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ISL6730 Datasheet, PDF (15/19 Pages) Intersil Corporation – Reduce component size requirements
ISL6730A, ISL6730B, ISL6730C, ISL6730D
VI
L
Q1
CF1
RCS
RSEN
ISEN
ICOMP
Ric
Cic
Cip
Gmi
VOUT
COUT
CURRENT
MIRROR
2:1
ICS
IREF
RIS
FIGURE 15. INDUCTOR CURRENT SENSING SCHEME
FROM DUTY TO
80
INDUCTOR CURRENT
40
20 COMPENSATION
GAIN
0
-20
FZ
FP
-40
-60
CURRENT GAIN
-80 MODULATOR GAIN
OPEN LOOP
GAIN
-100
100
1k
10k
FREQUENCY (Hz)
100k
FIGURE 16. ASYMPTOTIC BODE PLOT OF CURRENT LOOP GAIN
FZ=
-----------------1------------------
2π
•
Ri
c
•
C
ic
(EQ. 46)
FP= -2---π-----•----R-----i--c----•-1---C-C---------ii---pp---------+•---------CC---------i-i--c-c--
(EQ. 47)
Use the following guidelines for locating the poles and zeros of
the compensation network.
The cross over frequency of the current loop should be set
between 2kHz to 100kHz. At cross over frequency, the transfer
function from duty cycle to inductor current is well approximated
by Equation 48:
Gid(s)= L---V-B----OS----TU----T-⋅---s-
(EQ. 48)
It is recommended to set the cross over frequency from 1/10 to
1/6 of the switching frequency with phase margin of 60°. A high
frequency pole is set at 1/2 of the switching frequency for ripple
filtering. In this example, we set the cross over, FC at 1/6 of the
switching frequency.
FZ=
-------------------------F----C---------------------------
⎛
tan ⎜
⎝
atan
⎛
⎜
⎝
F-F---C-P--⎠⎟⎞
+
⎞
ΦM⎠⎟
(EQ. 49)
Where FC = FS/6 = 10.3kHz, ΦM is the phase margin, which is
60°. FP = FS/2 = 31kHz.
Thus, the current loop compensation zero is:
FZ=
---------------(---6---2----K----H-----z----)---⁄---6----------------
tan ⎝⎛
atan
⎛
⎝
26--⎠⎞
+
60 d e g⎠⎞
=
2.12 k H z
(EQ. 50)
The total compensation capacitance is calculated:
Cip + Cic=
⎛
⎜
⎜
⎝
⎛
⎜
⎝
------------V----O-----U----T-------------
LBST ⋅ (2πfc)2
⋅
A-----i-D----C---
Vm
⋅
R--R---S--C--E--S--N--⎠⎟⎞
⋅
1-1----+-+----((---ff--cc----⁄⁄--ff--pz---))---22-⎠⎟⎟⎞
(EQ. 51)
Cip + Cic= (19.8)nF
Cip
=
(Ci
p
+
Ci
C
)
-f--z-
fp
The value of the noise filtering capacitor is:
Cip
=
14.9 n
F
⋅
2----.--1---2----k---H-----z-
31 k H z
=
1.35 n F
(EQ. 52)
(EQ. 53)
(EQ. 54)
The value of Cic is:
Cic = 19.8nF – 1.35nF = 18.4nF
(EQ. 55)
The value of Ric is:
Ric = 2----π-----⋅---2---.--1---2----k----H1-----z----⋅---1---8----.-4----n----F-- = 4.11kΩ
(EQ. 56)
Select the RC value from the standard value, we have:
Ric = 4.02kΩ, Cic = 18nF, Cip = 1.2nF. Figure 17 shows the actual
bode plot of current loop gain.
15
FN8258.1
August 8, 2013