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In the presence of a resistor, we can reason that the voltage difference between the given battery voltage and breakdown will be given to resistor. My question is without any other devices just only a battery whose voltage can increase or decrease connected to a Zener diode in reverse bias mode. What will happen if the voltage is increased above tgebreakdown limit? Where will the extra voltage drop go? Will the diode be damaged?

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4 Answers 4

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With no resistor, a voltage source and zener diode is an invalid circuit. There is no extra voltage drop, and the zener will break.

Sometimes, if there is a fuse in the circuit, the zener could be used as an overvoltage protection device which breaks the fuse when voltage rises above zener voltage.

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  • \$\begingroup\$ Sir you mean it can behave like fuse ? Or something else meant by "breaks the fuse " \$\endgroup\$
    – Orion_Pax
    Apr 16 at 14:44
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    \$\begingroup\$ @Orion_Pax: If you include a fuse in the circuit, in place of the usual resistor, the fuse may blow when you increase the battery voltage above the Zener voltage. If the fuse rating is larger than the Zener maximum current rating, the Zener diode will be destroyed. \$\endgroup\$ Apr 16 at 16:07
  • \$\begingroup\$ I see thanks understood it \$\endgroup\$
    – Orion_Pax
    Apr 16 at 23:10
  • \$\begingroup\$ "Invalid circuit" - like, how? Will the Universe reboot if you assemble such a circuit? There are no invalid circuits. Just circuits that perform unexpectedly, but that's an easy problem to fix and has nothing to do with the circuit itself. \$\endgroup\$ Apr 17 at 8:09
  • \$\begingroup\$ @Kubahasn'tforgottenMonica the circuit performs as expected, it just has no real life value as either the zener does nothing or blows up depending on applied voltage. For what purpose would this circuit be valid? \$\endgroup\$
    – Justme
    Apr 17 at 8:46
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without any other devices just only a battery whose voltage can increase or decrease connected to zener diode in reverse bias mode . What will happen if the voltages gets increased from breakdown limit

This is a first order model of what you will get.

schematic

simulate this circuit – Schematic created using CircuitLab

Real batteries can be modeled as having an internal source resistance. For example, a Duracell MN1500 AA battery might have 71m Ohms of source resistance.

Real Zener diodes are also not perfect. The voltage changes with the applied current. Therefore the "breakdown voltage" of a Zener diode is often specified at a particular reverse current. At lower currents the voltage will be lower and at higher currents the voltage is higher. You can typically get a voltage vs current curve from the part datasheet.

Here is the curve for the BZX55C4V7 4.7V Zener diode.

enter image description here

Using a program like WebPlotDigitizer we can extract the following voltage vs current data.

enter image description here

Plotting the data we see that the output voltage of the zener over the current range 20mA to 80mA is a pretty good fit for a linear resistor model. The slope is about 9.6 ohms (0.0096V / mA). We would use 9.6 ohms for R_zener in our schematic above when attempting to approximate the output voltage.

enter image description here

Note that the current will increase with more applied voltage, and it can increase fairly quickly. If the applied wattage (voltage * current) across the device exceeds the ratings then the device will likely overheat and fail. Depending on the construction of the device it might fail open or short. Depending on how much current was applied the device might burn and small pieces might fly off.

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  • \$\begingroup\$ Real batteries do not have an internal source resistance; their behavior is modelled by inserting a resistor in series. This is only an electrical equivalent circuit helping understanding and calculating. See also my story Is the source internal resistance really a resistance?". \$\endgroup\$ Apr 16 at 20:51
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    \$\begingroup\$ I agree with circuit fantasist \$\endgroup\$
    – Orion_Pax
    Apr 16 at 23:11
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    \$\begingroup\$ @Circuitfantasist The cathode and anode of a battery (like all conductive materials) do in fact have some resistance. So does the electrolyte. If you are looking at a car battery for instance, the bulk resistivity of sulfuric acid is 4.81 ohm-cm. But since there are also chemical reactions going on which affect the voltage, I shall change the wording to say, "real batteries can be modeled as having an internal source resistance", since that might be a bit more accurate. \$\endgroup\$
    – user4574
    Apr 17 at 3:54
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    \$\begingroup\$ @Circuitfantasist Well, of course it's an electrical equivalent, but that's a reasonable model to use if all you want is a ballpark estimate of what will happen. Usually the battery's time constants are much longer than the survival time of the Zener, so treating the battery as an ideal voltage source with a series resistance is quite an accurate model. After a milliseconds or seconds the Zener emits the magic smoke, and then we don't care about any longer-term effects in the battery, since the circuit is mostly open :) \$\endgroup\$ Apr 17 at 8:11
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The zener diode has a characteristic that rapidly increases current with voltage above its zener voltage.

If you supply let's say 5 V, you'll get the (5 mA) zener voltage across the diode.

If you supply another 100 mV, or another 1 V, and your power supply has the power to deliver the current, the current will rise dramatically, probably releasing the magic smoke.

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To understand what is happening, imagine the Zener diode as a variable "resistor" that vigorously decreases its resistance when the applied voltage exceeds the diode threshold voltage. Figuratively speaking, from this point on, the zener diode shorts the source.

The problem of this connection is that both elements behave as voltage stabilizers. Each of them tries to set its voltage; as a result, a voltage conflict arises between them and the current vigorously changes as a response. This phenomenon is known as "current steering".

The conflict between the two constant-voltage elements can be "soften" by a resistor, which takes on the voltage difference across itself. The better solution is a current-stable "resistor".

Similarly, if you connect a current source to the collector-emitter part of a BJT, a "current conflict" appears and the voltage vigorously changes. Now you can imagine the transistor as a variable "resistor" that vigorously increases its resistance when the injected current exceeds the transistor threshold current (set by the base-emitter voltage). This arrangement is known as "dynamic load".

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    \$\begingroup\$ Nice explanation thanks \$\endgroup\$
    – Orion_Pax
    Apr 16 at 23:11

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