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ISL6442 Datasheet(PDF) 13 Page - Renesas Technology Corp

Part # ISL6442
Description  Dual (180째 Out-of-Phase) PWM and Linear Controller
PDF  17 Pages
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Manufacturer  RENESAS [Renesas Technology Corp]
Direct Link  http://www.renesas.com
Logo RENESAS - Renesas Technology Corp

ISL6442 Datasheet(HTML) 13 Page - Renesas Technology Corp

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ISL6442
FN9204 Rev 2.00
Page 13 of 17
Oct 31, 2008
The modulator transfer function is the small-signal transfer
function of VOUT/VCOMP. This function is dominated by a DC
gain, given by dMAXVIN/VOSC, and shaped by the output filter,
with a double pole break frequency at FLC and a zero at FCE.
For the purpose of this analysis, L and D represent the channel
inductance and its DCR, while C and E represent the total output
capacitance and its equivalent series resistance.
The compensation network consists of the error amplifier (internal
to the ISL6442) and the external R1-R3, C1-C3 components. The
goal of the compensation network is to provide a closed loop
transfer function with high 0dB crossing frequency (F0; typically
0.1 to 0.3 of FSW) and adequate phase margin (better than 45°).
Phase margin is the difference between the closed loop phase at
F0dB and 180°. The equations that follow relate the compensation
network’s poles, zeros and gain to the components (R1, R2, R3,
C1, C2, and C3) in Figure 14. Use the following guidelines for
locating the poles and zeros of the compensation network:
1. Select a value for R1 (1k
 to 5k, typically). Calculate
value for R2 for desired converter bandwidth (F0). If setting
the output voltage via an offset resistor connected to the FB
pin, Ro in Figure 14, the design procedure can be followed
as presented in Equation 12.
2. Calculate C1 such that FZ1 is placed at a fraction of the FLC, at
0.1 to 0.75 of FLC (to adjust, change the 0.5 factor to desired
number). The higher the quality factor of the output filter and/or
the higher the ratio FCE/FLC, the lower the FZ1 frequency (to
maximize phase boost at FLC).
3. Calculate C2 such that FP1 is placed at FCE.
4. Calculate R3 such that FZ2 is placed at FLC. Calculate C3 such
that FP2 is placed below FSW (typically, 0.5 to 1.0 times
FSW). FSW represents the switching frequency. Change the
numerical factor to reflect desired placement of this pole.
Placement of FP2 lower in frequency helps reduce the gain
of the compensation network at high frequency, in turn
reducing the HF ripple component at the COMP pin and
minimizing resultant duty cycle jitter.
It is recommended a mathematical model is used to plot the
loop response. Check the loop gain against the error
amplifier’s open-loop gain. Verify phase margin results and
adjust as necessary. The following equations describe the
frequency response of the modulator (GMOD), feedback
compensation (GFB) and closed-loop response (GCL):
where:
FIGURE 14. VOLTAGE-MODE BUCK CONVERTER
COMPENSATION DESIGN
-
+
E/A
VREF
COMP
C1
R2
R1
FB
C2
R3
C3
L
C
VIN
PWM
CIRCUIT
HALF-BRIDGE
DRIVE
OSCILLATOR
E
EXTERNAL CIRCUIT
ISL6442
VOUT
VOSC
D
UGATE
LGATE
Ro
PHASE
FLC
1
2
LC
---------------------------
=
(EQ. 10)
FCE
1
2
 CE

------------------------
=
(EQ. 11)
R2
VOSC R1 F0

dMAX VIN FLC

---------------------------------------------
=
(EQ. 12)
C1
1
2
 R2 0.5 F
LC

------------------------------------------------
=
(EQ. 13)
C2
C1
2
 R2 C1 F
CE
1
–

---------------------------------------------------------
=
(EQ. 14)
R3
R1
FSW
FLC
------------ 1
–
----------------------
=
C3
1
2
 R3 0.7 F
SW

-------------------------------------------------
=
(EQ. 15)
GMOD f
dMAX VIN
VOSC
------------------------------
1s f
 EC

+
1s f
 ED
+
 C

s
2 f LC

++
----------------------------------------------------------------------------------------
=
(EQ. 16)
GFB f
1s f
 R2 C1

+
sf
 R1 C1 C2
+


------------------------------------------------------
=
1s f
 R1 R3
+
 C3

+
1s f
 R3 C3

+
 1s f
 R2
C1 C2
C1 C2
+
----------------------



+


-----------------------------------------------------------------------------------------------------------------------------
(EQ. 17)
GCL f
GMOD f GFB f
=
(EQ. 18)
sf

2
 fj

=



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