The control-to-output transfer function is

The ESR zero frequency is:

The power stage double pole frequency is:

The corresponding Bode plots are shown in

Figure 5.

Since the ESR zero frequency is so low, it effectively cancels

the phase shift from one of the power stage poles. This limits

the total phase shift to 90%.

when gain = 0dB), it needs compensation to improve the DC

gain and cut off frequency (0dB frequency). Otherwise, the

low DC gain may cause a poor line regulation, and the low

cutoff frequency will hurt transient response performance.

The transfer function for the 2-pole-1-zero compensation

network shown in

Figure 4 is:

where

One of the poles is located at origin to help achieve the high-

est DC gain. So there are three parameters to determine, the

position of the zero, the position of the second pole, and the

constant A. To determine the cutoff frequency and phase

margin, the loop bode plots need to be generated. The loop

transfer function is:

TF = −TF1 x TF2

By choosing the zero close to the double pole position and

the second pole to half of the switching frequency, the closed

loop transfer function turns out to be very good.

That is, if f

−6

ΩF,

then the cutoff frequency will be 50 kHz, the phase margin

will be 72˚, and the DC gain will be that of the error amplifier.

See

Figure 6 below.

The compensation network component values can be deter-

mined by the following equations:

Notice there are three equations but four variables. So one

of the variables can be chosen arbitrarily. Since the current

driving capability of the error amplifier is limited to around

3 mA, it is a good idea to have a high impedance path from

the output of the error amplifier to the output of the converter.

From the above equations it can be told that a larger R

DS100119-13

FIGURE 5. Control-to-Output Bode Plots

DS100119-17

FIGURE 6. Loop Bode Plots

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