LT3507A
22
3507afa
For more information www.linear.com/LT3507A
The maximum allowed power dissipation by the LT3507A
can be determined by:
P
DISS(MAX)
=
T
J(MAX)
T
A
θ
JA
where T
JMAX
is the maximum die temperature of 125°C
(150°C for H-grade).
However, take care in determining T
A
since the catch di-
odes also dissipate power and must be located close to
the LT3507A. Another potential heat sour
ce is the LDO
pass transistor. In a compact layout the pass transistor
will be located close to the LT3507A. The inductors will
also dissipate some power due to their series resistance
and they must be close to the LT3507A. All of these heat
sources will increase the effective ambient temperature
seen by the LT3507A.
A thorough analysis of eight heat sources in a small PCB
area is beyond the scope of this data sheet, however a
number of thermal analysis programs are available to
calculate the temperature rise in each component (such as
PCAnalyze from K&K Associates or Flo Therm PCB from
Mentor). The power dissipation of each component will be
needed to accurately calculate the thermal characteristics
of the system.
The contributors to power dissipation inside the LT3507A
are switch DC loss, switch AC loss, boost current, quies
-
cent current and LDO drive current. The total dissipation
within the LT3507A can be expressed as:
P
DISS
= P
SWDCi
+P
SWACi
+P
BSTi
( )
+P
Q
+P
LDO
i=1
3
The switch DC and AC losses in channel i are:
P
SWDCi
=
R
SWi
I
OUTi
( )
2
V
OUTi
V
INi
P
SWACi
= 17ns I
OUTi
( )
V
INi
( )
f
( )
applications inForMation
where R
SWi
is the equivalent switch resistance (0.18Ω for
channel 1 and 0.22Ω for channels 2 and 3) and f is the
operating frequency.
The boost loss in channel i is:
P
BSTi
=
V
OUTi
V
BOOSTi
( )
I
OUTi
50
+0.02A
V
INi
The quiescent loss is:
P
Q
= V
IN1
(I
Q(VIN1)
) + V
BIAS
(I
Q(BIAS)
)
If the BIAS pin does not have a voltage of at least 3V ap-
plied, then V
IN1
must replace V
BIAS
in the equation. Also,
I
Q(VIN1)
can be reduced by 0.2mA (typ) if the LDO is shut
off (see the LDO section).
The LDO drive loss is:
P
LDO
=(V
BIAS
V
LDO(OUT)
0.7V)
I
OUT(LDO)
β
PASS
,
if V
BIAS
V
LDO(OUT)
+1.5V
or
P
LDO
=(V
IN1
V
LDO(OUT)
0.7V)
I
OUT(LDO)
β
PASS
,
if V
BIAS
<V
LDO(OUT)
+1.5V
where β
PASS
is the current gain of the external pass
transistor.
Next, the power in the external components must be taken
into account. The diode power is given by:
P
DIODE
=
V
F
V
IN
V
OUT
V
F
( )
I
OUT
V
IN
where V
F
is the forward drop of the diode at I
OUT
.
The inductor power is:
P
IND
= (I
OUT
)
2
ESR
IND
where ESR
IND
is the inductor equivalent series resistance.
LT3507A
23
3507afa
For more information www.linear.com/LT3507A
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 LT3507A 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 LT3507A 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 finally:
P
NPN
= 0.2A(3.3V – 3V) = 0.06W
Thus the total power dissipated by the LT3507A 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.
LT3507A
24
3507afa
For more information www.linear.com/LT3507A
typical applications
3.3V, 5V and 12V from a 24V Input with Ratiometric Tracking
V
IN1
BOOST1 UVLO
OVLO
BOOST3
SW3
FB3
V
C3
BIAS
DRIVE
FB4
PGOOD1
PGOOD2
PGOOD3
PGOOD1
PGOOD2
PGOOD3
SW1
FB1
V
C1
TRK/SS1
TRK/SS2
TRK/SS3
BOOST2
0.1µF
10µF
50V
22µF
D1
16.2k
1000pF
1.5nF
34.8k
L1 3.3µH
V
OUT1
3.3V
2A
V
IN
21V TO 27V
V
OUT2
5V
1.2A
L2 6.8µH
L3 10µH
BAT54
V
IN2
V
IN3
V
INSW
100k
49.9k
4.53k
OVLO = 29V
UVLO = 16V
100k
4.32k
V
OUT1
11k
41.2k
BAT54
BAT54
18.2k
SW2
FB2
V
C2
RT/SYNC TRK/SS4
GND
LT3507A
RUN1
RUN2
RUN3
0.1µF
0.1µF
22µF
10µF
D2
D3
24.3k
26.7k
SHDN
470nF
470pF
53.6k
68.1kNC
V
OUT1
D1: ON SEMI MBRS230LT3
D2, D3: ON SEMI MBRA130LT3
L1: COILCRAFT DO1813H-332ML
L2: COILCRAFT DO1813H-682ML
L3: COILCRAFT DO1813H-103ML
10k
3507A TA02
54.9k
f
SW
= 800kHz
10.2k
100k 100k
V
OUT3
12V
1A
140k
V
OUT1

LT3507AIFE#PBF

Mfr. #:
Manufacturer:
Analog Devices / Linear Technology
Description:
Switching Voltage Regulators 3x Mono Buck Reg w/ LDO
Lifecycle:
New from this manufacturer.
Delivery:
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