Electrical – remote sensing in power supply

Figure 1. Actual circuit from tradeofic.com. Note addition of 22 \$\Omega\$ resistor at (3).

There are odd things about the circuit apart from the differences in the schematic you posted and the one in the link. The one in the link I can understand to some extent and will use that in this answer.

• The 317 style regulators settle down with 1.25 V between Vout and ADJ. With the component values in Figure 1 that means there is a 1.25 V drop across the 121 Ω resistor so current = \$ \frac {1.25}{121} = 10~mA\$.
• With no load connected \$V_2 = 10m(365+25) = 3.9~V\$. This will result in an output voltage of 3.9 + 1.25 = 5.15 V at (1).
• Again, with no load, \$ V_4 = V_1 = V_3 \$ as the op-amp is configured as a voltage follower.

Now let's apply a load and let's say it causes a 100 mV drop across \$ R_{DROP} \$.

• \$ V_4 \$ drops by 100 mV. \$ V_3 \$ does the same. This causes a 100 mV drop across the 22 Ω resistor and \$ \frac {0.1}{22} = 4.5~mA \$ will flow into A1 which will have to sink the current.
• At this point we should note that A1's negative terminal is connected to GND via the 25 Ω resistor. This sink currrent (and any quiescent current of A1) will return to GND through the 25 Ω resistor raising \$V_5\$ by \$ 4.5m \cdot 25 = 112~mV\$.
• Raising \$V_5\$ will therefore raise \$V_2\$ and ultimately \$V_1\$ by the same amount thus compensating for the voltage drop across \$R_{DROP}\$.

Points yet to be figured out:

• The LM301 has a quiescent current of 1.8 mA at \$V_S\$ = ±15V. You'd need to work out what it is on a single rail supply at \$ V_{IN} \$.
• The circuit appears to stabilise at a voltage higher than 5 V. Why? The 301 quiescent current will make this even worse.
• Normally a voltage compensating circuit would compensate for voltage drop in the feed and return lines. This circuit doesn't show the return line clearly. I would expect that the 25 Ω resistor should terminate at the load with a separate return wire. The compensation circuit components should then be chosen to adjust for twice the drop detected across \$ R_{DROP} \$ assuming that feed and return conductors are the same gauge.

\$ V_{DROP} \$ limit of 300 mV

Figure 2. LM301 maximum current.

Figure 11 of the datasheet shows the current limit at two temperatures. We're unlikely to be operating at such a high voltage and may be further limited. Let's pick 12 mA as maximum sink current to give some safety margin.

Remember that the sink current goes through the 25 Ω resistor giving a voltage rise to compensate for the voltage drop across \$ R_{DROP} \$. Therefore the maximum voltage rise the opamp can generate is \$12m \cdot 25 = 300~mV \$. This is the (design) maximum compensation of this circuit. It may work up to 400 or 500 mV but it's not guaranteed.