PD or Iota for my upgrade?

Salvo wrote:
The battery has just one impedance, not two as you suggest.

I'm not saying the battery has two impedances. I'm saying that the converter sees a load that is indistinguishable from a resistive load as long as the voltage and current are constant steady state.

The resistance will have minor variations over frequency, but we're discussing source resistance at very low frequency. You say it's so elementary that there are two distinct impedances that are orders of magnitude apart, then there should be many links describing this unusual feature.

There's nothing about this that is an "unusual feature." It's just Ohm's Law. When current and voltage aren't changing you can always model the load as a pure resistor. I haven't done any searches, but I'll note that it's been posted that the manufacturer of the PD tested with a resistive load, others here have tested with a resistive load, and you can look up Ohms Law. The resistor used for those tests isn't the resistance of the battery model (50 milliohms). The resistance used in each of those cases is the resistance required for the converter to supply its maximum rated current at the rated output voltage. If we wanted to know if the converter could output 10A at 12.5V, we'd use a 1.25 ohm resistor. If we want to know if it will supply 60A at 14.4V, we'd use a 240 milliohm resistor.

Have you heard of Kirchoff's Voltage Law?

Yes, and Maxwell's equations and Ohm's Law. The latter says that when voltage and current aren't changing, the effective resistance is equal to the voltage divided by the current.

If you do a loop analysis of the circuit (perfect voltage source plus series resistor connected to a battery charging source), you would realize the internal resistance of the battery can not be 1.25 ohms.

I didn't say it was. I said the internal resistance of the battery is what you calculated (50 milliohms) in series with a perfect voltage source having zero impedance and I said that in steady state conditions a perfect voltage source having zero impedance looks exactly like a resistor having a value of the voltage divided by the current going through it. In other words, the perfect voltage source looks like 1.25 ohms minus 50 milliohms so that the sum is 1.25 ohms.

According to KVL, the sum of the voltage drops of the voltage source and series resistance must equal 12.5V when 10A are conducting into the battery.

We agree to here.

At 10A, the internal resistor creates a voltage drop of 12.5V. That means the pure voltage source (battery voltage minus internal resistance) has zero volts! That's impossible.

We're replacing the model of an internal resistor (50 milliohms) in series with a zero ohm perfect voltage source by a model comprising a single resistor (1.25 ohms), or if you prefer, as your 50 milliohm resistor in series with a 1.25ohm - 50milliohm resistor. The two are equivalent. In steady state the 1) pure resistor model and the 2) internal battery resistance (50 milliohm) in series with a perfect voltage source model are identical. The second model is better for changing conditions, but the first is fine for steady state. Since we are testing a steady state limit, the first model is better because you can't easily get a perfect voltage source needed for the second model.

The actual battery source voltage is
V = V_at_bat_terminals - R_internal_resistance * I
V = 12.5V - 0.05 ohm * 10A
V = 12.0V
The actual battery voltage is 12.0V, not zero.

In your model, I agree. But remember what we're doing here - we're testing the output capability of the converter. The converter doesn't know what the battery voltage is or what the internal resistance is. It just knows the voltage it is trying to hold and how much current the load is allowing to flow when that voltage is applied to it. It sees that at 12.5 volts, 10A flows. Whether that is because there is a real battery, a perfect voltage source in series with a 50 milliohm resistor, or just a plain old 1.25 ohm resistor doesn't matter to it. It just knows that when 12.5 volts is applied at the output, 10A flows into the load.

http://batteryuniversity.com/learn/article/how_to_measure_internal_resistance

This is fine for determining the internal battery resistance, but that's not of interest in steady state conditions if you just want to know how much current the converter will supply. If you had a perfect battery, you could do your tests with this internal resistance model resistor in series with your perfect battery, or you could simplify things by using a larger resistor and modeling the battery as a pure resistor. In steady state, according to Ohm's Law, they are equivalent, and that's why people use a large resistor ("load bank") to performance test devices that produce power.

I've used salt water tanks as a load bank to performance test generators when you have to dissipate a few kilowatts. The salt water tank is just a big resistor. I've used heating coil load banks to performance test voltage supplies I've built. That's all a converter is - a big voltage supply capable of producing 60A or whatever it's rated for at its rated output voltage. There are lots of articles about load banks, and that's all we're doing - putting a load on the converter to see how much current it can supply at the different voltages it operates at. It's a perfectly standard engineering technique for doing tests like this, and that's why the converter manufacturers test that way. It's a more controllable and reproducible test than using a battery.