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IP1206PBF Datasheet(PDF) 23 Page - International Rectifier |
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IP1206PBF Datasheet(HTML) 23 Page - International Rectifier |
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23 / 30 page ![]() 2/26/2008 www.irf.com 23 iP1206PbF Compensation for Current Loop (slave channel) The slave error amplifier is differential transconductance amplifier, in 2-phase configuration the main goal for the slave channel feedback loop is to control the inductor current to match the master channel inductor current as well provides highest bandwidth and adequate phase margin for overall stability. The following analysis is valid for both using external current sense resistors and using DCR of inductor. The transfer function of power stage is expressed by: Where: V in=Input voltage L 2=Output inductor V osc=Oscillator Peak Voltage As shown the G(s) is a function of inductor current. The transfer function for compensation network is given by equation (21), when using a series RC circuit as shown in Fig 23. The loop gain function is: ) ( V sL V V s I s G osc in e L 20 * ) ( ) ( 2 2 - - - - = = L2 L1 C2 R2 RS2 RS1 Ve IL2 IL1 Fb2 E/A2 Comp2 Vp2 ) ( sC R sC R R g R s V s D s s m s e 21 1 * * ) ( ) ( 2 2 2 2 1 2 - - - - ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = = [ ] 2 s R s D s G s H * ) ( * ) ( ) ( = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = osc 2 in 2 2 2 2 s 1 s m 2 s V sL V sC C sR 1 R R g R s H * * * * * ) ( Select a zero frequency for current loop (F o2) 1.5 times larger than zero cross frequency for voltage loop (F o1). From (22), R2 can be expressed as: V in=12V V osc=1.25V g m=2800umoh L 2=1uH R s1=DCR=2.4mOhm F o2=60kHz This results to : R 2=5.84K The power stage of current loop has a dominant pole (Fp) at frequency expressed by: Where Rds(on1) is the on-resistance of control FET, Rds(on2) is the on-resistance of synchronous FET, RL is the DCR of output inductance and D is the duty cycle Req=9.48mOhm Set the zero of compensator at 10 times the dominant pole frequency FP, the compensator capacitor, C2 can be expressed as: C2=1nF All designs should be tested for stability to verify the calculated values. 1 O 2 O F 5 1 F * % . ≅ ) ( V L F V R R g F H osc O in s m O 22 1 * * * 2 * * * ) ( 2 2 2 1 2 - - - - = = π ) ( V V L F R g R in osc O s m 23 * * * 2 * * 1 2 2 1 2 - - - - π = s L on ds eq R R R R + + = ) ( z 2 2 P z F R 2 1 C F 10 F * * * π = = Fig. 23: The Compensation network for current loop L 2 R F 2 eq P * π = L 2 on ds 1 on ds eq R D 1 R D R R + − + = ) ( * * ) ( ) ( |
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