I've realised that several ICs with simple components, such as the FDS6690, provide duplicate pins in order to comply with a standard IC package (e.g. SOT23). My question is - are these pins connected internally or one should do it on the PCB? Can I just use one of the Drain and Source pins?
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2\$\begingroup\$ Related: arduino.stackexchange.com/questions/3089/… \$\endgroup\$– jippieJun 28, 2014 at 14:36
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\$\begingroup\$ I have come accross chips where the pins were not connected internally, however in every case the datasheet was very explicit that they must be connected externally and they were given different names. Probably the main place people will encounter this nowadays is combined analog/digital chips that have independent analog and digital grounds, the reason they are not connected internally is so you can decide on the 'star-ground' point on your board where all your grounds meet. if every adc had its own connection ground loops would result. \$\endgroup\$– John MeachamFeb 11, 2015 at 22:48
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4 Answers
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They are commonly used to remove heat as well as increase conductivity so my advice is read the data sheet pertaining to the device in question: -
The picture above is recommending that all pins be used and, in order to improve the thermal dissipation of the device, various copper areas are suggested.
answered Jun 28, 2014 at 14:00
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Here is a representation of a part packaging similar to your FDS6690-
If you don't connect all the pins on the top together (and thence to a big pad of copper or stitched to a ground plane), the heat flow out of the package will be significantly compromised. This could be enough to cause failure.
If you don't connect all the source pins together, some of the wire bonds may carry more current than necessary, and the Rds(on) will be a bit higher, also heat flow out of the package will be somewhat reduced.
answered Jun 28, 2014 at 15:07
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2\$\begingroup\$ Lower parasitic inductance too. +1 for the drawing. \$\endgroup\$ Jun 28, 2014 at 15:42
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\$\begingroup\$ @Alexeev, yup- n inductors in parallel have 1/n inductance. \$\endgroup\$ Jun 28, 2014 at 15:55
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You don't have to connect both pins to the board if you don't have a need for higher current capability. If they're labeled as being the same pin, then they are connected internally. Some pins sometimes are no connects. If that's the case, you'll usually want them connected to ground just so you don't have a floating voltage in the chip.
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1\$\begingroup\$ Sometimes the "higher" current is normal operating current though, as the internal connects may (intentionally) not be large enough for supply and are merely for reference. Some MCUs/CPUs and other larger devices are designed this way. \$\endgroup\$ Jun 28, 2014 at 14:26
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\$\begingroup\$ @IgnacioVazquez-Abrams agreed. Everyone should still be reading the datasheets of their devices. \$\endgroup\$– hortaJun 28, 2014 at 16:27
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Other people have covered off the main reasons:
- Current handling increase,
- Heat conduction increase,
- Reduced inductance
To complete the list:
- Some military requirements require that there are redundancy in bond wires (and indeed bond wire strength is tested), there may not be sufficient space on the lead frame so another pin is used,
- There may not be space on die to run the signal across the die or doing so may incur too much delay/distortion (these by necessity would not be side-by-side), but sometimes there is local congestion so it is easier just to connect on the outside,
- In some designs this is done to provide backward compatibility to a previous suboptimal pin placement,
- In some designs in which the chips are designed to be used in a grid, the same signal is provided on different pins to provide easier routing on the PCB,
- and sometimes it is to provide wiring options (usually in the case when the package has too many pins).
answered Jun 28, 2014 at 17:10