I don't quite understand how a buck converter works in terms of maintaining a constant voltage.
Let's say you have a 5V, 3A buck converter, but you hook it up to a 1K resistor (load) in series to give you a current draw of 5mA.
Now, let's say you start adding more 1K resistors in parallel to increase the current draw of the load(s).
What changes in the buck converter to meet the increase current demand?
Based on my understanding, there are only two parameters to play it here - switching frequency and duty cycle.
However, I thought that many buck converters have a fixed switching frequency, which rules that out as a possible parameter to tweak?
Part of my confusion comes from this graph shown in TI's power topologies handbook, where is shows the Imin and I max of the inductor current:
So my new hypothesis is that there is a part of the graph here that is not shown, which is the initial current rise from 0A to whatever the necessary current is to maintain the specified voltage drop.
To answer my own question, what would happen if we started adding 1K resistors in parallel would be that there is a single instance/cycle where the switch stays closed for a longer time to allow the current through the inductor to ramp up to the necessary value in order to maintain the specified voltage drop across the load(s). See my crude drawing below: