参数资料
型号: ISL6740EVAL1
厂商: Intersil
文件页数: 18/28页
文件大小: 0K
描述: EVALUATION BOARD 1 ISL6740
标准包装: 1
主要目的: DC/DC,步降
输出及类型: 1,非隔离
输出电压: 12V
电流 - 输出: 8A
输入电压: 48V
稳压器拓扑结构: 降压
板类型: 完全填充
已供物品:
已用 IC / 零件: ISL6740
ISL6740, ISL6741
determined by measurement, calculation, estimate, or by some
combination of these methods.
V OUT is the output voltage
If we allow a current ramp, Δ I, of 5% of the rated output current,
t zvs ≈ ------------------------------------------------------------------
π L lk ? ( 2C oss + C xfrmr ) (EQ. 19)
2
V L ? T ON 0.25 ? 2.08
Δ I
L ≥ ---------------------- = ----------------------------- = 1.04
S
Device output capacitance, Coss, is non-linear with applied
the minimum inductance required is:
0.5
μ H
(EQ. 23)
voltage. To find the equivalent discrete capacitance, Cfet, a
charge model is used. Using a known current source, the time
required to charge the MOSFET drain to the desired operating
voltage is determined and the equivalent capacitance is
calculated.
An inductor value of 1.4 μ H, rated for 18A was selected.
With a maximum input voltage of 53V, the maximum output
voltage is about 13V. The closest higher voltage rated capacitor
is 16V. Under steady state operating conditions the ripple current
Ichg ? t
Cfet = -------------------
V
F
(EQ. 20)
in the capacitor is small, so it would seem appropriate to have a
low ripple current rated capacitor. However, a high rated ripple
current capacitor was selected based on the nature of the
Once the estimated transition time is determined, it must be
verified directly in the application. The transformer leakage
inductance was measured at 125nH and the combined
capacitance was estimated at 2000pF. Calculations indicate a
transition period of ~ 25ns. Verification of the performance
yielded a value of t D closer to 45ns.
The remainder of the switching half-period is the charge time, t C ,
and can be found from
intended load, multiple buck regulators. To minimize the output
impedance of the filter, a Sanyo OSCON 16SH150M capacitor in
parallel with a 22 μ F ceramic capacitor were selected.
Overcurrent Component Selection
There are two circuit areas to consider when selecting the
components for overcurrent protection, current limit and short
circuit shutdown. The current limit threshold is fixed at 0.6V while
the short circuit threshold is set to a fraction of the duty cycle the
t C = --------------- – t D = ----------------------------------- – 45 ? 10
2 ? F S
2 ? 235 ? 10
1 1
3
– 9
= 2.08
μ s
(EQ. 21)
designer wishes to define as a short circuit.
The current level that corresponds to the overcurrent threshold
must be chosen to allow for the dynamic behavior of an open
where F S is the converter switching frequency.
Using Figure 4, the capacitor value appropriate to the desired
oscillator operating frequency of 470kHz can be selected. A C T
value of 100pF, 220pF, or 330pF is appropriate for this
loop converter. In particular, the low inductor ripple current under
steady state operation increases significantly as the duty cycle
decreases.
14
frequency. A value of 220pF was selected.
To obtain the proper value for R TD , Equation 3 is used. Since
there is a 10ns propagation delay in the oscillator circuit, it must
be included in the calculation. The value of R TD selected is
8.06k Ω .
A similar procedure is used to determine the value of R TC using
Equation 2. The value of R TC selected is the series combination of
17.4k Ω and 1.27k Ω . See section “Overcurrent Component
Selection” on page 18 for further explanation.
13
12
11
10
9
V (L1:1)
I (L1)
Output Filter Design
8
0.9950
0.9960
0.9970
0.9980
0.9990
1.000
The output filter inductor and capacitor selection is simple and
straightforward. Under steady state operating conditions the
voltage across the inductor is very small due to the large duty
cycle. Voltage is applied across the inductor only during the
switch transition time, about 45ns in this application. Ignoring
the voltage drop across the SR FETs, the voltage across the
inductor during the ON time with V IN = 48V is:
TIME (ms)
FIGURE 8. STEADY STATE SECONDARY WINDING VOLTAGE
AND INDUCTOR CURRENT
V IN ? N S ? ( 1 – D )
V L = V S – V OUT = --------------------------------------------- ≈ 250
2N P
mV
(EQ. 22)
where
V L is the inductor voltage
V S is the voltage across the secondary winding
18
FN9111.6
December 2, 2011
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