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ADP1850ACPZ-R7 数据表(PDF) 21 Page - Analog Devices |
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ADP1850ACPZ-R7 数据表(HTML) 21 Page - Analog Devices |
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21 / 32 page  Data Sheet ADP1850 Rev. C | Page 21 of 32 LOOP COMPENSATION (SINGLE PHASE OPERATION) As with most current mode step-down controller, a transcon- ductance error amplifier is used to stabilize the external voltage loop. Compensating the ADP1850 is fairly easy; an RC compen- sator is needed between COMPx and AGND. Figure 33 shows the configuration of the compensation components: RCOMP, CCOMP, and CC2. Because CC2 is very small compared to CCOMP, to simplify calculation, CC2 is ignored for the stability compensation analysis. ADP1850 FBx CCOMP Gm 0.6V COMPx AGND RCOMP CC2 Figure 33. Compensation Components The open loop gain transfer function at angular frequency, s, is given by ) ( ) ( ) ( s Z s Z V V G G s H FILTER COMP OUT REF CS m × × × × = (1) where: Gm is the transconductance of the error amplifier, 500 µS. GCS is the tranconductance of the power stage. ZCOMP is the impedance of the compensation network. ZFILTER is the impedance of the output filter. VREF = 0.6 V. GCS with units of A/V is given by MIN DSON CS CS R A G _ 1 × = (2) where: ACS is the current sense gain of either 3 V/V, 6 V/V, 12 V/V, or 24 V/V set by the gain resistor between DLx and PGNDx. RDSON_MIN is the low-side MOSFET minimum on resistance. If a sense resistor, RS, is added in series with the low-side FET, then GCSbecomes ) ( 1 _ S MIN DSON CS CS R R A G + × = Because the zero produced by the ESR of the output capacitor is not needed to stabilize the control loop, assuming ESR is small the ESR is ignored for analysis. Then ZFILTER is given by OUT FILTER sC Z 1 = (3) Because CC2 is small relative to CCOMP, ZCOMP can be simplified to COMP COMP COMP COMP COMP COMP sC C sR sC R Z × + = + = 1 1 (4) At the crossover frequency, the open-loop transfer function is unity or 0 dB, H (fCROSS) = 1. Combining Equation 1 and Equation 3, ZCOMP at the crossover frequency can be written as × × × π = REF OUT OUT CS m CROSS CROSS COMP V V C G G f f Z 2 ) ( (5) The zero produced by RCOMP and CCOMP is COMP COMP ZERO C R f × π = 2 1 (6) At the crossover frequency, Equation 4 can be shown as CROSS ZERO CROSS COMP CROSS COMP f f f R f Z 2 ) ( 2 + × = (7) Combining Equation 5 and Equation 7 and solving for RCOMP gives × × × × π × + = REF OUT OUT CS m CROSS ZERO CROSS CROSS COMP V V C G G f f f f R 2 2 2 (8) Choose the crossover and zero frequencies as follows: 12 SW CROSS f f = (9) 48 4 SW CROSS ZERO f f f = = (10) Substituting Equation 2, Equation 9, and Equation 10 into Equation 8 yields × × × π × × = REF OUT OUT m CROSS DSON CS COMP V V C G f R A R 2 97 . 0 (11) where: Gm is the transconductance of the error amplifier, 500 µS. ACS is the current sense gain of 3 V/V, 6 V/V, 12 V/V, or 24 V/V. RDSON is on resistance of the low-side MOSFET. VREF = 0.6 V. And combining Equation 6 and Equation 10 yields CROSS COMP COMP f R C × π = 2 (12) Note that the previous simplified compensation equations for RCOMP and CCOMP yield reasonable results in fCROSS and phase margin assuming that the compensation ramp current is ideal. Varying the ramp current or deviating the ramp current from ideal can affect fCROSS and phase margin. And lastly, set CC2 to COMP C COMP C C C × ≤ ≤ × 10 1 20 1 2 (13) |
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