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When repairing or maintaining equipment, or even during design revisions, why would I, for electrical reasons (not interested in the potential financial aspects) not want to replace a PN junction with a Schottky diode? Three reasons I can come up with:

  • Leakage: Schottky diodes have, in general, larger leakage
  • Biasing: If the PN junction is used to bias other devices (such as is sometimes done in BJT output Class AB amplifiers)
  • Switching applications: We might desire to use the slow reverse-recovery in RF switches.

Are there any others? In my specific case, I am replacing diodes in a older HP 6253A powersupply. The powersupply has several 3A rated PN junctions in the signal path which look very corroded, and I have a number of 6A Schottky diodes I am looking to replace them with. Is there any reason why I wouldn't want to do so?

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  • \$\begingroup\$ Got a schematic? \$\endgroup\$ – The Photon Sep 13 '17 at 16:05
  • \$\begingroup\$ There should be a schematic in the link (I don't know/want to risk uploading the schematics directly). In addition, I'm seeking to ask this question as a more general "why wouldn't I want to do this", disconnected from just this specific situation. \$\endgroup\$ – Joren Vaes Sep 13 '17 at 16:06
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    \$\begingroup\$ Everyone answering so far seems to think you want a hard and fast rule. You've outlined all of the reasons I can personally think of for not substituting. I'm looking forward to the odd corner cases that the analog gurus here will mention. This is an excellent question, excellently asked. +1 from me. \$\endgroup\$ – akohlsmith Sep 13 '17 at 17:30
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    \$\begingroup\$ One more reason: In circuit where the diode's capacitance is important, substituting a Shottky (or even another PN diode) could cause the circuit to fail. \$\endgroup\$ – The Photon Sep 13 '17 at 19:00
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While Schottky diodes generally have a lower forward voltage drop and faster recovery time, they are also more susceptible to reverse surges. In my experience a Schottky diode will fail at a lower reverse voltage (or during a transient) that a standard silicon diode would handle without a problem. It all comes down to your application, what the diode is doing, and what sort of electrical events you expect it to see. There is no right or wrong answer here.

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If the design has to function over an extended temperature range, leakage becomes a very big issue.

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You may as well ask why would you not want to replace all 1n4001s with 1n4007s?

There is no boilerplate answer to this question, and you can never dismiss cost.

The answer, in reality, is just the same as every other part you might chose.. "If the shoe fits and you can afford them, then use them if you want to."

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  • \$\begingroup\$ I think there's a slight but important difference between your hypothetical question and the original question: your question refers to the same technology unlike the original question. \$\endgroup\$ – horta Sep 13 '17 at 17:22
  • \$\begingroup\$ @horta it doesn't matter. If the part meets the design requirements, and you can carry the cost, technology is irrelevant. \$\endgroup\$ – Trevor_G Sep 13 '17 at 17:26
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    \$\begingroup\$ Sure, but the question is, does a different technology meet the design requirements and in what way might it not meet design requirements where a PN junction was? \$\endgroup\$ – horta Sep 13 '17 at 17:27
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    \$\begingroup\$ He's not asking for boilerplate "one size fits all" answers. He's asking what reasons outside of the ones he's already mentioned might be a reason to not substitute a regular PN diode with a Schottky. Taking your own answer as an example, if the specific leakage or capacitance of the 4001 is important for the circuit, that might be a good reason. Poor design on the part of the engineer, but a reason to not substitute. That's what I think @jorenvaes is asking about. \$\endgroup\$ – akohlsmith Sep 13 '17 at 17:33
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    \$\begingroup\$ @Trevor I feel like we are missing eachothers arguments here. Ofcourse, all the specifications are important, but these general, high-level ideas give you a good starting point. These high-level rules are used all the time in designs to find a starting point. If I know I need a ultra-low input current opamp, I know I probably don't need to go looking for a bipolar-input device... \$\endgroup\$ – Joren Vaes Sep 13 '17 at 17:48

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