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LMR43620-Q1: How to calculate R_fbt and R_fbb

Chase Jeong
Mastermind 7810 points
Part Number: LMR43620-Q1
Other Parts Discussed in Thread: LMR43620

Tool/software:

Dear TI experts,

My customer considers LMR43620-Q1 for their next project.

I saw the part 9.2.2.2.1 FB for adjustable output, and I have some questions.

1. There is a range for R_fbt, just equal or smaller than 10kohm X (Vout/1V).

And in the table 9-6, I think that the value of R_fbt is just set to the closest value of equal.

for example, R_fbt for 5V output is -> 10kohm X (5V/1V) = 50kohm. so it is just exactly set to 49.9kohm.

Other values are used same formula.

Then What happens if I set R_fbt to 40kohm or 30kohm? I think these values are also avaiable for equation 6.

2. Second question is also about ranges of R_fbb.

It is also just presented as larger than 5kohm and smaller(or equal) than 10kohm.

but in the example of table 9.6, it is set to the value that is closest to 10kohm.

for example, for the case of 5V output, parallel combination of of R_fbt and R_fbb is 9.932Kohm, it is very close to 10kohm.

Them I have same question as no.1 -> What happens if I set R_fbb as 7kohm? then parallel comblnation of 49.9kohm and 7kohm is 6.14kohm, for 5V output. It is also meets the equation (5).

Conclusion;

I think the range of equation(5) and equation(6) are pretty wide, so it is very confused to determine the exact value of R_fbt and R_fbb.

and my own conclusion is as below;

Equation(5) -> R_fbt || R_fbb = 10kohm (closest value)

Equation(6) -> r_fbt = 10kohm X (Vout/1V)

Please confirm it, and also advise me to determine r_fbt and R-fbb for 30V output.

++ one more request, please check that there any calculator (excel or other files) for LMR43620.

Best regards,

Chase

  • 0
    TI__Mastermind 22300 points

    Hello,

    Equations 5 and 6 have acceptable ranges of resistance values within the "greater than" and "less than" limits shown in the equations.

    For example, equation 6 shows that Rfbt must be less than or equal to 10k*vout/vref, which means for a 5V output you could use lower resistance than 50k.

    Generally you want the largest possible resistor values for feedback to improve efficiency by limiting the current through the feedback path.

    As long as you meet the limits set within the equations, it should be OK.

    Table 8-4 represents configurations which TI has tested and recommends for this device. For anything outside of those recommendations, you must follow the datasheet recommendations as closely as possible when selecting feedback resistors.

    Webench can be used for recommended designs. Here is a design I made for your reference: https://webench.ti.com/appinfo/webench/scripts/SDP.cgi?ID=729D3D78FC8F89F0

    Best regards,

    Ridge

  • 0 in reply to Ridge Lahti
    Mastermind 7810 points

    Dear Ridge,

    Thank you for your support.

    Please see the table below. It is in the datasheet.

    If I want to 5V output voltage, recommended values of R_fbt and R_fbb are 49.9kohm and 12.4kohm each.

    Let me calculate the equation (5) and (6) with these values.

    Equation (5) ---> 5kohm < R_fbt || R_fbb <= 10kohm ---> 5kohm < 49.9k || 12.4k <= 10kohm ---> 5kohm < 9.932k <= 10k ---> it is fine.

    Equation (6) ---> R_fbt <= 10kohm X (Vout/1V) ---> 49.9k <= 10kohm X (5V/1V) ---> 49.9k <= 50k ---> it is fine.

    And this time I will calculate again equation (5) and (6) with the value of 33.2kohm(R_fbt) and 14.3kohm(R_fbb), this value is recommended to 3.3V output voltage, but i will remain Vout as 5V.

    Equation (5) ---> 5kohm < R_fbt || R_fbb <= 10kohm ---> 5kohm < 33.2k || 14.3k <= 10kohm ---> 5kohm < 9.995k <= 10k ---> it is also fine.

    Equation (6) ---> R_fbt <= 10kohm X (Vout/1V) ---> 33.2k <= 10kohm X (5V/1V) ---> 33.2k <= 50k ---> it is also fine.

    I hope that you understand what I want to say. Regarding to the equation (5) and (6), 49.9k and 12.4k is recommended for 5V output voltage. but 33.2k and 14.3k is also fit to the equation (5) and (6) for 5V output although it is the recommended values for 3.3V output.

    Furthermore, 24.9k and 16.5k are also fit for 5V output.

    That's why I think the ranges of 2 equations are too wide, and it is best to calculate equations as close as 10k.

    Please check this issue, and let me know if there are my misunderstandings.

    Best regards,

    Chase

  • 0 in reply to Chase Jeong
    TI__Mastermind 22300 points

    Hi Chase,

    I think I see the confusion now.

    For this device, the Vref voltage is Vref=1V. Vref is the voltage inside the IC which the FB node is compared to for proper regulation.

    So, your FB resistor divider must divide the output voltage down to 1V on the FB node. That equation is not explicitly written in that datasheet, but your resistor divider must step the output down to 1V and it must meet the conditions set by equations 5 and 6.

    You would use the resistor divider equation of: Vref = Vout *RFBB/(RFBT+RFBB). You would choose a value for RFBT then solve for RFBB.

    In the examples you calculated, those values are acceptable to equations 5 and 6 but they would set the output to voltages other than 5V.

    In summary, you need to calculate resistor values for a feedback divider which steps your desired output voltage down to 1V at the FB node, and you need to make sure that your calculated resistance values meet the requirements of equations 5 and 6. 

    Best regards,

    Ridge