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Solved Examples · Example 3

Q.Given ID=10I_D = 10 mA and VGS=−1V_{GS} = -1 V, determine the value of VPV_P if IDSS=20I_{DSS} = 20 mA.

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[!TLDR]

Solving Shockley's equation for VPV_P gives VP=−3.41V_P = -3.41 V.

Start from Shockley's equation:

ID=IDSS[1−VGSVP]2I_D = I_{DSS}\left[1-\frac{V_{GS}}{V_P}\right]^2

Substitute the given values:

10×10−3=20×10−3[1−−1VP]210\times10^{-3} = 20\times10^{-3}\left[1-\frac{-1}{V_P}\right]^2 …

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