नीचे दर्शाये गये चित्र के अनुसार किसी उत्सर्जक-अभिनति नेटवर्क के लिये, संग्राही धारा _______________है।

F1 Madhuri Engineering 19.12.2022 D3

This question was previously asked in
KVS TGT WET (Work Experience Teacher) 23 Dec 2018 Official Paper
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  1. 40.1 μA
  2. 2.01mA 
  3. 6.66 mA 
  4. 10 mA 

Answer (Detailed Solution Below)

Option 2 : 2.01mA 
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संकल्पना:

उत्सर्जक-अभिनति नेटवर्क का डी.सी. विश्लेषण;

आधार धारा→ \(I_{B} = \frac{V_{CC}-V_{BE}}{R_{B}+(1+β)R_{E}}\)

संग्राहक धारा→ \(I_{C} = β I_{B}\)

गणना:

दिया गया है;

VCC = 20 V

RB = 430 KΩ 

VBE = 0.7 V

β = 50

RE = 1 KΩ 

तब;

आधार धारा → \(I_{B} = \frac{V_{CC}-V_{BE}}{R_{B}+(1+β)R_{E}}= \frac{20-0.7}{430+(1+50)1}=0.04\, mA\)

संग्राहक धारा → \(I_{C} = β I_{B} = 50 \times 0.04=2.01\, mA\)

 

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