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Suppose you've got two 5V1 zeners in parallel. Due to slight differences between the two parts (tolerance on zener voltage), one of them will conduct more current.

schematic

simulate this circuit – Schematic created using CircuitLab

The one consuming most current will heat up more. This will probably affect its properties. What kind of equilibrium will be reached? Will the current distribute evenly between them?

I couldn't estimate myself what equilibrium would be reached because...

  1. I got confused by the temperature coefficients. Some zeners tend towards negative, other towards positive temperature coefficients.

  2. Negative temperature coefficients would be the worst-case scenario: the zener voltage would drop when the part heats up, drawing even more current. Nevertheless, while drawing more current the voltage over the diode will increase a little because the zener I-V-Curve is not ideal. This would push some current to the other zener(s) mounted in parallel. But would that be enough to overcome the detrimental effects of a negative temperature coefficient?

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  • \$\begingroup\$ I'm voting to close this question as off-topic because this smells a bit too much like a test or homework assignment. By itself that is OK but then we do expect that you show what you tried to answer the question yourself. The very minimum you could have done is look up the relations between current and voltage for zener diodes. \$\endgroup\$ – Bimpelrekkie Sep 21 '17 at 9:04
  • \$\begingroup\$ @Bimpelrekkie This is not a homework assignment. I need to know more about parallel zeners for my next circuit. In fact, I got confused because some zeners tend to have positive and other negative temperature coefficients. Also, the current will distribute to some extent, because the voltage over the zener diode does increase (slightly) with current going through. I just don't know what kind of equilibrium will be reached. \$\endgroup\$ – K.Mulier Sep 21 '17 at 9:13
  • \$\begingroup\$ Well, experienced circuit designers would never connect two zeners in parallel, it is asking for trouble. So that made me guess this is just a theoretical question, I wasn't expecting you'd actually want to do this. I would simply use a higher rated (higher power dissipation) zener diode or use two of half the voltage in series. \$\endgroup\$ – Bimpelrekkie Sep 21 '17 at 9:16
  • \$\begingroup\$ I agree with you. But the reason I contemplated doing it anyway is purely for practical reasons. It would make the BOM-list a bit shorter. I don't like adding yet another different component type to the list... (okay, let's not dive too deep into the reasons why. Let us just assume this is a valid practical issue). :-) \$\endgroup\$ – K.Mulier Sep 21 '17 at 9:19
  • \$\begingroup\$ +1 @Bimpelrekkie. Or you can use a zener + high power BJT i think. \$\endgroup\$ – next-hack Sep 21 '17 at 9:19
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Low voltage Zener diodes (about 5V) feature true Zener breakdown, i.e. band-to-band tunneling. The larger the temperature, the more likely the tunneling occurs because the larger the carrier energy (and the smaller also the energy gap).

Therefore, there could be a thermal run-away. In fact, unlike normal diodes (or LED diodes) in forward conduction, small mismatches can lead to large current variations as the curve is very steep in breakdown voltage. The Zener carrying more current will likely heat more (Which will decrease the Zener voltage etc.).

High-voltage Zener diodes (e.g. 10V or more), instead, feature avalanche breakdown. The carriers are accelerated by an electric field, and if they reach enough energy, they will produce more electron-hole pairs, due to impact ionization (which are in turn accelerated by the electric field, etc.). However, the larger the temperature, the larger the lattice vibration and thus the smaller the probability that an electron can gain enough energy between impacts, therefore the less likely they can ionize (i.e. produce another electron-hole pair).

In this way, there will be a negative feedback, that will make the diode initially carrying more current to be less conductive (therefore the current will spread more equally).

The datasheets shows in fact different temperature coefficient for different Zener values.

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    \$\begingroup\$ Very interesting. Which effect would be most prominent: the negative tempco (bad effect) or the slight voltage increase when more current flows through the part, due to non-ideal I-V-curve (good effect) ? \$\endgroup\$ – K.Mulier Sep 21 '17 at 9:22
  • \$\begingroup\$ This could be actually calculated, at least considering a constant dynamic zener resistance and constant temperature coefficient. One must consider the power dissipation, the thermal resistance from junction to ambient (the product will give you the \$\Delta T_j\$), the dynamic zener resistance and the temperature coefficient. \$\endgroup\$ – next-hack Sep 21 '17 at 9:27
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    \$\begingroup\$ don't waffle, it's specified in the data sheet \$\endgroup\$ – Neil_UK Sep 21 '17 at 10:58
  • \$\begingroup\$ Could you please explain your downvote? Is there anything in the answer wrong? (the vote should not be on the comments, by the way). @Neil_UK If you just look at the datasheet you'll only get the minimum Vz at the maximum operating temperature. But in this way you're implicitly assuming that the thermal runaway, (if any) will stop (reach equilibrium) once the diode has reached the maximum admissible operating temperature. That's an error. You must first check if the diode will actually STAY at that temperature. \$\endgroup\$ – next-hack Sep 21 '17 at 11:49
  • \$\begingroup\$ Low voltage zeners have one tempco, high voltage zeners have another, with them becoming equalish around 5v. Your answer was long and waffly, his short and clear, so I voted differentially to push his above yours. I'm making none of the assumptions you seem to think I am. \$\endgroup\$ – Neil_UK Sep 21 '17 at 12:56
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You could look at a datasheet and notice it mentions a temperature coefficient for zener voltage, which answers your question.

In the case of a 5V1 zener tempco can be positive or negative, so anything can happen...

For a 3V zener, tempco is negative, so you'll have current hogging.

For a 12V zener, tempco is positive, so they would tend to share the current.

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  • \$\begingroup\$ Very interesting. So it depends from one zener to another, right? \$\endgroup\$ – K.Mulier Sep 21 '17 at 9:10
  • \$\begingroup\$ Yeah, the physical process that makes a zener diode is dependent on the voltage. Below 5V it's an actual Zener, above it uses the avalanche effect. I presume both have different tempcos. \$\endgroup\$ – bobflux Sep 21 '17 at 9:12
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You didn't state whether you're using zeners for overvoltage (OV) protection or a voltage source.

I'm going to assume you want OV protection because, if you were making a power supply, you would want to minimize the power wasted by running a semiconductor in its active region. For the purpose of OV protection, a better circuit would be to put several resistor-zener circuits in parallel. Then you have control over both the sign of the coefficient and its magnitude. Think about it from a feedback perspective. As the current increases through a resistor, its voltage increases proportionally. In a resistor-zener series circuit, now that the resistor is eating up more of the total voltage, there is less voltage for the zener to eat, therefore its current will be lower. Using a series resistor, you can always force the system into negative feedback* regardless of the sign of the diode's tempco. Each diode should be in series with its own resistor and in parallel with nothing at all. Put as many pairs as necessary in parallel in order to achieve an acceptable risk profile.

*Do not confuse sign of feedback with sign of tempco.

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  • \$\begingroup\$ Good idea @JimStrieter :-) \$\endgroup\$ – K.Mulier Sep 22 '17 at 11:54

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