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LM3743 Datasheet(PDF) 21 Page - Texas Instruments |
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LM3743 Datasheet(HTML) 21 Page - Texas Instruments |
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21 / 38 page ![]() VIN(MAX) - VOUT 'IOUT = fSW x LACTUAL x DMIN L = 1.34 µH L = 5.5V - 1.8V 0.3 x 10A x 300 kHz x 1.8V 5.5V L = VIN-MAX - VOUT 'IOUT x fSW x DMIN LM3743 www.ti.com SNVS427E – SEPTEMBER 2006 – REVISED APRIL 2013 DESIGN CONSIDERATIONS The following is a design procedure for selecting all the components in the Typical Application circuit on the front page. This design converts 5V (VIN) to 1.8V (VOUT) at a maximum load of 10A with an efficiency of 90% and a switching frequency of 300 kHz. The same procedures can be followed to create many other designs with varying input voltages, output voltages, load currents, and switching frequency. Switching Frequency Selection of the operating switching frequency is based on trade-offs between size, cost, efficiency, and response time. For example, a lower frequency will require larger more expensive inductors and capacitors. While a higher switching frequency will generally reduce the size of these components, but will have a reduction in efficiency. Fast switching converters allow for higher loop gain bandwidths, which in turn have the ability to respond quickly to load and line transients. For the example application we have chosen a 300 kHz switching frequency because it will reduce the switching power losses and in turn allow for higher conduction losses considering the same power loss criteria, thus it is possible to sustain a higher load current. Output Inductor The output inductor is responsible for smoothing the square wave created by the switching action and for controlling the output current ripple ( ΔIOUT) also called the AC component of the inductor current. The DC current into the load is equal to the average current flowing in the inductor. The inductance is chosen by selecting between trade-offs in efficiency, size, and response time. The recommended percentage of AC component to DC current is 30% to 40%, this will provide the best trade-off between energy requirements and size, (read AN-1197 (SNVA038) for theoretical analysis). Another criteria is the ability to respond to large load transient responses; the smaller the output inductor, the more quickly the converter can respond. The equation for output inductor selection is: (12) (13) (14) Here we have plugged in the values for input voltage, output voltage, switching frequency, and 30% of the maximum load current. This yields an inductance of 1.34 µH. The output inductor must be rated to handle the peak current (also equal to the peak switch current), which is (IOUT + (0.5 x ΔIOUT)) = 11.5A, for a 10A design and a AC current of 3A. The Coiltronics DR125–1R5 is 1.5 µH, is rated to 13.8A RMS current, and has a direct current resistance (DCR) of 3 m Ω. After selecting the Coiltronics DR125–1R5 for the output inductor, actual inductor current ripple must be re-calculated with the selected inductance value. This information is needed to determine the RMS current through the input and output capacitors. Re-arranging the equation used to select inductance yields the following: (15) VIN(MAX) is assumed to be 10% above the steady state input voltage, or 5.5V at VIN = 5.0V. The re-calculated current ripple will then be 2.69A. This gives a peak inductor/switch current will be 11.35A. Output Capacitor The output capacitor in a switching regulator is selected on the basis of capacitance, equivalent series resistance (ESR), size, and cost. In this example the output current is 10A and the expected type of capacitor is an aluminum electrolytic, as with the input capacitors. An important specification in switching converters is the output voltage ripple ΔVOUT. At 300 kHz the impedance of most capacitors is very small compared to ESR, hence ESR becomes the main selection criteria. In this design the load requires a 2% ripple , which results in a ΔVOUT of 36 mVP-P. Thus the maximum ESR is then: Copyright © 2006–2013, Texas Instruments Incorporated Submit Documentation Feedback 21 Product Folder Links: LM3743 |
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