P

0.432 W

I

0.036 A

V

12 V

=

=

=

UCC27323, UCC27324, UCC27325

UCC37323, UCC37324, UCC37325

www.ti.com

SLUS492H – JUNE 2001 – REVISED MAY 2013

VDD

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

current and the programmed oscillator frequency. Total VDD current is the sum of quiescent VDD current and the

average OUT current. Knowing the operating frequency and the MOSFET gate charge (Qg), average OUT

current is calculated from Equation 1.

where

•

f is frequency

(1)

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

μF ceramic capacitor must be

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

μF) with relatively low

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

capacitors must present a low impedance characteristic for the expected current levels in the driver application.

Drive Current and Power Requirements

The UCC37323/4/5 family of 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 which 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 (see

References).

When a driver IC is tested with a discrete, capacitive load calculating the power that is required from the bias

supply is fairly simple. The energy that must be transferred from the bias supply to charge the capacitor is given

by Equation 2.

E = ½CV

2

where

•

C is the load capacitor

•

V is the bias voltage feeding the driver

(2)

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

power loss given by Equation 3.

P = 2 × ½CV

where

•

f is the switching frequency

(3)

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 helps to clarify this.

P = 10 nF × (12)

(4)

With a 12-V supply, this equates to a current of (see Equation 5):

(5)

current that is due to the IC internal consumption must 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, which 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.

Copyright © 2001–2013, Texas Instruments Incorporated

Submit Documentation Feedback

11

Product Folder Links: UCC27323 UCC27324 UCC27325 UCC37323 UCC37324 UCC37325