CS5127 datasheet(12/24 Pages) CHERRY

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Applications Information: continued

The zero frequency due to the output capacitor ESR is

given as

= 4.8 kHz.

The double pole frequency of the power output stage is

= 1.95 kHz.

The ESR zero approximately cancels one of the poles, and

the total phase shift is limited to 90. Bode plots are provid-

ed below.

Figure 7: Bode plot of gain response for VOUT/VCONTROL.

Figure 8: Bode plot of phase response for VOUT/VCONTROL.

This uncompensated system is stable, but the low gain will

result in poor DC accuracy, and the low cutoff frequency

will result in poor transient response. Note that we have

not yet included the gain factor from the feedback resistor

divider. This factor will further reduce the overall system

gain.

By adding the two pole, one zero compensation network

shown in figure 6, we can maximize the DC gain and push

out the crossover frequency. The transfer function for the

compensation network is

This can be rewritten in terms of pole and zero frequencies

and a gain constant A.

where

fZ =

fP =

and A = R1 C2

Note that, due to the first s term in the denominator, a pole

is located at f = 0. This will provide the maximum DC

gain.

The optimum performance can be obtained by choosing fZ

equal to the output double pole frequency and setting fP to

approximately half of the switching frequency. Gain fac-

tors can be chosen somewhat arbitrarily.

Values between

1E-6½F and 20E-6½F are practical. We then have a set of

equations that can be solved for component values:

C1 R1 =

[ - ],C1R2= ,C2=

Since there are only three equations, we must arbitrarily

choose one of the components. One option is to set the

value of R1 fairly large. This provides a high impedance

path between the VFB pin and the COMP pin.

For our design, we have fZ = the double pole frequency =

1.95 kHz and fP = fOSC/2 = 100kHz. LetÕs arbitrarily choose

R1 = 4.7K. Then we solve the first equation for C1 and

obtain C1 = 17nF. Use a standard value of 22 nF.

We next solve for R2. With C1 =22 nF, R2= 72½. Use a

standard value of 75½.

We can choose a gain factor from somewhere in the

middle of our range and solve for C2. If A = 10E-6½F, we

have

C2 = 2.1 nF. Use a standard value of 2.2 nF.

2¹fP

2¹

2¹ C1R2

(2¹ C1 (R1 + R2))

s/(2¹fZ + 1)

-A s ((s/2fP) + 1)

VCONTROL

VFB

s C1(R1 + R2) + 1

-s C2 R1(s C1 R2 + 1)

VCONTROL

VFB

-270.0

Frequency (Hz)

-90

-180

102

103

104

106

105

107

-60.0

Frequency (Hz)

-20

-40

102

103

104

106

105

107

R + R1

LC(R + RC)

(2¹)

(2¹CRC)

1 + s(3.3E-5)

s2(2.772E-9) + s(2.902E-5) + 0.42

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