LM2633 datasheet(34/40 Pages) NSC

34 / 40 page

Control Loop Design (Continued)

(45)

(46)

Back to the previous example. Let B = K, f

z1 =fp,fp2 =fz,fz2

n, then:

c_o = 5.1 x 310Hz = 1581Hz

The corresponding Bode plots of the compensation network

and the loop transfer function are shown in

Figure 12 and

Figure 13 respectively.

It can be seen from

Figure 13 that the crossover frequency is

20kHz, and the phase margin is about 84 degrees.

One thing that should be pointed out is this Bode plot is only

for the 0.4

Ω load. That is, when load current is 4A. If load

current is lower than 4A, the portion of the gain plot from the

corresponding f

p to 310Hz will be −40dB/dec. If load current

is higher than 4A, then the portion of the gain plot from

310Hz to fp will be flat. However, this usually does not have

much effect on the cross-over frequency and phase margin

because it happens at low frequencies.

If a shorter recovery time is desired during a load transient,

z1 can be increased so that the gain of the loop transfer

function becomes higher. However, try not to let f

z1 be higher

than the desired cross-over frequency, otherwise phase mar-

gin can be too low.

Figure 14 shows a situation where f

z1 is

placed at a higher frequency than the f

p, which results in a

−40 dB/dec section before the cross-over frequency. Notice

the phase margin is lower.

Sometimes the slow transient response is caused by the

current source and sink capability of the error amplifier.

Reducing the value of the compensation capacitor helps, but

make sure the small-signal loop is stable.

The power stage component selection can be significantly

different from the example values.

Figure 15 shows how the

two high frequency poles of a current-mode-control buck

regulator change with the Q value.

20000876

FIGURE 12. Example Compensation Transfer Function

20000877

FIGURE 13. Example Loop Transfer Function

200008D6

FIGURE 14. Higher Low Frequency Gain

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