PHT.301 Physics of Semiconductor Devices  

High frequency response of MOSFETsThe maximum speed at which a MOSFET can operate can be estimated from the formulas derived in the gradual channel approximation. Consider a MOSFET that has an AC current $\tilde{i}_{in}$ applied to its gate. The AC gate voltage will be $\tilde{v}_G = \tilde{i}_{in}/(2\pi f ZLC_{\text{ox}})$ where $f$ is the frequency of the AC signal, $Z$ is the width of the gate, $L$ is the length of the gate, and $C_{\text{ox}}$ is the specific capacitance. The output drain current will be given by the transconductance, $$\tilde{i}_{out}=g_m\tilde{v}_G,$$where the transconductance in the saturation regime is, $$g_m= \frac{dI_D}{dV_G}=\frac{Z}{L}\mu C_{\text{ox}}(V_GV_T).$$There will be gain if $\tilde{i}_{in}\lt \tilde{i}_{out}$. This results in the condition, $$f\lt f_T=\frac{g_m}{2\pi ZLC_{\text{ox}}}=\frac{\mu (V_GV_T)}{2\pi L^2},$$where $f_T$ is called the transit frequency. This formula seems to imply that the transit frequency increases as the size of the transistor decreases like $1/L^2$. However, the voltages are typically scaled down with $L$ to hold the electric field constant. The characteristic electric field is on the order of $E\approx V_DL \approx V_GL$ so the transit frequency is approximately, $$f_T=\frac{\mu E}{2\pi L}.$$The average electron velocity is $\mu E$ for low electric fields and saturates at a saturation velocity $v_s$ for high electric fields. If the MOSFET operates at high electric fields (which is typical) the transit frequency is approximately, $$f_T=\frac{v_s}{2\pi L}.$$ Here $t_T= L/v_s$ is the transit time for an electron travelling at the saturation velocity to cross the gate of length $L$. The maximum frequency at which a MOSFET exhibits gain is approximately one over the transit time. 