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I'm currently using a 5V current transducer that's connecting to an I/O device ADC input. I'm trying to protect my ADC input from going over 5V by using a zener diode . The reason this is coming about, is a current I'm trying to measure has a high inrush current on start up, and I don't want to damage the system. On the zener datasheet, it says \$I_z=5mA\$

Does this mean it will only starts clamping at 5mA or does it mean it will not work after 5mA?

And is it necessary to have a current limiting resistor? What will happened if there's no resistor?

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  • \$\begingroup\$ This is not a problem. The 5V transducer can't output more than 5V. \$\endgroup\$ – DKNguyen Oct 16 '20 at 14:14
  • \$\begingroup\$ I accidentally hookup the wrong current yesterday and it outputs a 13V tho. \$\endgroup\$ – cy1125 Oct 16 '20 at 14:21
  • \$\begingroup\$ Really? Are you sure you have the 5V version? \$\endgroup\$ – DKNguyen Oct 16 '20 at 14:21
  • \$\begingroup\$ yes. I have the CR5210-0.5 version \$\endgroup\$ – cy1125 Oct 16 '20 at 14:23
  • \$\begingroup\$ That's really weird. Does it actually measure correctly? Current limiting resistor not necessary if your diode is big enough or the drive current is weak enough. The zener just won't clamp to spec if current is <5mA. \$\endgroup\$ – DKNguyen Oct 16 '20 at 14:24
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They don't show the I/V curve for the breakdown region in the datasheet, but for a diode in zener mode, it breaks down very fast as the voltage is increased (Vz in diagram below)

enter image description here
Source: https://www.electronics-tutorials.ws/diode/diode_7.html

The Iz point means that they set the current to 5mA and then measured the voltage Vz at that point. The intended application is more for setting a voltage so if you wanted to get a specific Vz (for a specific model) then set the Iz to 5mA, increase the current beyond 5mA and you get a cliff. So use a current limiting resistor before the zener to keep the current below the absolute maximum ratings of the diode.

You don't necessarily need a resistor, if the highest potential current that the zener would ever see was low enough to keep the part from over heating ( like from an opamp that could source up to 80mA on the output) then a 5V diode would dissipate P = 0.08*5V = 0.4W which is just under the max dissipation for the part which is 0.5W

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