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Op-Amp Transient Response and Voltage Saturation (Electrical Engineering) Solution

1. Domain and Vertical Asymptote Analysis

  1. Name the Governing Formulas: The Rational Function Domain Restriction Rule and Vertical Asymptote Existence Criterion: (These principles are formalized in Vertical Asymptotes.)
  2. Find Singularities: Set the denominator to zero:
    t2+4t+3=0β€…β€ŠβŸΉβ€…β€Š(t+1)(t+3)=0β€…β€ŠβŸΉβ€…β€Št=βˆ’1andt=βˆ’3t^2 + 4t + 3 = 0 \implies (t + 1)(t + 3) = 0 \implies t = -1 \quad \text{and} \quad t = -3
    Algebraic Domain across R\mathbb{R}:
    (βˆ’βˆž,βˆ’3)βˆͺ(βˆ’3,βˆ’1)βˆͺ(βˆ’1,∞)(-\infty, -3) \cup (-3, -1) \cup (-1, \infty)
  3. Physical Domain Verification: The amplifier circuit is switched on at t=0t = 0. On the operational physical domain [0,∞)[0, \infty), both singular points (t=βˆ’1Β mst = -1\text{ ms} and t=βˆ’3Β mst = -3\text{ ms}) lie strictly in negative time prior to energization. Thus, no vertical asymptotes or discontinuities exist during circuit operation; V(t)V(t) is smooth and continuous for all tβ‰₯0t \ge 0.

2. Initial Voltage Condition (yy-Intercept)

Evaluate V(t)V(t) at energization time t=0t = 0:

V(0)=15(0)2(0)2+4(0)+3=03=0Β VoltsV(0) = \frac{15(0)^2}{(0)^2 + 4(0) + 3} = \frac{0}{3} = 0\text{ Volts}
The initial voltage across the capacitor starts at 0Β V0\text{ V}, reflecting zero initial stored charge.


3. Horizontal Asymptote & Steady-State Saturation

  1. Name the Governing Formula: The Horizontal Asymptote Degree Dominance Rule for Balanced Rational Functions:
    lim⁑tβ†’βˆžantn+…bmtm+…=anbm(whenΒ n=m)\lim_{t \to \infty} \frac{a_n t^n + \dots}{b_m t^m + \dots} = \frac{a_n}{b_m} \quad (\text{when } n = m)
    (This theorem is detailed in Horizontal Asymptotes.)
  2. Compute Limit as tβ†’βˆžt \to \infty:
    lim⁑tβ†’βˆžV(t)=lim⁑tβ†’βˆž15t2t2+4t+3=lim⁑tβ†’βˆž151+4t+3t2=151+0+0=15Β Volts\lim_{t \to \infty} V(t) = \lim_{t \to \infty} \frac{15t^2}{t^2 + 4t + 3} = \lim_{t \to \infty} \frac{15}{1 + \frac{4}{t} + \frac{3}{t^2}} = \frac{15}{1 + 0 + 0} = 15\text{ Volts}
  3. Engineering Interpretation: The horizontal asymptote is V=15Β VV = 15\text{ V}. In physical circuit hardware, this corresponds to the positive DC power supply voltage rail (+VCC=+15Β V+V_{CC} = +15\text{ V}), setting the upper physical ceiling for the operational amplifier output.

4. Monotonicity via First Derivative Sign Analysis

  1. Name the Governing Formula: The Quotient Rule for Differentiation:
    [uv]β€²=uβ€²vβˆ’uvβ€²v2\left[\frac{u}{v}\right]' = \frac{u'v - uv'}{v^2}
  2. Differentiate V(t)V(t): Here u(t)=15t2β€…β€ŠβŸΉβ€…β€Šuβ€²(t)=30tu(t) = 15t^2 \implies u'(t) = 30t, and v(t)=t2+4t+3β€…β€ŠβŸΉβ€…β€Švβ€²(t)=2t+4v(t) = t^2 + 4t + 3 \implies v'(t) = 2t + 4.
    Vβ€²(t)=(30t)(t2+4t+3)βˆ’(15t2)(2t+4)(t2+4t+3)2V'(t) = \frac{(30t)(t^2 + 4t + 3) - (15t^2)(2t + 4)}{(t^2 + 4t + 3)^2}
    Expand the numerator:
    Num=(30t3+120t2+90t)βˆ’(30t3+60t2)=60t2+90t=30t(2t+3)\text{Num} = (30t^3 + 120t^2 + 90t) - (30t^3 + 60t^2) = 60t^2 + 90t = 30t(2t + 3)
    Thus:
    Vβ€²(t)=30t(2t+3)(t2+4t+3)2V'(t) = \frac{30t(2t + 3)}{(t^2 + 4t + 3)^2}
  3. Sign Analysis on [0,∞)[0, \infty):
    • At t=0t = 0: Vβ€²(0)=0V'(0) = 0 (the response starts with zero initial slew rate).
    • For all t>0t > 0: 30t>030t > 0, 2t+3>02t + 3 > 0, and the denominator squared is strictly positive.
    • Therefore, Vβ€²(t)>0V'(t) > 0 for all t>0t > 0. By the Monotonicity Theorem, the output voltage is strictly increasing, guaranteeing monotonic transient charging with zero parasitic ringing or overshoot.

5. Amplifier 90% Settling Time

Target settling threshold:

Vtarget=0.90Γ—15Β V=13.5Β VV_{\text{target}} = 0.90 \times 15\text{ V} = 13.5\text{ V}
Set V(t)=13.5V(t) = 13.5:
15t2t2+4t+3=13.5β€…β€ŠβŸΉβ€…β€Š15t2=13.5(t2+4t+3)\frac{15t^2}{t^2 + 4t + 3} = 13.5 \implies 15t^2 = 13.5(t^2 + 4t + 3)
15t2=13.5t2+54t+40.515t^2 = 13.5t^2 + 54t + 40.5
1.5t2βˆ’54tβˆ’40.5=01.5t^2 - 54t - 40.5 = 0
Divide through by 1.51.5:
t2βˆ’36tβˆ’27=0t^2 - 36t - 27 = 0
Apply the Quadratic Formula:
t=βˆ’(βˆ’36)Β±(βˆ’36)2βˆ’4(1)(βˆ’27)2=36Β±1296+1082=36Β±14042t = \frac{-(-36) \pm \sqrt{(-36)^2 - 4(1)(-27)}}{2} = \frac{36 \pm \sqrt{1296 + 108}}{2} = \frac{36 \pm \sqrt{1404}}{2}
1404β‰ˆ37.47\sqrt{1404} \approx 37.47
Since time must be positive (tβ‰₯0t \ge 0):
tsettle=36+37.472=73.472β‰ˆ36.74Β mst_{\text{settle}} = \frac{36 + 37.47}{2} = \frac{73.47}{2} \approx 36.74\text{ ms}
The pre-amplifier settles to within 90%90\% of rail saturation in approximately 36.74Β milliseconds36.74\text{ milliseconds}.