PHT.301 Physics of Semiconductor Devices

## High frequency response of MOSFETs

The 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,

$$g_m= \frac{dI_D}{dV_G}=\frac{Z}{L}\mu C_{\text{ox}}(V_G-V_T).$$

There will be gain if $\tilde{i}_{in}\lt \tilde{i}_{out}$. This results in the condition,

$$f\lt f_T=\frac{g_m}{2\pi ZLC_{\text{ox}}}=\frac{\mu (V_G-V_T)}{2\pi L^2},$$

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.