LT1054L 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 require-
ments 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:
V_IN
FB/SHDN V⁺
RX
+
C_IN
CAP⁺ OSC
LT1054
GND V_REF
R1
R2
CAP ^– V_OUT
C1
V_OUT
C_OUT
LT1054 • F06
Figure 6
R_X = V_X/(4.4 I_OUT)
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 with-
out 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
dissipated 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 temperature 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 tempera-
tures. 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 pack-
age. 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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