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BD9104FVM 데이터시트(PDF) 13 Page - Rohm

부품명 BD9104FVM
상세설명  Output 1.5A or Less High Efficiency Step-down Switching Regulators with Built-in Power MOSFET
PDF  18 Pages
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제조업체  ROHM [Rohm]
홈페이지  http://www.rohm.com
Logo ROHM - Rohm

BD9104FVM 데이터시트(HTML) 13 Page - Rohm

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BD9102FVM, BD9104FVM, BD9106FVM
Technical Note
13/17
www.rohm.com
2009.05 - Rev.A
© 2009 ROHM Co., Ltd. All rights reserved.
4. Determination of RITH, CITH that works as a phase compensator
As the Current Mode Control is designed to limit a inductor current, a pole (phase lag) appears in the low frequency area
due to a CR filter consisting of a output capacitor and a load resistance, while a zero (phase lead) appears in the high
frequency area due to the output capacitor and its ESR. So, the phases are easily compensated by adding a zero to the
power amplifier output with C and R as described below to cancel a pole at the power amplifier.
Stable feedback loop may be achieved by canceling the pole fp (Min.) produced by the output capacitor and the load resistance
with CR zero correction by the error amplifier.
5. Determination of output voltage (for BD9106FVM only)
The output voltage VOUT is determined by the equation (7):
VOUT=(R2/R1+1)×VADJ・・・(7)
VADJ: Voltage at ADJ terminal (0.8V Typ.)
With R1 and R2 adjusted, the output voltage may be determined
as required.(Adjustable output voltage range: 1.0V~2.5V)
Use 1 kΩ~100 kΩ resistor for R1. If a resistor of the resistance
higher than100 kΩ is used, check the assembled set carefully for
ripple voltage etc.
Fig.56 Determination of output voltage
Gain
[dB]
Phase
[deg]
Fig.53 Open loop gain characteristics
A
0
0
-90
A
0
0
-90
fz(Amp.)
Fig.54 Error amp phase compensation characteristics
fp=
2π×RO×CO
1
fz(ESR)=
2π×ESR×CO
1
Pole at power amplifier
When the output current decreases, the load resistance Ro
increases and the pole frequency lowers.
fp(Min.)=
2π×ROMax.×CO
1
[Hz]←with lighter load
fp(Max.)=
2π×ROMin.×CO
1
[Hz]←with heavier load
Zero at power amplifier
Increasing capacitance of the output capacitor lowers the pole
frequency while the zero frequency does not change. (This
is because when the capacitance is doubled, the capacitor
ESR reduces to half.)
fz(Amp.)=
2π×RITH.×CITH
1
GND,PGND
SW
VCC,PVCC
EN
VOUT
ITH
VCC
VOUT
Cin
RITH
CITH
L
ESR
CO
RO
VOUT
Fig.55 Typical application
fz(Amp.)= fp(Min.)
2π×RITH×CITH
1
=
2π×ROMax.×CO
1
SW
6
1
ADJ
4.7μH
10μF
R2
R1
Output
fp(Min.)
fp(Max.)
fz(ESR)
IOUTMin.
IOUTMax.
Gain
[dB]
Phase
[deg]



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