SBAA767 January   2026 OPA206

 

  1.   1
  2.   Abstract
  3.   Trademarks
  4. 1Introduction
  5. 2General Transfer Function of the Second Order System
    1. 2.1 Damping Ratio
      1. 2.1.1 Underdamped (0 < ζ < 1)
      2. 2.1.2 Critically Damped (ζ = 1)
      3. 2.1.3 Overdamped (ζ>1)
  6. 3Modeling Op-Amp as a Second Order System
  7. 4Phase Margin vs Percent Overshoot
    1. 4.1 Phase Margin
    2. 4.2 Represent AOLβ as ΦPM
    3. 4.3 Represent ΦPM as Damping ratio
    4. 4.4 Phase Margin Represented by Percent Overshoot
    5. 4.5 Phase Margin Represented by Gain Peaking
  8. 5Simulation of Ideal Second Order System
    1. 5.1 Phase Margin: 30 Degrees
    2. 5.2 Phase Margin: 45 Degrees
    3. 5.3 Phase Margin: 60 Degrees
    4. 5.4 Phase Margin: 75 Degrees
    5. 5.5 Step Response with Different Phase Margin (Damping Ratio)
    6. 5.6 Gain Peaking with Different Phase Margin (Damping Ratio)
  9. 6Simulation Example Using an Op-Amp
    1. 6.1 OPA392 With Non-Inverting Amp Configuration
      1. 6.1.1 Step Response Simulation
      2. 6.1.2 Gain Peaking Simulation
      3. 6.1.3 Loop Gain Simulation
    2. 6.2 TLV9052 with Unity Gain Buffer Configuration
      1. 6.2.1 Step Response Simulation
      2. 6.2.2 Gain Peaking Simulation
      3. 6.2.3 Loop Gain Simulation
    3. 6.3 OPA206 with Unity Gain Buffer Configuration
      1. 6.3.1 Step Response Simulation
      2. 6.3.2 Gain Peaking Simulation
      3. 6.3.3 Loop Gain Simulation
  10. 7Causes of the Mismatch of Phase Margin Between Step Response and AC Analysis
    1. 7.1 The Transfer Function is not a Second Order System
    2. 7.2 Amplifier Showing Large-Signal Behavior
    3. 7.3 Noise Gain is Not Flat Within Crossover Frequency
  11. 8Summary
  12. 9References

2 General Transfer Function of the Second Order System

The general transfer function of a second order system can be described as below using DC gain K, damping ratio ζ and natural frequency ωn.

Equation 1. G s = K ω n 2 s 2 + 2 ζ ω n s + ω n 2

From denominator of Equation 1, obtain poles shown in Equation 3 by solving Equation 2

Equation 2. s 2 + 2 ζ ω n s + ω n 2 = 0
Equation 3. - ω n ζ ± ω n ζ 2 - 1