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Forum: Electrical & Computer Engineering :: Electronics :: BJTs  New Topic
Problem #18 Reply
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mohammadfawaz Problem #18 Sun 12 Jul 2009 6:41:00 PM
Student
Joined Sat 4 Apr 2009
Posts 642

Design the bias circuit in the figure below (find RB and RC) to get a collector current of 10µA and a collector voltage of 0.9V. Assume that b = 60.

Assume VBE(ACTIVE) = 0.7V


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Mohammad Fawaz
Student Rating3.5/5
mohammadfawaz Sun 12 Jul 2009 6:42:00 PM
Student
Joined Sat 4 Apr 2009
Posts 642

The BJT is in active region since VCE = VC = 0.9V > 0.3V (therefore not in SAT), and a current is flowing in the BJT (therefore not in cutoff).

IC = 10µA therefore, the base current is IB = IC/b = 10/60 = 0.167µA.

BJT in active: VBE = VB = 0.7 (since VE = 0 = ground)

 Therefore, the voltage across RB is VC – VB = 0.9 – 0.7 = 0.2V.

Now, RB = VRB/IB = 0.2/0.167µ = 1.2 MOhms.

The voltage across RC is VRC = 1.5 – VC = 1.5 – 0.9 = 0.6V

Using KCL, the current flowing through RC is IRC = IB + IC = 0.167 + 10 = 10.167 µA.

This gives RB = VRC/IRC = 0.6/10.167µ = 3.6 MOhms.





Mohammad Fawaz
Student Rating3.5/5
cim07 Sat 30 Jan 2010 8:13:00 AM
Student
Joined Wed 27 Jan 2010
Posts 1
This gives RC = VRC/IRC = 0.6/10.167µ = 59KOhms.



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