# Phase Margin Estimation Using the Rate-of-Closure

March 08, 2019 by Dr. Sergio Franco## How can we gauge stability in a negative-feedback circuit? For minimum-phase circuits, try rate-of-closure.

How can an engineer gauge stability in an analog negative-feedback circuit? For minimum-phase circuits, you can consider using the rate-of-closure.

In everyday analog design, engineers often need to quickly estimate the degree of stability (or lack thereof) of a negative-feedback circuit. A convenient tool that is applicable to *minimum-phase* circuits (so-called because all their poles and zeros are in the left half of the complex plane) is the *rate-of-closure *(ROC).

To prepare the background, consider the familiar block diagram of Figure *1a*:

*(a) (b)*

**Figure 1.** (a) Block diagram of a negative feedback circuit, and (b) visualizing the loop gain T.

**Figure 1.**(a) Block diagram of a negative feedback circuit, and (b) visualizing the loop gain T.

This diagram consists of an *error amplifier* with gain *a(**jf**)*, a *feedback network* with transfer function *β(**jf**)*, and a summing block generating the *error signal S _{e}*,

*Equation (1)*

*Equation (1)*

Collecting and solving for *S _{e}* gives

*Equation (2)*

*Equation (2)*

where *T = aβ* is called the *loop gain* because any signal entering the amplifier and going clockwise around the loop, is first amplified by *a* and then by *β*, for an overall gain of *aβ*. It is apparent that larger *T* leads to smaller *S _{e} *for a given input

*S*.

_{i}Rewriting as *T = a/(1/β)*, taking the logarithms, and multiplying by 20 to convert to decibels, gives

*Equation (3)*

*Equation (3)*

indicating that we can visualize the decibel plot of |*T*| as the *difference* between the decibel plots of |*a*| and |*1/β*|. This is depicted in Figure *1b* for the case of a constant gain-bandwidth product op-amp and a frequency-independent *β*.

The frequency* f _{x}* at which the two curves intersect, aptly called the

*crossover frequency*, plays an important role in the circuit’s stability (or lack thereof). At this frequency, we have |

*T(*

*jf*

_{x}*)*|

_{dB}= 0, or |

*T(*

*jf*)| = 1. Should the phase ph[

_{x}*T(*

*jf*

_{x}*)*] ever reach –180°, then we’d have

*T(*

*jf*

_{x}*)*= –1, which, substituted into Eq. (2), indicates that

*S*would blow up and lead to oscillation. (Note that even if we set

_{e}*S*, the intrinsic noise of the circuit would trigger the buildup of

_{i}= 0*S*.) To prevent oscillation, we must make sure that ph[

_{e}*T(*

*jf*

_{x}*)*] is

*away*from the dreaded value of –180° by a sufficient amount, aptly called the

*phase margin ɸ*,

_{m}

### * ɸ*_{m} = 180° + ph[T(jf_{x})]

_{m}= 180° + ph[T(jf

_{x})]

*Equation (4)*

*Equation (4)*

Phase margins of practical interest are *ɸ** _{m}* = 90°,

*ɸ*= 65.5° (which marks the onset of

_{m}*peaking*in the ac response),

*ɸ*= 76.3° (which marks the onset of

_{m}*ringing*in the transient response), and

*ɸ*= 45° (which we’ll see to lend itself to easy geometric visualization, though it results in a peaking of 2.4 dB and ringing with an overshoot of 23%).

_{m}In the situation depicted in Figure *1b*, *1/β* is a real number, so its phase is 0° and we have ph[*T(**jf _{x}*

*)*] = ph[

*a(*

*jf*

_{x}*)*] ≈ –90°, owing to the fact that

*f*>>

_{x}

*f*. Consequently,

_{p}*ɸ*≈ 180° + (–90°) = 90°. However, things are not always this rosy, so we turn to the more general cases of Figure 2 to estimate the phase margins of the various cases. To this end, we shall use the

_{m}*rate-of-closure*(ROC).

### What Is Rate of Closure?

Rate of closure or ROC is defined as the *difference* between the slopes of the ⎪*1/β*⎪and of the⎪*a*⎪ curves *right* at the crossover frequency:

*Equation (5)*

*Equation (5)*

Once we know the ROC, we estimate the *phase margin* as

*Equation (6)*

*Equation (6)*

In Figure 2, we assume an error amplifier with a more general gain profile than the constant gain-bandwidth product type of Figure *1a*. (For easy drawing, we are using straight segments, though we know that in practice the sharp angles are *rounded*, as the |*a*| curve in Figure *1b*.)

*(a)*

*(b)*

**Figure 2.** (a) Frequently encountered phase-margin situations with (a) frequency-independent and (b) frequency-dependent feedback factor β(jf).

**Figure 2.**(a) Frequently encountered phase-margin situations with (a) frequency-independent and (b) frequency-dependent feedback factor β(jf).

Now, past the first pole frequency *f _{p1}*, ⎪

*a*⎪ rolls off at a rate of –20 dB/dec, and past the second pole frequency

*f*it rolls off at a rate of –40 dB/dec. Right at

_{p2}*f*the slope must be the mean of the two slopes, or –30 dB/dec. Likewise, past

_{p2}*f*, ⎪

_{p3}*a*⎪ rolls off at a rate of –60 dB/dec, and the slope right at

*f*must be –50 dB/dec.

_{p3}- In Figure
*2a*the ⎪*1/β*⎪ curve has a slope of 0 and intercepts the⎪_{1}*a*⎪ curve in the region where ⎪*a*⎪ rolls off at a rate of –20 dB/dec, so Eq. (5) gives ROC = 0 –(–20) = 20 dB/dec, and Eq. (6) gives*ɸ*≈ 180 – 4.5×20 = 90°._{m1} - The ⎪
*1/β*⎪ curve has a slope of 0 and intercepts the⎪_{2}*a*⎪ curve at*f*, where the⎪_{p2}*a*⎪ curve has a slope of –30 dB/dec, so Eqs. (5) and (6) give ROC = 0 –(–30) = 30 dB/dec, and*ɸ*≈ 180 – 4.5×30 = 45°._{m2} - The ⎪
*1/β*⎪ curve has a slope of 0 and intercepts the⎪_{3}*a*⎪ curve in the region where ⎪*a*⎪ has a slope of –40 dB/dec, so Eqs. (5) and (6) give ROC = 0 –(–40) = 40 dB/dec, and*ɸ*≈ 180 – 4.5×40 = 0°._{m3} - In Figure
*2b*the slope of the ⎪*1/β*⎪ curve changes from 0 dB/dec below its crossover frequency to +20 dB/dec above its crossover frequency, so right at the crossover its slope must be +10 dB/dec. Then, Eqs. (5) and (6) give ROC = +10 –(–20) = 30 dB/dec, and_{4}*ɸ*≈ 180 – 4.5×30 = 45°._{m4} - The slope of the ⎪
*1/β*⎪ curve at its crossover frequency is +20 dB/dec, so Eqs. (5) and (6) give ROC = +20 –(–20) = 40 dB/dec, and_{5}*ɸ*≈ 180 – 4.5×40 = 0°._{m5} - The slope of the ⎪
*1/β*⎪ curve changes from +20 dB/dec immediately below its crossover frequency to 0 dB/dec above its crossover frequency, so right at the crossover its slope must be +10 dB/dec. Then Eqs. (5) and (6) give ROC = +10 –(–20) = 30 dB/dec, and_{6}*ɸ*≈ 180 – 4.5×30 = 45°._{m6} - The slope of the ⎪
*1/β*⎪ curve changes from 0 below its crossover frequency to –20 dB/dec above its crossover frequency, so right at the crossover its slope must be +10 dB/dec. Moreover, the slope of the⎪_{7}*a*⎪ curve is –40 dB/dec, so Eqs. (5) and (6) give ROC = –10 –(–40) = 30 dB/dec, and*ɸ*≈ 180 – 4.5×30 = 45°._{m7} - The slope of the ⎪
*1/β*⎪ curve at its crossover frequency is –20 dB/dec, and that of the⎪_{8}*a*⎪ curve is –40 dB/dec, so Eqs. (5) and (6) give ROC = –20 –(–40) = 20 dB/dec, and*ɸ*≈ 180 – 4.5×20 = 90°. It is interesting that even though crossover occurs in the region where ⎪_{m8}*a*⎪ rolls off at the steep rate of –40 dB/dec, just like the ⎪*1/β*⎪ curve does, the –20 dB/dec slope of the ⎪_{3}*1/β*⎪ curve ameliorates the phase margin dramatically compared to the ⎪_{8}*1/β*⎪ case._{3}

It pays to regard ROC as the *angle* between the ⎪*1/β*⎪and ⎪*a*⎪curves at the point where they intersect. As this angle becomes narrower, the circuit becomes more stable. Conversely, the wider this angle is, the closer the circuit is to instability.

The phase-margin estimate provided by the ROC method can be fairly good if all pole and zero frequencies are at *least a decade* away from the crossover frequency. Even if this is not the case, the ROC method still provides a reasonable starting point. After gaining enough experience, a designer will be able to figure out an improved estimate for the phase margin.

### A Real-Life Example of Rate-of-Closure

Let us apply the above considerations to a real-life example. Suppose we want to design a high-precision amplifier with (*a*) a DC closed-loop gain of 10^{3} V/V (or 60 dB), and (*b*) a DC loop gain T_{0} of at least 10^{6}. All we have at hand are op amps having a DC gain of 10^{5} V/V (or 100 dB) and a 1 MHz constant gain-bandwidth product (i.e., a transition frequency of f_{t} = 1 MHz).

After drawing the ⎪*a*⎪ and ⎪*1/β*⎪ curves as in Figure *3b*, we note that with a *single* op-amp we’d have a stable circuit (its situation is similar to the⎪ *1/β1*⎪ case of Figure *2a*), but with *T _{0} *= 100 – 60 = 40 dB, or 10

^{2}, which is insufficient. To jazz up

*T*let’s use

_{0}*two*op-amps in cascade, for an overall gain of

*a×a = a*. Now we do meet the

^{2}*T*requirement (

_{0}*T*= 200 – 60 = 140 dB, or 10

_{0}^{7}>10

^{6}), but the slope of |

*a*| at the crossover frequency is –40 dB/dec, indicating that we are in the situation of the ⎪

^{2}*1/β*⎪ case of Figure

_{3}*2a*, with

*ɸ*≈ 0°. We need to stabilize the circuit by suitably

_{m}

*(a) (b)*

**Figure 3.** (a) Composite amplifier intended for a DC closed-loop gain of 60 dB with a DC loop gain T_{0} ≥ 10^{6}. (b) Bode diagram showing that the composite has ɸ_{m} ≈ 0° and thus needs frequency compensation. Click to enlarge.

**Figure 3.**(a) Composite amplifier intended for a DC closed-loop gain of 60 dB with a DC loop gain T

_{0}≥ 10

^{6}. (b) Bode diagram showing that the composite has ɸ

_{m}≈ 0° and thus needs frequency compensation. Click to enlarge.

**Figure 4. **PSpice circuit to simulate the amplifier of Figure 3a.

**Figure 4.**PSpice circuit to simulate the amplifier of Figure 3a.

modifying its ⎪*1/β*⎪ curve in the vicinity of the crossover frequency* f _{x}*, which in our example can be seen to be the geometric mean of 1 kHz and 1 MHz, or

*f*= (10

_{x}^{3}×10

^{6})

^{1/2}= 31.6 kHz. The ⎪

*1/β*⎪ case of Figure

_{7}*2b*suggests that for

*ɸ*≈ 45° we need to bend the curve

_{m}*downwards*by establishing a breakpoint right at

*f*, a task that we accomplish by placing a capacitor

_{x}*C*in parallel with

_{f}*R*and imposing the condition

_{2}

*Equation (7)*

*Equation (7)*

Plugging in* f _{x}* = 31.6 kHz and

*R*= 99.9 kΩ gives

_{2}*C*= 50.36 pF. Running the PSpice circuit of Figure 4, we get the traces of Figure

_{f}*5a*, where we measure

*f*≈ 40.2 kHz and

_{x}*ɸ*≈ 51.6°. The ac response

_{m}*A = V*exhibits some peaking, and the transient response (not shown but easily verifiable) exhibits some ringing.

_{o}/V_{i}

*(a) (b)*

**Figure 5.** Plots of |a^{2}|, ⎪1/β⎪, and the closed-loop gain |A| for (a) C_{f} = 50.36 pF, and (b) C_{f} = 283.3 pF.

**Figure 5.**Plots of |a

^{2}|, ⎪1/β⎪, and the closed-loop gain |A| for (a) C

_{f}= 50.36 pF, and (b) C

_{f}= 283.3 pF.

To avoid peaking and ringing, let’s establish a situation of the type of the ⎪*1/β _{8}*⎪ case of Figure

*2b*. We achieve this by

*shifting*our ⎪

*1/β*⎪ curve

*leftwards*until it intercepts the |

*a*| curve at the point where |

^{2}*a*| drops to 30 dB, or 31.6. This occurs at a new crossover frequency of

^{2}*f*= (31.6×10

_{x}^{3}×10

^{6})

^{1/2}= 178 kHz. To find the

*new*breakpoint

*f*of the ⎪

_{0}*1/β*⎪ curve, we exploit the constancy of the gain-bandwidth product on the slanted portion of the ⎪

*1/β*⎪ curve and impose 31.6×178×10

^{3}= 10

^{3}×

*f*. This gives

_{0}*f*= 5.623 kHz, so we now need

_{0}*C*= 1/(2π × 99.9 × 10

_{f}^{3 }× 5.623 × 10

^{3}) = 283.3 pF.

Rerunning the PSpice circuit of Figure 4, but with this new value of *C _{f}*, we get the plots of Figure

*5b*, where we measure

*f*≈ 177.8 kHz and

_{x}*ɸ*≈ 86.4°. We got rid of peaking and ringing, but at the price of a lower closed-loop bandwidth of about 5.8 kHz.

_{m}### Conclusion

This article has presented a graphical, frequency-domain method that provides a straightforward way to assess the stability of a negative-feedback system. By examining the intersection of the curves representing the open-loop gain and the inverse of the feedback factor *β*, we can determine the rate of closure, and from the rate of closure we can calculate an approximate phase margin.

As you’re reviewing this material, give particular attention to Fig. 2, because most circuits encountered in practice conform to one of the cases depicted in the figure, at least in the proximity of the crossover frequency.

4 CommentsFor Fig 3 and Fig 4, how the 1/beta is bend downward? The added Cf not only introduces a pole of 1/2/pi/Cf/R2 but also a zero of 1/2/pi/Cf/(R2//R1) for 1/beta. So the final curve should be like 1/beta6 while a^2 has a slope of -40dB/decade. There should be 180-4.5*(10-(-40))=-45 degree, no phase margin at all.