Given figure shows a silicon transistor connected as a common emitter amplifier. The quiescent collector voltage of the circuit is approximately.

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VSSC (ISRO) Technical Assistant Previous Year Paper (Held on 8 Feb 2015)

Option 3 : 14 V

ISRO VSSC Technical Assistant Mechanical held on 09/06/2019

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80 Questions
320 Marks
120 Mins

__Analysis__**:**

The quiescent Base voltage (V_{b}) is determined by the potential divider network formed by the two resistors R_{1}, R_{2 }(connected to the base of the transistor), and the power supply voltage V_{cc} as shown, with the current flowing through both the resistors.

\({V_B} = \frac{{{V_{CC}}{R_2}}}{{{R_1} + {R_2}}}\)

__Calculation__**:**

\({V_B} = \frac{{20}}{{10\; + \;5}} \times 5 = \frac{{20}}{3}\)

The Thevenin equivalent resistance connected to the base of the circuit will be:

\({R_{BB}} = \frac{{5\; \times \;10}}{{15}} = \frac{{10}}{3}\)

Applying KVL from V_{B} to emitter ground, we get:

V_{B} – I_{B}R_{BB} – V_{BE} – I_{E}R_{E} = 0 ---(1)

With I_{E} = (β + 1)I_{B} ≈ β I_{B}:

\({I_B} = \frac{{{I_E}}}{\beta }\)

Equation (1) becomes:

\({V_B} - \frac{{{I_E}}}{\beta }{R_{BB}} - {V_{BE}} - {I_E}{R_E} = 0\)

\({I_E} = \frac{{\frac{{20}}{3} - 0.7}}{{10\; + \;\frac{{10}}{3}\; \times\; \beta }}\)

\({I_E} = \frac{{5.96\;mA}}{{10\; + \;\frac{{10}}{3}\; \times \;100}}\)

Assuming I_{E} ≈ I_{C}, the output voltage will be:

\({V_o} = 20 - {I_E}{R_C} = 14\;V\)