参数资料
型号: ISL6752/54EVAL1Z
厂商: Intersil
文件页数: 11/16页
文件大小: 0K
描述: BOARD DEMO FOR ISL6752 ISL6754
标准包装: 1
主要目的: 电源,ATX
输出及类型: 1,隔离
输出电压: 12V
电流 - 输出: 50A
输入电压: 350 ~ 450 V
稳压器拓扑结构: 全桥式
板类型: 完全填充
已供物品:
已用 IC / 零件: ISL6752,ISL6754
ISL6752
Since the peak current limit threshold is 1.00V, the total
current feedback signal plus the external ramp voltage must
1
16
sum to this value.
2
ISL6752
15
V e + V CS = 1
(EQ. 15)
3
4 CTBUF
14
13
N P ? N CT 1
R CS = ------------------------ ? ----------------------------------------------------
I O + -------- t SW ? --- + ---- ?
L ? π 2 ?
Substituting Equations 13 and 14 into Equation 15 and
solving for R CS yields Equation 16:
Ω (EQ. 16)
N S V O 1 D
O
For simplicity, idealized components have been used for this
R CS
R6
R9
C4
5
6
7
8 CS
12
11
10
9
discussion, but the effect of magnetizing inductance must be
considered when determining the amount of external ramp
to add. Magnetizing inductance provides a degree of slope
compensation to the current feedback signal and reduces
the amount of external ramp required. The magnetizing
inductance adds primary current in excess of what is
FIGURE 6. ADDING SLOPE COMPENSATION
Δ I P = -----------------------------
reflected from the inductor current in the secondary.
V IN ? Dt SW
A
L m
(EQ. 17)
Assuming the designer has selected values for the RC filter
placed on the CS pin, the value of R9 required to add the
appropriate external ramp can be found by superposition.
V e – Δ V CS = -------------------------------------------------------------------------------
where V IN is the input voltage that corresponds to the duty
( D ( V CTBUF – 0.4 ) + 0.4 ) ? R6
R6 + R9
V
(EQ. 20)
cycle D and L m is the primary magnetizing inductance. The
effect of the magnetizing current at the current sense
Rearranging to solve for R9 yields Equation 21:
Δ I P ? R CS
Δ V CS = --------------------------
( D ( V CTBUF – 0.4 ) – V e + Δ V CS + 0.4 ) ? R6
V e – Δ V CS
resistor, R CS , is expressed in Equation 18:
V
N CT
(EQ. 18)
R9 = -------------------------------------------------------------------------------------------------------------------
Ω
(EQ. 21)
If Δ V CS is greater than or equal to V e , then no additional slope
compensation is needed and R CS becomes Equation 19:
The value of R CS determined in Equation 16 must be
rescaled so that the current sense signal presented at the
CS pin is that predicted by Equation 14. The divider created
V IN ? Dt SW
N S ?
Dt SW ?
R CS = ----------------------------------------------------------------------------------------------------------------------------------
-------- ? ? I O + --------------- ? ? V IN ? ------- S - – V O ? ? + -----------------------------
R ′ CS = ---------------------- ? R CS
N P ? 2L O ? N P ? ? L m
N CT
N ? ?
(EQ. 19)
by R6 and R9 makes this necessary.
R6 + R9
R9
(EQ. 22)
Example:
If Δ V CS is less than V e , then Equation 16 is still valid for the
value of R CS , but the amount of slope compensation added
by the external ramp must be reduced by Δ V CS .
Adding slope compensation may be accomplished in the
ISL6752 using the CTBUF signal. CTBUF is an amplified
representation of the sawtooth signal that appears on the CT
pin. It is offset from ground by 0.4V and is 2x the peak-to-peak
amplitude of CT (0.4V to 4.4V). A typical application sums this
signal with the current sense feedback and applies the result
to the CS pin, as shown in Figure 6.
11
V IN = 280V
V O = 12V
L O = 2.0μH
Np/Ns = 20
Lm = 2mH
I O = 55A
Oscillator Frequency, F SW = 400kHz
Duty Cycle, D = 85.7%
N CT = 50
R6 = 499 Ω
Solve for the current sense resistor, R CS , using Equation 16.
R CS = 15.1 Ω .
FN9181.3
October 31, 2008
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