Answer to Question #247456 in Electric Circuits for Johannes Steven

Question

Question

The output characteristics of a transistor in common-emitter configuration can be regarded as straight lines connecting the following points

         

   IB =     20 µA        50 µA          80 µA       

 

VCE(V) 1.0 8.0 1.0 8.0 1.0 8.0

 

IC (mA) 1.2 1.4 3.4 4.2 6.1 8.1

 

Plot the characteristics and superimpose the load line for a 1 kΩ load, given that the supply voltage is 9V and the d.c. base bias is 50 μA. The signal input resistance is 800 Ω. When a peak input current of 30 μA varies sinusoidally about a mean bias of 50 μA, determine

(a) The quiescent collector current 

(b) The current gain 

(c) The voltage gain

(d) The power gain

Answer 1

From Common Emitter configuration, we can develop the load line analysis equation as follows

9 - I_c * 1 - V_CE = 0

V_CE = 9 - I_c * 1 ................................Equation(1)

We will plot equation(1) with the output characteristics of the transistor as shown in the image attached below.

(a) From graph we see that quiescent I_c = 4.2 mA

(b) Current gain $\beta$ = $\dfrac{}{}$ $\dfrac{Ic}{I_B}$ = $\dfrac{4.2 \ mA}{30\ \ uA}$ = 140

(c) Voltage gain A_V = $\beta\ *\ \dfrac{Output\ resistance}{Input\ Resistance}$ = 140 * $\dfrac{1000}{800}$ = 175

(d) Power gain = A_v * $\beta$ = 140 * 175 = 24500