Recalling this question: Choosing series resistors for zener diodes

I thought to open a new question related to this, because it is centered to a specific part of the old question.

It is said that, when a Zener of a given voltage is used with a lower current than the tested one, it provides a smaller zener value. I tried with the BZX84C3V3 and applying 5V with a 560Ω: I have a bit more of 3.3V, which is reduced when using 2.2kΩ. Does the zener cut the voltage better (or even to a low voltages than 3V3) when derated in current?

This is coherent with the datasheet, but I don't get the logic: doesn't the Zener shall be less opened to GND with lower current? And being less open, more zener resistance shall be present and according to the voltage divider, I shall see an higher voltage. Instead, it is happening the opposite.

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    \$\begingroup\$ Good datasheets contain an IV curve, have a look at them \$\endgroup\$ – PlasmaHH Aug 26 '16 at 11:36
  • \$\begingroup\$ The change in voltage due to the change in biasing current is not only caused by the Zener's series resistance. The Zener voltage also changes slightly with current. Dave did a video on the Zener, have a look here: eevblog.com/2016/08/07/eevblog-908-zener-diodes \$\endgroup\$ – Bimpelrekkie Aug 26 '16 at 11:53
  • \$\begingroup\$ I realize my mistake: we are talking about polarizing current, not to the applied voltage. In the eevblog video (in its last 60seconds), Dave said that simply "does not work" at low currents with no particular explanation: does simply depends on the change of the IV curve type? (not the V-I working point, but actually the curve type and thresholds) \$\endgroup\$ – thexeno Aug 26 '16 at 14:27

You are mixing up a couple of concepts. The zener dynamic resistance is the change in zener voltage divided by the change in zener current around an operating point. It's a linearized portion of the I-V curve.

The zener voltage vs. current curve is nonlinear and generally pretty miserably non-ideal for a 3.3V zener. Below is a typical set of curves from another answer:

enter image description here

You can see that at 10mA you get around 3.4V from the "3.3V" zener but at 0.1mA you get more like 1.9V. It's not easy to see the dynamic resistance on that set of curves because it's a log-linear graph, but the dynamic resistance will be much less at high current. It would be the reciprocal of the slope of the curve on a similar but linear-linear graph.

Higher voltage 'zener' diodes behave in a much more ideal fashion, with almost vertical I-V curves over a wide range of currents. So mostly we avoid use of low voltage zener diodes in favor of active circuits such as the TL431.

  • \$\begingroup\$ You are perfectly clear. But note that my point is due to the circuit configuration in which the zener is used: I ideally thought that with low currents the diode is almost shutoff (no avalanche/zener effect at all, like a reversed diode before the Vbd) and so it is like the diode is not present, so no regulation, so higher voltage, not lower. This is where my confusion came from. Confusion that I still have. \$\endgroup\$ – thexeno Aug 26 '16 at 14:43
  • \$\begingroup\$ It seems that with low currents, the diode let it flow all across it self (where if all current is absobed, when a very low one is provided, it acts almost like a ground), so very low voltage across with it. And not like an open circuit. \$\endgroup\$ – thexeno Aug 26 '16 at 14:46
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    \$\begingroup\$ That's right- it acts more like a short. To have the voltage increase as current decreases is rare- it's called negative resistance. Something like a neon bulb or a tunnel diode will exhibit negative resistance over a certain range of currents. \$\endgroup\$ – Spehro Pefhany Aug 26 '16 at 15:03
  • \$\begingroup\$ I have not tried yet, but you are saying that if I use a normal diode, it will happens the same? Or I will see 5V instead because the normal diode is almost electrically disconnnected, except for the leakage? \$\endgroup\$ – thexeno Aug 26 '16 at 15:10
  • \$\begingroup\$ A normal diode (in the forward direction) like 1N4148. 0.694V at 10mA and 0.480V at 0.1mA. So it's a better regulator than a 3.3V Zener. 5 in series would give you 3.47V at 10mA and 2.4V at 0.1mA. In reverse direction it might be more like a 150V or 200V zener. \$\endgroup\$ – Spehro Pefhany Aug 26 '16 at 15:17

enter image description here

Zener diode works in the reverse bias, so take a look at the lower half of the I-V curve. The voltage stays almost constant, but deviates slightly with increasing current. So, the lower the current, the closer the voltage to the rated Vz.

I do not think, "more or less opened" to the ground is an actually correct scientific expression.

  • \$\begingroup\$ It was a simplification that could make the correct idea of the concept. Whether or not it is a correct or wrong concept (understand the correctness was the goal of the question, btw). Thakns and see comments for the update. \$\endgroup\$ – thexeno Aug 26 '16 at 13:25

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