PM6670ASTR STMicroelectronics, PM6670ASTR Datasheet - Page 44

IC CTLR DDR2/3 MEM PS 24VFQFPN

PM6670ASTR

Manufacturer Part Number
PM6670ASTR
Description
IC CTLR DDR2/3 MEM PS 24VFQFPN
Manufacturer
STMicroelectronics
Datasheet

Specifications of PM6670ASTR

Applications
Memory, DDR2/DDR3 Regulator
Current - Supply
800µA
Voltage - Supply
4.5 V ~ 36 V
Operating Temperature
-40°C ~ 85°C
Mounting Type
Surface Mount
Package / Case
24-VFQFN, 24-VFQFPN
For Use With
497-8412 - BOARD EVAL PM6670AS DDR2/3497-8411 - BOARD EVAL PM6670S DDR2/3
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Other names
497-6305-2
PM6670ATR

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Quantity
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Quantity:
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Application information
44/53
where R
Switching losses are approximately given by:
Equation 41
where t
gate-driver current capability and the gate charge Q
low R
As general rule, the R
MOSFET.
Logic-level MOSFETs are recommended, as long as low-side and high-side gate drivers are
powered by V
greater than the maximum input voltage.V
Table 16
Table 16.
In buck converters the power dissipation of the synchronous MOSFET is mainly due to
conduction losses:
Equation 42
Maximum conduction losses occur at the maximum input voltage:
Equation 43
The synchronous rectifier should have the lowest R
MOSFET turns on, high d
through its gate-drain capacitance C
the choice of the low-side MOSFET is a trade-off between on resistance and gate charge; a
good selection should minimize the ratio C
Equation 44
Manufacturer
DSon
ON
P
DSon
ST
IR
lists tested high-side MOSFETs.
switching
. Unfortunately low R
and t
Evaluated high-side MOSFETs
is the drain-source on-resistance of the control MOSFET.
VCC
OFF
=
= +5 V. The breakdown voltage of the MOSFETs (V
V
are the turn-on and turn-off times of the MOSFET and depend on the
IN
DSon
STS12NH3LL
I (
IRF7811
LOAD
P
V
Type
conduction
x Q
/d
t
(max)
of the phase node can bring up even the low-side gate
Doc ID 14436 Rev 2
gate
DSon
2
product should be minimized to find out the suitable
=
P
R
MOSFETs have a great gate charge.
RRS
C
DLowSide
Δ
DSon
2
GS
I
L
, causing a cross-conduction problem. Once again,
)
=
R
INmax
(mΩ)
RSS
t
10.5
DSon
C
on
9
1
ISS
P
/ C
f
.
sw
V
conduction
V
IN
GS
OUT
C
+
,
MAX
gate
DSon
RSS
V
where
IN
. A greater efficiency is achieved with
as possible. When the high-side
I (
Gate charge
I
LOAD
LOAD
(nC)
12
18
,
MAX
(max)
2
2
+
BRDSS
Δ
2
I
L
)
Rated reverse
t
) must be
off
voltage (V)
PM6670AS
f
sw
30
30

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