Saturday, March 6, 2021

Audio Amplifier Feedback - Rate-of-Closure

This post is a part of the series on audio amplifier feedback. The contents of the series can be found here.

Earlier in this series of posts, I showed that the loop gain is the product of open loop gain $A$ and the feedback network gain $B$: $LG = A \times B$:

If one plots the magnitudes $|A|$ and $|{1 / B}|$ on a logarithmic scale, the difference between these two plots corresponds to the loop gain magnitude: ${{|A|}_{dB}} - {{|{1 / B}|}_{dB}} = {{|LG|}_{dB}}$:

The crossover point where the two plots meet, and where the magnitude of the loop gain becomes unity (or 0dB), is important for the circuit's stability: if the phase lag around the feedback loop reached 180°, the circuit becomes unstable. The difference between the phase lag and 180° is the phase margin $\phi_m$:

Here, the phase margin is negative and the circuit is unstable. One of the objectives of compensation is to keep the phase margin sufficiently positive. A practical requirement is to keep the phase margin of at least 30° in the range of frequencies where the magnitude of loop gain is between 10dB and -10dB.

Rate-of-closure or ROC is a tool to estimate the phase margin from the logarithmic plots of $|A|$ and $|{1 / B}|$. ROC is defined as the difference between the slopes of the $|{1 / B}|$ and of the $|A|$ curves at the crossover frequency. Once we know the ROC, we can estimate the phase margin as $$\phi_m=180° - 4.5 \times ROC$$ where $\phi_m$ is in degrees and ROC is in dB/decade. In the example below, ROC 1 is about 45 dB/decade, so $\phi_m$ is about $180° - 4.5 \times 45 = -25°$. ROC 1 is about 25 dB/decade, so $\phi_m$ is about $180° - 4.5 \times 25 = 65°$:

A more detailed discussion of rate-of-closure can be found in this excellent article by Sergio Franco.

How does this concept apply to the lead compensation discussed in my last post? Let's take the above example and add a lead compensation capacitor, which introduces a pole-zero pair into the feedback network transfer function $B$. If the crossover point falls between the pole and the zero, where the $|1/B|$ curve slopes down, ROC decreases, indicating a higher phase margin:

Here, ROC 1 is about 35 dB/decade, indicating $\phi_m$ about $180° - 4.5 \times 35 = 20°$ (compare to -25° without the compensation capacitor).

If the compensation capacitor is too small, $|1/B|$ curve slopes down at the frequencies higher than the crossover, the ROC remains high, and the phase margin remains low:

If the compensation capacitor is too large, $|1/B|$ curve slopes down at the frequencies lower than the crossover, returning to horizontal at the crossover point. Again, the ROC remains high, and the phase margin remains low:

That is, as stated in my previous post, the value of the lead compensation capacitor is critical. If it is too high or too low, it will not improve the phase margin. 

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