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I'm looking at the datasheet for a TPS6103 boost converter, and trying to figure out how to configure it for 5V 1A output. The datasheet is not making this easy to understand.

Datasheet: http://www.mouser.com/ds/2/405/tps61030-450662.pdf

It says this about setting the output voltage.

The output voltage of the TPS61030 dc/dc converter section can be adjusted with an external resistor divider. The typical value of the voltage on the FB pin is 500 mV. The maximum allowed value for the output voltage is 5.5 V. The current through the resistive divider should be about 100 times greater than the current into the FB pin. The typical current into the FB pin is 0.01 µA, and the voltage across R6 is typically 500 mV. Based on those two values, the recommended value for R4 should be lower than 500 kΩ, in order to set the divider current at 1 µA or higher. Because of internal compensation circuitry the value for this resistor should be in the range of 200 kΩ. From that, the value of resistor R3, depending on the needed output voltage (VO), can be calculated using equation 1:

enter image description here

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Given all this seems pretty hard for me to parse, I've come the following conclusions that I'm not sure are right. Can someone please tell me I've interpreted this correctly?

  • It's recommending that R4 be 180KΩ.
  • R3 would therefore be 1.62MΩ via: 180KΩ * (5V/0.5V - 1) = 180KΩ * 9 = 1620KΩ.
  • These two resistors would configure the boost converter for 5V output at up to 1A.
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  • \$\begingroup\$ Yes, that looks fine to me. \$\endgroup\$
    – markt
    Jan 18 '15 at 0:37
  • \$\begingroup\$ Congratulations. It looks that the datasheet has made it easy enough to understand for you to have understood it. \$\endgroup\$
    – Russell McMahon
    Jan 18 '15 at 3:11
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Congratulations! :-)
It seems that the datasheet has made this easy enough to understand for you to have understood it correctly. ie yes your chosen values are correct.

Their wording is verbose but would be useful were it not for an error in referring to R6 when they meant to indicate R4.

They say

  • " ... the voltage across R6 is typically 500 mV..."

but SHOULD say

  • " ... the voltage across R4 is typically 500 mV..."

This was probably caused by a cut and paste error during data sheet production.
R6 is related to the operation of the low battery comparator circuit and is only involved in the output considerations to the extent that the pullup voltage for R6 may not be more than Vout.

A less explanatory but more concise statement of the values of R3 & R4 are:

  • R3 <= 200K

  • R4 = (Vout/Vref -1) x R3

ie for Vout:Vref
= 5.0V:0.5V
= 10:1
then
R4:R3 = 9:1

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  • \$\begingroup\$ Ah that makes more sense. I was confused why R6 mattered at all for the output programming, since it looks like part of the low battery repoting circuit. But why R3 <= 200K? I mean where in the datasheet does it explain what values you should use? I only picked 180K because it used that in the example for solving the equation, and it was less than 200K. Thank you for helping me level up my datasheet reading skills :) \$\endgroup\$
    – Alex Wayne
    Jan 18 '15 at 6:51
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    \$\begingroup\$ @AlexWayne The <= 200k is buried in the fine print you quoted. I've added the BUT in the following. This really amounts to just a "do this" instruction hedged about with fever trees, er, pseudo reasons: They say "The typical current into the FB pin is 0.01 µA, and the voltage across R6 is typically 500 mV. Based on those two values, the recommended value for R4 should be lower than 500 kΩ, in order to set the divider current at 1 µA or higher. [BUT] Because of internal compensation circuitry the value for this resistor should be in the range of 200 kΩ." \$\endgroup\$
    – Russell McMahon
    Jan 18 '15 at 13:17

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