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):

Block diagram of µA723 from 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):

Negative voltage regulator circuit from µA723 datasheet

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.


2 Answers 2


Because of physics. Turns out that around 6-7 V is the voltage for which the temperature coefficient of Zener diodes is minimized. Much lower voltage Zener diodes also operate on a different principle and have very soft 'knees', so the regulation as well as the temperature coefficient is inferior.

Discrete ~6.2 V Zener references were once popular, for example, the 1N829. They're still available, but not inexpensive.

Modern circuits tend to use bandgap references, which require lower minimum operating voltage but are noisier and perhaps not as stable. They're very cheap and good enough for many applications. There are also some references that operate on the principle of a floating gate with a charge applied at the factory.

The best available Zener references (and some of the finest voltage references available, period) are 'buried' ovenized Zener diodes in the 6-7 V range. The crystal-looking cans in the image are ultra-precision resistors.

enter image description here

  • \$\begingroup\$ This makes a lot of sense for the Zener that creates \$V_{ref}\$ but I'm not sure how critical it is for the output Zener to be so insensitive to temperature, since it's in the feedback loop. I guess that another way of putting the question is: if I stick a 2V Zener on \$V_{out}\$ and don't use the built-in Zener, will I cause a problem? \$\endgroup\$
    – Tony
    Nov 27, 2023 at 3:58
  • \$\begingroup\$ I imagine it's the same zener junction design as the reference. You could use a TL431 or something like that, 2V Zeners are pretty terrible. Even an LED would be better most likely. \$\endgroup\$ Nov 27, 2023 at 4:51
  • \$\begingroup\$ Point well made. I"m not actually thinking of doing this, just wondering if it would cause problems with, e.g., drop-out/common-mode limitation on the error amplifier. \$\endgroup\$
    – Tony
    Nov 27, 2023 at 5:02
  • \$\begingroup\$ It should only affect the output voltage range if it's just in series with the output. The output is supposed to be able to go down to 2.0V worst case. \$\endgroup\$ Nov 27, 2023 at 5:05

Here's a more mechanical and internal-design explanation, if you like:

Zeners around 6V are typical for vintage IC processes (and still today, I suppose, in applicable bipolar or BiCMOS processes). The sequence of doping steps used was (in part):

  • Start with lightly P-doped ("p-") substrate.
  • Deposit a layer of n- via epitaxy; this p-, n- junction isolates VCC from GND.
  • Diffuse p into specific locations; this forms anodes of diodes, and bases of NPN transistors.
  • Diffuse n+ into n- to form "degenerate" schottky contacts, and into p to form NPN emitter junctions.

"Doping" means to make a subtle chemical modification -- adding trace elements such as B, Al, Ga for p-type, P or As for n-type.

Insulation (SiO2 glass) is applied, openings etched out, and contacts made with aluminum metallization. Aluminum can't be deposited just anywhere, lest it make a rectifying junction or other weirdness, but it does make well-behaved resistive junctions with p-type and highly-doped Si. So the p+ and n+ regions are applied for this reason.

Breakdown voltage scales inversely with doping strength, so the total supply voltage handling might be over 30V, NPN VCBO about 30V, and VEBO merely 7V.

Conveniently, the increasing doping strength per step also gives good hFE for BJTs, so the NPNs in this process performs rather well; they're essentially a general-purpose BJT like you would buy. But with a couple extra process steps, instead of single transistors you can have potentially thousands together on a single chip.

A zener diode might simply be constructed as an NPN transistor, with the collector left unconnected, or shorted to base. This gives a cromulent quality zener, often used for start-up / bias circuitry, limiting signal amplitude, etc.

The µA723, and related parts, took this one step further, going out of their way (additional process steps) to make what's called a "buried zener", which, as the name suggests, lies under the surface -- away from the quirks of surface states and contamination. (The NPN emitter is not just a planar junction, but has a bowl shape, with emission taking place all around its surface; but in zener breakdown, the characteristics are different around the edges than at the bottom.)

Once you have a buried zener in your process, you can use it pretty much anywhere at no additional cost; it's just more openings in the masks. Although I think the second zener on this device is just a regular one (maybe to save space?).

(As it happens, most variants since the original (LM723, etc.), have used plain (NPN E-B junction) zeners, or even bandgap references, for the internal reference. Here's a thread photographing every one they can find: LM723 die pictures | EEVblog Forum)

Zeners of other voltages could be constructed, but at significant cost -- additional process steps. Most often, multiples of ~6.2V were made by connecting zeners in series, ~30V might be made using C-B breakdown, and lower voltages might be constructed with (forward) diode junctions, or more complicated (active) circuitry -- bandgap reference, active clamping circuit, etc.

Presumably, they could've done the latter here (chained a bunch of diodes) -- but, this would take up quite some space, since each diode occupies about a transistor's worth of area.

  • \$\begingroup\$ Neat answer thank you. An interesting thing to me was that the old application note suggested also using 6.2V for an external output zener (e.g. on the tin-can part, which didn't have an internal one), even though that wouldn't make any difference to the fabrication processes. If it's just "some voltage drop" to make the error amp work properly then anything would do, although I guess stability is still important and if 6.2V was the magic number for a stable zener back that would explain much. \$\endgroup\$
    – Tony
    Nov 27, 2023 at 6:12

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