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BD9104FVM 데이터시트(PDF) 13 Page - Rohm |
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BD9104FVM 데이터시트(HTML) 13 Page - Rohm |
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13 / 18 page ![]() BD9102FVM, BD9104FVM, BD9106FVM Technical Note 13/17 www.rohm.com 2009.05 - Rev.A © 2009 ROHM Co., Ltd. All rights reserved. 4. Determination of RITH, CITH that works as a phase compensator As the Current Mode Control is designed to limit a inductor current, a pole (phase lag) appears in the low frequency area due to a CR filter consisting of a output capacitor and a load resistance, while a zero (phase lead) appears in the high frequency area due to the output capacitor and its ESR. So, the phases are easily compensated by adding a zero to the power amplifier output with C and R as described below to cancel a pole at the power amplifier. Stable feedback loop may be achieved by canceling the pole fp (Min.) produced by the output capacitor and the load resistance with CR zero correction by the error amplifier. 5. Determination of output voltage (for BD9106FVM only) The output voltage VOUT is determined by the equation (7): VOUT=(R2/R1+1)×VADJ・・・(7) VADJ: Voltage at ADJ terminal (0.8V Typ.) With R1 and R2 adjusted, the output voltage may be determined as required.(Adjustable output voltage range: 1.0V~2.5V) Use 1 kΩ~100 kΩ resistor for R1. If a resistor of the resistance higher than100 kΩ is used, check the assembled set carefully for ripple voltage etc. Fig.56 Determination of output voltage Gain [dB] Phase [deg] Fig.53 Open loop gain characteristics A 0 0 -90 A 0 0 -90 fz(Amp.) Fig.54 Error amp phase compensation characteristics fp= 2π×RO×CO 1 fz(ESR)= 2π×ESR×CO 1 Pole at power amplifier When the output current decreases, the load resistance Ro increases and the pole frequency lowers. fp(Min.)= 2π×ROMax.×CO 1 [Hz]←with lighter load fp(Max.)= 2π×ROMin.×CO 1 [Hz]←with heavier load Zero at power amplifier Increasing capacitance of the output capacitor lowers the pole frequency while the zero frequency does not change. (This is because when the capacitance is doubled, the capacitor ESR reduces to half.) fz(Amp.)= 2π×RITH.×CITH 1 GND,PGND SW VCC,PVCC EN VOUT ITH VCC VOUT Cin RITH CITH L ESR CO RO VOUT Fig.55 Typical application fz(Amp.)= fp(Min.) 2π×RITH×CITH 1 = 2π×ROMax.×CO 1 SW 6 1 ADJ 4.7μH 10μF R2 R1 Output fp(Min.) fp(Max.) fz(ESR) IOUTMin. IOUTMax. Gain [dB] Phase [deg] |
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