Discussion
Consider an n-channel depletion mode MOSFET having following parameters :
\(V_{TN} = -2.5\) V and \(K_{n} = 1.1\) mA/V\(^{2}\)
If \(V_{GS} = 0\)V and \(V_{DS} = 0.5\) V, then current \(I_{D}\), is
Answer: C
The drain current is 2.48 mA, option (C). With \(V_{GS} = 0\) and \(V_{TN} = -2.5\) V, the overdrive is \(V_{GS} - V_{TN} = 2.5\) V. As \(V_{DS} = 0.5\) V is less than 2.5 V, the device works in the triode region, where
\[ I_{D} = K_{n}\left[2(V_{GS}-V_{TN})V_{DS}\right.\]
\[ \left. -\,V_{DS}^{2}\right] \]
\[ I_{D} = 1.1\left[2(2.5)(0.5) - 0.25\right] \]
\[ = 1.1\times2.25 = 2.48\ \text{mA} \]
Options (A), (B) and (D) are not given by the triode equation. The saturation formula \(K_{n}(V_{GS}-V_{TN})^{2}\), which would give 6.9 mA, applies only when \(V_{DS}\) is at least 2.5 V, which is not the case here.
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