P + C

V2

f + Qg

f

I +

P

V

+

0.432 W

12 V

+ 0.036 A

P + 2

1

2

CV2f

UCC27423, UCC27424, UCC27425

SLUS545D – NOVEMBER 2002 – REVISED MAY 2013

www.ti.com

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:

, where f is the switching frequency.

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.

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

(1)

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

0.0027A. 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 10nF 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

specified conditions. Using the gate charge Qg, one can determine the power that must be dissipated when

charging a capacitor. This is done by using the equivalence Qg = CeffV to provide the following equation for

power:

(2)

This equation allows a power designer to calculate the bias power required to drive a specific MOSFET gate at a

specific bias voltage.

ENABLE

UCC27423/4/5 provides dual Enable inputs for improved control of each driver channel operation. The inputs

for active high operation. When ENBA and ENBB are driven high, the drivers are enabled and when ENBA and

ENBB are low, the drivers are disabled. The default state of the Enable pin is to enable the driver and therefore

can be left open for standard operation. However, if the enable pin is left open, it is recommended to terminate

any PCB traces to be as short as possible to limit noise. If large noise is present due to non-optimal PCB layout,

it is recommended to tie the Enable pin to Vcc or to add a filter capacitor (0.1 µF) to the Enable pin. The output

states when the drivers are disabled is low regardless of the input state. See the truth table of Table 1 for the

operation using enable logic.

Enable input are compatible with both logic signals and slow changing analog signals. They can be directly

driven or a power-up delay can be programmed with a capacitor between ENBA, ENBB and AGND. ENBA and

ENBB control input A and input B respectively.

10

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