lm27241 National Semiconductor Corporation, lm27241 Datasheet - Page 17
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lm27241
Manufacturer Part Number
lm27241
Description
Synchronous Buck Regulator Controller For Mobile Systems
Manufacturer
National Semiconductor Corporation
Datasheet
1.LM27241.pdf
(27 pages)
Application Information
Only in the case of the input capacitor, the situation may be
different as is explained next.
INPUT CAPACITOR
In a buck regulator, the input capacitor provides most of the
pulsed current waveform demanded by the switch. However
the DC (average) value of the current through a capacitor in
steady state must be zero. Otherwise, the capacitor would
start accumulating charge every cycle, and that would
clearly not represent a ‘steady state’ by definition.
The equation for the RMS current through the input capacitor
is then
The function D(1-D) has a maxima at D = 0.5. This would
correspond to an input voltage of 5V/0.5 = 10V. And the input
capacitor current at this worst case input voltage would be
FIGURE 13. Evaluation Board Transient Response
Iout step = 2.5A to 5A, and 5A to 2.5A Load Step.
V
OUT
= 1.50VDC, V
IN
= 14V
(Continued)
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17
The input capacitors must be positioned physically close to
the power stage.
MOSFETs
Selection of FETs for the controller must be done carefully
taking into account efficiency, thermal dissipation and drive
requirements. Typically the component selection is made
according to the most efficient FET for a given price.
When looking for a FET, it is often helpful to compose a
spreadsheet of key parameters. These parameters may be
summarized as ON resistance (R
rise and fall times (t
device may then be calculated according to the following
equations:
High-side FET:
P = P
Where
P
P
P
Low-side FET:
P = P
Where
P
P
One will note that the gate charge requirements should be
low to ensure good efficiency. However, if a FET’s gate
charge requirement is too low (less than 8nC), the FET can
turn on spuriously. A good starting point for a 10A load is to
use a high-side and low-side FET each with an on resistance
of 5mΩ (FET on resistance is a function of temperature,
therefore it is advisable to apply the appropriate correction
factor provided in the FET datasheet), gate to source charge
of 8nC (total gate charge of 36 nC), t
temperature coefficient of 1.4. For a 5V input and 1.2V/10A
output (f = 300kHz), this yields a power dissipation of 0.62 W
(high-side FET) and 0.54 W (low-side FET). The efficiency is
then 91%. While the same FET may be used for both the
high side and low side, optimal performance may not be
realized.
C
GC
SW
C
GC
= D x (I
= (1 - D) x (I
= 5V x Q
= 5V x Q
= 0.5 x V
C
C
+ P
+ P
OUT
GC
GC
GS
GS
IN
+ P
2
OUT
x R
x I
x f
x f
SW
r
OUT
2
DS_ON
and t
x R
x (t
DS_ON
f
). The power dissipated in a given
)
r
+ t
f
)
) x f
DS_ON
r
= 11ns, t
), gate charge (Q
f
= 47ns, and
www.national.com
GS
),