Question
Download Solution PDFBJT में कलेक्टर-टू-बेस बायस परिपथ, स्थिर बायस परिपथ की तुलना में बेहतर स्थिरता रखता है क्योंकि __________।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFBJT में कलेक्टर-टू-बेस बायस परिपथ, स्थिर बायस परिपथ की तुलना में बेहतर स्थिरता रखता है, क्योंकि इसका स्थिरता कारक स्थिर बायस की तुलना में कम होता है, जो (β + RB / RE) के गुणांक से कम होता है।
स्थिर बायस के लिए: \(S(V_{BE}) = \frac{\partial I_C}{\partial V_{BE}} = -\frac{β}{R_E}\)
कलेक्टर-टू-बेस बायस के लिए: \(S(V_{BE}) = \frac{\partial I_C}{\partial V_{BE}} = -\frac{β}{R_B+β R_E}= -\frac{β/R_E}{β + R_B/R_E}\)
विभिन्न विन्यासों के लिए स्थिरता कारक
| बायसिंग विन्यास | बायसिंग परिपथ | VBE के सापेक्ष स्थिरता कारक |
| स्थिर बायस |
\(I_C = β I_B = β \left(\frac{V_{CC}-V_{BE}}{R_B}\right)\) \(S(V_{BE}) = \frac{\partial I_C}{\partial V_{BE}} = -\frac{β}{R_E}\) |
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| कलेक्टर-टू-बेस बायस |
मान लें कि IE = β IB,
\(V_{CC} - I_BR_B-V_{BE} - β I_BR_E=0\) \(I_C = β I_B = β\left(\frac{V_{CC}-V_{BE}}{R_B+β R_E}\right) \) \(S(V_{BE}) = \frac{\partial I_C}{\partial V_{BE}} = -\frac{β}{R_B+β R_E}= -\frac{β/R_E}{β + R_B/R_E}\) |
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| वोल्टेज डिवाइडर |
मान लें कि IE = β IB, \(V_{TH}-I_BR_{TH}-V_{BE}-β I_BR_E=0\) \(I_C = β I_B = β\left(\frac{V_{TH}-V_{BE}}{R_{TH}+β R_E}\right)\) \(S(V_{BE}) = \frac{\partial I_C}{\partial V_{BE}} = -\frac{β}{R_{TH}+β R_E}= -\frac{β/R_E}{β + R_{TH}/R_E}\) |
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| प्रतिक्रिया बायस |
\(V_{CC}-β I_BR_C-I_BR_B-V_{BE}=0\) \(I_C = β I_B = β\left(\frac{V_{TH}-V_{BE}}{R_{B}+β R_C}\right)\) \(S(V_{BE}) = \frac{\partial I_C}{\partial V_{BE}} = -\frac{β}{R_B+β R_C}= -\frac{β/R_C}{β + R_B/R_C}\)
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Last updated on Jun 7, 2025
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