Question
Download Solution PDFFor the circuit shown below, the value of IC is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
DC Analysis of BJT Amplifier
During the calculation of the DC parameter of a BJT amplifier, the capacitor is open-circuited.
\(I_BR_B+V_{BE}+I_ER_E=0\)
VBE = 0.7 V for NPN transistor
and \(I_E=I_B(1+\beta)=I_C+I_B\)
\(240I_B+0.7+2I_E-20=0\)
\(240I_B+0.7+2I_B(1+\beta)-20=0\)
\(240I_B+0.7+2I_B(1+90)-20=0\)
\(I_B={19.3\over 422}=0.04 \space mA\)
The collector current is given by:
\(I_E=(1+\beta) I_B\)
\(I_E=(1+90) 0.04=4.16 \space mA\)
\(I_C=I_E-I_B\)
\(I_C=4.16-0.04=4.12\space mA\)
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