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I have an N-MOSFET array IC and I don't need to use all of them, but I do need to use a diode. Is it OK to use the body diode of one of the MOSFETs as a regular diode to decrease component count and footprint? Can I just tie source to gate and use the MOSFET in place of a diode in the circuit?

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    \$\begingroup\$ I think you should insert more details about the application of this diode. What are you going to do with it? \$\endgroup\$ – gstorto Feb 19 '15 at 18:48
  • \$\begingroup\$ One use would be a flyback diode for 3V 40mA motor. Another just a diode to isolate a tactile switch from digital output \$\endgroup\$ – Cano64 Feb 19 '15 at 19:19
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If the MOSFETs in the array are isolated from each other and the body diode rating is not exceeded, the voltage rating of the MOSFET is suitable, and the specs (current rating and reverse recovery time, especially) of the body diode are acceptable it should be fine.

It might not be a good choice in something like a switching regulator where trr is important.

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The below figure shows a typical diode-connected MOSFET (for NMOS). It performs a rectifying function similar to a diode. It is not necessary to connect bulk and source as in the figure, it is an artifact of the lack of a true 4-terminal device in the schematics editor for electronics.stackexchange. The source terminal which is tied with bulk in the figure could be at a voltage higher than bulk.

Bulk is typically shared among many devices and the lowest voltage is used for bulk. In bulk CMOS and NMOS technologies the NMOS transistor body is made of P-type material (excess positive charges), while source and drain are made of N-type material (excess negative charges). There are therefore also some "parasitic" PN-junctions you could exploit: Bulk(P)-Source(N) and Bulk(P)-Drain(N) are real PN junctions - however this is normally not done! For ICs the body material is usually shared by many transistors, and fixed (by specification) to 0 Volt. It then only makes sense if the voltage you need on bulk (for the P-side of the diode) is 0 Volt. Source implants usually have very high number of dopants. Therefore the diodes are somewhat weak - you need a higher forward voltage to send the same amount current through them - compared to a device with more moderate doping.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ The unedited circuit does not work as a single diode. It will work as 2 diodes in antiparallel: the first one is the body-drain diode as you showed. The second one is not technically a diode and will not show an exponential I-V behavior, but a quadratic behavior (and its "anode" is on the top). When Vth is exceeded, the current will be K*(Vgs-Vt)^2, as the mosfet will be always in saturation. To make your circuit work, you should short the gate to the source. (I have requested edits to the schematics) \$\endgroup\$ – next-hack Sep 11 '17 at 19:50
  • \$\begingroup\$ Thanks for the attention. In my post I was trying to point the asker to a diode-connected transistor, rather than the drain-body junctions (which are normally weak devices with forward bias, with strong limitations). I have updated the post! \$\endgroup\$ – HKOB Sep 21 '17 at 21:13
  • \$\begingroup\$ I see now :) But however, let me point out that discrete MOSFET arrays (I mean chips containing more than one MOSFET. And I think that the OP was referring to these) are made with vertical structure, and the source is always connected to the bulk (like in discrete MOSFETs). In analog and digital ICs, instead, MOSFET are planar and the bulk is just connected to the rails (gnd or VDD), unless triple well is adopted, therefore your diode works (still one must ensure that the body is at the lowest - in nMOSFET, highest for pMOSFET- potential on the circuit). \$\endgroup\$ – next-hack Sep 21 '17 at 21:18

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