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參數(shù)資料
| 型號(hào)： | LM2637M/NOPB |
| 廠商： | NATIONAL SEMICONDUCTOR CORP |
| 元件分類： | 穩(wěn)壓器 |
| 英文描述： | SWITCHING CONTROLLER, 1000 kHz SWITCHING FREQ-MAX, PDSO24 |
| 封裝： | SOIC-24 |
| 文件頁數(shù)： | 4/17頁 |
| 文件大?。?/td> | 803K |
| 代理商： | LM2637M/NOPB |
Applications Information (Continued)
(7)
The corresponding Bode plots are shown in Figure 7.
Notice since the ESR zero frequency is so low that the phase
doesn’t even go beyond 90. This makes the compensation
easier to do.
Since the DC gain and cutoff frequency (0 dB frequency) are
too low, some compensation is needed. Otherwise the low
DC gain will cause a poor line regulation, and the low cutoff
frequency may hurt transient response performance.
The transfer function for the 2-pole-1-zero compensation
network shown in Figure 6 is:
(8)
where
(9)
One of the poles is located at origin to help achieve the
highest DC gain. So there are three parameters to deter-
mine, 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 gener-
ated. The loop transfer function is:
TF = TF1 x TF2
(10)
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
z = 1.32 kHz, fp = 153 kHz, and A = 4.8 x 10
6
F,
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 8.
The compensation network component values can be deter-
mined by Equation (9), since the values of f
z,fp and A are
now known. To more conveniently calculate the values,
Equation (9) can be rearranged as follows:
(11)
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
EAO to FB. From Equation (11) it can be told that a larger R
2
will result in a smaller C
1,C2 and a larger R1. Calculations
show that the following combination is a good one: R
2 =51
,
C
1 = 0.022 f, R1 = 5.6 k
,C
2 = 820 pF.
For a different application or different type of output capaci-
tors, a different compensation scheme may be necessary.
The user can either follow the steps above to figure the
appropriate component values or contact National for help.
10084817
FIGURE 6. Buck Converter from a Control Viewpoint
10084818
FIGURE 7. Control-to-Output Bode Plots
LM2637
www.national.com
12
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