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BD9104FVM Datasheet(PDF) 23 Page - Rohm

Part # BD9104FVM
Description  4.0V(or 4.5V) to 5.5V, 0.8A 1ch Synchronous Buck Converter Integrated FET
PDF  29 Pages
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Manufacturer  ROHM [Rohm]
Direct Link  http://www.rohm.com
Logo ROHM - Rohm

BD9104FVM Datasheet(HTML) 23 Page - Rohm

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BD9102FVM BD9104FVM BD9106FVM
Datasheet
02.MAR.2012 Rev.001
www.rohm.com
© 2011 ROHM Co., Ltd. All rights reserved.
TSZ22111・15・001
TSZ02201-0J3J0AJ00080-1-2
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 to 2.5V)
Use 1 kΩ to 100 kΩ resistor for R1.
If a resistor of the
resistance higher than10
0 kΩ is used, check the assembled set
carefully for ripple voltage etc.
Fig.57 Determination of output voltage
Gain
[dB]
Phase
[deg]
Fig.54 Open loop gain characteristics
A
0
0
-90
A
0
0
-90
fz(Amp.)
Fig.55 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×R
OMax.
×CO
1
[Hz]←with lighter load
fp(Max.)= 2π×R
OMin.
×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×R
ITH.
×CITH
1
GND,PGND
SW
VCC,PVCC
EN
VOUT
ITH
VCC
VOUT
Cin
RITH
CITH
L
ESR
CO
RO
VOUT
Fig.56 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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