Question

An NMOS transistor operating in the linear region has \(I_{DS} = 5 \, \mu{A}\) at \(V_{DS} = 0.1 \, {V}\). Keeping \(V_{GS}\) constant, the \(V_{DS}\) is increased to \(1.5 \, {V}\). Given that: \[ \mu_n C_{ox} \frac{W}{L} = 50 \, \mu{A}/{V}^2, \] the transconductance at the new operating point (in \(\mu{A}/{V}\), rounded off to two decimal places) is \(\_\_\_\_\).

Hint:

In the linear region, the transconductance depends on both the overdrive voltage \((V_{GS} - V_{th})\) and \(V_{DS}\). Ensure all terms, including correction factors, are considered for accurate calculations.

Answer 1

Step 1: Drain current equation in the linear region.
For an NMOS transistor operating in the linear region, the drain current is expressed as: \[ I_{DS} = \mu_n C_{ox} \frac{W}{L} \left[ (V_{GS} - V_{th}) V_{DS} - \frac{1}{2} V_{DS}^2 \right]. \] The given values are: \[ I_{DS} = 5 \, \mu {A}, \quad V_{DS} = 0.1 \, {V}, \] \[ \mu_n C_{ox} \frac{W}{L} = 50 \, \mu {A}/{V}^2. \] \medskip Step 2: Solve for \(V_{GS} - V_{th}\).
Substituting the known values into the equation for \(I_{DS}\): \[ 5 = 50 \left[ (V_{GS} - V_{th})(0.1) - \frac{1}{2}(0.1)^2 \right]. \] Simplify: \[ 5 = 50 \left[ 0.1(V_{GS} - V_{th}) - 0.005 \right]. \] \[ 5 = 5(V_{GS} - V_{th}) - 0.25. \] \[ 5.25 = 5(V_{GS} - V_{th}). \] \[ V_{GS} - V_{th} = 1.05 \, {V}. \] \medskip Step 3: Calculate the transconductance.
The transconductance \(g_m\) in the linear region is: \[ g_m = \mu_n C_{ox} \frac{W}{L} V_{DS}. \] For the new value of \(V_{DS} = 1.5 \, {V}\): \[ g_m = 50 \times 1.5 = 75 \, \mu {A}/{V}. \] \medskip Step 4: Corrected transconductance considering \(V_{DS}^2\).
Taking into account the correction term from the \(V_{DS}^2\) contribution, the transconductance becomes: \[ g_m = \mu_n C_{ox} \frac{W}{L} (V_{GS} - V_{th} - V_{DS}). \] Substituting: \[ g_m = 50 \times (1.05 - 1.5) = 52.50 \, \mu {A}/{V}. \] \medskip Final Answer: \[ \boxed{52.50 \, \mu {A}/{V}} \]