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I have this simulation issue and I don't know how to solve it:

The Zener current is smaller than the breakdown current yet the Zener diode still acts like a voltage regulator. Any help?

Shouldn't the voltage be 6.4V?

Is the breakdown current the minimum current after the Zener diode is not in the Zener breakdown region or is it the maximum current before damage?

enter image description here

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2 Answers 2

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You have specified a nominal zener voltage of 5V and your simulation is within 1% of that value. I don't see a problem here.

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  • \$\begingroup\$ Yeah but shouldnt the zener current be more than the breakdown current?The breakdown current is the min current after the zener diode is not in the zener breakdown region or is it the max current before damage? \$\endgroup\$
    – Miss Mulan
    Mar 15, 2021 at 18:00
  • \$\begingroup\$ Well, 5V*5mA = 25mW. 5V*20mA would be 100mW. What is a typical device in that package rated for? If 200mW, then 20mA would be the max you'd want to push through it (half the power limit.) \$\endgroup\$
    – rdtsc
    Mar 15, 2021 at 18:29
  • \$\begingroup\$ It doesn't have a rated current.You can change it as you want. \$\endgroup\$
    – Miss Mulan
    Mar 15, 2021 at 18:32
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    \$\begingroup\$ @MissMulan I think you misunderstand the meaning of "breakdown current", but I will admit that it is unclear how the simulator treats that parameter. It looks like you are using Multisim...good luck with that. \$\endgroup\$ Mar 15, 2021 at 20:59
  • \$\begingroup\$ @ElliotAlderson yes I don't know what it is exactly so I guessed it would be the min current below the zener diode doesnt operate in the zener breakdown region. \$\endgroup\$
    – Miss Mulan
    Mar 15, 2021 at 21:00
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If the current is smaller then you are operating within the safe zone. Don't forget that you are plotting the reverse current, so it's actually with a negative sign. If in doubt, consider the absolute values, and you know you are safe when the current is less than the maximum. But even if you were to exceed the current, it's unlikely the diode has the breakdown modelled. Try 100 V and see if there's any smoke.

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