LT1054 Datasheet, PDF (9/16 Pages) Linear Technology – Switched-Capacitor Voltage Converter with Regulator

LT1054/LT1054L
APPLICATIONS INFORMATION
where dV = peak-to-peak ripple and f = oscillator frequency.
For output capacitors with significant ESR a second term
must be added to account for the voltage step at the switch
transitions. This step is approximately equal to:
(2I_OUT)(ESR of C_OUT)
Power Dissipation
The power dissipation of any LT1054 circuit must be
limited such that the junction temperature of the device
does not exceed the maximum junction temperature rat-
ings. The total power dissipation must be calculated from
two components, the power loss due to voltage drops
in the switches and the power loss due to drive current
losses. The total power dissipated by the LT1054 can be
calculated from:
P ≈ (V_IN – |V_OUT|)(I_OUT) + (V_IN)(I_OUT)(0.2)
where both V_IN and V_OUT are referred to the ground pin
(Pin 3) of the LT1054. For LT1054 regulator circuits, the
power dissipation will be equivalent to that of a linear
regulator. Due to the limited power handling capability of
the LT1054 packages, the user will have to limit output
current requirements or take steps to dissipate some power
external to the LT1054 for large input/output differentials.
This can be accomplished by placing a resistor in series
with C_IN as shown in Figure 6. A portion of the input
voltage will then be dropped across this resistor without
affecting the output regulation. Because switch current is
approximately 2.2 times the output current and the resistor
will cause a voltage drop when C_IN is both charging and
discharging, the resistor should be chosen as:
R_X = V_X/(4.4 I_OUT)
V_IN
FB/SHDN V⁺
R_X
+
C_IN
CAP⁺ OSC
LT1054
GND V_REF
R1
R2
CAP ^– V_OUT
C1
V_OUT
C_OUT
Figure 6
LT1054 • F06
where:
V_X ≈ V_IN – [(LT1054 Voltage Loss)(1.3) + |V_OUT|]
and I_OUT = maximum required output current. The factor
of 1.3 will allow some operating margin for the LT1054.
For example: assume a 12V to – 5V converter at 100mA
output current. First calculate the power dissipation without
an external resistor:
P = (12V – |– 5V|)(100mA) + (12V)(100mA)(0.2)
P = 700mW + 240mW = 940mW
At θ_JA of 130°C/W for a commercial plastic device this
would cause a junction temperature rise of 122°C so that
the device would exceed the maximum junction tempera-
ture at an ambient temperature of 25°C. Now calculate the
power dissipation with an external resistor (R_X). First find
how much voltage can be dropped across R_X. The maxi-
mum voltage loss of the LT1054 in the standard regulator
configuration at 100mA output current is 1.6V, so:
V_X = 12V – [(1.6V)(1.3) + |– 5V|] = 4.9V and
R_X = 4.9V/(4.4)(100mA) = 11Ω
This resistor will reduce the power dissipated by the
LT1054 by (4.9V)(100mA) = 490mW. The total power dis-
sipated by the LT1054 would then be (940mW – 490mW)
= 450mW. The junction temperature rise would now be
only 58°C. Although commercial devices are guaranteed
to be functional up to a junction temperature of 125°C, the
specifications are only guaranteed up to a junction tem-
perature of 100°C, so ideally you should limit the junction
temperature to 100°C. For the above example this would
mean limiting the ambient temperature to 42°C. Other
steps can be taken to allow higher ambient temperatures.
The thermal resistance numbers for the LT1054 packages
represent worst-case numbers with no heat sinking and
still air. Small clip-on type heat sinks can be used to lower
the thermal resistance of the LT1054 package. In some
systems there may be some available airflow which will
help to lower the thermal resistance. Wide PC board traces
from the LT1054 leads can also help to remove heat from
the device. This is especially true for plastic packages.
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