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Question 18.4: Perform the design of Example 18.3 numerically rather than u......

Perform the design of Example 18.3 numerically rather than using graphs.

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As before

I_{D(quiescent)} = \frac{V_{R}}{R_{D}} = \frac{5  V}{2.5  kΩ} = 2  mA

From Equation 18.3

I_{D} = I_{DSS}\left(1  –  \frac{V_{GS}}{V_{P}} \right)^{2}

or, by rearranging,

V_{GS} = V_{P} \left(1  –  \sqrt{\frac{I_{D}}{I_{DSS}}} \right) = -6 \left(1  –  \sqrt{\frac{2}{8}} \right) = -3  V

as before. Thus, R_{S} is 1.5 kV, as above.

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