Question
Download Solution PDFएक कॉमन बेस कनेक्शन में, यदि एमिटर धारा 2 mA और कलेक्टर धारा 1.95 mA है, तो बेस धारा कितनी होगी?
Answer (Detailed Solution Below)
Detailed Solution
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1.) उभयनिष्ठ उत्सर्जक धारा लाभ (β)
\(\beta={I_C\over I_B}\)
2.) उभयनिष्ठ आधार धारा लाभ (α)
\(\alpha={I_C\over I_E}\)
3.) उभयनिष्ठ संग्राहक धारा लाभ (γ)
\(\beta={I_E\over I_B}\)
साथ ही, \(I_E=I_C+I_B\)
जहाँ, IC = संग्राहक धारा
IB = आधार धारा
IE = उत्सर्जक धारा
गणना
दिया गया है, IE = 2 mA
IC = 1.95 mA
\(2=1.95+I_B\)
IB = 0.05 mA
Last updated on May 29, 2025
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