参数资料
型号: NCP1583DR2GEVB
厂商: ON Semiconductor
文件页数: 8/16页
文件大小: 0K
描述: EVAL BOARD FOR NCP1583DR2G
产品变化通告: Product Obsolescence 24/Jan/2011
设计资源: NCP1583 EVB BOM
NCP1583DR2GEVB Gerber Files
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: 0.8V
输入电压: 4.5 ~ 12 V
稳压器拓扑结构: 降压
频率 - 开关: 350kHz
板类型: 完全填充
已供物品:
已用 IC / 零件: NCP1583
其它名称: NCP1583DR2GEVBOS
NCP1582, NCP1582A, NCP1583
APPLICATION SECTION
Input Capacitor Selection
The input capacitor has to sustain the ripple current
produced during the on time of the upper MOSFET, so it
must have a low ESR to minimize the losses. The RMS value
of this ripple is:
The above calculation includes the delay from comp
rising to when output voltage becomes valid.
To calculate the time of output voltage rising to when it
reaches regulation; D V is the difference between the comp
voltage reaching regulation and 1.1 V.
IinRMS + IOUT
D
(1 * D) ,
Output Capacitor Selection
Iinrush +
COUT VOUT
tSS
D IOUT 2 LOUT
D VOUT ? DISCHARGE + ,
where D is the duty cycle, Iin RMS is the input RMS current,
& I OUT is the load current. The equation reaches its
maximum value with D = 0.5. Losses in the input capacitors
can be calculated with the following equation:
PCIN + ESRCIN IinRMS 2 ,
where P CIN is the power loose in the input capacitors &
ESR CIN is the effective series resistance of the input
capacitance. Due to large d I /d t through the input capacitors,
electrolytic or ceramics should be used. If a tantalum must
be used, it must be surge protected. Otherwise, capacitor
failure could occur.
Calculating Input Start ? up Current
To calculate the input start up current, the following
equation can be used.
,
where I inrush is the input current during start ? up, C OUT is the
total output capacitance, V OUT is the desired output voltage,
and t SS is the soft start interval.
If the inrush current is higher than the steady state input
current during max load, then the input fuse should be rated
accordingly, if one is used.
Calculating Soft Start Time
To calculate the soft start time, the following equation can
be used.
The output capacitor is a basic component for the fast
response of the power supply. In fact, during load transient,
for the first few microseconds it supplies the current to the
load. The controller immediately recognizes the load
transient and sets the duty cycle to maximum, but the current
slope is limited by the inductor value.
During a load step transient the output voltage initial
drops due to the current variation inside the capacitor and the
ESR. (neglecting the effect of the effective series inductance
(ESL)):
D VOUT ? ESR + D IOUT ESRCOUT,
where V OUT ? ESR is the voltage deviation of V OUT due to the
effects of ESR and the ESR COUT is the total effective series
resistance of the output capacitors.
A minimum capacitor value is required to sustain the
current during the load transient without discharging it. The
voltage drop due to output capacitor discharge is given by
the following equation:
2 COUT (VIN D * VOUT)
where V OUT ? DISCHARGE is the voltage deviation of V OUT
due to the effects of discharge, L OUT is the output inductor
value & V IN is the input voltage.
It should be noted that Δ V OUT ? DISCHARGE and V OUT ? ESR
are out of phase with each other, and the larger of these two
voltages will determine the maximum deviation of the
output voltage (neglecting the effect of the ESL).
tSS +
(CP ) CC) * D V
ISS
Inductor Selection
Both mechanical and electrical considerations influence
Where C C is the compensation as well as the soft start
capacitor,
C P is the additional capacitor that forms the second pole.
I SS is the soft start current
D V is the comp voltage from zero to until it reaches
regulation.
the selection of an output inductor. From a mechanical
perspective, smaller inductor values generally correspond to
smaller physical size. Since the inductor is often one of the
largest components in the regulation system, a minimum
inductor value is particularly important in
space ? constrained applications. From an electrical
perspective, the maximum current slew rate through the
output inductor for a buck regulator is given by:
SlewRateLOUT + IN * OUT .
V comp
V out
D V
1.1 V
V V
LOUT
This equation implies that larger inductor values limit the
regulator ’s ability to slew current through the output
inductor in response to output load transients. Consequently,
output capacitors must supply the load current until the
inductor current reaches the output load current level. This
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