Alternative fix to: How to calculate a Joule Thief

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I understand that the voltage rate will depend on the end-unit consuming the power (eg., an LED chip).

I don't believe there's such thing as a "voltage rate" in the context you're using and, in the same context, the power supply will generally be indifferent as to what the load looks like.

What happens if the output from the PSU is divided into two parallel connections, one lead connected to an end-unit needing to use 27v, and another using 30v?

The voltage across the loads will be the single voltage the supply generates in order to force a total of 2.7 amperes through the parallel loads.

What happens internally? What's the reason why for the end result?

The supply senses the current in its output lead and adjusts its output voltage so that its output current will remain constant at 2.7 amperes, regardless of the load resistance.

As an aside, your:

I see that some power supplies provide a wide range of output voltages meanwhile maintaining the current.

For example: 2.7 A, 20-36 V output.

Indicates that since - from Ohm's law - R = E/I, with 20 volts across a load and 2.7 amperes through it, the load's resistance must be 7.41 ohms.

Similarly, at the high end of the supply's compliance, with 36 volts across a load and 2.7 amperes through it, the load's resistance must be 13.33 ohms.

Note that the range of load resistances the supply can accommodate is from 7.41 ohms through 13.33 ohms, regardless of whether the load is a single resistor, a couple of parallel diodes, or whatever, as long as the compliance of the supply is maintained within its limits.

Next, re. your diode example:

Example:

[[80W 2.7 A PSU with 18-30 V]] |--> (21-24) V, 1 A Diode |--> (27-31) V, 2 A Diode How would it regulate the voltage in each case?

It CANNOT regulate the voltage, it can only increase it until it either pushes 2.7 amperes through the load or until it runs out of headroom.

In the first instance, assume a diode with a 4 volt drop when there's 1 ampere through it.

If it's running from a 23 volt constant-voltage supply it'll need a ballast resistor to drop the remaining 19 volts, that resistor being: Rs = (Vsupply - Vled)/Iled = (23V - 4V)/1A = 19 ohms

Notice though, that since it's not a constant voltage supply, but a constant current supply it'll raise its output voltage in an effort to get 2.7 amperes through the load. In this case its output voltage can only rise to 30 volts, so when it gets there the current into the load will be: Iled = (Vsupply - Vled)/Rs = (30V - 4V)/19R = 1.53 amperes.

In the instance where you have two LEDs in parallel, the output voltage of the supply will rise until the total current out of the supply is 2.7 amperes, or until its output voltage rises as high as it can if it can't output 2.7 amperes.

With a 1 ampere LED in parallel with a 2 ampere LED and each LED's current limited by a ballast resistor, the voltage of the supply will rise until the current into the parallel combination equals 2.7 amperes.