Question

Kindly Don't Spàm!
Thank uh !:)

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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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Answer 2

Answer:

• Ibq ≈ 47.1 μA
• Icq ≈ 2.35 mA
• Vcq ≈ 6.82 V

Explanation:

Q-point values will be determined by the base current and the transistor gain.

Base Current

The current supplied to the base of the transistor is the current through resistor Rb. That current is given by the equation ...

  $I_B=\dfrac{V_{CC}-V_B}{R_B}=\dfrac{12-0.7}{.24\cdot10^6}\approx 47.08\quad\mu\text{A}$

Collector Current

The collector current is the base current multiplied by the transistor beta.

  $I_C=\beta\times I_B=50\times47.08\text{ \mu A}=2.354\text{ mA}$

Collector Voltage

The collector voltage will be the supply voltage less the voltage drop across the collector resistor due to the collector current.

  $V_C=V_{CC}-I_C\times R_C=12-2.354\cdot10^{-3}\times2.2\cdot10^3\\\\V_C=12-5.179\approx 6.82\quad\text{V}$

The Q-point values are approximately ...

• Ibq ≈ 47.1 μA
• Icq ≈ 2.35 mA
• Vcq ≈ 6.82 V