frequency dependence and step responsebrew/ece304spr07/pdf/step response.pdf · bode phase plot...

ECE 304: Feedback, Frequency Response and Step Response

Frequency response So far we have looked at feedback for the case of frequencies at midband, where the open-loop gain has no frequency dependence. However, the open-loop gain does have frequency dependence at frequencies above midband. We have to deal with this dependence in order to tailor the transient step response of the feedback amplifier, and to insure that the amplifier is stable.

Single-pole response The simplest case is an open loop gain with a single pole, as shown in EQ. 1 below. EQ. 1

( )

1

0

ffj1

AfA+

= .

We know when this amplifier is put into a feedback amplifier, the closed loop gain is given by EQ. 2

β++

β+=

β++=

β+=

10FB0FB

0

0FB1

0FB

FB

f)A1(fj1

1A1

A

Affj1

A)f(A1

)f(A)f(A .

EQ. 2 shows that the corner frequency of the FB amplifier is moved out from f1 to a new value f1FB given by EQ. 3

f1FB = (1+βFBA0) f1 = PF × f1,

where PF = performance factor. That is, FB increases the bandwidth. The trade-off is that the gain has decreased from A0 (when there is no feedback) to A0/PF. Example Select the feedback resistor RF in the CE amplifier below to double the bandwidth, and sketch the Bode plots for the amplifier with and without feedback.

VP

OUT

VP

IN

I_DC

+

10uF

0

0

.model Q_n NPN (Is=I_S Bf=B_F Cjc=C_JC Cje=C_JE Tf=T_F)

0

+

10_F

15.000V

Sweep

+

-

ACV_AC

1V

+

R_F

PARAMETERS:

B_F = 100I_S = 10fAC_JC = 2pFC_JE = 2pFT_F = 400ps

TSF = 1msVSF = -5/126

FIRST_NPAIRS = 0,0, 0.01u,1, 100,1

V_PWL_ENH

-50.000mV

PARAMETERS:R_F = 100MegR_C = 1k

I_DC = 10.1mAR_S = 500

V_CC = 15V

+

R_S

+

-

V_CC

10.000mA +

R_C

5.0000V

Q_nQ1

10.000mA

-764.65mV

0

FIGURE 1

CE amplifier with shunt-shunt feedback resistor RF A PSPICE solution to sizing RF is shown in Figure 2. The corresponding change in step response is shown in Figure 3. The wider bandwidth corresponds to a faster step response.

Unpublished work © 4/19/05 J R Brews 5/7/2005

R_F

1.0K 10K 100K 1.0M 10M 100MV(OUT)

0V

100V

200V

(67.119K,65.863)

(100.000M,131.726)

no FB (open loop) FIGURE 2

At RF = 67.12 kΩ the gain of the small-signal gain of the amplifier is half its value at large RF (open-loop)

Time

0s 100ns 200ns 300ns 400ns 500nsV(OUT)

4V

6V

8V

10V

(42.052n,8.0000)

(78.927n,8.0000)

FIGURE 3

The rise time is approximately halved by the feedback of Figure 2

Frequency

1.0KHz 10KHz 100KHz 1.0MHz 10MHz 100MHzdB(V(OUT))

25.0

37.5

50.0

(1.6738M,39.393)

(3.3402M,33.381)

FIGURE 4

Bandwidth is doubled at RF = 67.12k

Frequency

1.0Hz 100Hz 10KHz 1.0MHz 100MHz 10GHz 1.0THzP(V(OUT))

-500d

0d

500d(3.3085M,135.000)

(1.6678M,-225.000)

FIGURE 5

Bode phase plot showing poles at approximately the same locations as Figure 4


3.38E+06

1.64E+06

-360-315-270-225-180-135

1.E+05 1.E+06 1.E+07 1.E+08 1.E+09 1.E+10Frequency (Hz)

Phas

e (d

eg)

RF=67.12kRF=100M

FIGURE 6

Bode phase plot with confusing extra 360° subtracted out; frequency values are a bit different from Figure 5 because no interpolation between data points was used here

The PSPICE phase plot has the disadvantage that PSPICE has added 360° to the phase for one of the choices for RF. Of course, a 360° phase shift is a full circle, and is equivalent to the phase without this addition, but it confuses the plot. In Figure 6 the same data is plotted without this 360°. We can see in Figure 6 that there are other poles in this amplifier, higher in frequency than the lowest pole, which is the one we have shifted. However, the gain drops off at 20 dB/decade above the corner and, so far as the corner frequency is concerned, this is a one-pole amplifier design problem.

Two-pole frequency and step response The previous example shows even this simple amplifier has multiple poles. Let’s go to a two-pole system where the open-loop gain is given as EQ. 4

( )( )210

j1j1A)f(A

ωτ+ωτ+= .

AGENDA The two-pole system is easy to analyze, and is used to find the pole positions necessary to get good step response from the amplifier. A real amplifier has more poles, but we still use the two-pole system that approximates the real amplifier to design the real system. The design consists of setting the poles the correct distance apart. This pole separation is found here using the two-pole model.

ANALYSIS: FINDING THE STEP RESPONSE USING LAPLACE TRANSFORM OF THE TRANSFER FUNCTION In EQ. 4, τ1 and τ2 are the two time constants for the circuit. Plugging this open-loop gain into the gain expression for the feedback amplifier, we find the frequency dependence of the FB amp gain is given by EQ. 5 below. EQ. 5

( )( )

+ττ

ω++

τ+τω++

=+ωτ+ωτ+

=

0FB212

0FB210FB

00FB21

0FB

Aβ1)j(

Aβ1j1

1Aβ1

AAβj1j1

A)f(A .

The step response of a system with the above gain function can be understood by switching to the Laplace parameter s = jω. The denominator of EQ. 5 in terms of s becomes EQ. 6

0FB212

0FB21

0FB212

0FB21

Aβ1s

Aβ1s1

Aβ1)j(

Aβ1j1

+ττ

++

τ+τ+→

+ττ

ω++

τ+τω+ .

To obtain the step response, we have to factor this expression. That is, we want to express the denominator as EQ. 7 below.


EQ. 7

( )( )

+ττ

−+++=

+ττ

+

τττ+τ

+ττ

+=

+ττ

++

τ+τ+

0FB21

0FB212

2121

210FB

0FB212

0FB21

Aβ1jbasjbas

Aβ1ssAβ1

Aβ1s

Aβ1s1

Once we have the denominator in the form of EQ. 7, the inverse Laplace transform is easy. To find the values of a and b in EQ. 7, we use the formula for the zeros of the quadratic in s, namely, EQ. 8

2/1

210FB

2

2121

Aβ114111

21s

ττ−

τ

−τ

±

τ

+τ

−= .

The interesting thing about this expression is that for large enough feedback βFB, the argument of the square root becomes negative, so the square root itself becomes imaginary and EQ. 8 becomes EQ. 9

2/12

21210FB

21

1141Aβj11

21s

τ−

τ−

ττ±

τ+

τ−= ,

= –a ± j b. Therefore, for big enough βFB, the denominator factors as (s+a+jb) ( s+a-jb). The inverse transform of such a denominator involves a damping factor times oscillatory sin and cos functions. The unit step response from Laplace inversion of EQ. 5 using EQ. 9 is (see Kuo1) EQ. 10

VOUT (t) =

ζζ−

+ζ−ωζ−

τ+

τ−

−+

2/122/12

N2/1221

0FB0 )1(tanat)1(sin

)1(

t1121exp

1Aβ1

A ,

where the parameter ζ is called the damping factor and is defined as EQ. 11

2/1

210FB21

)A1(1

2

ττβ+

τ+τ=ζ ,

while ωN is called the natural frequency of the system, and is defined as EQ. 12

τ+

τζ=

ττ+

=ω21

2/1

210FB

N11

21Aβ1 .

Although called the damping factor, ζ does not determine the exponential damping factor in EQ. 10. Parameter ζ depends on βFBA0, but the exponential damping factor depends only on the time constants τ1 and τ2, and depends on neither the feedback factor βFB nor the gain A0.

Although not determining the leading exponential damping factor, the parameter ζ does determine the amount of overshoot and ringing in the step response. These terms are illustrated shortly (Figure 7 - Figure 10).

In terms of ζ and ωN, the denominator of EQ. 5 is given by EQ. 13 below.

1 Benjamin C. Kuo, Automatic Control Systems, Fifth Edition, Prentice-Hall, 1987, pp. 314-316 and pp. 591-596.


EQ. 13

2

NN0FB212

0FB21 j2j1

Aβ1)j(

Aβ1j1

ωω

+ζ

ωω

+=+

ττω+

+τ+τ

ω+

Because EQ. 13 depends on only ζ and ωN, it is clear that all the properties of this two-pole system can be described using ζ and ωN. Defining the normalized frequency variable u as u = ω/ωN, we can find the closed loop gain M(u) as EQ. 14

M(u) = AFB =

−ζ+

β+ 20FB0

uuj211

A1A

,

and the magnitude of the closed loop gain as: EQ. 15

M(u) = |AFB | =

( )2/1

2220FB0

u2u1

1A1

A

ζ+

−

β+ .

Using EQ. 15, the closed-loop bandwidth frequency ωBW as defined by the gain dropping by 3dB is found as EQ. 16

2/12/1242

21

2/12/1242NBW 2442111

2124421

+ζ−ζ+ζ−

τ+

τζ=

+ζ−ζ+ζ−ω=ω .

EXAMPLE SIMULATIONS OF STEP RESPONSE

Often we don’t want a step response with oscillatory time dependence. So why not keep βFBA0 small enough that the square root in EQ. 8 is always a real number? The reason is that such a step response is very slow. So we want to live dangerously and allow designs where oscillatory response is possible, but not so bad that it actually is counterproductive. We undertake to determine how big ζ can be, and to design for that case. This decision can be done in either the frequency or the time domain if we understand the relation between these two approaches. Below are some example time domain step-responses.

6.28 2.08

0

12

3

4

0 1 2 3 4 5 6 7Time (µs)

Step

Res

pons

e

8

T_1=1000us T_2=1us Zeta=0.7071

Damping factor

FIGURE 7 Step response of two-pole system with τ1 = 1 ms and τ2 = 1 µs - this is the “Butterworth” case with ζ=1/ √2; the amplifier uses βFB = 0.5 and A0 = 103 V/V; time at maximum is labeled

Figure 7 shows the choice that we will examine further, with ζ=1/√2. This response exhibits only a small overshoot above the final value of 2 V/V and a fairly fast rise time. Any faster rise is


associated with ringing of the response, that is, oscillation around the final value, as shown below.

1.44

0

1

2

3

4

0 1 2 3 4 5 6 7Time (µs)

Step

Res

pons

e

8

T_1=100us T_2=1us Zeta=0.2256

FIGURE 8

Step response of same two-pole system with τ1 = 100 µs

0.45

0

1

2

3

4

0 1 2 3 4 5 6 7Time (µs)

Step

Res

pons

e

8

T_1=10us T_2=1us Zeta=0.0777

FIGURE 9

Step response of same two-pole system with τ1 = 10 µs

0.14

0

1

2

3

4

0 1 2 3 4 5 6 7Time (µs)

Step

Res

pons

e

8

T_1=1us T_2=1us Zeta=0.0447

FIGURE 10

Step response of two-pole system with τ1 = τ2

= 1µs Figure 7 - Figure 10 show step responses of the system for various values of ζ. The value of ζ was changed by moving the pole corresponding to τ1 while τ2 = 1 µs. The thing to notice is that the closer τ1 comes to τ2, the more rapid the ringing, the larger the maximum deviation from the long-time asymptotic value of 2 V/V, but the faster it settles to the final value.2

2 By the way, if the amplifier output swing was designed based upon the steady-state output, ringing may mean that transistors are driven out of the active mode, and the amplifier will not work as predicted from a linear model.


FREQUENCY DOMAIN RESPONSE OF TWO-POLE SYSTEM The magnitude plots of the gain also show peaking, this time as a function of frequency, with the peaks occurring somewhat below the corner frequency of the amplifier. Figure 11 shows the case ζ=1/√2, which corresponds to the largest ζ-value for which there is no peak in the frequency domain. This condition is referred to as “maximally flat”, or as the “Butterworth” condition, often used for filter design.3

1.13E+05

00.5

11.5

22.5

1.0E+04 1.0E+05 1.0E+06 1.0E+07Frequency (Hz)

Mag

nitu

de o

f gai

n

Zeta=0.7071T_1=1000us T_2=1us

FIGURE 11

Magnitude of gain of two-pole system with τ1 = 1 ms and τ2 = 1 µs - this is the “Butterworth” case with ζ=1/ √2; the amplifier uses βFB = 0.5 and A0 = 103 V/V; 3 dB frequency is labeled

Below are the gain plots for the other pole positions looked at for the step response.

5.33E+05

012345


Mag

nitu

de o

f gai

n

Zeta=0.2256T_1=100us T_2=1us

1.74E+06

0

5

10

15


Mag

nitu

de o

f gai

n

Zeta=0.0777T_1=10us T_2=1us

FIGURE 12

More magnitude plots corresponding to the step response curves already shown; the 3dB corner frequency is labeled

3 For example, see Chapter 6: Butterworth Lowpass Filters in M.E. Van Valkenburg, Analog Filter Design, pp. 157-167, Holt, Rinehart and Winston, 1982.


5.53E+06

05

10152025


Mag

nitu

de o

f gai

n

Zeta=0.0447T_1=1us T_2=1us

FIGURE 13

Remaining magnitude plot corresponding to the step response curves already shown Comparing Figure 11-Figure 13 with the step responses of Figure 7-Figure 10, we see that higher peaks in the frequency domain translate into more ringing and overshoot in the step response. So we can design for less ringing and overshoot by reducing peaking in the f-domain, or by direct reduction of the ringing in the time domain. The underlying design choice is the pole positions.

FINDING THE TIME CONSTANT RATIO FOR ζ=1/√2: POLE POSITIONING Every design has its own requirements. As a general-use amplifier, where we aren’t too

sure what it will be used for, the ζ=1/√2 design is a good compromise, so let’s look at that case. The bandwidth is then, for ζ=1/√2, from EQ. 16, EQ. 17

N

2/12/1242NBW 24421 ω=

+ζ−ζ+ζ−ω=ω (ζ=1/√2) =

τ+

τ 212/111

2

1 .

For ζ = 1/√2, we now show that the time constants are widely separated. We find from EQ. 11 EQ. 18

( ) 2/10FB

2/1

12

2/1

21

2/1

210FB21

A1

121

)A1(1

2 β+

ττ

+

ττ

=

ττβ+

τ+τ=ζ .

Let the time-constant ratio K be defined as EQ. 19

2/1

21K

ττ

≡

with τ1 the larger, longer time, so K ≥1. Ratio K tells us how far the poles are apart. Then EQ. 18 is a quadratic equation for K, namely: EQ. 20

( ) 01Aβ1K2K 2/10FB

2 =++ζ− ,

with the solution EQ. 21

( )2/1

0FB222/1

0FB A1A1K

βζ+−ζ±β+ζ= .

Setting ζ=1/√2, we find from EQ. 21 EQ. 22

( ) ( )

−β±β+= 2/1

0FB2/1

0FB2/11AA1

2

1K .

For the case τ1 > τ2, K >1, so we take the plus sign. As βFBA0 is normally much larger than 1, we have approximately the time constant ratio of EQ. 23 below.


EQ. 23

0FB2

21 A2K β≈=

ττ ,

As advertised, EQ. 23 says the poles are far apart. EQ. 23 agrees with the time constant ratio in Figure 7.

Using the open-loop Bode plot to set the pole positions An easy method to design the amplifier for a particular step response is to use the Bode plots for the open-loop amplifier. The open-loop Bode plots are easy to sketch using the open-loop gain function, for example, EQ. 4. To use these plots we need to know how to connect the gain and phase of the closed-loop amplifier to the open-loop Bode plots. For example, how do we know when the Bode plots of the open loop amplifier correspond to ζ=1/√2? The simple answer is: if the poles are spaced according to EQ. 23. Gain plot and location of the corner frequency: For large βFBA0, the closed loop amplifier gain is 1/βFB at frequencies below the closed loop corner frequency. As frequency increases, feedback amplifier gain will stay at 1/βFB as long as the closed loop gain is large enough to keep A(f)/(1+βFBA(f)) ≈ 1/βFB. It’s clear that this large-gain requirement for 1/βFB behavior will collapse for frequencies large enough that βFBA(f) << 1. For these frequencies, A(f)/(1+βFBA(f)) ≈ A(f) because βFBA(f) << 1. Then the feedback amplifier and the open-loop amplifier will have the same behavior. The transition from βFBA(f) >> 1 to βFBA0 << 1 might be said to be the frequency where βFBA(f) = 1. On that basis, we tentatively assign the frequency f1/β defined by βFBA(f1/β) = 1 as the corner frequency of the feedback amplifier. An example is shown in Figure 14.

Frequency

1.0Hz 100Hz 10KHz 1.0MHz 100MHz 10GHzDB(V(OUT))

-50

0

50

100

(Corner,66.6548M,58.2931)

(f1/B,52.7129M,61.3031)

(1.00000,61.3031)

FIGURE 14

Example comparison of gain curves for an open-loop amplifier (top) and a closed-loop feedback amplifier (bottom); the corner of the feedback amplifier is at 66.65MHz, while the frequency f1/β where the open-loop amplifier gain is the same as the zero-frequency feedback amplifier gain is f1/β = 52.71 MHz

Behavior like that in Figure 14 suggests that the bandwidth of the feedback amplifier can be estimated by drawing a Bode plot for the open-loop amplifier and finding where this gain plot crosses the line 1/βFB. For a Butterworth step response, we take the frequency at this crossing as ωBW. Then we adjust the design (set the lowest pole) to make the spacing of the poles satisfy EQ. 23, or to make the phase φ(ωBW) = –125.3° (see EQ. 38).

Practical recipe for design In fact, because it is easier, we usually will locate the corner frequency at the location of the second pole, that is ωBW = 2πf2, and take φ(ωBW) = –135°, which is approximately the phase of the


open-loop amplifier at the location of the second pole φ(f2) when poles are widely spaced. This choice leads to ζ = 0.583.

Plots of the closed-loop step response and gain for ζ=0.583 are shown in Figure 15 - Figure 16.

4.51 2.20

0

12

3

4

0 1 2 3 4 5 6 7Time (µs)

Step

Res

pons

e

8

T_1=681.17us T_2=1us Zeta=0.5839

FIGURE 15

Closed-loop step response for open loop phase at ωBW of –135°

1.60E+05

00.5

11.5

22.5


Mag

nitu

de o

f gai

n

Zeta=0.5839

T_1=681.17us T_2=1us

FIGURE 16

Closed-loop gain plot for the case of open loop phase at ωBW of –135°

Below are some example Bode plots for an amplifier designed for ζ = 0.5839, or an open-loop phase of –135° at the corner frequency of the closed-loop amplifier.

Frequency

1.0Hz 100Hz 10KHz 1.0MHz 100MHzVDB(OUT)

-100

0

100(233.6342Hz,56.99dB)(60dB)

(f_1/B=99.3144KHz,6.00325dB)

(fBW=159.6103KHz,2.99325dB)

(6.00325dB)

FIGURE 17

Bode gain plots for open-loop amplifier (top) and closed loop amplifier with ζ=0.5839 Figure 17 compares the Bode gain plots for the open-loop amplifier (top) and the closed-loop amplifier. The closed loop amplifier has a gain of 1/βFB = 6 dB. The 1/βFB = 6 dB line crosses the


open-loop gain curve at f1/β = 99 kHz , rather roughly locating the open-loop corner frequency, which is located at fBW = 159.6 kHz. Later, we look carefully at the relation between f1/β and fBW.

Frequency

1.0Hz 100Hz 10KHz 1.0MHz 100MHzP(V(OUT))

-200d

-100d

0d

(232.9731,-45.0000)

(f2=159.1550KHz,-134.9141)

(f_1/B=99.3144KHz,-121.8290)

(fBW=159.6103KHz,-105.0491)

(fBW=159.6103KHz,-134.9961)

Open loop amplifier

FIGURE 18 Bode phase plot for both amplifiers; care must be taken to include enough frequency points to get accurate phases 101 points/decade works OK

Figure 18 compares the phase plots of the two amplifiers. As is typical, the phase of the closed-loop amplifier drops very steeply near the corner frequency 159.6 kHz from 0° to –180°, passing through –90° at ω=ωΝ (see EQ. 14), which is not ωBW unless ζ=1/√2 (see EQ. 16). The phase of the open-loop amplifier is –135° at fBW, as expected. The slight departure of fBW from f2 results from the poles being too close together to treat entirely separately. Figure 17 and Figure 18 show that the low-frequency pole is properly located to get φ (fBW) = –135°, that is, ζ = 0.5839.

Example We are given an open-loop amplifier with poles at f1 = 105 Hz and f2 = 106 Hz. We wish to construct a feedback amplifier with a closed loop gain of 2 V/V, and a step response corresponding to ζ = 0.5839 (that is, φ = –135°at corner frequency). What should be the new pole frequency for f1?

1/βFB (dB)

FIGURE 19 Bode gain and phase plots for example


Answer: The intersection of the 1/βFB line with the open-loop gain shows the corner frequency of the closed loop amp is presently 7.04 MHz. We want the open-loop phase at this frequency to be –135°, but it is much lower: –171.1°. At the location of the second pole the open-loop phase is –129.3°, a bit higher than the value –135° expected for widely separated poles. If we move the first pole lower in frequency, it will be further from the second pole, and the phase there will drop to –135°. We want to move the first pole low enough that the 1/βFB line intersects the open-loop gain curve at the second pole frequency of f2 = 106 Hz. In that way , the closed loop amp will have a corner frequency of 106 Hz, and the open-loop amp will have a phase of –135° at this frequency. These conditions correspond to ζ = 0.5839.

GRAPHICAL METHOD We could do this by trial and error. However, we can use a graphical method. We place a

20dB/decade line at f2 = 106 Hz on the 1/βFB-line, as shown in Figure 20. If the first pole is moved to the intersection of the 20dB-line with A0, the 20dB-line imitates the Bode plot with the pole at the new position. So, we would expect moving the first pole to this point will result in a Bode plot that follows our 20 dB line.

f2

f1 (new)

FIGURE 20 Placing a 20 dB/dec line to imitate the Bode plot with the moved pole

20 dB/dec

54 dB

54/20 =2.7 dec Because A0 = 60 dB and 1/βFB = 6 dB, our line rises 54 dB at a slope of 20 dB/dec. That is, 20dB/dec = slope = 54dB/ number of decades in frequency → number of decades = 54/20 = 2.7 dec, or f2/f1 = 102.7, f1 = f2/102.7 = 1995 Hz.


FIGURE 21

Calculated new pole frequency of 1995 Hz at intersection of 20 dB/dec line with A0

FIGURE 22

Bode plots with pole moved to new position at 1995 Hz; phase at 106Hz is φ = –135° as wished, but closed-loop 3dB corner is at 1.27 × 106 Hz, where φ = –141.7°, a bit off


FIGURE 23

Redesigned amplifier using GOAL SEEK in EXCEL; new pole is now at 1468 HZ; closed loop 3dB corner at 106 Hz, and open-loop phase φ = –135° at 106 Hz

Figure 22 shows the amplifier as designed, and Figure 23 shows the design using GOAL SEEK in EXCEL. The approximate procedure locates the new pole within about 35%. The approximate method uses a straight-line 1/βFB approximation to the closed loop gain instead of the actual closed loop gain curve M(u) from EQ. 15.

ALGEBRAIC METHOD BASED ON TWO-POLE FORMULAS

An even faster approach that is completely unintuitive, is to use the pole ratio of EQ. 21 in EQ. 19. For ζ = 0.5839, EQ. 21 provides K=26.1. Then EQ. 19 provides τ1 = K2 τ2 = 108.4 µs, or f1 = 1/(2πτ1) = 1468 Hz. This is the same design shown in Figure 23.

This method is fast, but it depends heavily upon use of the two-pole model. In a real amplifier there are always more than two poles. A graphical method is more likely to keep your intuition active and keep you on top of what is happening with these poles when you move the lowest pole around.

Rise time to first maximum of step response and bandwidth The two-pole model provides a simple connection between the time domain and the frequency domain. Looking at the cases we want to design, where there is very little or no ringing, the step response approximately has reached its final value at the first maximum in the output response. We can find the time at the first maximum by setting the derivative of EQ. 10 to zero. This time is tMAX, EQ. 24

2/12N

MAX1

t

ζ−ω

π= .

However, the bandwidth is given by EQ. 16, repeated below:


EQ. 25

2/12/1242NBW 24421

+ζ−ζ+ζ−ω=ω .

Hence, the rise time is related to the bandwidth as EQ. 26

2/1

2

2/1242

BWMAX

1

24421t

ζ−

+ζ−ζ+ζ−

ωπ

= .

For example, for the Butterworth design ζ=1/√2 and ωBW = ωN. Using EQ. 12 for ωN, tMAX becomes EQ. 27

τ+

τ

π=

21

MAX 112t , (Butterworth design).

The utility of this estimate is that it puts a limit on how fast the step response can be for an amplifier with known τ1 and τ2. Because τ1 >> τ2 for a Butterworth design, the amplifier speed is limited by the faster time constant τ2, and tMAX ≈ 6.3 τ2. Finding the feedback amplifier bandwidth from open-loop Bode plots

Above we have used a simple method for finding the corner frequency ωBW of a feedback amplifier: namely, assume it is located where a horizontal line at the zero-frequency gain of the feedback amplifier intersects the open-loop Bode plot. Here we digress to find whether this method is accurate for a two-pole system.

If we substitute ω = ωBW, by definition, the closed-loop gain satisfies M = 1/√2 M(0). But what is the value of the open-loop gain at ωBW? Rewrite EQ. 16 as EQ. 28 below.

EQ. 28

)(hA124421A12/1

210FB

2/12/12422/1

210FB

BW ζ

ττβ+

=

+ζ−ζ+ζ−

ττβ+

=ω ,

where EQ. 12 for ωN was used and the square-root function is short-handed as h(ζ), that is, EQ. 29

2/12/1242 24421)(h

+ζ−ζ+ζ−≡ζ .

We evaluate the open loop gain at ω = ωBW as shown in EQ. 30 below. EQ. 30

( )( )( ) ( )

ττ

ζβ++

ττ

ζβ++

=τω+τω+

=ω

122/1

0FB212/1

0FB

02BW1BW

0BW

)(hA1j1)(hA1j1

Aj1j1

A)(A

.

Assume βFB A0 >> 1. Then, approximately, if βFBA0 >> 1, EQ. 30 becomes EQ. 31 below.


EQ. 31

( ) ( )2/1

122

0FB

2/1

212

0FB

0BW

)(hA1)(hA1

A)(A

ττ

ζβ+

ττ

ζβ+

=ω

( ) ( )

2/1

20FB

122/1

20FB

212

FB

)(hAβ/1

1

)(hAβ/1

1h

1

ζ

ττ+

ζ

ττ+

β=

)(

1FB ζαβ

= .

The second form is found by factoring out the nonunity term in both factors of the denominator. The parameter α for the case τ1 >> τ2 is, from EQ. 31, EQ. 32

2/1

20FB12Aβ

)(h)(h)(

ττ

+ζζ=ζα .

The phase of the open-loop gain from EQ. 30 is given by EQ. 33

( ) ( )

ττ

ζ−

ττ

ζ−=φ122/1

0FB212/1

0FB )(hAβtana)(hAβtana .

EQ. 31 suggests an adjustment of the simple method for finding the corner frequency of the closed loop amplifier using the Bode plot for the open-loop amplifier. Draw the open-loop Bode plot, then draw the line 1/(βFBα(ζ)) and find where it crosses the open-loop gain plot. That crossing is where ω = ωBW. How big is α? For example, in designs where we want a step response that is fast but with no oscillations around final value, we arrange that τ1 = 2βFBA0τ2 (a “maximally flat” or Butterworth op amp design, see EQ. 23). Then according to EQ. 11, ζ is given by EQ. 34

21

Aβ2)Aβ1(

12

)1Aβ2()A1(

12

2/1

220FB0FB

20FB2/1

210FB21 ≈

+

τ+=

ττβ+

τ+τ=ζ

τ,

where the approximation βFBA0 >> 1 is made. Then, according to EQ. 29, h(ζ) is given by EQ. 35 below. EQ. 35

124421)(h2/12/1242 =

+ζ−ζ+ζ−≡ζ .

The closed-loop corner frequency from EQ. 16 is now EQ. 36

22121BW

2111

21)(h11

21

τ≈

τ

+τ

=ζ

τ

+τζ

=ω , or 2

ff 2BW = .

That is, the closed-loop corner frequency is lower than the pole frequency f2. Substituting τ2/τ1 = 2 βFBA0 into EQ. 32, the gain of EQ. 31 is found in EQ. 37 below.


EQ. 37

FB2/1

2

2FB

BWβ31

)(h21

1)(hβ

1)(A =

ζ+

ζ=ω .

EQ. 37 suggests that, instead of plotting 1/βFB, we should plot the line 1/(1.73 βFB) on the open-loop Bode plot to find the bandwidth of the closed loop amplifier. The corresponding phase is found from EQ. 33 as EQ. 38

( ) ( ) oo 26.12521tana90

A2AtanaA2Atana

20FB22/1

0FB2

20FB2/10FB −=−−≈

τβτ

β−

ττβ

β−=φ .

As another common example, op amps are frequently designed to be “stable” 4, by which is meant that the open –loop gain drops at 20 dB/decade from the first pole all the way down to the frequency f1/β where A(f1/β) =1/βFB. For this behavior to be possible, there can be no pole frequency between the first one at f1 and the frequency f1/β. That is, the largest that f1 can be is f1 = f2/(βFBA0), or equivalently, τ1 = βFBA0τ2. See Figure 24.

12

0FB

12

10

0FB10

110210

FB10010

ffA

,so

;

ff

log

Alog20

)f(log)f(log)/1(log20Alog20

dec/dB20

=β

β=

−

β−=

20 log10 A0

20 log10 1/βFB

log10 f1 log10 f2

20 dB/dec

FIGURE 24

Slope calculation to determine frequency ratio from Bode plot; first pole located so f2 = f1/β

We choose f1 at this maximum value to get the most bandwidth possible for a “stable” design. According to EQ. 29, with τ1 = βFBA0τ2 the damping factor ζ is given by EQ. 39

21

Aβ)Aβ1(

12

)1Aβ()A1(

12

2/1

220FB0FB

20FB2/1

210FB21 ≈

+

τ+=

ττβ+

τ+τ=ζ

τ,

where the approximation βFBA0 >> 1 is made. Using this value for ζ, according to EQ. 29, h(ζ) is given by EQ. 40

.272.12

512141

21124421)(h

2/12/12/12/12/1242 =

+=

+−+−=

+ζ−ζ+ζ−≡ζ

The actual closed-loop corner frequency from EQ. 16 is now given by EQ. 41 below.

4 See Sedra and Smith, p. 849. This terminology mustn’t be misinterpreted as meaning the phase margin is zero; its closer to 45°. If the 20 dB/dec behavior is followed down to 0 dB (that is βFB=1) the terminology is “unconditionally stable” or “stable for unity β”.


EQ. 41

22121BW

272.1272.111)(h1121

τ≈

τ

+τ

=ζ

τ

+τζ

=ω , or f . 2BW f272.1=

That is, for τ1 = βFBA0τ2 the closed-loop corner frequency actually is higher than the second pole frequency. Substituting EQ. 40 into EQ. 31, we find the open-loop gain at fBW is given by EQ. 42 below. EQ. 42

FB2/1

2

2FB

BW 82.11

)(h11

1)(h

1)(Aβ

=

ζ+

ζβ=ω

EQ. 40 suggests we should plot the line 1/(1.82 βFB) on the open-loop Bode plot to find the bandwidth of the closed loop amplifier. The corresponding open-loop phase at fBW is found from EQ. 33 as EQ. 43

( ) ( ) .8.141272.1tana90A2

1hAtanaA

21hAtana

20FB22/1

0FB2

20FB2/10FB

oo −=−−≈

τβτ

β−

ττβ

β−=φ

In practice, an amplifier is likely to have more than two poles, and the fudge factor α will be different than for a two-pole system. Therefore, we usually just take α=1.

FIGURE 25

Comparison of closed-loop bandwidth fBW with frequency f2 of second pole for the “stable” and Butterworth designs; for the “stable” design fBW = 1.27 f2, and for the Butterworth design fBW = f2/1.41

FIGURE 26

Phase plots corresponding to Figure 25; phase at fBW agrees with EQ. 43 and EQ. 38 Figure 25 illustrates the behavior of the open- and closed-loop gains near the closed-loop

corner frequency fBW . It is clear that fBW switches from being greater than f2 (see EQ. 41) to being less than f2 (see EQ. 36) depending upon the choice of design. In Figure 25, τ2 = 1 µs and


τ1 = 500 µs (“stable” design) or τ1 = 1000µs (Butterworth design). Open-loop gain is A0 =1000 V/V and βFB = 0.5 V/V (1/βFB = 6 dB). Figure 26 shows the corresponding phase shifts. Bandwidth for the closed-loop amplifier is much lower for the Butterworth design, as expected because the Butterworth design has a slower step response with no ringing.

PSPICE plots To simulate the behavior of the two-pole system, the circuit of Figure 27 was used.

TSF = 1msVSF = 1V

FIRST_NPAIRS = 0,0, .1u,0, 0.11u,1, 100,1

+

C1C

+

C2C

VDB

0

0

-+

+-

E2

GAIN = 1

-+

+-

E1

GAIN = 1E3

B_FB

-+

+-

E4

GAIN = 1

+R1

R_1

-+

+-

E3

GAIN = 1

0

0

PARAMETERS:T_1 = 1000*T_2T_2 = 1E-6

R_1 = T_1/CR_2 = T_2/C

C = 10nF

B_FB = 0.5

+R2

R_2

Sweep

+

-

ACV21V

0

IN OUT

FIGURE 27 PSPICE circuit for simulating a two-pole amplifier with feedback βFB

In Figure 27 the feedback network is realized by two PSPICE E-parts (VCVS’s) and a GAIN part (Triangular Outline). This arrangement is equivalent to a single VCVS with gain βFB = B_FB. The complicated set-up allows a PARAM box to set the feedback βFB: E-parts cannot use gain variables, only numbers.

Summary Feedback controls step response and bandwidth. However, there is a trade-off between very slow response and a response with too much ringing and overshoot. These properties depend on the time constant ratio, τ1/τ2, with a more oscillatory response corresponding to closer spacing of the poles.

A common approach to determining the closed-loop bandwidth is plotting 1/βFB on the open-loop Bode plot and finding the frequency of intersection with the open-loop gain curve. For a two-pole system we’ve seen that this approach is approximately correct, although analysis suggests a “fudge factor” modifying 1/βFB to 1/(αβFB) would be more accurate. Usually it is not worth the trouble to use an α different from one, especially for a multiple pole system where there is no easy formula for α. The bottom line is that with α=1 we will not obtain a bandwidth exactly as specified, but will have to tweak the design using simulation. Nothing new about that, I guess.

For a two-pole system, an open-loop phase at fBW of φ(fBW) = –125.3° corresponds to the “maximally flat” or “Butterworth” design with damping factor ζ=1/√2 and τ1 = 2βFBA0 τ2. This design goal may not be suitable for every situation, but it serves as conservative design that exhibits the fastest step response without ringing.

A somewhat more adventurous design (the so-called “stable” design) selects the poles so the open-loop amplifier has a 20 dB/dec slope from the first pole down to the frequency f1/β defined by A(f1/β) = 1/βFB. The design of this type with largest bandwidth results if the first pole is located so f1/β = f2, which makes τ1 = βFBA0 τ2 . In a two-pole system, this choice makes the closed-loop corner frequency fBW = 1.27 f2 and the phase φ(fBW) = –141.8°.


A simple procedure that implements the “stable” design selects the intersection of the 1/βFB line with the open-loop Bode plot to occur at the frequency of the second pole. Then the first pole is placed at f1 = f2/(βFBA0) using graphical manipulations. Again, this position for f1 is the highest frequency allowed for f1 within the “stable” design scenario, providing the best bandwidth for this type of design. This design provides fast step response with only a little ringing. More precise ringing or settling-time specifications can be arranged by adjusting the first pole position f1 using PSPICE.

Appendix: Phase margin of two-pole system Step response determines the pole locations of the amplifier, which also determines its stability. Of course, a two-pole system is always stable because its phase always is > –180°. However, for a many-pole system approximated by a two-pole system, the phase margin calculated based upon the approximate two-pole system is an indicator of the stability of the system. So it is handy to see what the stability indications are for a desired step response.

The phase margin of a design is defined as EQ. 44

))f(Aarg(180 /1M β−=φ o ,

where f1/β is defined as EQ. 45

|A(f1/β)| = 1/βFB,

and the function arg denotes the angle of its argument. For a two-pole system given by EQ. 4, we find ω1/β using EQ. 46

( )( ) FB2/11/1

0β

1j1j1

A=

τω+τω+ ββ.

Explicitly finding the magnitudes we obtain EQ. 47

( ) ( 221

2/1

222

21

2/1

20FB 1Aβ τ+τω+

ττω−= ββ ) .

Solving for ω1/β we find EQ. 48

( ) 22

21

20FB

222

21

22

21

22

21

2/1 Aβ42 ττ+

τ−τ+

τ+τ−=ττω β .

EQ. 48 can be simplified using the assumption that τ1 >> τ2. We find EQ. 49

( )

τ

τ++−τ≈ττω β 2

1

222

0FB21

22

21

2/1 Aβ4112 ,

or, EQ. 50

2/12

120FB

2/1

Aβ21121

ττ

++−τ

=ω β .

For the case of a Butterworth design, τ1 = 2βFBA0τ2, and EQ. 50 provides ω1/β as EQ. 51 below.


EQ. 51

( )2/1

2

2/1

2/1 2

1211111

21

−

τ=++−

τ=ω β ; Butterworth.

For the case of a “stable” design, τ1 = βFBA0τ2, and EQ. 50 provides EQ. 52

( )2/1

2

2/1

2/1 2

15141121

−τ

=++−τ

=ω β ; Stable design.

The phase margin of the gain is then EQ. 53

( ) ( )

τ

τ++−−≈τω−τω−=φ ββ

2/12

120FB

2/11/1MAβ211

21tana90tanatana180 oo .

For the Butterworth design τ1 = 2 βFBA0τ2 and EQ. 53 provides a margin of EQ. 54

( ) ( ) oooooo 5.655.11418021

21tana90180tanatana180

2/1

2/11/1M =−=

−−−≈τω−τω−=φ ββ

and for the “stable” design τ1 = βFBA0τ2 and EQ. 53 provides a margin of EQ. 55

( ) ( ) oooooo 8.512.1281802

15tana90180tanatana1802/1

2/11/1M =−=

−−−≈τω−τω−=φ ββ .

From these results we see that both these designs have substantial phase margins. Whether they are adequate depends on other factors, for example, manufacturing variations in components and operating conditions.


frequency dependence and step responsebrew/ece304spr07/pdf/step response.pdf · bode phase plot...

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