In the circuit shown below, device X has the characteristics shown in the i_x–v_x plot. The x-axis division is 0.3 V, and the y-axis division is 20 μA. Assume that V_A = 9 V and R = 20 KΩ. Use the graphical method to plot the load line.

a.       What is the x-axis intercept of the load line?

b.      What is the slope of the load line?

c.       What is the bias voltage V_x of the device?

a.       To find the x-axis intercept, we need to find the load line from circuit. If we find the Thevenin equivalent of the voltage source together with the two resistors, we can easily find the relation between v_x and i_x.

R_Th = R//3R = 0.75R (if we short the source).

V_Th = [R/(3R+R)].V_A = 0.25V_A (voltage divider).

Now, we can apply Ohms law for the new resistor: 0.75R.

Get: 0.25V_A – v_x = 0.75Ri_x. Therefore, the load line equation is i_x = -(4/3R)v_x +V_A/3R.

Therefore, i_x = -[4/(3´20)]v_x + 9/(3´20) i.e. i_x = -(1/15)v_x + 3/20. (see figure)

x-axis interception, so i_x = 0, i.e. -(1/15)v_x + 3/20 = 0 therefore, v_x = 2.25

b.      The slope of the load line is -1/15 mA/V as found above in the equation.

c.       To find the bias voltage, we just need to read the v_x component of the meeting point of the load line and the current voltage characteristics of the device X. We can clearly see that the v_x component of this point is at 5 units. But each unit represents 0.3V, then the bias voltage is 5´0.3 = 1.5V.