CS5212

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12

Boost Component Selection for Upper and Lower

FET Gate Drive

The boost (BST) pin provides for application of a higher

voltage to drive the upper FET. This voltage may be

provided by a fixed higher voltage or it may be generated

with a boost capacitor and charging diodes, as shown in

Figure 1. The voltage in the boost configuration would be

the summation of the voltage from the charging diodes and

the output voltage swing. Care must be taken to keep the

peak voltage with respect to ground less than 20 V peak. The

capacitor value should be ten times larger than the

capacitance of the top FET. The boost circuit requires a

modification to achieve startup. See Rpull–up Selection for

boost circuit startup.

Rpull–up Value Selection for Boost Circuit Startup

The CS5212 application circuit incorporates a pull–up

resistor, R7, into the boost circuitry. This resistor is essential

to achieving startup of the boost circuit. At startup, the

GATE(H) output may be limited to 0.8 V, due to internal Vbe

drops. Until the boost circuitry charges up, the GATE(H)

output cannot provide sufficient Vgs to turn on the

MOSFET. A resistor from GATE(H) to BST allows

bypassing of the GATE(H) driver until the boost circuitry is

charged. The time constant, set by the pull–up resistor and

the Cin of the top MOSFET, must be fast enough to turn on

the MOSFET during the switching period. The following

equation is used to determine Rpull–up:

Rpull–up

t

1

(Cin

fSW)

Choosing components according to this equation will

insure that approximately 63% of the boost voltage will be

applied to GATE(H) within one switching period. To start

charge pumping, the control MOSFET must pull up the

switching node above 0.6 V, two Schottky drops, which will

by GATE(H) must be 0.6 V greater than Vth of the top FET.

Both high– and low–side switches must be sublogic level

ensure proper up.