0
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\begingroup\$ You can find the datasheet here \$\endgroup\$ Apr 26, 2021 at 19:00

2 Answers 2

1
\$\begingroup\$

This technique is usually found on regulator evaluation boards. It allows to break the feedback loop and do stability testing by injecting signals while still keeping the feedback divider and forward compensation capacitor in the path of injected signal. That is why the name is Rinj, and it is 50 ohms as typically lab equipment and coaxial cables use 50 ohms impedance. 50 ohms is very little compared to the impedance of the feeback divider so it has almost no effect during normal operation.

\$\endgroup\$
2
  • \$\begingroup\$ Ok thanks ! So I don't really need to put it in my design right ? \$\endgroup\$ Apr 26, 2021 at 19:24
  • \$\begingroup\$ No, typically these are not put on final product using the regulator. \$\endgroup\$
    – Justme
    Apr 26, 2021 at 19:26
0
\$\begingroup\$

Inserting a resistance in series with the upper divider network in a switching regulator is a very common way of measuring the open-loop gain of a power supply without physically opening the loop. This method has been introduced by Dr. Middlebrook in a paper published in 1975, Measurement of loop gain in feedback systems. As shown in the below picture, a resistance is inserted in series at a point where impedance matching is favorable (low output resistance at point B and high input resistance at point A). Because of its low value - usually 10-20 ohms - it does not bother the divider network.

enter image description here

By injecting a sinusoidal waveform via an injector - it can be a solid-state injector or a simple transformer - it is possible to extract the control-to-output transfer function or the compensated loop gain of a converter without opening the loop. It is usually wise to include this resistance during the prototype stage with two pins allowing easy probing later on when the assembly is ready.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.