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ADP3192A Datasheet(PDF) 20 Page - Analog Devices |
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ADP3192A Datasheet(HTML) 20 Page - Analog Devices |
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20 / 32 page ![]() ADP3192A Rev. 0 | Page 20 of 32 If the resulting ripple voltage is less than what is designed for, the inductor can be made smaller until the ripple value is met. This allows optimal transient response and minimum output decoupling. The smallest possible inductor should be used to minimize the number of output capacitors. For this example, choosing a 320 nH inductor is a good starting point and gives a calculated ripple current of 11 A. The inductor should not saturate at the peak current of 35.5 A and should be able to handle the sum of the power dissipation caused by the average current of 30 A in the winding and core loss. Another important factor in the inductor design is the dc resistance (DCR), which is used for measuring the phase currents. A large DCR can cause excessive power loss, though too small a value can lead to increased measurement error. A good rule is to have the DCR (RL) be about 1 to 1½ times the droop resistance (RO). This example uses an inductor with a DCR of 1.4 mΩ. Designing an Inductor Once the inductance and DCR are known, the next step is to either design an inductor or find a standard inductor that comes as close as possible to meeting the overall design goals. It is also important to have the inductance and DCR tolerance specified to control the accuracy of the system. Reasonable tolerances most manufacturers can meet are 15% inductance and 7% DCR at room temperature. The first decision in designing the inductor is choosing the core material. Several possibilities for providing low core loss at high frequencies include the powder cores (from Micrometals, Inc., for example, or Kool Mu® from MAGNETICS®) and the gapped soft ferrite cores (for example, 3F3 or 3F4 from Philips®). Low frequency powdered iron cores should be avoided due to their high core loss, especially when the inductor value is relatively low, and the ripple current is high. The best choice for a core geometry is a closed-loop type, such as a potentiometer core (PQ, U, or E core) or toroid. A good compromise between price and performance is a core with a toroidal shape. Many useful magnetics design references are available for quickly designing a power inductor, such as • Intusoft Magnetics Designer Software • Designing Magnetic Components for High Frequency dc-dc Converters , by Colonel Wm. T. McLyman, Kg Magnetics, Inc., ISBN 1883107008 Selecting a Standard Inductor The following power inductor manufacturers can provide design consultation and deliver power inductors optimized for high power applications upon request: • Coilcraft® • Coiltronics® • Sumida Corporation® CURRENT SENSE AMPLIFIER Most designs require the regulator output voltage, measured at the CPU pins, to drop when the output current increases. The specified voltage drop corresponds to a dc output resistance (RO), also referred to as a load line. The ADP3192A has the flexibility of adjusting RO, independent of current-limit or compensation components, and it can also support CPUs that do not require a load line. For designs requiring a load line, the impedance gain of the CS amplifier (RCSA) must be to be greater than or equal to the load line. All designs, whether they have a load line or not, should keep RCSA ≥ 1 mΩ. The output current is measured by summing the voltage across each inductor and passing the signal through a low-pass filter. This summer filter is the CS amplifier configured with resistors RPH(X) (summers), and RCS and CCS (filter). The impedance gain of the regulator is set by the following equations, where RL is the DCR of the output inductors: () L X PH CS CSA R R R R × = (6) CS L CS R R L C × = (7) The user has the flexibility to choose either RCS or RPH(X). However, it is best to select RCS equal to 100 kΩ, and then solve for RPH(X) by rearranging Equation 6. Here, RCSA = RO = 1 mΩ because this is equal to the design load line. () () Ω k 140 Ω k 100 mΩ 0 . 1 Ω m 4 . 1 = × = × = X PH CS CSA L X PH R R R R R Next, use Equation 7 to solve for CCS. nF 8 2 . 2 Ω k 100 Ω m 4 . 1 nH 320 = × = CS C |
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