In an op-amp, what does connecting the negative supply to the feedback do?

Question

I'm trying to understand how the LM317 works? I've found that a good simplification is the following:

So a practical example would be

^{simulate this circuit – Schematic created using CircuitLab}

Where R2 corresponds to a load.

What I don't have clear is the effect of connecting the negative voltage supply of the op-amp to the emmiter, instead of earth.

I guess that this is done so there is no need for another pin in the LM317, but would make any difference? The simulation, doesn't show any difference, but could it affect the stability or something like that?

Answer 1

It basically makes the LM317 a "floating" type voltage regulator. This provides the advantage that the regulator only sees the difference between the voltage input voltage and output voltage as compared to the usual style regulator that sees the voltage difference between input and ground. What this means is that the LM317 does not have a maximum input voltage, but rather a maximum Input-to-output differential voltage (see datasheet). This makes the LM317 suitable in pre-regulation application where you might have like a 48V system and you want to drop the voltage down to 20V and then use a traditional regulator to come down to 12V. This way your traditional regulator doesn't require a input voltage tolerance of more than 48V and reduces the power dissipation across it by loading it off to the LM317. So more of less, it basically reduces the voltage seen by the opamp by basically "floating" the negative supply.

Answer 2

As you say it reduces the number of pins but has the disadvantage that the supply current for the amplifier flows to the output pin and so imposes a minimum output current. (around 10mA is typical for this type of device).

For this design the current through the reference diode also has to be made constant. So resistor as shown wouldn't work well.