Topic: Electronic Devices
The transconductance $g_{m}$ of an FET in the saturation region equals
- A. $\frac{-2I_{DSS}}{V_{p}}[1-\frac{V_{GS}}{V_{p}}]$
- B. $\frac{-2I_{DSS}}{V_{p}}[1-\frac{V_{GS}}{V_{p}}]^{2}$
- C. $\frac{-2I_{DSS}}{V_{p}}[1-\frac{V_{GS}}{V_{p}}]^{1/2}$
- D. $\frac{1}{V_{p}}[I_{DSS}\times I_{DS}]^{1/2}$
Answer: A. $\frac{-2I_{DSS}}{V_{p}}[1-\frac{V_{GS}}{V_{p}}]$
Explanation: $I_{DS}=I_{DSS}(1-\frac{V_{GS}}{V_{P}})^{2}$. $g_{m}=\frac{\partial I_{D}}{\partial V_{GS}}=\frac{-2I_{DSS}}{V_{p}}(1-\frac{V_{GS}}{V_{p}})$.