Here is the excerpt of the schematic again for reference:
Since the author has expressed considerable doubt about the grounding problem, here is a more detailed explanation. The error amplifier amplifies the voltage difference between the In+ and In- terminals and it has a lot of gain. Any difference will result in the output changing and it only takes a difference measured in microvolts to swing the output to the rail.
The key to this is Kirchhoff's Voltage Law which is simply that the sum of voltages in a closed loop is zero. Here I am going to take advantage of the virtual short circuit concept in opamps. That is where because of feedback the voltages at the In+ and In- inputs are the same. Just as if they were shorted together. Using that I can then sum the voltages around the loop between them.
First is the voltage across R7 (assume fixed resistors) then the wires and finally Vref between V- and In+. Being careful about the signs:
$$ V(R7) + IR - Vref = 0 $$
The IR losses in the wires are neglibile except in the connection between the filter capacitor and the output terminal. The resistance is low but the current is high and worse, variable. IR is the voltage drop in that wire where I is theload current and R is the wire resistance.
The voltage at R7 is calculated simply because of the voltage divider on Vout:
$$ V(R7) = Vout (\frac{R7}{R6+R7}) $$
If the output voltage is set to 13.8V at no load then R7/(R6+R7) = Vref/13.8V. Or:
$$ Vref = Vout (\frac{Vref}{13.8V}) + IR $$
Solving for Vout:
$$ Vout = (Vref - IR) \frac{13.8V}{Vref} $$
Or:
$$ Vout = 13.8V - IR \frac{13.8V}{Vref} $$
Vref is approximately 7.15V so the ratio of 13.8V/7.15V is 1.93. So the voltage at the output terminals will drop about twice the IR drop in the wire between the filter capacitor and the output terminal. The article stated that the output voltage dropped by 80mV with a 10A load. That is the voltage drop expected with a 4 milliohm resistance in this connection.
Sure the output voltage drop is small but by making a very simple change that costs nothing it can be made nearly nonexistent.
This could be taken a bit further by extending the sense point. If, for example, you have a power distribution box that you plug your equipment into, there will some voltage drop in the cable connecting to it. If you sense the voltage at that bus instead of at the output of the supply, that voltage drop goes away.
Doing that is not without hazard. You must take care to never turn on the supply unless the sense connection is complete. The sense leads must be a shielded twisted pair to protect it from noise. Even with that it is possible that the regulator will oscillate. That can be fixed and methods for doing that are in the articles at Walt Jung's blog. Look for the References and Regulators pages.
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