Question
Download Solution PDFIn a fixed bias circuit silicon NPN transistor, common emitter configuration with β = 50 is used. Calculate VCE at quiscent point when RB = 106 Ω, RC = 5 kΩ and VCC = 10V.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept
The circuit of the fixed bias circuit is shown below:
The value of collector to emitter voltage is given by:
\(V_{CE}=V_{CC}-I_CR_C\)
where, the value of IC is given by:
\(I_C = β I_B = β \left(\frac{V_{CC}-V_{BE}}{R_B}\right)\)
Calculation
Given, β = 50
RB = 106 Ω, RC = 5 kΩ and VCC = 10V
\(I_C = 50 \left(\frac{10-0.7}{10^3}\right)\) mA
\(V_{CE}=10- 50 \left(\frac{10-0.7}{10^3}\right)\times 5\)
VCE = 7.67 V
Last updated on May 29, 2025
-> SSC JE Electrical 2025 Notification will be released on June 30 for the post of Junior Engineer Electrical/ Electrical & Mechanical.
-> Applicants can fill out the SSC JE application form 2025 for Electrical Engineering from June 30 to July 21.
-> SSC JE EE 2025 paper 1 exam will be conducted from October 27 to 31.
-> Candidates with a degree/diploma in engineering are eligible for this post.
-> The selection process includes Paper I and Paper II online exams, followed by document verification.
-> Prepare for the exam using SSC JE EE Previous Year Papers.