数据搜索系统,热门电子元器件搜索
  Chinese  ▼
ALLDATASHEETCN.COM

X  

LTC1148CS 数据表(PDF) 12 Page - Linear Technology

部件名 LTC1148CS
功能描述  High Efficiency Synchronous Step-Down Switching Regulators
PDF  20 Pages
Scroll/Zoom Zoom In 100%  Zoom Out
制造商  LINER [Linear Technology]
网页  http://www.linear.com
标志 LINER - Linear Technology

LTC1148CS 数据表(HTML) 12 Page - Linear Technology

Back Button LTC1148CS Datasheet HTML 8Page - Linear Technology LTC1148CS Datasheet HTML 9Page - Linear Technology LTC1148CS Datasheet HTML 10Page - Linear Technology LTC1148CS Datasheet HTML 11Page - Linear Technology LTC1148CS Datasheet HTML 12Page - Linear Technology LTC1148CS Datasheet HTML 13Page - Linear Technology LTC1148CS Datasheet HTML 14Page - Linear Technology LTC1148CS Datasheet HTML 15Page - Linear Technology LTC1148CS Datasheet HTML 16Page - Linear Technology Next Button
Zoom Inzoom in Zoom Outzoom out
 12 / 20 page
background image
12
LTC1148
LTC1148-3.3/LTC1148-5
APPLICATIO S I FOR ATIO
LTC1148 DC supply current is 160
µA for no load, and
increases proportionally with load up to a constant
1.6mA after the LTC1148 series has entered continu-
ous mode. Because the DC bias current is drawn from
VIN, the resulting loss increases with input voltage. For
VIN = 10V the DC bias losses are generally less than 1%
for load currents over 30mA. However, at very low load
currents the DC bias current accounts for nearly all of
the loss.
2. MOSFET gate charge current results from switching the
gate capacitance of the power MOSFETs. Each time a
MOSFET gate is switched from low to high to low again,
a packet of charge dQ moves from VIN to ground. The
resulting dQ/dt is a current out of VIN which is typically
much larger than the DC supply current. In continuous
mode, IGATECHG = f (QN + QP). The typical gate charge
for a 0.1
Ω N-channel power MOSFET is 25nC, and for
a P-channel about twice that value. This results in
IGATECHG = 7.5mA in 100kHz continuous operation, for
a 2% to 3% typical mid-current loss with VIN = 10V.
Note that the gate charge loss increases directly with
both input voltage and operating frequency. This is the
principal reason why the highest efficiency circuits
operate at moderate frequencies. Furthermore, it ar-
gues against using larger MOSFETs than necessary to
control I
2R losses, since overkill can cost efficiency as
well as money!
3. I
2R losses are easily predicted from the DC resistances
of the MOSFET, inductor, and current shunt. In continu-
ous mode the average output current flows through L
and RSENSE, but is “chopped” between the P-channel
and N-channel MOSFETs. If the two MOSFETs have
approximately the same RDS(ON), then the resistance of
one MOSFET can simply be summed with the resis-
tances of L and RSENSE to obtain I2R losses. For
example, if each RDS(ON) = 0.1Ω, RL = 0.15Ω, and
RSENSE = 0.05Ω, then the total resistance is 0.3Ω. This
results in losses ranging from 3% to 12% as the output
current increases from 0.5A to 2A. I2R losses cause the
efficiency to roll-off at high output currents.
Figure 5 shows how the efficiency losses in a typical
LTC1148 series regulator end up being apportioned.
Figure 5. Efficiency Loss
OUTPUT CURRENT (A)
0.01
90
95
1
LTC1148 • F05
85
80
0.03
0.1
0.3
3
100
GATE CHARGE
LTC1148 IQ
I2R
The gate charge loss is responsible for the majority of
the efficiency lost in the mid-current region. If Burst
Mode operation was not employed at low currents, the
gate charge loss alone would cause efficiency to drop to
unacceptable levels. With Burst Mode operation, the
DC supply current represents the lone (and unavoid-
able) loss component which continues to become a
higher percentage as output current is reduced. As
expected, the I2R losses dominate at high load currents.
Other losses including CIN and COUT ESR dissipative
losses, MOSFET switching losses, Schottky conduction
losses during dead time, and inductor core losses, gener-
ally account for less than 2% total additional loss.
Design Example
As a design example, assume VIN = 12V (nominal),
VOUT = 5V, IMAX = 2A, and f = 200kHz; RSENSE, CT and L
can immediately be calculated:
RSENSE = 100mV/2 = 0.05Ω
tOFF = (1/200kHz)[1 – (5/12)] = 2.92µs
CT = 2.92µs/[(1.3)(10
4)] = 220pF
LMIN = 5.1(10
5)0.05
Ω(220pF)5V = 28µH
Assume that the MOSFET dissipations are to be limited to
PN = PP = 250mW.
If TA = 50°C and the thermal resistance of each MOSFET
is 50
°C/W, then the junction temperatures will be 63°C



Html Pages

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20


数据表 下载

Go To PDF Page


链接网址



ALLDATASHEET是否为您带来帮助?  [ DONATE ] 

关于 Alldatasheet   |   广告服务   |   联系我们   |   隐私政策   |   数据表链接    |   链接交换   |   制造商名单
All Rights Reserved©Alldatasheet.com


Mirror Sites
English : Alldatasheet.com  |   English : Alldatasheet.net  |   Chinese : Alldatasheetcn.com  |   German : Alldatasheetde.com  |   Japanese : Alldatasheet.jp
Russian : Alldatasheetru.com  |   Korean : Alldatasheet.co.kr  |   Spanish : Alldatasheet.es  |   French : Alldatasheet.fr  |   Italian : Alldatasheetit.com
Portuguese : Alldatasheetpt.com  |   Polish : Alldatasheet.pl  |   Vietnamese : Alldatasheet.vn
Indian : Alldatasheet.in  |   Mexican : Alldatasheet.com.mx  |   British : Alldatasheet.co.uk  |   New Zealand : Alldatasheet.co.nz
Family Site : ic2ic.com  |   icmetro.com