ncp5381a ON Semiconductor, ncp5381a Datasheet - Page 30

• Manufacturer Part Number: ncp5381a
• Description: 2/3/4 Phase Buck Controller For Vr10 And Vr11 Pentium Iv Processor Applications
• Manufacturer: ON Semiconductor

Inductor Current Sense Compensation
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
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
The NCP5381A uses the inductor current sensing
The demoboard inductor measured 350 nH and 0.75 mW
−
+
VRamp_min
1.3 V
+
−
1E3
−
+
0
Rsense(T) +
Vin
12
Figure 18.
BW = 15 MHz
4
Unity
Gain
0
4.3 k
C3
10.6 n
RF
12
0
22 p
CH
R6
1 k
1.5 n
CF
0
E1
0.47 · mF · DCR 25C · (1 ) 0.00393 · C −1 · (T−25 · C))
E
GAIN = 6
1
+
−
(250e−9/4)
Voff
+ −
+
−
RDRP
5.11 k
L
2
Voff
http://onsemi.com
CFB1
680 p
(0.85e−3/4)
Figure 19.
RFB
1 k
DCR
RFB1
30
100
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 mF capacitor for C and then
solve for R.
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.
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
will keep this offset at the droop pin below 2.5 mV.
Simple Average PSPICE Model
used to determine a stable solution and provide insight into
the control system.
2
ESLBulk
(3.5e−9/10)
1
ESRBulk
(7e−3/10)
1.3
1
CBulk
(560e−6*10)
It is desirable to keep the Rsense resistor value below
A simple state average model shown in Figure 19 can be
L
+
−
100 p
LBRD
Voffset
0
2
0.75 m
RBRD
+
−
−
+
VDAC
1.25 V
0
2
1
ESRCer
(1.5e−3/18)
ESLCer
(1.5e−9/18)
CCer
(22e−6*18)
1Aac
0Adc
+
−
I1 = 10
I2 = 110
TD = 10u
TR = 50n
TF = 50n
PW = 40u
PER = 80u
Vout
+
−
I2
0
(eq. 8)