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| 型号： | IRU3048CFTR |
| 厂商： | International Rectifier |
| 文件页数： | 9/18页 |
| 文件大小： | 0K |
| 描述： | IC REG CTRLR BUCK PWM VM 16TSSOP |
| 标准包装： | 1 |
| PWM 型： | 电压模式 |
| 输出数： | 3 |
| 频率 - 最大： | 220kHz |
| 占空比： | 90% |
| 电源电压： | 4.2 V ~ 25 V |
| 降压： | 是 |
| 升压： | 无 |
| 回扫： | 无 |
| 反相： | 无 |
| 倍增器： | 无 |
| 除法器： | 无 |
| Cuk： | 无 |
| 隔离： | 无 |
| 工作温度： | 0°C ~ 70°C |
| 封装/外壳： | 16-TSSOP（0.173"，4.40mm 宽） |
| 包装： | 剪切带 (CT) |
| 产品目录页面： | 1384 (CN2011-ZH PDF) |
| 其它名称： | *IRU3048CFTR IRU3048CFCT |
IRU3048
For:
V IN1 = 12V
F LC1 = 2.8KHz
Z IN
V OUT
C 12
V OSC = 1.25V
F O1 = 30KHz
F ESR1 = 26.5KHz
R 8 = 1K
R 6 = 1.64K
g m = 600 m mho
C 10
R 8
R 6
R 7
C 11
Z f
This results to R 9 =46.4K V ; Choose R 9 =46.4K V
To cancel one of the LC filter poles, place the zero be-
R 5
Fb
E/A
Comp
Ve
fore the LC filter resonant frequency pole:
F Z ? 75%F LC1
Gain(dB)
V REF
F Z ? 0.75 3
2 p
1
L 3 3 C O
---(13)
H(s) dB
For:
L 3 = 10.2 m H
Co = 300 m F
F Z 1
F Z 2
F P 2
F P 3
Frequency
Fz = 2.1KHz
R 9 = 46.4K V
Using equations (11) and (13) to calculate C 9 , we get:
C 9 = 1630pF; Choose C 9 = 1800pF
Figure 7 - Compensation network with local
feedback and its asymptotic gain plot.
In such configuration, the transfer function is given by:
Using equations (11),(12) and (13) for Ch2, where:
V e 1-
V OUT = 1 +
g m Z f
g m Z IN
V IN2 = 5V
V OSC = 1.25V
F O2 = 30KHz
F LC2 = 3.5KHz
R 15 = 1K
R 14 = 442 V
The error amplifier gain is independent of the transcon-
ductance under the following condition:
F ESR2 = 26.5KHz
g m = 600 m hmo
g m Z f >> 1
and
g m Z IN >>1
---(14)
We get:
R 11 = 38.9K V ; Choose R 11 = 39.2K V
By replacing Z IN and Z f according to figure 7, the trans-
former function can be expressed as:
[ ( )]
3 (1+sR 8 C 10 )
1+sR 7
C 19 = 1554pF; Choose C 19 = 1800pF
One more capacitor is sometimes added in parallel with
C 9 and R 4 . This introduces one more pole which is mainly
H(s)=
1
3
sR 6 (C 12 +C 11 )
(1+sR 7 C 11 ) 3 [1+sC 10 (R 6 +R 8 )]
C 12 C 11
C 12 +C 11
used to supress the switching noise. The additional pole
is given by:
As known, transconductance amplifier has high imped-
ance (current source) output, therefore, consider should
2 p 3 R 9 3
F P =
1
C 18 3 C POLE
C 18 + C POLE
be taken when loading the E/A output. It may exceed its
source/sink output current capability, so that the ampli-
fier will not be able to swing its output voltage over the
necessary range.
The pole sets to one half of switching frequency which
results in the capacitor C POLE:
The compensation network has three poles and two ze-
C POLE =
1
p 3 R 9 3 f S -
1
C 18
?
1
p 3 R 9 3 f S
ros and they are expressed as follows:
1
F P1 = 0 F P2 =
2 p3 R 8 3 C 10
( C C 3 +C C )
f S
For F P <<
2
For a general solution for unconditionally stability for any
type of output capacitors, in a wide range of ESR values
we should implement local feedback with a compensa-
tion network. The typically used compensation network
F P3 =
F Z1 =
1
2 p3 R 7 3
1
2 p3 R 7 3 C 11
12 11
12 11
?
1
2 p3 R 7 3 C 12
F Z2 = 2 p3 C 10 3 (R 6 + R 8 ) ? 2 p3 C 10 3 R 6
for voltage-mode controller is shown in Figure 7.
1 1
Rev. 1.7
09/12/02
www.irf.com
9
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