BD9140MUV-E2 Rohm Semiconductor, BD9140MUV-E2 Datasheet - Page 10

• Manufacturer Part Number: BD9140MUV-E2
• Description: IC SWITCH REG 2A W/FET VQFN020
• Manufacturer: Rohm Semiconductor
• Series: -r
• Type: Step-Down (Buck), PWM - Current Moder

Specifications of BD9140MUV-E2

• Internal Switch(s): Yes
• Synchronous Rectifier: Yes
• Number Of Outputs: 1
• Voltage - Output: 2.5 V ~ 6 V
• Current - Output: 2A
• Frequency - Switching: 500kHz
• Voltage - Input: 4.5 V ~ 13.2 V
• Operating Temperature: -40°C ~ 105°C
• Mounting Type: *
• Package / Case: *
• Lead Free Status / Rohs Status: Lead free / RoHS Compliant

© 2009 ROHM Co., Ltd. All rights reserved.
BD9141MUV
Phase
Phase
[deg]
[deg]
Gain
Gain
www.rohm.com
[dB]
[dB]
3. Selection of input capacitor (Cin)
4. Determination of RITH, CITH that works as a phase compensator
-90
-90
V
A low ESR 22μF/25V ceramic capacitor is recommended to reduce ESR dissipation of input capacitor for better
efficiency.
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.
A
A
0
0
0
0
CC
Fig.29 Input capacitor
Fig.31 Error amp phase compensation characteristics
I
OUT
fp(Min.)
fz(Amp.)
L
Min.
Cin
Fig.30 Open loop gain characteristics
fp(Max.)
Co
I
OUT
Max.
V
OUT
fz(ESR)
Input capacitor to select must be a low ESR capacitor of the capacitance
sufficient to cope with high ripple current to prevent high transient voltage. The
ripple current IRMS is given by the equation (5):
If V
I
< Worst case > I
When Vcc is twice the V
I
RMS
RMS
CC
=I
=2×
=8V, V
OUT
×
OUT
10/15
fp=
fz
Pole at power amplifier
Zero at power amplifier
√
√
(ESR)
V
=5V, and I
RMS(max.)
When the output current decreases, the load resistance Ro
increases and the pole frequency lowers.
Increasing capacitance of the output capacitor lowers the pole
frequency while the zero frequency does not change.
because when the capacitance is doubled, the capacitor ESR
reduces to half.)
OUT
5(8-5)
3.3
2π×R
=
(V
V
2π×E
CC
CC
OUT
1
-V
O
OUTmax.=
fp
fp
fz
, I
×C
=0.97[A
OUT
1
(Amp.)
(Min.)
(Max.)
RMS
SR
O
)
×C
=
=
=
=
2A, (BD9140MUV)
[A]・・・(5)
RMS
O
2π×R
2π×R
2π×R
I
OUT
2
]
OMax.
OMin.
ITH
1
1
1
×C
×C
×C
ITH
O
O
[Hz]←with lighter load
[Hz] ←with heavier load
Technical Note
2009.09 - Rev.B
(This is