Let's break down and solve the problem step by step: ## **Given:** - **Circuit:** Source and drain are **AC grounded**. - **Components:** Two capacitances, \( C_{GS} \) (gate-source) and \( C_{GD} \) (gate-drain). - **Input:** \( i_{in} \) is injected at the gate. - **Output:** \( i_{out} \) is taken from the source (which is also AC grounded). - **Definition:** Transit frequency (\( f_T \)) is the frequency at which the current gain drops to 1: \[ \left| \frac{i_{out}}{i_{in}} \right| = 1 \quad \text{at} \quad f = f_T \] --- ## **Step 1: Small-Signal Equivalent Circuit** For this configuration, the small-signal model at high frequencies is dominated by the gate-source and gate-drain capacitances. The **source** and **drain** are both AC grounded. - The input current \( i_{in} \) splits between \( C_{GS} \) and \( C_{GD} \). --- ## **Step 2: Current Division** ### **Impedance of Capacitances:** \[ Z_{C_{GS}} = \frac{1}{j\omega C_{GS}}, \quad Z_{C_{GD}} = \frac{1}{j\omega C_{GD}} \] ### **Current through \( C_{GS} \):** The total input current \( i_{in} \) splits between \( C_{GS} \) and \( C_{GD} \) (both to ground): \[ i_{out} = \text{current through } C_{GS} = i_{in} \frac{Z_{C_{GD}}}{Z_{C_{GS}} + Z_{C_{GD}}} \] But since both capacitors are to ground and in parallel, the current division is: \[ i_{in} = i_{GS} + i_{GD} \] \[ i_{GS} = \frac{C_{GS}}{C_{GS} + C_{GD}} i_{in} \] \[ i_{GD} = \frac{C_{GD}}{C_{GS} + C_{GD}} i_{in} \] But **transconductance (\( g_m \))** comes into play: The output current is \( i_{out} = g_m v_{gs} \). --- ## **Step 3: Relating Input Current to Gate-Source Voltage** \[ i_{in} = (j \omega C_{GS} + j \omega C_{GD}) v_{gs} \] \[ i_{in} = j \omega (C_{GS} + C_{GD}) v_{gs} \] \[ v_{gs} = \frac{i_{in}}{j \omega (C_{GS} + C_{GD})} \] --- ## **Step 4: Output Current** \[ i_{out} = g_m v_{gs} = g_m \left( \frac{i_{in}}{j \omega (C_{GS} + C_{GD})} \right) \] --- ## **Step 5: Current Gain** \[ \frac{i_{out}}{i_{in}} = \frac{g_m}{j \omega (C_{GS} + C_{GD})} \] Magnitude: \[ \left| \frac{i_{out}}{i_{in}} \right| = \frac{g_m}{\omega (C_{GS} + C_{GD})} \] --- ## **Step 6: Solve for Transit Frequency (\( f_T \))** At \( f = f_T \): \[ \left| \frac{i_{out}}{i_{in}} \right| = 1 \] \[ \frac{g_m}{2\pi f_T (C_{GS} + C_{GD})} = 1 \] \[ f_T = \frac{g_m}{2\pi (C_{GS} + C_{GD})} \] --- ## **Final Expression:** \[ \boxed{ f_T = \frac{g_m}{2\pi (C_{GS} + C_{GD})} } \] --- ### **Summary Table** | Parameter | Expression | |--------------------|-----------------------------------| | Transit Frequency | \( f_T = \frac{g_m}{2\pi (C_{GS} + C_{GD})} \) | **This is the required expression for the MOSFET transit frequency (\( f_T \)) for the given circuit.**