V( 1) 7.94 k (81)(0.68 k )Q

Th BEB

Th E

E VIR Rβ

− −= =

+ + Ω + Ω = 21.42 μA

(b) ICQ = β IBQ = (80)(21.42 μA) = 1.71 mA

(c) VCEQ = VCC − ICQ (RC + RE) = 16 V − (1.71 mA)(3.9 kΩ + 0.68 kΩ) = 8.17 V

(d) VC = VCC − ICRC = 16 V − (1.71 mA)(3.9 kΩ) = 9.33 V

(e) VE = IERE ≅ ICRE = (1.71 mA)(0.68 kΩ) = 1.16 V (f) VB = VE + VBE = 1.16 V + 0.7 V = 1.86 V (g) ICsat = 16 V

3.9 k 0.68 kCC

C E

VR R

=+ Ω + Ω

= 16 V4.58 kΩ

= 3.49 mA

For β = 120 (h) ETh = 2.05 V, RTh = 7.94 kΩ IB = 2.05 V 0.7 V

( 1) 7.94 k + (121)(0.68 k )Th BE

Th E

E VR Rβ

− −=

+ + Ω Ω= 14.96 μA

IC = βIB = (120)(14.96 μA) = 1.8 mA VCE = VCC − IC(RC + RE) = 16 V − (1.8 mA)(3.9 kΩ + 0.68 kΩ) = 7.76 V 1.8 mA 1.71 mA%

1.71 mACI −Δ = ×100 = 5.26% & 7.76 V 8.17 V%

8.17 VCEV −Δ = ×100 = 5.02%

11c 11f 20c %ΔIC 49.83% 34.59% 5.26% %ΔVCE 48.70% 46.76% 5.02% Fixed-bias Emitter

Feedback Voltage-divider

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(e) It is obviously the voltage-divider configuration is the least sensitive to changes in β.

Question #4.10 (Txt 4.13) For the voltage-divider bias configuration shown in Fig. 4.10, if VCC=18 V, VC=12 V, R2=5.6 kΩ, RE=1.2 kΩ and RC=4.7 kΩ, determine:

(a) IC (b) VE (c) VB

(d) R1

Fig. 4.10 Voltage-divider bias network Solution (a) IC = 18 V 12 V

4.7 kCC C

C

V VR− −

=Ω

= 1.28 mA

(b) VE = IERE ≅ ICRE = (1.28 mA)(1.2 kΩ) = 1.54 V

(c) VB = VBE + VE = 0.7 V + 1.54 V = 2.24 V

(d) R1 = 1

1

:R

R

VI

1RV = VCC − VB = 18 V − 2.24 V = 15.76 V

1RI ≅

22

2.24 V5.6 k

BR

VIR

= =Ω

= 0.4 mA

R1 = 1

1

R

R

VI

= 15.76 V0.4 mA

= 39.4 kΩ

Question #4.11 (Txt 4.14) For the voltage-divider bias configuration shown in Fig. 4.11, if VC=10.6 V, IB=20 μA, R2=8.2 kΩ, RE=1.2 kΩ, RC=2.7 kΩ, RE=0.68 kΩ and β = 100 determine:

(a) IC (b) VE

(c) VCC (d) VCE

(e) VB (f) R1

Solution Fig. 4.11 Voltage-divider bias network (a) IC = βIB = (100)(20 μA) = 2 mA

(b) IE = IC + IB = 2 mA + 20 μA = 2.02 mA VE = IERE = (2.02 mA)(1.2 kΩ) = 2.42 V

(c) VCC = VC + ICRC = 10.6 V + (2 mA)(2.7 kΩ) = 10.6 V + 5.4 V = 16 V

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(d) VCE = VC − VE = 10.6 V − 2.42 V = 8.18 V

(e) VB = VE + VBE = 2.42 V + 0.7 V = 3.12 V (f)

1 2R R BI I I= + = 3.12 V8.2 kΩ

+ 20 μA = 380.5 μA + 20 μA = 400.5 μA

R1 = 1

16 V 3.12 V400.5 A

CC B

R

V VI μ− −

= = 32.16 kΩ

Question #4.12 (Txt 4.32) Determine RC and RB for a fixed-bias configuration if VCC = 12 V, β = 80, ICQ = 2.5 μA and VCEQ = 6 V. Use the standard values.

Solution IB = 2.5 mA

80CIβ= = 31.25 μA

RB = 12 V 0.7 V31.25 A

BR CC BE

B B

V V VI I μ

− −= = = 361.6 kΩ

RC = 12 V 6 V 6 V2.5 mA 2.5 mA

QC

Q

CC CER CC C

C C C

V VV V VI I I

−− −= = = = = 2.4 kΩ

Standard values: RB = 360 kΩ RC = 2.4 kΩ Question #4.13(Txt 4.33) Design an emitter-stabilized bias network at ICQ = 0.5 ICsat and VCEQ = 0.5 VCC. Use VCC = 20 V, ICsat = 10 mA, β = 120 and RC = 4 RE. Use the standard values.

Solution

satCI = CC

C E

VR R+

= 10 mA

20 V4 E ER R+

= 10 mA ⇒ 20 V5 ER

= 10 mA ⇒ 5RE =

20 V10 mA

= 2 kΩ

RE = 2 k5Ω = 400 Ω

RC = 4RE = 1.6 kΩ IB = 5 mA

120CIβ

= = 41.67 μA

RB = VRB/IB = 20 V 0.7 V 5 mA(0.4 k ) 19.3 2 V

41.67 A 41.67 Aμ μ− − Ω −

=

= 415.17 kΩ Standard values: RE = 390 Ω,

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RC = 1.6 kΩ, RB = 430 kΩ Question #4.14 (Txt 4.34) Design a voltage-divider bias network using the supply voltage of 24 V, a transistor with beta of 110, and an operating point of ICQ = 4 mA and VCEQ = 8 V. Chose VE =(1/8) VCC. Use the standard values. Solution RE = 3 V

4 mAE E

E C

V VI I

≅ = = 0.75 kΩ

RC = ( )

QC CC CE ER CC C

C C C

V V VV V VI I I

− +−= =

= 24 V (8 V + 3 V) 24 V 11 V 13 V4 mA 4 mA 4 mA− −

= = = 3.25 kΩ

VB = VE + VBE = 3 V + 0.7 V = 3.7 V

VB = 2 2

2 1 2 1

(24 V)3.7 V = CCR V RR R R R

⎫⇒ ⎬+ + ⎭

2 unknowns!

∴ Use βRE ≥ 10R2 for increased stability

(110)(0.75 kΩ) = 10R2

R2 = 8.25 kΩ

Choose R2 = 7.5 kΩ

Substituting in the above equation: 3.7 V =

1

7.5 k (24 V)7.5 k R

ΩΩ+

R1 = 41.15 kΩ Standard values: RE = 0.75 kΩ RC = 3.3 kΩ R2 = 7.5 kΩ, R1 = 43 kΩ

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Question #4.15 (Txt 4.35) Using the characteristics of Fig. 4.15, design a voltage-divider bias configuration to have a saturation level of 10 mA and a Q-point one-half the distance between the cutoff and saturation. The available supply voltage is 28 V and VE is to be one-fifth of VCC. Use the standard values.

Fig. 4.15 Transistor characteristics

Solution VE = 1

5 CCV = 1 (28 V)5

= 5.6 V

RE = 5.6 V5 mA

E

E

VI

= = 1.12 kΩ (use 1.1 kΩ)

VC = 28 V2 2CC

EV V+ = + 5.6 V = 14 V + 5.6 V = 19.6 V

CRV = VCC − VC = 28 V − 19.6 V = 8.4 V

RC = 8.4 V5 mA

CR

C

VI

= = 1.68 kΩ (use 1.6 kΩ)

VB = VBE + VE = 0.7 V + 5.6 V = 6.3 V VB = 2

2 1

CCR VR R+

⇒ 6.3 V = 2

2 1

(28 V)RR R+

(2 unknowns)

β = 5 mA37 A

C

B

II μ

= = 135.14

βRE = 10R2 then (135.14)(1.12 kΩ) = 10(R2) and R2 = 15.14 kΩ (use 15 kΩ) Substituting: 6.3 V =

1

(15.14 k )(28 V)15.14 k R

ΩΩ+

Solving, R1 = 52.15 kΩ (use 51 kΩ) Standard values:

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RE = 1.1 kΩ, RC = 1.6 kΩ, R1 = 51 kΩ and R2 = 15 kΩ

Assigment #4 Solution DC Biasing- BJTs

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