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CS5305 Datasheet(PDF) 22 Page - ON Semiconductor |
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CS5305 Datasheet(HTML) 22 Page - ON Semiconductor |
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22 / 33 page ![]() CS5305 http://onsemi.com 22 Share Bus Components Five external components are required to implement the module−to−module current share function. These components set the current sense load line, provide the share bus pull−down and compensate the share adjust amplifier. The share current sense amplifier monitors the total module current and provides a DC voltage output proportional to that current. The share sense amplifier gain is programmable and allows the user to set the share bus transconductance. It is important that all modules in a system have a share current load line that approximates that of all the other modules to ensure accurate module−to−module current sharing. Let us arbitrarily set the share bus maximum voltage for full load at 2 V. If a module is designed to provide 81 A at full load, the module share transconductance should be 81 A/2 V or 40.5 A/V. Two resistors and a capacitor set the share current sense amplifier gain. The resistors set the DC gain while the capacitor provides a zero to minimize errors due to noise. The total module current is measured as described in the section dealing with the OCSET current limit function. That is, each phase within a module generates a voltage between the CSx and CSREF leads that is proportional to the current flow in the output inductor and the inductor’s ESR: VCSx * VCSREF + (IL)(ESRL) This signal is amplified by a factor of 3.7 for each phase and then summed for all three phases. This signal is provided as input to the share current sense amplifier. If we assume that all three phases are sharing current equally within a single module, the input to the share current sense amplifier can be expressed as: VIN(SENSE) + 11.1(VCSx * VCSREF) + 11.1(IL)(ESRL) + 3.7(IOUT)(ESRL) If we set IL equal to the maximum per phase current at full load, and if we know the value of ESR for our inductors, we can calculate the required share current sense amplifier gain as: AV(SHARESENSE) + share maximimum voltage VIN(SENSE) As an example, let us again consider the case for a module providing full load current of 81 A. Each output phase is conducting 27 A. If we assume ESR = 1.5 mΩ then input to the share current sense amplifier is (11.1)(27 A)(1.5 mΩ) = 0.45 V. The required share current sense amplifier gain is then 2 V/0.45 V = 4.44. + − RIOUT CIOUT IFB RIFB VIN(SENSE) IOUT Figure 38. From the schematic in Figure 38, we derive the DC gain as: AV(SHARESENSE) + 4.44 + RIOUT RIFB ) 1 This specifies that RIOUT should be 3.44 times greater than RIFB. Another important consideration is the type of resistor selected for RIFB. The thermal performance of RIFB must match that of whatever sense element is being used to monitor module current. Inductive sensing has been shown to be reasonably accurate, but copper’s thermal coefficient of resistivity is approximately +4000 parts per million per °C (0.4% per °C). In order to maintain accurate control of the share bus over temperature, RIFB must have a similar thermal coefficient. This requires a positive temperature coefficient element such as the KOA−Speer LT73. If a standard sense resistor is used in series between the inductor and the load, there is no need to use special resistors for sensing, but efficiency will suffer due to power dissipation in the sense resistor. As regards the value of CIOUT, it should be noted that the complete transfer function for the share current sense amplifier in Figure 38 is: AV(SHARESENSE) + RIOUT RIFB(1 ) sCIOUT RIOUT) ) 1 CIOUT causes the gain for high frequency noise to decrease, thus quieting the share bus. The share resistor provides a passive pull−down on the SHARE lead. This allows the share bus voltage to be pulled all the way down to ground. The share resistor is selected to satisfy a number of criteria. First, the resistor cannot be made too small. The SHARE lead source current is guaranteed to be above 1 mA and must be capable of driving the SHARE lead voltage to 3 V. The share bus of one module serves as master to all and drives the total resistance of all SHARE leads. Thus, the total impedance of all share resistors should be made greater than or equal to 3 kΩ. That is, |
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