The µA723 voltage regulator has a 6.2 V Zener diode on its output. I am trying to understand why the Zener voltage is as high as 6.2 V. (LM723 or LM723C are more or less equivalent parts, raising the same question.)
Here is the block diagram from the µA723 datasheet (\$V_z\$ is the Zener voltage, which is specified as 6.2 V later in the datasheet):
The Zener is used in circuits for a negative regulated voltage, or for a general floating regulated voltage. To make things more concrete, here is the negative voltage regulator circuit from the datasheet (\$V_I\$ is some fairly negative voltage):
Without the Zener diode, the regulated output would be only a diode drop (2N5001 base to emitter) above \$V_{CC-}\$. I can see this being a problem for at least the error amplifier, which might need a bit more space between its negative rail (\$V_{CC-}\$) and output. But I don't understand why it needs 6.2 V. In the standard low-voltage positive supply circuit in the datasheet, for example, the output voltage may be as low as 2 V above ground, when \$V_{CC-}\$ is also ground.
There is a very old application note which has a couple of tantalising mentions of the Zener diode, but it doesn't really indicate why 6.2 V is the correct value:
Diode \$D_2\$ used to shift the \$V_{OUT}\$ terminal of the \$\mu\$A723 up to provide sufficient operating voltage to \$Q_{12}\$. \$D_2\$ can be eliminated when using the DIP package by using the \$V_{Z}\$ terminal instead of the \$V_{OUT}\$ terminal. [page 2.10]
The zener provides the necessary level shifting required to maintain biasing of the regulator. (\$\mu\$A723's in the metal can package do not have a \$V_{zener}\$ output. When using these devices, it would be necessary to add an external 6.2 volt zener between \$V_{out}\$ and the switch transistors). [page 4.6]
I know this is an old part and newer, shinier options are available. Nevertheless, I'm trying to understand this one.