4
\$\begingroup\$

For the circuits I've studied thus far involving diodes (which admittedly are not that many), they have been nominally used in the forward-bias mode. For example, the LED only lights up when it's operated in the forward-biased region, and is not designed for reverse bias, let alone the burn-out breakdown region.

However, I recently read about the Zener diode, and I found that this particular diode is predominantly used in the reverse-biased, breakdown region, with the following regulator circuit being a popular example:

enter image description here

Although this circuit works, why can we not achieve the same voltage regulation functionality by operating the diode in the forward-biased mode, like this:

enter image description here

This is I-V curve I am assuming for the diode:
enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ Ypur assumed I-V curve is very wrong. In the forward direction zeners act much like std silicon diodes with somewhat higher fwd voltage drop. In the reverse direction they exhibit reverse breakdown at modestly well defined voltages which is what makes them useful as regulators. If you need other than a very poor 0.8V reference zeners in the fwd direction are not suitable. \$\endgroup\$ – Russell McMahon Jun 16 '16 at 23:38
9
\$\begingroup\$

Three reasons:

First, operating in the forward orientation only allows operation at a single voltage, nominally about 0.7 volts for a silicon diode. Diode construction can be tailored to produce a wide range of breakdown voltages, with a consequent choice of different regulator outputs.

Second, your V-I curve overstates the sharpness of a forward-biased junction. There is no relatively flat portion other than in the vicinity of zero, and that's not very useful.

Third, with an exponential V-I curve, the forward-biased junction cannot be operated at useful current levels with good regulation.

\$\endgroup\$
  • \$\begingroup\$ For actual curves, take this diode's datasheet that happened to be a result on Digikey. Definitely a difference in sharpness. \$\endgroup\$ – user2943160 Jun 17 '16 at 1:47
5
\$\begingroup\$

The obvious answer is ... that they were designed to do that. Not being facetious here, rectifying diodes are designed for forward operation and high withstand in reverse, regulating diodes are designed for a relatively accurate reverse breakdown voltage to be used in regulation circuits. If we want to continue, you could say that Photodiodes are optimized for reverse operation without breakdown too ...

In short an diode is not a diode is not a diode, the type really matters.

Interestingly, there is a very precise mechanism that dictates the operational voltage of a Zener diode. There are other regulator diodes that operate similarly to a Zener but use a different effect.

\$\endgroup\$
  • 1
    \$\begingroup\$ For OP's benefit, the "other regulator diodes" mentioned in this answer are also commonly called "zener diodes", even though they don't actually work on the zener effect. \$\endgroup\$ – The Photon Jun 16 '16 at 21:52
3
\$\begingroup\$

Because there's nothing special about forward-biased operation. It still has the 0.6-0.7V drop typical of most silicon rectifiers. It's only when reverse-biased that it exhibits its nominally high voltage drop.

\$\endgroup\$
  • 1
    \$\begingroup\$ Followup question if you don't mind: if we happen to want 0.6-0.7 V as the regulating voltage, we could conceivably operate in the forward-bias region (as my second schematic shows), right? \$\endgroup\$ – Tosh Jun 16 '16 at 22:04
  • 1
    \$\begingroup\$ Well... sure, but then there'd be no point in using a zener diode. You may as well use a more stable rectifier instead. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 16 '16 at 22:06
  • 1
    \$\begingroup\$ @Tosh as you learn more about electronics you will see plenty of example where the 0.6-0.7 V forward bias of a conventional diode is used as a regulated voltage, for example to Figs 5.5.3 and 5.5.4 here: learnabout-electronics.org/Amplifiers/amplifiers55.php - but don't try to understand those circuits until you have studied the basics of how transistors work. \$\endgroup\$ – alephzero Jun 17 '16 at 3:01
3
\$\begingroup\$

Because they are manufactured to have very specific and sharp curve in reverse polarity around this specific voltage value so while it may have typical diode forward voltage it can for example have around -5V of this reverse one. And that value is the one we use. It is all about desired properties of your diode.

\$\endgroup\$
2
\$\begingroup\$

It depends on the application but I'll give two examples, including the zener:

Zener Diode

This is operated in reverse bias because, simply put, that's where the interesting, 'Zener' property is. In forward bias, Zeners look like regular diodes. That is to say that their voltage changes a lot with current and isn't very tune-able. The reverse-bias breakdown is both tune-able (which is why you can buy Zeners in all manner of voltages) and sharp. That means you can put a lot of current through it and the voltage remains relatively constant.

Photodiode

These are often operated in reverse when used to measure light input. That's because the IV curve is extremely flat at negative voltages. That means that the negative voltage you apply to the thing can be noisy and unstable and you will still get a steady current for a given light level.

\$\endgroup\$
0
\$\begingroup\$

The rectifying and LED diodes break down thermally and their breakdown is irreversible.

On the other hand Zener diodes break down due to quantum tunnelling effect and this breakdown is reversible.

That's why Zener diodes can be used in both current directions and the standard diodes cannot. The actual values of breakdown and opening voltages, conductivities etc. define the application of the diode - Zeners are used for stabilisation of voltages near 5 V, Avalanche diodes are used to stabilise higher voltages, power (standard) diodes are used in rectifiers and LEDs are used as light sources.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.