Microchip Transistor Thermal Cycle Profiling (Electrical & Computer Engineering) Solution

Problem Overview

We analyze the junction temperature $T(t) = t^3 - 9t^2 + 15t + 100$ ($^\circ\text{C}$) of a power GaN transistor during a $6$-second cycle ($t \in [0, 6]$). We locate critical numbers, apply the First Derivative Test, determine absolute extrema using the Closed Interval Method, and audit hardware thermal safety and firmware calibration timing.

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Step-by-Step Mathematical Solution

Step 1: Differentiate and Locate Critical Numbers

The governing formulation is the definition of Critical Numbers:

(This concept is defined in Critical Numbers and Critical Points.)

1. Compute the first derivative using the Power Rule:
   $$T'(t) = \frac{d}{dt}[t^3 - 9t^2 + 15t + 100] = 3t^2 - 18t + 15$$
2. Set $T'(t) = 0$ to find stationary points:
   $$3(t^2 - 6t + 5) = 0$$
   Factoring the quadratic:
   $$3(t - 1)(t - 5) = 0 \implies t = 1\text{ s} \quad \text{and} \quad t = 5\text{ s}$$
3. Check singular points: Since $T'(t)$ is a polynomial, it is defined for all $t$.
4. Both $t = 1\text{ s}$ and $t = 5\text{ s}$ lie strictly within the interior interval $(0, 6)$.

Step 2: Classify Critical Numbers via First Derivative Test

Construct a sign chart for $T'(t) = 3(t - 1)(t - 5)$ across the interval $[0, 6]$:

• Interval $(0, 1)$: Test $t = 0.5 \implies T'(0.5) = 3(-0.5)(-4.5) = +6.75 > 0 \implies (+)$. Temperature is increasing ($\nearrow$).
• Interval $(1, 5)$: Test $t = 2 \implies T'(2) = 3(1)(-3) = -9 < 0 \implies (-)$. Temperature is decreasing ($\searrow$).
• Interval $(5, 6)$: Test $t = 5.5 \implies T'(5.5) = 3(4.5)(0.5) = +6.75 > 0 \implies (+)$. Temperature is increasing ($\nearrow$).

Classification:

• At $t = 1\text{ s}$: $T'(t)$ changes sign from positive to negative ($+ \to -$). By the First Derivative Test, $t = 1\text{ s}$ is a local maximum.
• At $t = 5\text{ s}$: $T'(t)$ changes sign from negative to positive ($- \to +$). By the First Derivative Test, $t = 5\text{ s}$ is a local minimum.

Step 3: Evaluate Candidate Values for Absolute Extrema

Using the Closed Interval Method Algorithm:

1. Left Boundary ($t = 0\text{ s}$):
   $$T(0) = (0)^3 - 9(0)^2 + 15(0) + 100 = 100^\circ\text{C}$$
2. First Critical Point ($t = 1\text{ s}$):
   $$T(1) = (1)^3 - 9(1)^2 + 15(1) + 100 = 1 - 9 + 15 + 100 = 107^\circ\text{C}$$
3. Second Critical Point ($t = 5\text{ s}$):
   $$T(5) = (5)^3 - 9(5)^2 + 15(5) + 100 = 125 - 225 + 75 + 100 = 75^\circ\text{C}$$
4. Right Boundary ($t = 6\text{ s}$):
   $$T(6) = (6)^3 - 9(6)^2 + 15(6) + 100 = 216 - 324 + 90 + 100 = 82^\circ\text{C}$$

Comparison:

• Absolute Maximum Temperature Value: $107^\circ\text{C}$ at location $t = 1\text{ second}$.
• Absolute Minimum Temperature Value: $75^\circ\text{C}$ at location $t = 5\text{ seconds}$.

Step 4: Engineering Analysis & Diagnostic Implications

1. Safety Margin Audit: The component thermal ceiling is $T_{\text{fail}} = 125^\circ\text{C}$. The peak temperature is $T_{\max} = 107^\circ\text{C}$.
   $$\text{Safety Margin} = T_{\text{fail}} - T_{\max} = 125^\circ\text{C} - 107^\circ\text{C} = 18^\circ\text{C}$$
   Since the margin is positive ($18^\circ\text{C} > 0$), the transistor operates safely below its thermal breakdown threshold throughout the duty cycle.
2. Firmware Calibration Scheduling: Johnson-Nyquist electronic thermal noise in semiconductor junctions is proportional to absolute temperature ($v_n^2 = 4k_B T R \Delta f$). The global minimum temperature of $75^\circ\text{C}$ occurs at $t = 5\text{ seconds}$. Firmware engineers should program the microcontroller to trigger high-precision ADC calibration sequences during the window surrounding $t = 5\text{ seconds}$ to ensure maximum signal-to-noise ratio (SNR).