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UCC27423-Q1 Datasheet(PDF) 17 Page - Texas Instruments |
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UCC27423-Q1 Datasheet(HTML) 17 Page - Texas Instruments |
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17 / 28 page  UCC27423-Q1 UCC27424-Q1 UCC27425-Q1 www.ti.com 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 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 can be calculated from: IOUT = Qg × f, where f is frequency 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 = ½CV2, 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 × ½CV2f, 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 × (12)2 × (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. 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: P = C × V2 × f = Qg × f Copyright © 2008–2011, Texas Instruments Incorporated 17 |
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