The BZY97C47 data sheet (I found no direct PDF link) confuses me on several points.

I fail to find straight forward information about the maximum allowed forward current from the data sheet.

Also, what is the parameter "Admis. Zener Current" supposed to mean?

And I am confused that additionally to the "zener voltage" (45-50V) there is a "reverse voltage" (25V) specified. Is that supposed the be the breakdown boundary where leakage current starts to increase?

  • \$\begingroup\$ Your link returns some "temporarily not available" pdf for me \$\endgroup\$ – PlasmaHH Mar 20 '15 at 13:59

It's a 1.5 watt zener so to generate 1.5 watts at 47 volts requires a current of 31.9 mA. This is the maximum allowable zener (EDITED) current but, there's a little sting in the tail. If the device is actually a 50 volt zener then this current is of course a little less. Ditto if the tolerance forces it to be a 43 volt zener but this time the current is slightly higher.

This is largely confirmed by the term "Admis. Zener Current" being 30mA. A 1.5 watt 50V zener can only be driven with 30mA. I'm assuming "Admis." means admissible meaning "acceptable" or "valid".

Zeners don't switch off immediately below their zener voltage. In the data sheet that I found (because your link doesn't work for me), it quotes 24V as being the point where 1uA of reverse current flows. Above that voltage, current will increase and below that voltage current will get less.

It's also worth pointing out that if the zener self-heats and takes the local temperature above 60degC then the zener power rating reduces. Here are two graphs, the left graph shows the non-instantanous rise in current as the zener voltage is approached. The right graph shows the maximum power curve just mentioned and ultimately also governs current flow into the device i.e. if you are putting 30mA in and the temperature rises above 60 then you are exceeding the power rating of the device: -

enter image description here

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  • \$\begingroup\$ Maybe I misinterpret your answer, but I was actually asking about the maximal allowed forward current when the diode is used in forward biased direction. I intend to use it for a capacitive power supply and thus, every negative half wave, the Z-diode is forward biased. I was trying to figure out a adequate series resistor with diode current in mind. However, I could probably indeed use the maximum power divided by forward voltage approach. I was just confused that there is no definitive value about that like we know it from regular diodes. \$\endgroup\$ – Rev1.0 Mar 20 '15 at 14:10
  • \$\begingroup\$ If you are referring to peak instantaneous current as opposed to average current then the last column in the data sheet suggest 0.31 amps for 10ms. Note that I edited my answer because it was confusing with the word "forward" incorrectly used. \$\endgroup\$ – Andy aka Mar 20 '15 at 14:45
  • \$\begingroup\$ The continuous forward current was what I was looking for, but I guess in the end it comes down to maximum allowed power dissipation. You answer covers some good points, thanks. \$\endgroup\$ – Rev1.0 Mar 20 '15 at 14:49
  • \$\begingroup\$ Yeah tricky to find and certainly not mentioned on the data sheet but maybe if you knwo the forward volt drop and used "1 watt" as the limiting case you might get somewhere. Are there any zener data sheets that give this parameter I wonder? \$\endgroup\$ – Andy aka Mar 20 '15 at 14:52

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