LTC3553-2_15 Datasheet, PDF (31/36 Pages) Linear Technology – Micropower USB Power Manager with Li-Ion Charger, Always-On LDO and Buck Regulator

LTC3553-2
OPERATION
part. For high charge currents the LTC3553-2 power dis-
sipation is approximately:
P_D = (V_BUS–BAT) • I_BAT + P_D(REGS)
where P_D is the total power dissipated, V_BUS is the supply
voltage, BAT is the battery voltage, and I_BAT is the battery
charge current. P_D(REGS) is the sum of power dissipated
on chip by the step-down switching regulators.
The power dissipated by the buck regulator can be esti-
mated as follows:
P_D(BUCK) = (B_OUTx • I_OUT) • (100 - Eff)/100
Where B_OUTx is the programmed output voltage, I_OUT is
the load current and Eff is the % efficiency which can
be measured or looked up on an efficiency table for the
programmed output voltage.
The power dissipated by the LDO regulator can be esti-
mated using:
P_D(LDO) = (V_INLDO – V_LDO) • I_LDO
where V_INLDO is the LDO input supply voltage, V_LDO is
the LDO regulated output voltage, and I_LDO is the LDO
load current.
Thus the power dissipated by all regulators is:
P_D(REGS) = P_D(BUCK) + P_D(LDO)
It is not necessary to perform any worst-case power dissi-
pation scenarios because the LTC3553-2 will automatically
reduce the charge current to maintain the die temperature
at approximately 110°C. However, the approximate ambi-
ent temperature at which the thermal feedback begins to
protect the IC is:
T_A = 110°C – P_D • θ_JA
Example: Consider the LTC3553-2 operating from a wall
adapter with 5V (V_BUS) providing 400mA (I_BAT) to charge a
Li-Ion battery at 3.3V (BAT). Also assume P_D(REGS) = 0.3W,
so the total power dissipation is:
P_D = (5V – 3.3V) • 400mA + 0.3W = 0.98W
The ambient temperature above which the LTC3553-2
will begin to reduce the 400mA charge current, is ap-
proximately:
T_A = 110°C – 0.98W • 70°C/W = 41.4°C
The LTC3553-2 can be used above 41.4°C, but the charge
current will be reduced below 400mA. The charge current
at a given ambient temperature can be approximated by:
P_D = (110°C – T_A) / θ_JA = (V_BUS – BAT) • I_BAT + P_D(REGS)
Thus:
I_BAT
=
[(110°C – T_A) / θ_JA –P_D(REGS)]
(V_BUS –BAT)
Consider the above example with an ambient tem-
perature of 60°C. The charge current will be reduced to
approximately:
I_BAT = [(110°C – 60°C) / 70°C/W – 0.3W] / (5V – 3.3V)
I_BAT = (0.71W – 0.3W) / 1.7V = 241mA
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