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ECE 3040 Quiz 7 - BJT Bias Circuit Solutions, Quizzes of Electrical and Electronics Engineering

Georgia Institute of Technology - Main CampusElectrical and Electronics Engineering

The solutions to quiz 7 of ece 3040, a university-level electrical engineering course focusing on bjt (bipolar junction transistor) bias circuits. Students are required to determine the value of v- for a given current, and verify the bjt is operating in the active mode using the given circuit parameters and the provided formula. The calculations and the final results.

Typology: Quizzes

Pre 2010

Uploaded on 08/05/2009

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ECE3040Quiz7—July6,2005
Professor Leach Printed Name
Instructions. Place a box around your answers. Express numerical answers as a decimal number.
Place a box around all numerical answers. Honor Code Statement: I have neither given nor
received help on this quiz. Initials
The figure shows a BJT bias circuit. It is given that V+=18V,RE=7.5kΩ,R1= 110 kΩ,
R2=36kΩ,RC=2.4kΩ,VBE =0.7V,andβ=94.
(a) Solve for V−for IE=2mA.
(b) Verify that the BJT is biased in the active mode.
Vp18 R E7500 R 1110000 R236000 R C2400 VBE 0.7
β
94
αβ
1
β
Vm12
VBB Vp
R2
R1R2
.
VBB 4.438
=
RBB RpR1R2
,
RBB 2.712 104
=
IC
VBB VmVBE
RE
α
RBB
β
IC1000
.
2
=
VCB VpICRC
.
Vm
IC
α
RE
.
VBE VCB 9.338
=
1

Partial preview of the text

Download ECE 3040 Quiz 7 - BJT Bias Circuit Solutions and more Quizzes Electrical and Electronics Engineering in PDF only on Docsity!

ECE 3040 Quiz 7 — July 6, 2005

Professor Leach Printed Name

Instructions. Place a box around your answers. Express numerical answers as a decimal number.

Place a box around all numerical answers. Honor Code Statement: I have neither given nor

received help on this quiz. Initials

The figure shows a BJT bias circuit. It is given that V

= 18 V, R

E

= 7.5 kΩ, R

1

= 110 kΩ,

R

2

= 36 kΩ, R

C

= 2.4 kΩ, V

BE

= 0.7 V, and β = 94.

(a) Solve for V

−

for IE = 2 mA.

(b) Verify that the BJT is biased in the active mode.

V

p

18 R

E

7500 R

110000 R

36000 R

C

2400 V

BE

0.7 β 94

α

β

1 β

V

m

V

BB

V

p

R

R

R

V

BB

= 4.438 R

BB

R

p

R

R

, R

BB

I

C

V

BB

V

m

V

BE

R

E

α

R

BB

β

I

C

V

CB

V

p

I

C

R

C

V

m

I

C

α

R

E

V

BE

V

CB