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Schottky Diode Case Connection - Why?

fatman57

May 27, 2013
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I am looking to buy some TO-220AC Schottky Diodes, but when I look at the schematics I find something strange. It seems the case is live! Please see image below taken from the datasheet:
kak66e.png

I was going to put several diodes on the same heatsink, but it looks as if I cannot do this now. Will the heatsink be live to touch? I have also seen 3 pin dual diodes but they also have a connection to the case. Is there a special use case for the dual diodes?

Much obliged.
 

Harald Kapp

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This connection minimizes the thermal resistance between diode and case which improves cooling when mounted on a heatsink.
It has, unfortunately, the side effect that the heat sink becomes "live", too and that the heatsink creates unwanted short circuits between components mounted onto the same heatsink.
The remedy is isolated mounting using thermally conductive insulators and plastic screws or plastic sleeves to insulate metal screws. A well intelligible description is e.g. in this document.
 

73's de Edd

Aug 21, 2015
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Sometimes they also offer the device as a "R" version, in which case the diode polarity would be reversed.
It was very common on stud mounted diode cases.
 

fatman57

May 27, 2013
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Thanks very much for all your replies. Thanks very much for the document Harold! I guess I will have to revise my approach to take this into account.

Is there any difference between dual and single diodes apart from the extra input?
 

Harald Kapp

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Dual diodes have 2 diodes, single diodes have 1 diode. You use what you need.
 

fatman57

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This connection minimizes the thermal resistance between diode and case which improves cooling when mounted on a heatsink.

Out of interest - do these things get quite warm/hot? I presume the heat generated is proportional to the resistance of the device?
 

Harald Kapp

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The power dissipation is equal to P=I*V where I = current through the diode, V = voltage across the diode.
The current depends on the way the diode is used and which curent flows through it. The voltage you take from teh datasheet for teh given current. A Schottky has ~0.3 V voltage drop.
When using fast switching currents/voltages, add the dynamiv losses which are not that easy to calculate.

From the thermal resistance (given in K/W) of the heat sink and the diode and the power dissipated you can calculate the heating of the diode. Here's some background in case you are willing to dig through 36 pages of documentation :D

do these things get quite warm/hot?
It depends on the power dissipation, see above. They can get hot.
 

fatman57

May 27, 2013
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The power dissipation is equal to P=I*V where I = current through the diode, V = voltage across the diode.
The current depends on the way the diode is used and which curent flows through it. The voltage you take from teh datasheet for teh given current. A Schottky has ~0.3 V voltage drop.
When using fast switching currents/voltages, add the dynamiv losses which are not that easy to calculate.

From the thermal resistance (given in K/W) of the heat sink and the diode and the power dissipated you can calculate the heating of the diode. Here's some background in case you are willing to dig through 36 pages of documentation :D


It depends on the power dissipation, see above. They can get hot.
Thanks Harold, once again the documentation is much appreciated.
 
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