ncp5389 ON Semiconductor, ncp5389 Datasheet - Page 23

• Manufacturer Part Number: ncp5389
• Description: 2/3 Phase Buck Controller For Vr11 Pentium Iv Processor Applications
• Manufacturer: ON Semiconductor

Inductor Current Sense Compensation
This method uses an RC filter to cancel out the inductance
of the inductor and recover the voltage that is the result of
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 NCP5389 uses the inductor current sensing method.
The demoboard inductor measured 350 nH and 0.75 mW
-
+
VRamp_min
1.3 V
+
-
1E3
-
+
0
Rsense(T) +
Vin
12
Figure 16.
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-9/3)
Voff
+ -
+
-
RDRP
5.11 k
L
2
Voff
http://onsemi.com
CFB1
680 p
(0.85e-3/3)
Figure 17.
NCP5389
RFB
1 k
DCR
RFB1
23
100
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 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 17 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)