参数资料
型号: ISL6442EVAL1Z
厂商: Intersil
文件页数: 13/16页
文件大小: 0K
描述: EVAL BOARD FOR ISL6442
标准包装: 1
主要目的: DC/DC,LDO 步降
输出及类型: 3,非隔离
输出电压: 1.8V,3.3V,5V
电流 - 输出: 3A,3A,300mA
输入电压: 6V
稳压器拓扑结构: 降压
频率 - 开关: 1.4MHz
板类型: 完全填充
已供物品:
已用 IC / 零件: ISL6442
ISL6442
to the FB pin, Ro in Figure 14, the design procedure can
C2
be followed as presented in Equation 12.
V OSC ? R1 ? F 0
d MAX ? V IN ? F LC
COMP
R2
C1
R3
C3
R2 = ---------------------------------------------
(EQ. 12)
E/A
-
+
FB
Ro
R1
2. Calculate C1 such that F Z1 is placed at a fraction of the F LC ,
at 0.1 to 0.75 of F LC (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 F CE /F LC , the lower the F Z1
VREF
frequency (to maximize phase boost at F LC ).
C1 = ------------------------------------------------
1
2 π ? R2 ? 0.5 ? F LC
(EQ. 13)
C2 = ---------------------------------------------------------
PWM
CIRCUIT
OSCILLATOR
V OSC
V IN
V OUT
3. Calculate C2 such that F P1 is placed at F CE .
C1
2 π ? R2 ? C1 ? F CE – 1
(EQ. 14)
UGATE
L
D
4. Calculate R3 such that F Z2 is placed at F LC . Calculate C3
HALF-BRIDGE
DRIVE
PHASE
LGATE
C
E
such that F P2 is placed below F SW (typically, 0.5 to 1.0
times F SW ). F SW represents the switching frequency.
Change the numerical factor to reflect desired placement
of this pole. Placement of F P2 lower in frequency helps
reduce the gain of the compensation network at high
frequency, in turn reducing the HF ripple component at
ISL6442
EXTERNAL CIRCUIT
the COMP pin and minimizing resultant duty cycle jitter.
R3 = ----------------------
F SW
C3 = -------------------------------------------------
FIGURE 14. VOLTAGE-MODE BUCK CONVERTER
COMPENSATION DESIGN
R1
------------ – 1
F LC
1
2 π ? R3 ? 0.7 ? F SW
(EQ. 15)
The modulator transfer function is the small-signal transfer
function of V OUT /V COMP . This function is dominated by a DC
gain, given by d MAX V IN /V OSC , and shaped by the output
filter, with a double pole break frequency at F LC and a zero at
F CE . 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.
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 (G MOD ), feedback
compensation (G FB ) and closed-loop response (G CL ):
F LC = ---------------------------
G MOD ( f ) = ------------------------------ ? ----------------------------------------------------------------------------------------
V OSC
1 + s ( f ) ? ( E + D ) ? C + s ( f ) ? L ? C
F CE = ------------------------
1
2 π ? L ? C
1
2 π ? C ? E
(EQ. 10)
(EQ. 11)
d MAX ? V IN 1 + s ( f ) ? E ? C
2
(EQ. 16)
G FB ( f ) = ------------------------------------------------------ ?
? -----------------------------------------------------------------------------------------------------------------------------
( 1 + s ( f ) ? R3 ? C3 ) ? ? 1 + s ( f ) ? R2 ? ? ----------------------
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 (F 0 ; typically 0.1 to 0.3 of F SW ) and adequate phase
margin (better than 45°). Phase margin is the difference
between the closed loop phase at F 0dB 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 Ω to 5k Ω , typically). Calculate
value for R2 for desired converter bandwidth (F 0 ). If
setting the output voltage via an offset resistor connected
1 + s ( f ) ? R2 ? C1
s ( f ) ? R1 ? ( C1 + C2 )
1 + s ( f ) ? ( R1 + R3 ) ? C3
? ? C1 + C2 ? ?
G CL ( f ) = G MOD ( f ) ? G FB ( f )
where:
s ( f ) = 2 π ? f ? j
C1 ? C2 ? ?
(EQ. 17)
(EQ. 18)
13
FN9204.2
October 31, 2008
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