SLAAET1 March   2025 ADC08DJ3200 , ADC08DJ5200RF , ADC09DJ1300 , ADC09DJ800 , ADC09QJ1300 , ADC09QJ800 , ADC09SJ1300 , ADC09SJ800 , ADC12DJ1600 , ADC12DJ2700 , ADC12DJ3200 , ADC12DJ4000RF , ADC12DJ5200RF , ADC12DJ800 , ADC12QJ1600 , ADC12QJ800 , ADC12SJ1600 , ADC12SJ800 , ADC32RF52 , ADC32RF54 , ADC32RF55 , ADC32RF72 , ADC34RF52 , ADC34RF55 , ADC34RF72 , ADC3548 , ADC3549 , ADC3568 , ADC3569 , ADC3648 , ADC3649 , ADC3664 , ADC3668 , ADC3669 , ADC3683

 

  1.   1
  2.   Abstract
  3.   Trademarks
  4. 1Introduction
  5. 2Active (Amplifier) vs. Passive (Transformer/Balun) Tradeoffs
  6. 3Amplifiers vs. Baluns: The Advantages
  7. 4Amplifiers vs. Baluns: The Disadvantages
  8. 5Understanding the Importance of Phase Imbalance
  9. 6Phase Balance with Amplifiers and Baluns
  10. 7Summary
  11. 8References
  12.   A Appendix A

A Appendix A

Magnitude and Phase Imbalance Derivation

Mathematical model of two input signals to represent the differential interface into the ADC:

Equation 8. x1t=k1sinωtx2t=k2sinωt-180°+ρ=-k2sinωt+ρ

Mathematical model of the ADC, as a symmetrical third-order transfer function:

Equation 9. hxt=a0+a1 xt+a2x2t+a3x3t

Putting the two together, this can be represented as follows:

Equation 10. yt=hx1t-hx2tyt=a1x1t-x2t+a2x12t-x22t+a3x13t-x23t

In the case, where there is no imbalance to the two input signals, the transfer function of the ADC is modeled as follows, where magnitude is k1 = k2 = k and phase is exactly 180° out of phase (φ = 0°):

Equation 11. x 1 t = k sin ω t x 2 t = - k sin ω t y t = 2 a 1 k sin ω t + 2 a 3 k 3 sin 3 ω t

With simplification of the model, this can be seen that the even harmonics cancel and the odd harmonics do not or:

Equation 12. y t = 2 a 1 k + 3 a 3 k 3 4 sin ω t - a 3 k 3 2 sin 3 ω t

Looking at the case where the magnitude is imbalanced, where k1 ≇ k2 and φ = 0 and phase is balanced. The two inputs signals look as follows:

Equation 13. x1t = k1sinωtx2t = -k2sinωt

With some substitution, the following can be found:

Equation 14. yt=a22×k12-k22=2+a1k1+k2+3a34×k13-k23sinωt-a22×k12-k22cos2ωt-a34×k13-k23sin3ωt

As shown from the equation that the second harmonic is proportional to the difference of the squares of the magnitude terms, or:

Second harmonic is α k12 - k22

Looking at the case where the phase is imbalanced, where k1 = k2 and φ ≇ 0 and the magnitude is balanced. The two inputs signals look as follows:

Equation 15. x 1 t = k 1 sin ω t x 2 t = - k 1 sin ω t + ρ

With some substitution, the following can be found:

Equation 16. y t = a 1 k 1 + 3 a 3 k 1 3 4 × sin ω t + sin ω t × cos ρ + cos ω t × sin ρ - a 2 k 1 2 2 × cos 2 ω t - cos 2 ω t × cos 2 ρ + sin 2 ω t × sin 2 ρ - a 3 k 1 3 4 × sin 3 ω t + sin 3 ω t × cos 3 ρ + cos 3 ω t × sin 3 ρ

As shown from the equation that the second harmonic is proportional to the square of the magnitude term, or:

Second harmonic is α k12

In summary, the second harmonic is influenced by the phase imbalance more so than the magnitude imbalance. Therefore, the phase imbalance and second harmonic is proportional to the square of k1. While for magnitude imbalance, the second harmonic is proportional to the difference of squares of k1 and k2. Since k1 and k2 are typically and approximately equal, the difference of k12 and k22 is small. Thus, the second harmonic is generally not affected as much do to magnitude imbalance.