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LT3579 Datasheet(PDF) 29 Page - Linear Technology |
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LT3579 Datasheet(HTML) 29 Page - Linear Technology |
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29 / 40 page ![]() LT3579/LT3579-1 29 35791f feedback loop. As with any feedback loop, identifying the gain and phase contribution of the various elements in the loop is critical. Figure 18 shows the key equivalent elements of a boost converter. Because of the fast current control loop, the powerstageoftheIC,inductoranddiodehavebeenreplaced by a combination of the equivalent transconductance amplifier gmp and the current controlled current source (which converts IVIN to ηVIN VOUT IVIN ). Gmp acts as a current APPENDIX 1.215V REFERENCE IVIN • VIN VOUT • IVIN VOUT COUT CPL RESR RL RO RC CC CF R1 FB R2 R2 – + 3579 F18 gmp gma CC: COMPENSATION CAPACITOR COUT: OUTPUT CAPACITOR CPL: PHASE LEAD CAPACITOR CF: HIGH FREQUENCY FILTER CAPACITOR gma: TRANSCONDUCTOR AMPLIFIER INSIDE IC gmp: POWER STAGE TRANSCONDUCTANCE AMPLIFIER RC: COMPENSATION RESISTOR RL: OUTPUT RESISTANCE DEFINED AS VOUT/ILOADMAX RO: OUTPUT RESISTANCE OF gma R1, R2; FEEDBACK RESISTOR DIVIDER NETWORK RESR: OUTPUT CAPACITOR ESR : CONVERTER EFFICIENCY (~90% AT HIGHER CURRENTS) Figure 18. Boost Converter Equivalent Model source where the peak input current, IVIN, is proportional to the VC voltage. Note that the maximum output currents of gmp and gma are finite. The output of the gmp stage is limited by the minimum switch current limit (see Electrical Specifications) and gma is nominally limited to about ±12μA. From Figure 18, the DC gain, poles and zeros can be calculated as follows: DC Gain Ag R g V V RR R DC ma O mp IN OUT L : . =• • • • • • η 2 05 2 1+ + = •• • 05 1 2 2 2 . : R Output Pole P RC Error Amp Po L OUT π lle P RR C Error Amp Zero Z OC C : : 2 1 2 1 1 2 = •• + ()• = •• π π R RC ESR Zero Z RC RHP Zero Z V CC ESR OUT • = •• • = : : 2 1 2 3 π IIN L OUT S R VL High Frequency Pole P f P 2 2 2 3 3 • •• • > π : h hase Lead Zero Z RC Phase Lead Pole P PL : : 4 1 2 1 = •• • π 4 4 1 2 05 05 12 12 = •• • + • π RR RR C Error Amp Filter P PL . . o ole P RR RR C C C CO CO F F C : , 5 1 2 10 = •• • + < π |
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