LT1172MJ8 Linear Technology, LT1172MJ8 Datasheet - Page 8

• Manufacturer Part Number: LT1172MJ8
• Description: SP-SWREG/Monolithic, 100kHz 5A 2.5A And 1.25A High Efficiency Switching Regulato
• Manufacturer: Linear Technology

LT1 170/LT1 171/LT1172
OPERATIO
*See note under block diagram.
function; when pulled low with an external resistor, it
programs the LT1170/LT1171/LT1172 to disconnect the
main error amplifier output and connects the output of the
flyback amplifier to the comparator input. The LT1170/
LT1171/LT1172 will then regulate the value of the flyback
pulse with respect to the supply voltage.* This flyback
pulse is directly proportional to output voltage in the
traditional transformer coupled flyback topology regula-
tor. By regulating the amplitude of the flyback pulse, the
output voltage can be regulated with no direct connection
between input and output. The output is fully floating up to
the breakdown voltage of the transformer windings. Mul-
tiple floating outputs are easily obtained with additional
windings. A special delay network inside the LT1170/
LT1171/LT1172 ignores the leakage inductance spike at
the leading edge of the flyback pulse to improve output
regulation.
The error signal developed at the comparator input is
brought out externally. This pin (V
functions. It is used for frequency compensation, current
limit adjustment, soft starting, and total regulator shut-
down. During normal regulator operation this pin sits at a
voltage between 0.9V (low output current) and 2.0V (high
output current). The error amplifiers are current output
(g
adjusting current limit. Likewise, a capacitor coupled
external clamp will provide soft start. Switch duty cycle
goes to zero if the V
diode, placing the LT1170/LT1171/LT1172 in an idle mode.
Pulling the V
shutdown, with only 50 A supply current for shutdown
circuitry biasing. See AN19 for full application details.
Extra Pins on the MiniDIP and Surface Mount Packages
The 8- and 16-pin versions of the LT1172 have the
emitters of the power transistor brought out separately
from the ground pin. This eliminates errors due to ground
pin voltage drops and allows the user to reduce switch
current limit 2:1 by leaving the second emitter (E2) discon-
nected. The first emitter (E1) should always be connected
to the ground pin. Note that switch “on” resistance doubles
when E2 is left open, so efficiency will suffer somewhat
8
m
) types, so this voltage can be externally clamped for
C
pin below 0.15V causes total regulator
U
C
pin is pulled to ground through a
C
) has four different
when switch currents exceed 300mA. Also, note that chip
dissipation will actually increase with E2 open during
normal load operation, even though dissipation in current
limit mode will decrease . See “Thermal Considerations”
next.
Thermal Considerations When Using the MiniDIP and
SW Packages
The low supply current and high switch efficiency of the
LT1172 allow it to be used without a heat sink in most
applications when the TO-220 or TO-3 package is se-
lected. These packages are rated at 50 C/W and 35 C/W
respectively. The miniDIPs, however, are rated at 100 C/W
in ceramic (J) and 130 C/W in plastic (N).
Care should be taken for miniDIP applications to ensure
that the worst case input voltage and load current condi-
tions do not cause excessive die temperatures. The follow-
ing formulas can be used as a rough guide to calculate
LT1172 power dissipation. For more details, the reader is
referred to Application Note 19 (AN19), “Efficiency Calcu-
lations” section.
Average supply current (including driver current) is:
Switch power dissipation is given by:
Total power dissipation is the sum of supply current times
input voltage plus switch power:
In a typical example, using a boost converter to generate
12V at 0.12A from a 5V input, duty cycle is approximately
60%, and switch current is about 0.65A, yielding:
I
I
DC = switch duty cycle
P
R
P
I
P
P
IN
SW
IN
SW
D(TOT)
SW
D(TOT)
SW
= 6mA + 0.65(0.004 + DC/40) = 18mA
= switch current
= (I
= LT1172 switch “on” resistance (1 maximum)
= (0.65)
6mA + I
= (I
= (5V)(0.018A) + 0.25 = 0.34W
SW
)
IN
2
2
SW
)(V
• (R
• (1 )(0.6) = 0.25W
(0.004 + DC/40)
IN
SW
) + P
)(DC)
SW