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Problem #40 Reply
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mohammadfawaz Problem #40 Sun 12 Jul 2009 7:29:00 PM
Student
Joined Sat 4 Apr 2009
Posts 642

In the circuit shown below, device X has the characteristics shown in the ixvx plot. The x-axis division is 0.3 V, and the y-axis division is 20 μA. Assume that VA = 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 Vx of the device?


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Mohammad Fawaz
Student Rating3.5/5
mohammadfawaz Sun 12 Jul 2009 7:30:00 PM
Student
Joined Sat 4 Apr 2009
Posts 642

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 vx and ix.

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

VTh = [R/(3R+R)].VA = 0.25VA (voltage divider).

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

Get: 0.25VA – vx = 0.75Rix. Therefore, the load line equation is ix = -(4/3R)vx +VA/3R.

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

x-axis interception, so ix = 0, i.e. -(1/15)vx + 3/20 = 0 therefore, vx = 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 vx component of the meeting point of the load line and the current voltage characteristics of the device X. We can clearly see that the vx component of this point is at 5 units. But each unit represents 0.3V, then the bias voltage is 5´0.3 = 1.5V. 


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Mohammad Fawaz
Student Rating3.5/5
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