Electronic – DC to DC converter: How to convert 110 Vdc to 5 Vdc with 2.5 Amps current

Basically what you're asking is, do current controlled switching convertes exist. The answer is yes, as you can see here:

Yes, that looks complicated ! But it's not (for me at least). The MAX649 is a DCDC Buck converter controller IC, it does not have a build in switch, PMOS Q1 is the switch. If you would want a constant output VOLTAGE, you would need a voltage divider (2 resistors) and feed a divided value of the output voltage back to the FB (feedback) pin.

But in this circuit, has a constant output CURRENT, to charge a battery ! Instead of sensing the output voltage, now the output current is sensed by taking the voltage across R10, amplifying this voltage (the opamp does this, with a feedback network to set the gain) and where does ths feedback signal go ? Straight back in to the FB pin !

The simple buck converter in your schematic is too simple to be considered having a voltage or a current output. If you change the load then voltage and current change, it is not regulated in any way. Yes you can change the PWM of the switch signal but that PWM signal needs te be determined somehow. This is done with voltage or current feedback.

input current \$I_o < I_i\$

This equation refers to the average input and output currents.

During the switching cycle the input current will increase and decrease, roughly following a triangle wave shape.

When the switch M1 is closed, the current drawn from V1 increases, building up energy in the magnetic field of L1. Meanwhile the load draws current from output capacitor C2. When M1 is opened, current drawn from V1 decreases as the built up energy in L1 is delivered to C2.

At all other points, is current simply \$I_o = \frac{V}{R}\$ at B, C, D?

At B, it will be roughly \$\frac{V}{R}\$, but will still have some ripple because \$V_o\$ will have some ripple.

At C, you will have the combined return current of the load current (\$\frac{V_o}{R}\$) plus the current through C2, which will be positive in some parts of the cycle and negative in other parts.

At D you will have the return current from the load, the return current from C2, and return current from the switch, all adding up to exactly the same value as the inductor current \$I_i\$, which you could see using KCL.