Ncert Exercises & Solutions

Consider the circuit arrangement shown in the figure for studying input and output characteristics of an n-p-n transistor in CE configuration.
Select the value of R_B and R_C for a transistor whose V_BE = 0.7 V, so that the transistor is operating at point Q as shown in the characteristics (see figure).

Given that the input impedance of the transistor is very small and $V_{CC} = V_{BB} = 16 V$ , also find the voltage gain and power gain of the circuit making appropriate assumptions.

Hint: Power gain = current gain x voltage gain

Step 1: Find the collector resistance.

Given,

V_{BE} = 0.7 V, V_{CC} = V_{BB} = 16 V

V_{CE} = 8 V (from graph)

I_{C} = 4 mA = 4 \times 10^{- 3} A

I_{B} = 30 μA = 30 \times 10^{- 6} A

For the output characteristic at Q,

V_{CC} = I_{C} R_{C} + V_{CE}

R_{C} = \frac{V_{CC} - V_{CE}}{I_{C}} = \frac{16 - 8}{4 \times 10^{- 3}} = \frac{8 \times 1000}{4} = 2 kΩ

Step 2: Find the base resistance.

Using the relation,

V_{BB} = I_{B} R_{B} + V_{BE}

R_{B} = \frac{V_{BB} - V_{BE}}{I_{B}} = \frac{16 - 0.7}{30 \times 10^{- 6}}

= 510 \times 10^{3} Ω = 510 kΩ

Step 3 : Fidn the power gain .

β = \frac{I_{C}}{I_{B}} = \frac{4 \times 10^{- 3}}{30 \times 10^{- 6}} = 133

Voltage gain = β \frac{R_{C}}{R_{B}} = \frac{133 \times 2 \times 10^{3}}{510 \times 10^{3}} = 0.52

Power gain = β \times Voltage gain = 133 \times 0.52 = 69

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