
MAX1778/MAX1880–MAX1885
Quad-Output TFT LCD DC/DC
Converters with Buffer
where I MAIN includes the primary load current and the
[ ]
P LDO ( INT ) = LDO V SUPL - ( V LDO + 0 . 7 V )
input supply currents for the charge pumps (see the
Charge-Pump Input Power and Efficiency
Considerations section), linear regulator, and VCOM
buffer.
The linear regulator generates an output voltage by dis-
sipating power across an internal pass transistor, so
the power dissipation is simply the load current times
the input-to-output voltage differential:
P LDO ( INT ) = I LDO ( V SUPL - V LDO )
When driving an external transistor, the internal linear
regulator provides the base drive current. Depending
on the external transistor’s current gain ( β ) and the
maximum load current, the power dissipated by the
internal linear regulator can still be significant:
I
β
= I LDOOUT ( V SUPL - V LDOOUT )
The charge pumps provide regulated output voltages
by dissipating power in the low-side n-channel
MOSFET, so they could be modeled as linear regula-
tors followed by unregulated charge pumps. Therefore,
their power dissipation is similar to a linear regulator:
P MAX = (T J(MAX ) - T A )/( θ JB + θ BA )
where T J - T A is the temperature difference between
the controller’s junction and the surrounding air, θ JB (or
θ JC ) is the thermal resistance of the package to the
board, and θ BA is the thermal resistance from the PCB
to the surrounding air.
Design Procedure
Main Step-Up Converter
Output-Voltage Selection
Adjust the output voltage by connecting a voltage-
divider from the output (V MAIN ) to FB to GND (see the
Typical Operating Circuit). Select R2 in the 10k ? to
50k ? range. Calculate R1 with the following equations:
R1 = R2 [(V MAIN /V REF ) - 1]
where V REF = 1.25V. V MAIN can range from V IN to 13V.
Inductor Selection
Inductor selection depends upon the minimum required
inductance value, saturation rating, series resistance, and
size. These factors influence the converter’s efficiency,
maximum output load capability, transient response time,
P POS = I POS [ ( V SUPP DIODE ) N +
P NEG = I NEG [ ( V SUPN - 2 V DIODE ) N - V NEG ]
- 2 V V SUPD - V POS
]
and output-voltage ripple. For most applications, values
between 4.7μH and 22μH work best with the controller’s
switching frequency (Tables 1 and 2).
The inductor value depends on the maximum output
? V IN ( MIN ) ? ? V MAIN IN ( MIN ) ? ? 1 ?
L MIN = ? ? ? I f ? ? ? η
where N is the number of charge-pump stages, V DIODE
is the diodes ’ forward voltage, and V SUPD is the
positive charge-pump diode supply (Figure 4).
The VCOM buffer’s power dissipation depends on the
capacitive load (C LOAD ) being driven, the peak-to-
peak voltage change (V P-P ) across the load, and the
load’s switching rate:
P BUF = V P - P C LOAD f LOAD V SUPB
To find the total power dissipated in the device, the
power dissipated by each regulator and the buffer must
be added together:
P TOTAL = P STEP - UP + P LDO ( INT )
+ P NEG + P POS + P BUF
The maximum allowed power dissipation is 975mW (24-
pin TSSOP)/879mW (20-pin TSSOP) or:
Maxim Integrated
load the application must support, input voltage, output
voltage, and switching frequency. With high inductor
values, the MAX1778/MAX1880–MAX1885 source high-
er output currents, have less output ripple, and enter
continuous conduction operation with lighter loads;
however, the circuit’s transient response time is slower.
On the other hand, low-value inductors respond faster
to transients, remain in discontinuous conduction oper-
ation, and typically offer smaller physical size for a
given series resistance and current rating. The equa-
tions provided here include a constant LIR, which is the
ratio of the peak-to-peak AC inductor current to the
average DC inductor current. For a good compromise
between the size of the inductor, power loss, and
output-voltage ripple, select an LIR of 0.3 to 0.5. The
inductance value is then given by:
2
- V
? V MAIN ? ? MAIN ( MAX ) OSC ? ? LIR ?
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