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UCC27325P Datasheet(PDF) 10 Page - Texas Instruments |
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UCC27325P Datasheet(HTML) 10 Page - Texas Instruments |
10 / 15 page UCC27323, UCC27324, UCC27325 UCC37323, UCC37324, UCC37325 SLUS492B – JUNE 2001 – REVISED SEPTEMBER 2002 10 www.ti.com APPLICATION INFORMATION 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. 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 + 1 2 CV2, where C is the load capacitor and V is the bias voltage feeding the driver. 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 1 2 CV2f, 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 VDD = 12 V, CLOAD = 10 nF, and f = 300 kHz, the power loss can be calculated as: P = 10 nF x (12)2 x (300 kHz) = 0.432 W With a 12-V supply, this would equate to a current of: I + P V + 0.432 W 12 V + 0.036 A The actual current measured from the supply was 0.037 A, and is very close to the predicted value. But, the IDD 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. |
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