TPS541620 Buck converter IC - Feedback loop

I've been designing a POL supply using TI's new TPS541620. This seemed to be a good choice but it being new, it's quite hard to get more support than the datasheet can provide. And unfortunately some things are missing from it...

Looking at the Dual Independent Outputs application information, there is a 49.9kohm resistor called $$\R_{inj}\$$ in series with the $$\R_{fbt}\$$ resistor in the feedback loop.

There is no mention of the $$\R_{inj}\$$ resistor in the datasheet, and having it here doesn't make sense. According to the datasheet, to select the output voltage, you need to put a voltage divider with a 10k bottom resistor. The top resistor value can be chosen with the following equation (where $$\V_{fb} = 0.5V\$$ is the reference voltage) :

$$R = 10k * \frac{V_{out} - V_{fb}}{V_{fb}}$$

In their example, $$\V_{out}=1V\$$, so they take a value of $$\R=10k\$$ which makes sense. But if you add the 49.9kohm to it, then the output voltage would be $$\V_{out}=3.495V\$$.

I am far from being an expert in designing power supplies, so I'm listening if anyone has an explanation.

Thanks !

PS: I would have asked on the E2E forum but apparently my account doesn't allow me to do so.

EDIT:

I've just realized that $$\R_{inj}\$$ is 49.9 ohms and not 49.9kohms, so it is indeed negligeable.

• You can find the datasheet here Apr 26, 2021 at 19:00