参数资料
型号: NCP5382MNR2G
厂商: ON Semiconductor
文件页数: 29/33页
文件大小: 0K
描述: IC BUCK CTLR 2-6PHASE 48-QFN
产品变化通告: NCP5371/72/82 Discontinuation 12/Aug/2008
标准包装: 2,500
应用: 控制器,Intel Pentium? IV
输入电压: 10.8 V ~ 13.2 V
输出数: 6
工作温度: 0°C ~ 70°C
安装类型: 表面贴装
封装/外壳: 48-VFQFN 裸露焊盘
供应商设备封装: 48-QFN(7x7)
包装: 带卷 (TR)
NCP5382
Inductor Current Sense Compensation
The NCP5382 uses the inductor current sensing method.
This method uses an RC filter to cancel out the inductance
of the inductor and recover the voltage that is the result of
the current flowing through the inductor ’s DCR. This is
done by matching the RC time constant of the current sense
filter to the L/DCR time constant. The first cut approach is
to use a 0.47 m F capacitor for C and then solve for R.
Rsense(T) +
L
0.47 · m F · DCR25C · (1 ) 0.00393 · C ? 1 · (T ? 25 · C))
(eq. 8)
inductor temperature final selection of R is best done
experimentally on the bench by monitoring the Vdroop pin
and performing a step load test on the actual solution.
It is desirable to keep the Rsense resistor value below
1.0 k whenever possible by increasing the capacitor values
in the inductor compensation network. The bias current
flowing out of the current sense pins is approximately
100 nA. This current flows through the current sense
resistor and creates an offset at the capacitor which will
appear as a load current at the Vdroop pin. A 1.0 k resistor
0 GAIN = 6
Figure 19.
The demoboard inductor measured 350 nH and 0.75 m W
at room temp. The actual value used for Rsense was 953 W
which matches the equation for Rsense at approximately
50C. Because the inductor value is a function of load and
E1
+
+
? ?
E
will keep this offset at the droop pin below 2.5 mV.
Simple Average PSPICE Model
A simple state average model shown in Figure 20 can be
used to determine a stable solution and provide insight into
the control system.
+
2
+
?
+
?
VRamp_min
1.3 V
?
+
Vin
12
0
4
12
0
1
L
(250e ? 9/6)
Voff
DCR
(0.85e ? 3/6)
LBRD
1 2
100 p
CBulk
(560e ? 6*10)
ESRBulk
(7e ? 3/10)
2
ESLBulk
(3.5e ? 9/10)
RBRD
0.75 m
CCer
(22e ? 6*18)
ESRCer
(1.5e ? 3/18)
2
ESLCer
(1.5e ? 9/18)
1Aac
0Adc
+
?
I1 = 10
I2 = 110
TD = 10u
TR = 50n
TF = 50n
PW = 40u
PER = 80u
I2
?
RDRP
1
1
CH
5.11 k
0
RF
22 p
CF
4.3 k
1.5 n
CFB1
680 p
RFB1
100
?
1E3
Unity
Gain
BW = 15 MHz R6
C3 1k
10.6 n
0
? +
Voff
RFB
1k
1.3
+
?
Voffset
0
+
?
+
VDAC
1.25 V
0
Vout
Figure 20.
http://onsemi.com
29
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