Electronic – Replacing resistors with equivalent resistor

1. For a black-box type of converter like these linear regulator replacements, black-box empirical analysis of the transfer functions is about all you'll be able to do. Gain-phase analyzers are expensive but allow you to directly measure things like loop stability, plant response, input impedance and audio susceptibility. I wouldn't advocate $8k or more to characterize a 3-terminal regulator. You've already hit on one of the potential gotchas - lack of headroom caused by the input sagging under load.

2. With buck converters, paralleling involves some external OR-ing circuitry (be it a diode or a MOSFET) to prevent backfeed. Also, if load sharing is required, some form of balancing (forced current sharing or droop sharing) is also needed. Adding diodes will work but there's always the possibility that there will be no load sharing until one of the converters starts dropping out. I doubt you'd be able to make these 3-terminal devices share well given the low current rating without wasting lots of power implementing external droop (and even then, variations in the OR-ing diode drop could swamp out everything).

If you wish to solve the circuit using node voltage analysis, you would not bother to write a KCL equation at node 2.

Remember, when doing node voltage analysis, one is solving for the node voltages.

But, the voltage at node 2 is given: \$V_2 = V_{ab}\$

So, you might think that you must write a KCL equation for node 1 but, in fact, you don't because there is a voltage source connected there too.

Simply use KVL to write:

$$V_x + 0.01V_x = V_{ab} \rightarrow V_x = \dfrac{V_{ab}}{1.01}$$

Now, you know the node voltages so you can find the resistor currents. Can you take it from here to find \$I_a\$?

Finally, about Ia. I am also confused by the presence of 0.01Vx. Would applying KCL only means finding current between node 1 and 2 or do we have to involve 0.01Vx too?

Since you know the node voltages, you know the currents through the resistors connected to node 1. Thus, if you write a KCL equation there, the only unknown is the current through the dependent source so use this KCL equation to solve for the dependent source current.

Now that you've found the dependent source current, KCL at node 2 involves only one unknown current, the current \$I_a\$.

The reason why I applied node analysis is because I am studying it these days, and wanted to apply it correctly. Did I?

To correctly apply node voltage analysis, you must enclose the dependent voltage source and parallel resistor inside a supernode. The KCL equation for the supernode is:

$$\dfrac{V_x - V_s}{25} + \dfrac{V_x}{150} = I_a $$

There are two unknowns so you need another equation which is the KVL equation I wrote above.

Note that the 50 ohm resistor is not a factor in the equation. This is due to the fact that it is in parallel with a voltage source which means that the only circuit variable the 50 ohm resistor affects is the current through the dependent source.