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 |
\(I_C = β I_B = β \left(\frac{V_{CC}-V_{BE}}{R_B}\right)\) \(S(V_{BE}) = \frac{\partial I_C}{\partial V_{BE}} = -\frac{β}{R_E}\) |
|
कलेक्टर-टू-बेस बायस |
मान लें कि IE = β IB,
\(V_{CC} - I_BR_B-V_{BE} - β I_BR_E=0\) \(I_C = β I_B = β\left(\frac{V_{CC}-V_{BE}}{R_B+β R_E}\right) \) \(S(V_{BE}) = \frac{\partial I_C}{\partial V_{BE}} = -\frac{β}{R_B+β R_E}= -\frac{β/R_E}{β + R_B/R_E}\) |
|
वोल्टेज डिवाइडर |
मान लें कि IE = β IB, \(V_{TH}-I_BR_{TH}-V_{BE}-β I_BR_E=0\) \(I_C = β I_B = β\left(\frac{V_{TH}-V_{BE}}{R_{TH}+β R_E}\right)\) \(S(V_{BE}) = \frac{\partial I_C}{\partial V_{BE}} = -\frac{β}{R_{TH}+β R_E}= -\frac{β/R_E}{β + R_{TH}/R_E}\) |
|
प्रतिक्रिया बायस |
\(V_{CC}-β I_BR_C-I_BR_B-V_{BE}=0\) \(I_C = β I_B = β\left(\frac{V_{TH}-V_{BE}}{R_{B}+β R_C}\right)\) \(S(V_{BE}) = \frac{\partial I_C}{\partial V_{BE}} = -\frac{β}{R_B+β R_C}= -\frac{β/R_C}{β + R_B/R_C}\)
|
Last updated on Jun 7, 2025
-> RRB JE CBT 2 answer key 2025 for June 4 exam has been released at the official website.
-> Check Your Marks via RRB JE CBT 2 Rank Calculator 2025
-> RRB JE CBT 2 admit card 2025 has been released.
-> RRB JE CBT 2 city intimation slip 2025 for June 4 exam has been released at the official website.
-> RRB JE CBT 2 Cancelled Shift Exam 2025 will be conducted on June 4, 2025 in offline mode.
-> RRB JE CBT 2 Exam Analysis 2025 is Out, Candidates analysis their exam according to Shift 1 and 2 Questions and Answers.
-> The RRB JE Notification 2024 was released for 7951 vacancies for various posts of Junior Engineer, Depot Material Superintendent, Chemical & Metallurgical Assistant, Chemical Supervisor (Research) and Metallurgical Supervisor (Research).
-> The selection process includes CBT 1, CBT 2, and Document Verification & Medical Test.
-> The candidates who will be selected will get an approximate salary range between Rs. 13,500 to Rs. 38,425.
-> Attempt RRB JE Free Current Affairs Mock Test here
-> Enhance your preparation with the RRB JE Previous Year Papers.