SC480EVB View Datasheet(PDF) - Semtech Corporation
Part Name
Description
MFG CO.
'SC480EVB' PDF : 25 Pages View PDF
SC480
POWER MANAGEMENT
Application Information (Cont.)
Z TOP
RBOT x
0.015
V RIPPLE_VBAT (MIN) 0.015 Ohms
Secondly calculating the value of CTOP required to achieve
this:
at its highest. The capacitance required for smaller tran-
sient steps my be calculated by substituting the desired
current for the Ifinal term. In this case Ifinal is set for 5A.
For our example:
COUT(MIN) = 392μF.
C TOP
¨¨©§
1
ZTOP
1
RTOP
¸¸¹·
F
2 x Ư x f SW _VBAT (MIN)
Since our example uses internal feedback ,this method
cannot be used, however the voltage seen at the internal
FB point is already greater than 15mV.
We will select 440μF, using two 220μF, 12mΩ capacitors
in parallel.
Next we calculate the RMS input ripple current, which is
largest at the minimum battery voltage:
I IN ( RMS )
V OUT x VBAT (MIN) VOUT
x
IOUT
VBAT _ MIN
ARMS
Next we need to calculate the minimum output capaci-
tance required to ensure that the output voltage does not
exceed the transient maximum limit, POSLIMTR, starting
from the actual static maximum, V , OUT_ST_POS when a load
release occurs:
V OUT_ST_POS V OUT ERR DC V
For our example:
For our example:
IIN(RMS) = 4.17ARMS
Input capacitors should be selected with sufficient ripple
current rating for this RMS current, for example a 10μF,
1210 size, 25V ceramic capacitor can handle approxi-
mately 3ARMS. Refer to manufacturer’s data sheets and
derate appropriately.
VOUT_ST_POS = 1.818V,
POSLIM TR
V OUT x TOL TR V
Where TOLTR is the transient tolerance. For our example:
POSLIMTR = 1.944V,
Finally, we calculate the current limit resistor value. As de-
scribed in the current limit section, the current limit looks
at the “valley current”, which is the average output cur-
rent minus half the ripple current.
I VALLEY
IOUT
IRIPPLE_VBAT(MIN)
2
A
The minimum output capacitance is calculated as fol-
lows:
Iinit
IOUT (MAX )
IRIPPLE_VBAT (MAX )
2
A
The ripple at low battery voltage is used because we want
to make sure that current limit does not occur under nor-
mal operating conditions.
and,
C OUT(MIN)
Lx
Iinit 2 Ifinal 2
POSLIMTR2 VOUT_ST_POS2
F
R ILIM
IVALLEY
x 1.2
x
RDS (ON) x 1.4
10 x 106
Ohms
For our example:
This calculation assumes the condition of a full-load to no-
load step transient occurring when the inductor current is
IVALLEY = 8.31A, RDS(ON) = 4mΩ, giving RILIM = 5.62kΩ
© 2006 Semtech Corp.
17
www.semtech.com
Share Link: 
