In other words, why don't we always use Schottky diodes if they are so much better? What diode properties do Schottky diodes have that makes them unfit for certain applications?


They cost more, have higher reverse leakage current, and are physically larger according to a quick search. Of course they're much faster though :)

Looks like in a same size comparison they can't dissipate as much power as a typical power diode. Also with larger currents you lose that Vfw advantage. Oh and wiki says they normally have lower reverse voltage rating on the order of 50V.

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    \$\begingroup\$ +1 for reverse leakage current, which is highly temperature-sensitive. Some people use some pretty complicated work-arounds to avoid that leakage current ( a ). \$\endgroup\$ – davidcary Oct 17 '12 at 17:26

Far from a comprehensive list:

  • Schottky diodes of a comparable rating are generally more expensive than PN silicon diodes. I've seen price differences of 20% - 200% depending on the rating.
  • Schottky diodes have a lower maximum reverse voltage rating than is possible with PN diodes.

For essentially the same reason that schottkys have low forward drop, they have large reverse currents.

From the diode equation:

\$I_f = I_s\cdot e^{\frac{-qV_f}{kT}}, V_f = \frac{kT} {q}\cdot \ln\frac{I_f}{I_s}\$

-- having a large Is term is what makes Vf small. However, the reverse leakage current is also equal to the Is value.

From their structure, silicon schottkys can only withstand about -30 V alone. Higher voltage ones are created, but basically these have an internal JFET in series with them -- this is what actually withstands most of the reverse voltage.

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    \$\begingroup\$ Wow +1 ,I did not know about the internal parasitic JFET \$\endgroup\$ – Autistic Mar 12 '17 at 0:14
  • \$\begingroup\$ +6(0) for the gist behind most other answers on this page. It would perhaps make your answer even better if you expanded why schottkys cannot exceed -30V on their own (although, granted, that could also be a separate question). \$\endgroup\$ – Fizz Jan 1 '18 at 4:00

Here's one that may sound a bit strange, but is important in some uses: low forward voltage drop.

Sometimes it is useful to distribute the heat dissipation among the components in a device. Take for example the traditional linear voltage source: You have a transformer, a full wave rectifier, large capacitor and a voltage regulator plus some smaller capacitors near it.

Let's say the transformer has nominal output voltage of 12 V AC. Once we rectify that and fill up the capacitor, we have around 17 V DC on the capacitor in the case of ideal diodes with no voltage drop. If we want to power a device regulated by for example LM7812, we'll need to somehow dissipate away 5 extra volts. The typical dropout voltage for the regulator is 2 V, so we're left with around 3 V to get rid of. That would go into the regulator's heatsink and will increase the amount of heat the regulator dissipates. On the other hand, if we take a look at say 1N4007's datasheet, we can see that the forward voltage drop is between 0.7 V and 1 V in the forward current region that would be interesting to the users of LM7812. So with low current consumption, those 3 remaining volts would turn into at most 1.6 V (since we have two diodes conducting in the rectifier at any one time) that need to be dissipated into the heatsink of the regulator. At higher currents, the remaining 3 V would turn into 1 V which isn't as big problem and gives us some margin if the drop-out voltage of the regulator is higher than the typical 2 V.

If we used Shottky type 1N5819 diodes for the bridge rectifier, we'd have voltage drop on the diodes of around 1.2 V, leaving us with much more heat to dissipate on the regulator itself.

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    \$\begingroup\$ One could right away use a lower rated transformer and not dissipate that heat at all (while using shottkys) \$\endgroup\$ – PlasmaHH Jan 20 '15 at 13:37
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    \$\begingroup\$ @PlasmaHH One could! One could also use a cheap SMPS or something else. What if you don't have an appropriately rated transformer? What if you have an overstock of inappropriate transformers that need to be used up? What if transformers with higher voltage are cheaper? When I actually wrote this answer, I was working on a project where I had a rectifier made out of Shottky diodes and a transformer. I didn't have another transformer in my parts bin, so the diodes had to go. \$\endgroup\$ – AndrejaKo Jan 20 '15 at 15:53

Silicon schottkys can be found at 250 Volt easily but at 250V there is a VERY limited selection.Manufacturers via their sales reps state that they can not make them above 250 V.There is the problem of increased reverse leakage current that can upset some circuits AND cause thermal runaway at elevated temps below Tjmax at voltages below Vrmax .This runaway can occur at low voltages when using low voltage devices just as easily as in high voltages . OK keep them cool unless you really know what you are doing. SiC schottkys are available at high voltages and are fast and expensive but the foward drop can be worse than a normal diode at realistic currents .These Sic devices have significant bulk resistance .The overall tempco of the foward volts is positive at modest currents meaning that thermal runaway is possible despite the low leakage of the high band gap semiconductor.

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    \$\begingroup\$ Shottkies generaly have more capacitance than the standard diode they replaced which may be bad ,good ,or indifferent. \$\endgroup\$ – Autistic Sep 3 '15 at 3:23

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