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For the circuit shown in Fig. 4.38 (i), draw the d.c. load line.

Answers

The collector-emitter voltage V_CE is given by ;
V_CE = V_CC - I_CR_C ...(i)

When I_C = 0, then,
V_CE = V_CC = 12.5 V
This locates the point B of the load line on the collector-emitter voltage axis.

When V_CE = 0, then,
I_C = V_CC/R_C = 12.5 V/2.5 kilo-ohms = 5 mA
This locates the point A of the load line on the collector current axis. By joining these two points, we get the d.c. load line AB as shown in Fig. 4.38 (ii).

Wilfykil answered the question on August 15, 2019 at 06:27

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