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Forum: Electrical & Computer Engineering :: Electronics :: BJTs  | |||||||||
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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   Mohammad Fawaz |
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| mohammadfawaz | Sun 12 Jul 2009 6:42:00 PM | |||||||||||
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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 |
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| cim07 | Sat 30 Jan 2010 8:13:00 AM | |||||||||||
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Student Joined Wed 27 Jan 2010 Posts 1 |
This gives RC = VRC/IRC
= 0.6/10.167µ = 59KOhms. Chadi |
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