Questions about MAX20751 and VT1697SB

Hello,

25A is the lowest OCP can be selected for MAX20751, while i think 5A is designed output current for those two rails.

PMBus has 7 bits, lowest 3 bits are selectable by R_SEL0 and R_SEL1, the upper 4 bits are constant at the value 1110b. 

RDES value is set by R_SEL0

Hello,

25A is the lowest OCP can be selected for MAX20751, while i think 5A is designed output current for those two rails.

PMBus has 7 bits, lowest 3 bits are selectable by R_SEL0 and R_SEL1, the upper 4 bits are constant at the value 1110b. 

RDES value is set by R_SEL0

How it is 5 A that the main question. I have attached the two pages of schematic where it says 5A but my calculations mentioned in the post above does not match. How these power supplies are 5 A ? 

25A is the lowest OCP can be selected for MAX20751 with 1-phase, which gives Ioutmax = 0.85x25=21.25A. Any selected designed Iout value below 21.25A, can only choose OCP value as 25A in the design. In another word, in Xilinx design, Iout = 5A, minimum OCP in theory will be 5/0.85 = 5.88A, however, MAX20751 does not have 5.88A OCP option for 1-phase, in the design, it can only be selected as 25A with RDES = 604ohm

Thank you very much for reply. Yes, that is true that the lowest value of the OCP is 25 A that gives Ioutmax = 0.85x25=21.25A.

Question on the RDES is still there. I only see the resistors R_SEL0, R_SEL1, and R_SEL2 physically present in the schematic and placed in the layout as physical components but RDES having a value 604 Ohm from Table 5, where do we place it in the circuit ? It look like that RDES is only a value in theory and will be used in Table 7 to find the resistors R_SEL0, R_SEL1, and R_SEL2 for a given voltage. In other words, that RDES is not a physical resistor in the schematic, right ? 

Given the following values of the following resistors in Table 7 to get Vout = 0.030 + 0.970 = 1 V, how do we found three bits in the address PMAD [3:1] ? 

RSEL2 = 402 Ohm (Vout = 0.970 V) 
RSEL1 = 576 Ohm (Vout = 0.030 V) 
RSEL0 = 0 Ohm (RDES = 604 Ohm) 

My finding PMAD [3:1] = 010, is that correct ? 

Given the following values of the following resistors in Table 7 to get Vout = 0.030 + 1.170 = 1.2 V, how do we found three bits in the address PMAD [3:1] ? 

RSEL2 = 576 Ohm (Vout = 1.170 V)
RSEL1 = 1020 Ohm (Vout = 0.030 V)
RSEL0 = 0 Ohm (RDES = 604 Ohm)

My finding PMAD [3] = 0, is that correct but what about PMAD [2:1] ?

That is right, R_DES is not a physical resistor in the schematic.

PMAD[3] is determined by R_SEL0, if R_SEL0 = 0 ohm, then PMAD[3] =0. PMAD[2:1] is determined by R_SEL1, if R_SEL1 = 1020ohm, then PMAD[2:1] will be 11. You can refer to Table 7 in the datasheet.

yes, it is correct.

Alright. Thanks for reply once again. According to datasheet "R_SEL1 and R_SEL0 are the configuration resistors used to set the lowest 3 bits (PMAD[3:1])"

This means that only R_SEL1 and R_SEL0 are used to set the lowest 3 bits (PMAD[3:1])

R_SEL2 is NOT selecting the any of the bit (PMAD[3:1]). 

However, R_SEL2 and R_SEL1 are the configuration resistors used to set the output voltage. 

According to datasheet "R_SEL1 and R_SEL0 are the configuration resistors used to set the lowest 3 bits (PMAD[3:1]) of the entire PMBus address (PMAD[7:1]). The upper 4 bits (PMAD[7:4]) are constant at the value 1110b."

lowest three bits (PMAD[3:1]) ? or (PMAD[2:0) ? 

what about PMAD [3] ? if lowest three bits are selected by R_SEL1 and R_SEL0