Datasheet

D =
V
OUT
+ V
SW_BOT
+ I
OUT
x R
DC
V
IN
+ V
SW_BOT
- V
SW_TOP
I
irrms
= I(I2d)I(Id1)I(I
21
2
av2
2
av1
-++-+- 3d)I
2
av
LM26420
SNVS579F –FEBRUARY 2009–REVISED MARCH 2013 www.ti.com
with a saturation current limit of > 2.3A. There is no need to specify the saturation or peak current of the inductor
at the 3.25A typical switch current limit. The difference in inductor size is a factor of 5. Ferrite based inductors
are preferred to minimize core losses when opperating with the frequencies used by the LM26420. This presents
little restriction since the variety of ferrite-based inductors is huge. Lastly, inductors with lower series resistance
(R
DCR
) will provide better operating efficiency. For recommended inductors see Example Circuits.
INPUT CAPACITOR SELECTION
The input capacitors provide the AC current needed by the nearby power switch so that current provided by the
upstream power supply does not carry a lot of AC content, generating less EMI. To the buck regulator in
question, the input capacitor also prevents the drain voltage of the FET switch from dipping when the FET is
turned on, therefore providing a healthy line rail for the LM26420 to work with. Since typically most of the AC
current is provided by the local input capacitors, the power loss in those capacitors can be a concern. In the case
of the LM26420 regulator, since the two channels operate 180° out of phase, the AC stress in the input
capacitors is less than if they operated in phase. The measure for the AC stress is called input ripple RMS
current. It is strongly recommended that at least one 10µF ceramic capacitor be placed next to each of the VIND
pins. Bulk capacitors such as electrolytic capacitors or OSCON capacitors can be added to help stabilize the
local line voltage, especially during large load transient events. As for the ceramic capacitors, use X7R or X5R
types. They maintain most of their capacitance over a wide temperature range. Try to avoid sizes smaller than
0805. Otherwise significant drop in capacitance may be caused by the DC bias voltage. See OUTPUT
CAPACITOR SELECTION section for more information. The DC voltage rating of the ceramic capacitor should
be higher than the highest input voltage.
Capacitor temperature is a major concern in board designs. While using a 10µF or higher MLCC as the input
capacitor is a good starting point, it is a good idea to check the temperature in the real thermal environment to
make sure the capacitors are not over heated. Capacitor vendors may provide curves of ripple RMS current vs.
temperature rise, based on a designated thermal impedance. In reality, the thermal impedance may be very
different. So it is always a good idea to check the capacitor temperature on the board.
Since the duty cycles of the two channels may overlap, calculation of the input ripple RMS current is a little
tedious. Use the following equation.
(10)
I
1
is Channel 1's maximum output current. I
2
is Channel 2's maximum output current. d1 is the non-overlapping
portion of Channel 1's duty cycle D
1
. d2 is the non-overlapping portion of Channel 2's duty cycle D
2
. d3 is the
overlapping portion of the two duty cycles. I
av
is the average input current. I
av
= I
1
·D
1
+ I
2
·D
2
. To quickly determine
the values of d1, d2 and d3, refer to the decision tree in Figure 36. To determine the duty cycle of each channel,
use D = V
OUT
/V
IN
for a quick result or use the following equation for a more accurate result.
(11)
R
DC
is the winding resistance of the inductor. R
DS
is the ON resistance of the MOSFET switch.
Example:
V
IN
= 5V, V
OUT1
= 3.3V, I
OUT1
= 2A, V
OUT2
= 1.2V, I
OUT2
= 1.5A, R
DS
= 170mΩ, R
DC
= 30mΩ. (I
OUT1
is the same as
I
1
in the input ripple RMS current equation, I
OUT2
is the same as I
2
).
First, find out the duty cycles. Plug the numbers into the duty cycle equation and we get D1 = 0.75, and D2 =
0.33. Next, follow the decision tree in Figure 36 to find out the values of d1, d2 and d3. In this case, d1 = 0.5, d2
= D2 + 0.5 - D1 = 0.08, and d3 = D1 - 0.5 = 0.25. I
av
= I
OUT1
·D1 + I
OUT2
·D2 = 1.995A. Plug all the numbers into
the input ripple RMS current equation and the result is I
irrms
= 0.77A.
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