Datasheet

LT3507
23
3507fa
The LDO pass transistor power is:
P
NPN
= I
OUTLDO
(V
C
– V
OUTLDO
)
where V
C
is the collector voltage on the NPN pass tran-
sistor.
Example: An LT3507 design requirements are:
V
IN
= 8V, f= 500kHz
V1 = 2.5V at I1 = 1.6A
V2 = 3.3V at I2 = 0.8A (used for boost, bias and V4)
V3 = 1.2V at I3 = 1A
V4 = 3V at I4 = 0.2A (from 3.3V output)
T
A
= 50°C, T
JMAX
= 125°C
θ
JA
= 34°C/W
Schottky V
F
= 0.45V and Inductor ESR = 0.05Ω
P
DISS(MAX)
=
125°C–50°C
34°C/W
= 2.2W
P
SWDC1
=
0.18Ω 1.6A
()
2
2.5V
8V
= 0.14W
P
SWAC1
=17ns 1.6A
()
8V
()
500k
()
= 0.11W
P
BST1
=
2.5V 3.3V
()
1.6A
50
+ 0.02A
8V
= 0.06W
Similarly, P
SWDC2
= 0.09W, P
SWAC2
= 0.07W, P
BST2
=
0.06W, P
SWDC3
= 0.03W, P
SWAC3
= 0.07W and P
BST2
=
0.03W. Remember, the total current from channel 2 is I2
+ I4 since the LDO pass transistor draws from V2. Ignore
bias and boost currents.
P
Q
= 8V 3.5mA
()
+ 3.3V 7.5mA
()
= 0.05W
P
LDO
= 8V
0.2A
100
= 0.02W
APPLICATIONS INFORMATION
The total dissipation on the LT3507 is the sum of all these
and is equal to 0.73W. Note that this is less than half of
P
DISS(MAX)
. Next, the power dissipation of the external
components are:
P
DIODE1
=
0.45V 8V 2.5V 0.45
()
1.6A
8V
= 0.46W
P
IND1
= 1.6A
()
2
0.05Ω=0.13W
Similarly, P
DIODE2
= 0.24W, P
IND2
= 0.05W, P
DIODE3
=
0.36W and P
IND3
= 0.05W. And fi nally:
P
NPN
= 0.2A(3.3V – 3V) = 0.06W
Thus the total power dissipated by the LT3507 and external
components is 2.08W. The thermal analysis will use these
power dissipations to calculate the internal component
temperatures. Make sure that none of the components
exceed their rated temperature limits.
RELATED LINEAR TECHNOLOGY PUBLICATIONS
Application Notes 19, 35, 44, 76 and 88 contain more
detailed descriptions and design information for buck
regulators and other switching regulators. The LT1375
data sheet has a more extensive discussion of output
ripple, loop compensation, and stability testing. Design
Note 318 shows how to generate a dual polarity output
supply using a buck regulator.