Semiconductor Transistor Current Fluctuation (Electrical Engineering) Solution

In integrated circuit design, an advanced field-effect transistor (FET) regulates output current based on gate-to-source voltage ($V_g$) and drain-to-source voltage ($V_d$). In the saturation regime, the electrical drain current $I$ (in milliamperes, $\text{mA}$) is modeled by the non-linear constitutive equation:

The transistor's nominal operating bias point (Q-point) is established at $V_g = 2.0\text{ V}$ and $V_d = 4.0\text{ V}$, providing a baseline current of:

During high-frequency switching, power rail switching noise causes the terminal voltages to shift slightly to an off-nominal state of $V_g = 2.02\text{ V}$ and $V_d = 3.95\text{ V}$.

1. State the multivariable total differential formula $dI$ for the transistor current in terms of voltage differentials $dV_g$ and $dV_d$.
2. Compute the analytical partial derivatives $\frac{\partial I}{\partial V_g}$ and $\frac{\partial I}{\partial V_d}$, and evaluate them at the nominal bias point $(2, 4)$.
3. Use the linear approximation $L(V_g, V_d)$ centered at $(2, 4)$ to estimate the perturbed current $I(2.02, 3.95)\text{ mA}$.
4. If manufacturer power rail tolerances guarantee $|dV_g| \le 0.05\text{ V}$ and $|dV_d| \le 0.10\text{ V}$, compute the maximum worst-case current fluctuation $|dI|_{\max}$.

Step-by-Step Solution:

1. State the Total Differential Formula:

   • As defined in Differentials and Error Propagation, the total differential $dI$ is:
     $$dI = \frac{\partial I}{\partial V_g} dV_g + \frac{\partial I}{\partial V_d} dV_d$$

2. Compute Analytical Partial Derivatives:

   • Let $u(V_g, V_d) = V_g^3 + V_d^2$. Applying the Generalized Power Rule:
     $$\frac{\partial I}{\partial V_g} = \frac{1}{2}(V_g^3 + V_d^2)^{-1/2}(3V_g^2) = \frac{3V_g^2}{2\sqrt{V_g^3 + V_d^2}}$$
     $$\frac{\partial I}{\partial V_d} = \frac{1}{2}(V_g^3 + V_d^2)^{-1/2}(2V_d) = \frac{V_d}{\sqrt{V_g^3 + V_d^2}}$$
   • Evaluate at $(V_g, V_d) = (2, 4)$, where $\sqrt{V_g^3 + V_d^2} = \sqrt{24} = 2\sqrt{6}$:
     $$\left.\frac{\partial I}{\partial V_g}\right|_{(2, 4)} = \frac{3(2^2)}{2\sqrt{24}} = \frac{12}{2\sqrt{24}} = \frac{6}{\sqrt{24}} = \frac{3}{\sqrt{6}} \approx 1.2247\text{ mA/V}$$
     $$\left.\frac{\partial I}{\partial V_d}\right|_{(2, 4)} = \frac{4}{\sqrt{24}} = \frac{2}{\sqrt{6}} \approx 0.8165\text{ mA/V}$$

3. Construct Linear Approximation and Estimate Current:

   • Formulate the linearization function centered at $(2, 4)$:
     $$L(V_g, V_d) = I(2, 4) + \left.\frac{\partial I}{\partial V_g}\right|_{(2, 4)}(V_g - 2) + \left.\frac{\partial I}{\partial V_d}\right|_{(2, 4)}(V_d - 4)$$
     $$L(V_g, V_d) = \sqrt{24} + \frac{6}{\sqrt{24}}(V_g - 2) + \frac{4}{\sqrt{24}}(V_d - 4)$$
   • Calculate offsets for target point $(2.02, 3.95)$:
     $$\Delta V_g = 2.02 - 2.0 = +0.02\text{ V}$$
     $$\Delta V_d = 3.95 - 4.0 = -0.05\text{ V}$$
   • Substitute offsets:
     $$L(2.02, 3.95) = \sqrt{24} + \frac{6(0.02) + 4(-0.05)}{\sqrt{24}} = \sqrt{24} + \frac{0.12 - 0.20}{\sqrt{24}} = \sqrt{24} - \frac{0.08}{\sqrt{24}}$$
     $$L(2.02, 3.95) \approx 4.89898 - \frac{0.08}{4.89898} \approx 4.89898 - 0.01633 \approx 4.8827\text{ mA}$$

4. Compute Worst-Case Current Fluctuation:

   • Applying the Triangle Inequality for maximum error bound:
     $$|dI|_{\max} \le \left|\frac{\partial I}{\partial V_g}\right| |dV_g| + \left|\frac{\partial I}{\partial V_d}\right| |dV_d|$$
     $$|dI|_{\max} \le (1.2247)(0.05) + (0.8165)(0.10) = 0.06124 + 0.08165 = 0.1429\text{ mA}$$
   • The maximum current deviation due to power rail noise is bounded by approximately $\pm 0.143\text{ mA}$ (or $\frac{0.1429}{4.8990} \times 100\% \approx 2.92\%$).