参数资料
| 型号： | NCP1650EVB |
| 厂商： | ON Semiconductor |
| 文件页数： | 24/31页 |
| 文件大小： | 0K |
| 描述： | BOARD EVAL NCP1650 PFC CTLR |
| 设计资源： | NCP1650EVB Schematic NCP1650EVB Bill of Materials NCP1650EVB Test Procedure NCP1650EVB Gerber Zip |
| 标准包装： | 1 |
| 主要目的： | 电源管理，功率因数校正 |
| 嵌入式： | 否 |
| 已用 IC / 零件： | NCP1650 |
| 已供物品： | 板 |
| 其它名称： | NCP1650EVB-ND NCP1650EVBOS |
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NCP1650
For a first approach, the following formula will give the
inductance value that will cause the peak current to be a
fraction of the peak line frequency current.
Using the ON Semiconductor spreadsheet, a value of
250 m H allows for continuous mode operation at full load
and most input voltages. At the high line value of 265 vac,
?
?
L =
T · Vin2
2 · I% · Pout
1 ?
? 2 · V in
Vout
the unit will operate in the continuous mode from 30 ? to
150 ? , and discontinuous when the input voltage is near zero.
Using information from the ON Semiconductor
I max =
Where:
L is the inductance ( m H)
T is the switching period ( m s)
V in is the minimum rms line voltage (v)
I% is the percent switching current ripple relative to the line
current (.xx)
P out is the maximum output power (w)
V out is the output voltage (v)
So for the following unit:
V in = 85 vrms
V out = 400 VDC
P max = 1000 watts
T = 10 m s (100 kHz)
I% = .30
the inductance would be 84 m H.
? 2 · P out
Vin
The maximum low frequency line current would be
determined at full load and low line, or:
where the definitions of P out and V in are as in the above
equation. For the above conditions, I max would be
16.6 amps. The peak current in the inductor at full load and
low line would be 30% greater than this, or 21.6 amps.
For thermal calculations the transformer will have to pass
11.8 amps rms, and not saturate with a peak current of
21.6 amps.
There are several options available for the design of
inductors. You can contact a magnetics manufacturer, such
as Coiltronics (cooperet.com) or inductor designs can be
made simply with the use of programs such as the DC
inductor design program from Magnetics Inc. This software
is free at their website, www.mag--inc.com.
Using the equation provided, and the following variables:
T = 10 m s (f = 100 kHz)
V rms = 265 v
V o = 400 VDC
P max = 1000 watts
I% = 30
the inductance would be 74 m H.
spreadsheet the inductor can either be specified to a
magnetics company to design, or can be designed by the
Magnetics Inc. software. In either case, the critical
information for the inductor design, (inductance, maximum
average current, peak--to--peak ripple current, and switching
frequency) can be obtained from the spreadsheet.
If a secondary winding is desired to provide a bias supply,
it should provide a minimum of 11.8 volts (to exceed the
UVLO spec) and a maximum of 18 volts. The secondary
should be connected such that it conducts when the power
switch is off. This will create an output voltage that varies
with the input voltage, and near the zero crossings of the line
frequency will have a peak voltage equal to the regulated
output voltage divided by the turns ratio. The filter cap on the
Vcc pin needs to be of sufficient size to hold the voltage up
over between the zero crossings.
Oscillator
The relationship between the frequency and timing
capacitor is:
CT = 47, 000 ∕ f
Where C T is in pF and f is in kHz.
AC Voltage Divider
The voltage divider from the input rectifiers to ground is
a simple but important calculation. For this calculation it is
necessary to know the maximum line that the unit can
operate at. The peak input voltage will be:
Vinpeak = 1.414 × Vrms max
The maximum voltage at the AC input (pin 5) is 3.75 volts
(this is true for both multipliers).
If the maximum line voltage is 265 vac, the peak input
voltage is:
Vinpeak = 1.414 × 265 Vrms = 375 Vpk
To keep the power dissipation reasonable for a ? watt
resistor (R ac1 ), it should dissipate no more than ? watt.
Depending on environmental conditions, further derating
may be required. The power in this resistor is:
PRac1 = (375 v ? 3.75 v)2 ∕ Rac1 = .25 watts
so : Rac1 = 551 kOhms
To minimize dissipation, use the next largest standard
value, or 560 kOhms.
Then, Rac2 = 3.75 v ∕ ((375 v ? 3.75 v) ∕ 560 k)
= 5.6 kOhms
http://onsemi.com
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