The is a section of power supply schematics. I have a -12V power supply with a parallel LM4040.
I know there is a voltage drop over the 1K resistor . What is the trick of the LM4040 diode which allows to control the current through the 1K resistor through its neighbor?
The LM4040 is not a simple diode but a sophisticated integrated circuit incorporating a bandgap reference to make a shunt regulator. As the datasheet linked above shows, this is the simplified internal diagram:
The overlapping circles represent current sources and sinks. The triangle represents a differential amplifier.
When the voltage across the IC from "cathode" to "anode" gets larger than 2.5V the transistor to the right is turned on to draw current. The internal circuitry draws about 80uA maximum so you need to have a net current at least that high flowing through the IC to get it to regulate, and it can sink a maximum of 25mA without losing regulation.
As to how the (temperature-voltage) curvature-corrected reference works, the answers, including mine here have further information.
I suggest revising the symbol and pin numbers for the LM4040. The circuit will work as shown, but only because there are two errors that cancel out.
