Whenever we load a circuit, there's a drop from the open circuit voltage \$v_{oc}\$ due to the internal source resistance or Thevenin's resistance\$R_{th}\$. So we need to have an \$R_{load}\$ that is very large compared to \$R_{th}\$, (i.e) $$R_{load}>>R_{th}$$ so that almost all the voltage drop occurs across \$R_{load}\$ . But according to maximum power transfer theorem we need to have an \$R_{load}\$ that is equal to the \$R_{th}\$ to obtain maximum power transfer. Aren't these two statements contradictory, maybe contradictory isn't the right word, but you get it. So, which one do we need to follow while constructing a circuit.
Electronic – Contradiction:Maximum Power Transfer and High resistance of load
circuit analysiscircuit-designinternal-resistancepowerthevenin
Related Solutions
The text doesn't deny that the efficiency is 50% at maximum power transfer conditions. It is stating that maximum power transfer conditions are not the same as maximum efficiency conditions, so there may be a different point (namely, when Rs is much less than RL) where efficiency is higher but power transfer is lower.
See Wikipedia for more details: Maximizing power transfer versus power efficiency
As your quotation says, linear power supplies are called linear because of the transistor's (BJT's) operating region, not because it's a linear circuit.
To find a Thevenin equivalent for a nonlinear circuit, you can do the same things you do for any transistor amplifier:
- Make the AC components of the input conditions small.
- Use a nonlinear, large-signal model to find the characteristics of the circuit at the DC bias point.
- Make a linearizing approximation around the bias point (a so-called "small-signal model").
- Plug your small AC signal into the small-signal model to get the AC output.
The typical small-signal model for BJTs is the hybrid-pi model. So starting with a linear regulator:
simulate this circuit – Schematic created using CircuitLab
You would end up with this:
That's a linear circuit, so it has a Thevenin equivalent. This model only applies when changes in the voltages and currents are small, though. That's the price you pay for linearity.
You can't do this directly with an SMPS -- the switching can't be linearized. For a constant duty cycle, you can use a transformer to model the switching action. In the comments, Tom mentions a source for linearizing the control loop.
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Best Answer
You are in good company. The maximum power transfer concept frequently causes confusion. Take note of the fact that maximum power transfer does not mean maximum power efficiency. I think this may be the main point that trips people up.
In fact, when Rload = Rsource, then the maximum possible efficiency is 50%, since the load and source will be consuming equal power. If the goal is to extract power from the source at the highest possible rate, then the load and source should be matched.
But often, it is more important to protect the source from getting hot, or to maximize power efficiency from the standpoint of DC input power to the amplifier. In these cases, Rsource should be much lower than Rload. You see this in audio amplifiers, for example. Also, usually when you are discharging a battery, you make sure Rload is much greater than Rbattery to avoid over-heating the battery.
Another point is that when signals need to pass through transmission lines (such as coaxial cable or similar) the load must be matched to the source (and line) to prevent reflections or other signal degradation.
Key points