Question

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Answer 1

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Collector Current at Saturation :

• $\sf \large {I_{C(SAT)} = \frac{ V_{CC} }{R_C} }$

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• $\sf \large {I_{C(SAT)} = \frac{12}{2.2 \times 10 ^{3} } }$

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• $\bold{ \bf \large {I_{C(SAT)} =5.45mA }}$

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Value Of Cut - off Voltage :

• $\sf \large \: V_{CS(cut-off)} = V_{CC}$

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Therefore ,

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• $\bf \large \: V_{CS(cut-off)} = \: 12v$

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${ \bf\large \: Base \: Current , I_B = \frac{V_{CC}}{R_C} }$

• $\sf \large \: I_B ={ \frac{12}{2.2 \times 10 ^{3} } }$

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• $\bf \large \: I_B = 50μA$

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Collector Current ,

• $\sf \large \: I_C = β * I_B$

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• $\sf \large \: I_C = 50 * 50 * 10^{-6}$

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• $\bf \large \: I_C = 2.5mA$

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Collector to emitter Voltage :

• $\sf \large \: V_{CE} = V_{CC} - ( I_C * R_C )$

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• $\sf \large \: V_{CE} = 12 - ( 2.5 * 10-³ * 2.2 * 10³ )$

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• $\bf \large \: V_{CE} = 6.5v$

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Q - point are :

• $\sf \large \: I_{CEQ} = 2.5mA$

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• $\sf \large \: V_{CEQ} = 6.5v$

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Q - point located on the DC load line as shown in fig ~

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Hope Helps!:)

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