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I think I get the idea of barrier or cut of voltage (e.g., 0.7 V for Si). So diodes start conducting after cut voltage. I have few questions for the cut off voltage

  1. What happens if I increase the voltage to say 1 Volt? My understanding is the current will exponentially increase and may damage the diode. Is that correct?

  2. If 1 above is true then is it fair to say that diodes are operated at around their cut off voltage?

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  • \$\begingroup\$ 1. Depends how big the diode is and how well it is heatsinked. 2. Ehhhhh that entire question is too vague and therefore the answer is too vague to have any real meaning. \$\endgroup\$
    – DKNguyen
    Commented Sep 2, 2021 at 12:33
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    \$\begingroup\$ "cut off voltage" I've never heard it being called that. The forward voltage (~0.7 V for a silicon diode) is either called forward voltage or knee voltage (the I-V characteristik has a knee-shape). 1) correct, never apply 1 V directly to a diode unless you want to break it 2) no, forward voltage or knee voltage. \$\endgroup\$ Commented Sep 2, 2021 at 12:33
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    \$\begingroup\$ @Bimpelrekkie Me neither but I Googled it and it's a thing. Maybe regional? Or "archaic"? \$\endgroup\$
    – DKNguyen
    Commented Sep 2, 2021 at 12:33
  • \$\begingroup\$ I would say that "cut off voltage" meanings the voltage at which a battery being charged is considered full so charging is cut off: en.wikipedia.org/wiki/Cutoff_voltage \$\endgroup\$ Commented Sep 2, 2021 at 12:36
  • \$\begingroup\$ Yes, what you say is true for "ideal" diode. People call that 0.7V (approx.) thingee "forward (breakdown) voltage". \$\endgroup\$
    – jay
    Commented Sep 2, 2021 at 13:36

3 Answers 3

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Your statements are correct.

Except for your first sentence, which assumes an ideal diode. Real diode conduct even if the voltage across them is lower than their forward voltage, it's just that the current will be low. The relationship between voltage and current for a diode is given by the Shockley diode equation and looks like this: Voltage-current graph for a diode

As the graph shows, a small increase in voltage results in a huge increase in current. For this reason, diodes are not voltage-driven but current-driven: your design circuits to have a desired current through a diode.
Also, notice that even below 0.5V, the current is not really equal to 0, it is just very small (probably in the order of tens microamps around 0.4V).

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is it fair to say that diodes are operated at around their cut off voltage

Diodes are operated at significantly less than their: -

  • maximum continuous current
  • peak instantaneous current
  • values in between (or less) depending on the amount of heatsinking provided.
  • maximum peak reverse voltage

How significantly less that is depends on how reliable you want them to be.

What happens if I increase the voltage to say 1 Volt? My understanding is the current will exponentially increase and may damage the diode. Is that correct?

Some diodes will be fine (like the bulk diode inside most SiC MOSFETs; others will be destroyed. Anywhere between those two extremes lies the reality for any particular diode.

Take the 1N4148: -

enter image description here

  • at around 0.8 volts there might be 15 mA of current flow
  • at around 0.7 volts there might be 3 mA flowing
  • at around 0.6 volts about 0.8 mA will flow
  • at around 0.5 volts about 0.1 mA flows

Diodes conduct at voltages lower than 0.7 volts. Some more than others. A lot of diodes conduct mA even when reverse biased when the temperature is high (especially Schottky devices).

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  • \$\begingroup\$ how do you find the max continuous current.? \$\endgroup\$
    – user31058
    Commented Sep 2, 2021 at 15:40
  • \$\begingroup\$ You read the data sheet. \$\endgroup\$
    – Andy aka
    Commented Sep 2, 2021 at 21:55
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  1. The magic smoke comes out, and you need to buy a new diode.

The current through a diode increases rapidly as you increase the voltage (see the graph in the post by Sacha). So once you go much above 0.7V for a silicon diode, the current will be so large as to blow the diode.

  1. Yes. Because that's the voltage you get for any sensible current, with the diode forward biased.
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