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zener circuit

I'm learning how Zener diodes can regulate voltage. Can you help me check if my understanding of its transient behaviour is correct?

Say that the Zener diode in the diagram is in its conducting state, where the voltage across it exceeds the Zener voltage of 9V. It basically becomes a short circuit, which will prevent current from flowing to R2.

But when the current goes above 0.5A, the voltage drop across R1 becomes greater than 0.5V, which will shut off the Zener diode.

Then current can only flow through the two resistors as a series circuit, but R1 can only draw around 9.5 / 1M = 9.5uA with a voltage drop of 9.5uV. This will exceed the Zener voltage and make the Zener diode conducting again.

Is this oscillating behaviour really how the circuit behaves, instead of constantly dropping voltage?

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    \$\begingroup\$ No. It will find a stable point where the current is divided between the diode and the 1M resistor. \$\endgroup\$
    – Hearth
    Jul 6, 2022 at 14:52
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    \$\begingroup\$ A 9V "zener" is more likely operating in avalanche mode (lower-voltage zener diodes have different characteristics). For an explanation of avalanche operation, try: vishay.com/docs/85966/thenoiseofavalanchebreakdown.pdf \$\endgroup\$
    – glen_geek
    Jul 6, 2022 at 16:50
  • \$\begingroup\$ @glen_geek Wow this is really interesting. Thanks for the link! \$\endgroup\$
    – coulombs
    Jul 7, 2022 at 0:27
  • \$\begingroup\$ Compare: "A normal diode turns on when the voltage is 0.7V so when the voltage is more than 0.7V it turns into a short circuit and then starts conducting. When it is conducting it acts as a short circuit. The voltage across a short circuit is 0V so it turns off. Does this mean every forward-biased diode oscillates between on and off?" \$\endgroup\$
    – user253751
    Jul 7, 2022 at 11:15
  • \$\begingroup\$ @user253751: The voltage across a forward-biased silicon diode will be about 0.7 volts, increasing slightly with increasing current. It does not "turn into a short circuit" and oscillate between Zero and 0.7 volts. \$\endgroup\$ Jul 7, 2022 at 15:44

3 Answers 3

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That mental model of a Zener diode is slightly too simplistic. A Zener diode isn’t a simple switch. In all operating regions it has shunt capacitance. With “small” reverse bias voltage, it is a soft voltage-controlled current source. In active state, it’s a voltage source with nonlinear internal series resistance (ESR).

The active state means reverse bias current above some minimum, say 10uA or more.

The shunt capacitances of the resistors are much smaller than that of the diode and can be ignored unless you’re looking at the behavior of really tiny Zener diodes at frequencies past 1GHz.

  1. When the circuit is turned on, the diode is inactive, and acts like a capacitor. It starts to charge through R1, with a bit of current shunted away via R2.

  2. Once the diode charges enough, it becomes “mildly conductive” as the operating point passes through the knee of the I-V characteristic curve. The diode begins to act like a progressively smaller resistor in parallel to R2, or like a very soft voltage controlled current source.

  3. Once the diode enters the active region, it starts to act like a voltage source with nonlinear (“stiffening”) series resistance. The more current you push into it, the lower the voltage gets, although the behavior is nonlinear, so the rate of voltage change drops as the current grows. The ESR “stiffens” or gets progressively smaller with growing current, up to a certain point.

So, in general, assuming no lead inductances, there’s no oscillation. As the voltage on the diode capacitance grows, so does the current, until an equilibrium is reached.

Also: real Zener diodes won’t take kindly to 1 ohm series resistance. Compared to typical Zener series impedance, 1 ohm is a short, and the diode is driven hard. You’ll quickly exceed the package dissipation limits.

For stable voltage, a Zener needs a fixed current source as a load - typically in the 5-15mA range, depending on the diode. A resistor driven from a fixed voltage can do this job, but current sources are the best for this purpose. They have infinite impedance, and there’s no interaction between that impedance and the Zener’s series impedance. The Zener series impedance depends on the current only, so you end up with a pretty good voltage source with a series resistor driven by a fixed current.

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  • \$\begingroup\$ Thanks, your mentioning of the I-V characteristic curve definitely pointed me in the right direction. I wasn't sure how to read one at first, but I found this post that explains it in detail: electronics.stackexchange.com/questions/372611/… \$\endgroup\$
    – coulombs
    Jul 7, 2022 at 0:11
  • \$\begingroup\$ Can I take it that the Zener diode is designed to release significant current around its Zener voltage, and R1 balances this current by reducing the voltage across the Zener diode, so the Zener diode will always be in active mode if euqilibrium point is below the knee? Also, is shunt capacitance an actual physical property or is it used only for analogy? \$\endgroup\$
    – coulombs
    Jul 7, 2022 at 0:14
  • \$\begingroup\$ @coulombs Equilibrium point should be above the knee, or else it's not regulating voltage. Above the knee, a tiny voltage change would cause a large current change which (because of the series resistor) would push the voltage back to equilibrium. Below the knee, current is basically zero, so any voltage change causes basically zero change in current. \$\endgroup\$
    – user253751
    Jul 7, 2022 at 13:07
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A Zener diode does not become a short circuit when reverse biased. Instead it attempts to maintain its advertised voltage across itself.

Your R1 at 1 Ohm is much too low a value for proper operation - you should use a few hundred Ohms at least.

With your circuit, as Vcc rises from zero, the voltage across the Zener will also rise and the Zener will conduct a small leakage current, until it approaches 9 volts. As the voltage rises above 9 volts, the Zener current will increase, limited by R1, as the Zener attemps to hold the voltage across itself and R2 at 9 volts (the voltage will rise slightly as the Zener current increases).

R1 should be selected so that the current through the Zener is near the recommended current shown in its datasheet at the normal Vcc voltage and with the load (R2) drawing its normal current.Zener

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  • \$\begingroup\$ Thanks I think I get the gist of it. But when the Zener diode is in reverse bias above the Zener voltage, isn't its exponential current increase the same as a short? (assuming the diode is directly connected to power and ground with no resistors) \$\endgroup\$
    – coulombs
    Jul 7, 2022 at 0:21
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    \$\begingroup\$ Depends on your definition of "same as a short". If you attempt to apply more than the Zener breakdown voltage to the Zener diode, it will attempt to draw enough current to hold the voltage near its breakdown voltage - but there will still be significant voltage across the diode, where there would be almost no voltage across a short circuit. \$\endgroup\$ Jul 7, 2022 at 0:39
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The reverse VI curve slope is related to the forward Vf vs If diode current in Zener mode.

You must not use a 1 Ohm series R for a current limiter but rather limit that to just include the load max plus a small mA for full load to regulate.

They are only used for low power applications. Pay attention to the datasheets threshold and knee resistance .

Then ensure that you do not exceed the rated power at 125’C

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  • \$\begingroup\$ Thanks I'll keep that in mind. What is a knee resistance? \$\endgroup\$
    – coulombs
    Jul 7, 2022 at 0:24
  • \$\begingroup\$ the incremental V/I slope at the corner of the curve where the slope changes like the leg of a "knee" . All diodes have this knee shape but different slopes and voltages incl. LEDs when saturated due to bulk resistance and power rating \$\endgroup\$ Jul 7, 2022 at 3:07
  • \$\begingroup\$ electronics.stackexchange.com/questions/624794/… \$\endgroup\$ Jul 7, 2022 at 3:12

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