Texas Instruments LMZ14201 SIMPLE SWITCHER, Руководство пользователя

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500 mA/Div 2.00
P
s/Div
500 mA/Div 2.00
P
s/Div
LMZ14201
www.ti.com SNVS649E – JANUARY 2010 – REVISED MARCH 2013
Following is a comparison pair of waveforms of the showing both CCM (upper) and DCM operating modes.
Figure 33. CCM and DCM Operating Modes
V
IN
= 12V, V
O
= 3.3V, I
O
= 1 A / 0.25 A
The approximate formula for determining the DCM/CCM boundary is as follows:
I
DCB
≊ V
O
*(V
IN
–V
O
)/(2*10 μ H*f
SW(CCM)
*V
IN
) (16)
Following is a typical waveform showing the boundary condition.
Figure 34. Transition Mode Operation
V
IN
= 24V, V
O
= 3.3V, I
O
= 0.29 A
The inductor internal to the module is 10
μ
H. This value was chosen as a good balance between low and high
input voltage applications. The main parameter affected by the inductor is the amplitude of the inductor ripple
current (I
LR
). I
LR
can be calculated with:
I
LR P-P
=V
O
*(V
IN
- V
O
)/(10µH*f
SW
*V
IN
) (17)
Where V
IN
is the maximum input voltage and f
SW
is determined from Equation 10 .
If the output current I
O
is determined by assuming that I
O
= I
L
, the higher and lower peak of I
LR
can be
determined. Be aware that the lower peak of I
LR
must be positive if CCM operation is required.
POWER DISSIPATION AND BOARD THERMAL REQUIREMENTS
For the design case of V
IN
= 24V, V
O
= 3.3V, I
O
= 1A, T
AMB(MAX)
= 85°C , and T
JUNCTION
= 125°C, the device must
see a thermal resistance from case to ambient of less than:
θ
CA
< (T
J-MAX
— T
AMB(MAX)
) / P
IC-LOSS
- θ
JC
(18)
Given the typical thermal resistance from junction to case to be 1.9 °C/W. Use the 85°C power dissipation curves
in the
Typical Performance Characteristics section to estimate the P
IC-LOSS
for the application being designed. In
this application it is 0.52W.
θ
CA
= (125 — 85) / 0.52W — 1.9 = 75
To reach
θ
CA
= 75, the PCB is required to dissipate heat effectively. With no airflow and no external heat, a good
estimate of the required board area covered by 1 oz. copper on both the top and bottom metal layers is:
Board Area_cm
2
= 500°C x cm
2
/W /
θ
JC
(19)
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