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*:

*(a) (b)*

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

**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 *Î² (**jf**)*, and an adder block that generates the *Smi error signal*,

*Equation (1)*

*Equation (1)*

Collect and solve for *Smi* gives

*Equation (2)*

*Equation (2)*

where *T = aÎ²* 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 *Î²*, for a total gain of *aÎ²*. Obviously bigger *T* leads smaller *Smi*for a given input *Syo*.

Rewriting as *T = a / (1 / Î²)*, taking the logarithms, and multiplying by 20 to convert to decibels, gives

*Equation (3)*

*Equation (3)*

which indicates that we can visualize the decibel graph of |*T*| As the *difference* between the decibel plots of |*a*| and |*1 / Î²*|. This is represented in the figure *1 B* for the case of an op amp with constant bandwidth gain and independent frequency *Î²*.

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 â€“180 Â°, then we would have *T (**jfX**)* = â€“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 â€“180 Â° for a sufficient quantity, appropriately called *phase margin Ï†meter*,

### * Ï†meter = 180 Â° + ph[T(jf[T(jfX)] *

*Equation (4)*

*Equation (4)*

The phase margins of practical interest are *Ï†meter* = 90 Â°, *Ï†meter* = 65.5 Â° (which marks the beginning of *in peak hours* in the answer to c), *Ï†meter* = 76.3 Â° (which marks the start of *buzz* in the transient response), and *Ï†meter* = 45 Â° (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 / Î²* is a real number, so its phase is 0 Â° and we have ph

The[*T (**jfX**)*]= ph

The[*una(**jfX**)*]â‰ˆ â€“90 Â°, because *FX* >>* **Fpag*. Therefore, *Ï†meter* â‰ˆ 180 Â° + (â€“90 Â°) = 90 Â°. 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 âŽª*1 / Î²*And of the*a*âŽª 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 *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*.)

*(a) (b)*

**Figure 2.** (a) Frequent phase margin situations with (a) feedback factor Î² (jf) independent of frequency and (b) dependent of frequency.

**Figure 2.**(a) Frequent phase margin situations with (a) feedback factor Î² (jf) independent of frequency and (b) dependent of frequency.

Now, past the first pole frequency. *Fp1*, âŽª*a*âŽª outputs at a rate of â€“20 dB / dec, and passes the second pole frequency* Fp2* takes off at a rate of â€“40 dB / dec. On the right in *Fp2* the slope must be the average of the two slopes, or â€“30 dB / dec. Likewise, past* Fp3*, âŽª*a*âŽª takes off at a rate of â€“60 dB / dec, and the slope to the right at *Fp3*it should be â€“50 dB / dec.

- In figure
*2nd*the âŽª*1 / Î²1*The curve has a slope of 0 and is intercepted*a*âŽª curve in the region where âŽª*a*âŽª takes off at a rate of â€“20 dB / dec, so Eq. (5) gives ROC = 0 – (- 20) = 20 dB / dec, and Eq. (6) gives*Ï†m1*â‰ˆ 180 – 4.5 Ã— 20 = 90 Â°. - The âŽª
*1 / Î²2*The curve has a slope of 0 and is intercepted*a*âŽª curve in*Fp2*, where he*a*The curve has a slope of â€“30 dB / dec, so the equations. (5) and (6) give ROC = 0 – (- 30) = 30 dB / dec, and*Ï†m2*â‰ˆ 180 – 4.5 Ã— 30 = 45 Â°. - The âŽª
*1 / Î²3*The curve has a slope of 0 and is intercepted*a*âŽª curve in the region where âŽª*a*âŽª has a slope of â€“40 dB / dec, so equations. (5) and (6) give ROC = 0 – (- 40) = 40 dB / dec, and*Ï†m3*â‰ˆ 180 – 4.5 Ã— 40 = 0 Â°. - In figure
*2b*the slope of the âŽª*1 / Î²4*The 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*Ï†m4*â‰ˆ 180 – 4.5 Ã— 30 = 45 Â°. - The slope of the âŽª
*1 / Î²5*The curve at its crossover frequency is +20 dB / dec, so equations. (5) and (6) give ROC = +20 – (- 20) = 40 dB / dec, and*Ï†m5*â‰ˆ 180 – 4.5 Ã— 40 = 0 Â°. - The slope of the âŽª
*1 / Î²6*The 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*Ï†m6*â‰ˆ 180 – 4.5 Ã— 30 = 45 Â°. - The slope of the âŽª
*1 / Î²7*The changes curve changes from 0 below its crossover frequency to â€“20 dB / dec above its crossover frequency, so right at the crossover its slope should be +10 dB / dec. Furthermore, the slope of the*a*The curve is â€“40 dB / dec, so the equations. (5) and (6) give ROC = â€“10 – (- 40) = 30 dB / dec, and*Ï†m7*â‰ˆ 180 – 4.5 Ã— 30 = 45 Â°. - The slope of the âŽª
*1 / Î²8*The curve of at at its crossover frequency is â€“20 dB / dec, and that of âŽª*a*The curve is â€“40 dB / dec, so the equations. (5) and (6) give ROC = â€“20 – (- 40) = 20 dB / dec, and*Ï†m8*â‰ˆ 180 – 4.5 Ã— 20 = 90 Â°. It is interesting that although the crossing occurs in the region where*a*âŽª drifts at a steep rate of â€“40 dB / dec, as does âŽª*1 / Î²3*The curve âŽª makes, the slope of â€“20 dB / dec of the âŽª*1 / Î²8*Curve âŽª improves the phase margin compared to âŽª*1 / Î²3*âŽª case.

ROC is worth considering as the *angle* between the âŽª*1 / Î²*âŽª and âŽª*a*âŽªCurves 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 âŽª*a*âŽª and âŽª*1 / Î²*âŽª 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 âŽª *1 / Î²1*âŽª 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 Ã— a = a2*. Now we meet him *T0* requirement*T0* = 200 – 60 = 140 dB, or 107> 106), but the slope of |*a2*| at the crossover frequency it is â€“40 dB / dec, which indicates that we are in the situation of âŽª*1 / Î²3*âŽª case figure *2nd*, with *Ï†meter* â‰ˆ 0 Â°. We need to stabilize the circuit properly.

*(a) (b)*

**Figure 3.** (a) Composite amplifier intended for 60 dB DC closed-loop gain with DC loop gain T0 â‰¥ 106. (b) Bode plot showing that the compound has Ï† meter Ï† 0 Â° and therefore needs compensation of frequency.

**Figure 3.**(a) Composite amplifier intended for 60 dB DC closed-loop gain with DC loop gain T0 â‰¥ 106. (b) Bode plot showing that the compound has Ï† meter Ï† 0 Â° and therefore needs compensation of frequency.

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

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

modifying your âŽª*1 / Î²*âŽª 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 Ã— 106) 1/2 = 31.6 kHz. The âŽª*1 / Î²7*âŽª case figure *2b* suggests that for* Ï†meter* â‰ˆ 45 Â° we have to bend the curve *down* setting a breaking point right at *FX*, a task we perform by placing a capacitor. *doF*in parallel with *R2* and imposing the condition

*Equation (7)*

*Equation (7)*

Plugging in* FX* = 31.6 kHz and* R2*= 99.9 kÎ© gives *doF* = 50.36 pF. Running the PSpice circuit of Figure 4, we obtain the traces of Figure *5th*where we measure *FX* â‰ˆ 40.2 kHz and *Ï†meter* â‰ˆ 51.6 Â°. The answer ac *A = Vo / Vyo* it exhibits some peaks, and the transient response (not shown but easily verifiable) exhibits some timbre.

*(a) (b)*

**Figure 5.** Plots of | a2 |, âŽª1 / Î²âŽª, and the closed-loop gain | A | for (a) CF = 50.36 pF, and (b) CF = 283.3 pF.

**Figure 5.**Plots of | a2 |, âŽª1 / Î²âŽª, 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 of*1 / Î²8*âŽª case figure *2b*. We did this *Changing* our âŽª*1 / Î²*âŽª 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 Ã— 103 Ã— 106) 1/2 = 178 kHz. To find the *new* breaking point *F0* of the âŽª*1 / Î²*In the curve we, we exploit the constancy of the gain bandwidth product on the sloping part of the âŽª*1 / Î²*âŽª curve and impose 31.6 Ã— 178 Ã— 103 = 103 Ã—*F0*. Is all* F0* = 5.623 kHz, so now we need *doF* = 1 / (2Ï€ Ã— 99.9 Ã— 103 Ã— 5.623 Ã— 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. *5b*where we measure *FX* â‰ˆ 177.8 kHz and *Ï†meter* â‰ˆ 86.4 Â°. We got rid of the peaks and tones, but at the price of a lower closed-loop bandwidth of about 5.8 kHz.

In this article,