How can an engineer measure stability in an analog negative feedback circuit? For minimum phase circuits, you may consider using the closing rate.
In everyday analog design, engineers often need to quickly estimate the degree of stability (or lack thereof) of a negative feedback loop. A convenient tool that is applicable to minimum phase The circuits (so called because all their poles and zeros are in the left half of the complex plane) are the closing rate (ROC).
To prepare the background, consider the familiar block diagram in Figure 1st:
Figure 1. (a) Block diagram of a negative feedback circuit, and (b) display of the loop gain T.
This diagram consists of a error amplifier with profit a(jf), a feedback network with transfer function β (jf), and an adder block that generates the Smi error signal,
Collect and solve for Smi gives
where T = aβ it’s called the loop gain because any signal going into the amplifier and rotating clockwise around the loop will be amplified first a and then by β, for a total gain of aβ. Obviously bigger T leads smaller Smifor a given input Syo.
Rewriting as T = a / (1 / β), taking the logarithms, and multiplying by 20 to convert to decibels, gives
which indicates that we can visualize the decibel graph of |T| As the difference between the decibel plots of |a| and |1 / β|. This is represented in the figure 1 B for the case of an op amp with constant bandwidth gain and independent frequency β.
The frequency FX in which the two curves intersect, aptly called the crossover frequency, plays an important role in the stability of the circuit (or lack thereof). At this frequency, we have |T (jfX)| dB = 0, or |T (jfX) | = 1. If the phase ph
The[T (jfX)]ever reach –180 °, then we would have T (jfX) = –1, which, substituted in Eq. (2), indicates that Smi It would explode and lead to oscillation. (Note that even if you set Syo = 0, the intrinsic noise of the circuit would cause the accumulation of Smi.) To avoid oscillation, we must ensure that ph
The[T (jfX)]it is far from the dreaded value of –180 ° for a sufficient quantity, appropriately called phase margin φmeter,
φmeter = 180 ° + ph[T(jf[T(jfX)]
The phase margins of practical interest are φmeter = 90 °, φmeter = 65.5 ° (which marks the beginning of in peak hours in the answer to c), φmeter = 76.3 ° (which marks the start of buzz in the transient response), and φmeter = 45 ° (which we will see to lend itself to easy geometric visualization, although it results in a 2.4 dB peak and a 23% excess timbre).
In the situation represented in the figure. 1 B, 1 / β is a real number, so its phase is 0 ° and we have ph
The[T (jfX)]= ph
The[una(jfX)]≈ –90 °, because FX >> Fpag. Therefore, φmeter ≈ 180 ° + (–90 °) = 90 °. However, things are not always that attractive, so we turn to the more general cases in Figure 2 to estimate the phase margins of the various cases. For this we will use the closing rate (ROC).
What is the closing fee?
The rate of closure or ROC is defined as the difference between the slopes of the ⎪1 / βAnd of thea⎪ curves Right at the crossover frequency:
Once we know the ROC, we estimate the phase margin as
In Figure 2, we assume an error amplifier with a more general gain profile than the constant-gain bandwidth product type of Figure 1st. (For an easy drawing, we are using straight segments, although we know that in practice acute angles are rounded, like the |a| curve in figure 1 B.)
Figure 2. (a) Frequent phase margin situations with (a) feedback factor β (jf) independent of frequency and (b) dependent of frequency.
Now, past the first pole frequency. Fp1, ⎪a⎪ outputs at a rate of –20 dB / dec, and passes the second pole frequency Fp2 takes off at a rate of –40 dB / dec. On the right in Fp2 the slope must be the average of the two slopes, or –30 dB / dec. Likewise, past Fp3, ⎪a⎪ takes off at a rate of –60 dB / dec, and the slope to the right at Fp3it should be –50 dB / dec.
- In figure 2nd the ⎪1 / β1The curve has a slope of 0 and is intercepteda⎪ curve in the region where ⎪a⎪ takes off at a rate of –20 dB / dec, so Eq. (5) gives ROC = 0 – (- 20) = 20 dB / dec, and Eq. (6) gives φm1 ≈ 180 – 4.5 × 20 = 90 °.
- The ⎪1 / β2The curve has a slope of 0 and is intercepteda⎪ curve in Fp2, where heaThe curve has a slope of –30 dB / dec, so the equations. (5) and (6) give ROC = 0 – (- 30) = 30 dB / dec, and φm2 ≈ 180 – 4.5 × 30 = 45 °.
- The ⎪1 / β3The curve has a slope of 0 and is intercepteda⎪ curve in the region where ⎪a⎪ has a slope of –40 dB / dec, so equations. (5) and (6) give ROC = 0 – (- 40) = 40 dB / dec, and φm3 ≈ 180 – 4.5 × 40 = 0 °.
- In figure 2b the slope of the ⎪1 / β4The 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 should be +10 dB / dec. So, ecs. (5) and (6) give ROC = +10 – (- 20) = 30 dB / dec, and φm4 ≈ 180 – 4.5 × 30 = 45 °.
- The slope of the ⎪1 / β5The curve at its crossover frequency is +20 dB / dec, so equations. (5) and (6) give ROC = +20 – (- 20) = 40 dB / dec, and φm5 ≈ 180 – 4.5 × 40 = 0 °.
- The slope of the ⎪1 / β6The 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 should be +10 dB / dec. So the equations. (5) and (6) give ROC = +10 – (- 20) = 30 dB / dec, and φm6 ≈ 180 – 4.5 × 30 = 45 °.
- The slope of the ⎪1 / β7The changes curve changes from 0 below its crossover frequency to –20 dB / dec above its crossover frequency, so right at the crossover its slope should be +10 dB / dec. Furthermore, the slope of theaThe curve is –40 dB / dec, so the equations. (5) and (6) give ROC = –10 – (- 40) = 30 dB / dec, and φm7 ≈ 180 – 4.5 × 30 = 45 °.
- The slope of the ⎪1 / β8The curve of at at its crossover frequency is –20 dB / dec, and that of ⎪aThe curve is –40 dB / dec, so the equations. (5) and (6) give ROC = –20 – (- 40) = 20 dB / dec, and φm8 ≈ 180 – 4.5 × 20 = 90 °. It is interesting that although the crossing occurs in the region wherea⎪ drifts at a steep rate of –40 dB / dec, as does ⎪1 / β3The curve ⎪ makes, the slope of –20 dB / dec of the ⎪1 / β8Curve ⎪ improves the phase margin compared to ⎪1 / β3⎪ case.
ROC is worth considering 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. On the contrary, the wider this angle, the closer the circuit will be to the instability.
The phase margin estimate provided by the ROC method can be quite good if all the pole and zero frequencies are at less 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 calculate an improved estimate for the phase margin.
A real life example of the closing rate
Let’s apply the above considerations to a real life example. Suppose we want to design a high-precision amplifier with (a) a DC loop gain of 103 V / V (or 60 dB), and (second) a DC T0 loop gain of at least 106. All we have on hand are op amps that have a DC gain of 105 V / V (or 100 dB) and a constant gain bandwidth product of 1 MHz (that is, a transition frequency of ft = 1 MHz).
After drawing the ⎪a⎪ and ⎪1 / β⎪ curves as in the figure 3b, we note that with a single op-amp we would have a stable circuit (your situation is similar to ⎪ 1 / β1⎪ case figure 2nd), but with T0= 100 – 60 = 40 dB, or 102, which is insufficient. To make jazz T0 let’s use two cascaded op-amps, for an overall gain of a × a = a2. Now we meet him T0 requirementT0 = 200 – 60 = 140 dB, or 107> 106), but the slope of |a2| at the crossover frequency it is –40 dB / dec, which indicates that we are in the situation of ⎪1 / β3⎪ case figure 2nd, with φmeter ≈ 0 °. We need to stabilize the circuit properly.
Figure 3. (a) Composite amplifier intended for 60 dB DC closed-loop gain with DC loop gain T0 ≥ 106. (b) Bode plot showing that the compound has φ meter φ 0 ° and therefore needs compensation of frequency.
Figure 4. PSpice circuit to simulate the amplifier of figure 3a.
modifying your ⎪1 / β⎪ curve in the vicinity of the crossover frequency FX, which in our example can be seen as the geometric mean of 1 kHz and 1 MHz, or FX = (103 × 106) 1/2 = 31.6 kHz. The ⎪1 / β7⎪ case figure 2b suggests that for φmeter ≈ 45 ° we have to bend the curve down setting a breaking point right at FX, a task we perform by placing a capacitor. doFin parallel with R2 and imposing the condition
Plugging in FX = 31.6 kHz and R2= 99.9 kΩ gives doF = 50.36 pF. Running the PSpice circuit of Figure 4, we obtain the traces of Figure 5thwhere we measure FX ≈ 40.2 kHz and φmeter ≈ 51.6 °. The answer ac A = Vo / Vyo it exhibits some peaks, and the transient response (not shown but easily verifiable) exhibits some timbre.
Figure 5. Plots of | a2 |, ⎪1 / β⎪, and the closed-loop gain | A | for (a) CF = 50.36 pF, and (b) CF = 283.3 pF.
To avoid spikes and hums, let’s establish a situation of the type of1 / β8⎪ case figure 2b. We did this Changing our ⎪1 / β⎪ curve toward the left until it intercepts the |a2| curve at the point where |a2| drops to 30 dB, or 31.6. This occurs at a new crossover frequency of FX = (31.6 × 103 × 106) 1/2 = 178 kHz. To find the new breaking point F0 of the ⎪1 / βIn the curve we, we exploit the constancy of the gain bandwidth product on the sloping part of the ⎪1 / β⎪ curve and impose 31.6 × 178 × 103 = 103 ×F0. Is all F0 = 5.623 kHz, so now we need doF = 1 / (2π × 99.9 × 103 × 5.623 × 103) = 283.3 pF.
Execution of the PSpice circuit of Figure 4, but with this new value of doF, we obtain the plots of the figure. 5bwhere we measure FX ≈ 177.8 kHz and φmeter ≈ 86.4 °. We got rid of the peaks and tones, but at the price of a lower closed-loop bandwidth of about 5.8 kHz.
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