UCC27423-Q1

UCC27424-Q1

UCC27425-Q1

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SGLS274D

– SEPTEMBER 2008 – REVISED AUGUST 2011

VDD

Although quiescent VDD current is very low, total supply current is higher, depending on OUTA and OUTB

current can be calculated from:

For the best high-speed circuit performance, two VDD bypass capacitors are recommended tp prevent noise

problems. The use of surface mount components is highly recommended. A 0.1-

μF ceramic capacitor should be

located closest to the VDD to ground connection. In addition, a larger capacitor (such as 1-

μF) with relatively low

ESR should be connected in parallel, to help deliver the high current peaks to the load. The parallel combination

of capacitors should present a low impedance characteristic for the expected current levels in the driver

application.

Drive Current and Power Requirements

The UCC2742x drivers are capable of delivering 4 A of current to a MOSFET gate for a period of several

hundred nanoseconds. High peak current is required to turn the device ON quickly. Then, to turn the device OFF,

the driver is required to sink a similar amount of current to ground. This repeats at the operating frequency of the

power device. A MOSFET is used in this discussion because it is the most common type of switching device

used in high frequency power conversion equipment.

References 1 and 2 discuss the current required to drive a power MOSFET and other capacitive-input switching

devices. Reference 2 includes information on the previous generation of bipolar IC gate drivers.

When a driver IC is tested with a discrete, capacitive load it is a fairly simple matter to calculate the power that is

required from the bias supply. The energy that must be transferred from the bias supply to charge the capacitor

is given by:

E =

There is an equal amount of energy transferred to ground when the capacitor is discharged. This leads to a

power loss given by the following:

P = 2

This power is dissipated in the resistive elements of the circuit. Thus, with no external resistor between the driver

and gate, this power is dissipated inside the driver. Half of the total power is dissipated when the capacitor is

charged, and the other half is dissipated when the capacitor is discharged. An actual example using the

conditions of the previous gate drive waveform should help clarify this.

P = 10 nF

With a 12-V supply, this would equate to a current of:

I = P / V = 0.432 W / 12 V = 0.036 A

current that is due to the IC internal consumption should be considered. With no load, the IC current draw is

0.0027 A. Under this condition, the output rise and fall times are faster than with a load. This could lead to an

almost insignificant, yet measurable, current due to cross-conduction in the output stages of the driver. However,

these small current differences are buried in the high-frequency switching spikes and are beyond the

measurement capabilities of a basic lab setup. The measured current with 10-nF load is reasonably close to that

expected.

The switching load presented by a power MOSFET can be converted to an equivalent capacitance by examining

the gate charge required to switch the device. This gate charge includes the effects of the input capacitance plus

the added charge needed to swing the drain of the device between the ON and OFF states. Most manufacturers

provide specifications that provide the typical and maximum gate charge, in nC, to switch the device under

power:

P = C

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