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.
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:
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
R3would therefore be
180KΩ * (5V/0.5V - 1) = 180KΩ * 9 = 1620KΩ.
- These two resistors would configure the boost converter for
5Voutput at up to