JFET symmetry

Discussion in 'General Electronics Chat' started by tindel, Feb 27, 2013.

  1. tindel

    Thread Starter Active Member

    Sep 16, 2012
    568
    193
    I have some JFET's that I'm trying to put into a design of mine, but I'm curious about JFET symmetry. I am familiar with the JFET's basic operation, assuming I know which input is the source.

    In a symmetrical JFET what determines which pin is the source and which pin is the drain? The current direction through the JFET? In theory you can put the FET in circuit in either orientation, and it will continue to operate properly, meaning the source and drain can switch for some reason. What is that reason?
     
  2. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    It's because of the symmetry you've mentioned.

    Think of a large brick of whichever doping you'd prefer silicon. This has a given resistivity, given by the bulk properties of the doped silicon. This device is a silicon resistor, nothing much special about it, attaching either end to a source will work the same as a resistor, no polarity. Now, add in a means to deplete an area of charge carriers and you have a means to effectively narrow the conductive region of that silicon resistor. Doped wells of the opposite type and a voltage applied to those wells will give you that control. The amount of reverse biasing will dictate the depletion region around the wells, removing charge carriers and the conductive path.

    If you think about it as a silicon resistor with a voltage controlled conductance, it doesn't matter which side is source.

    The ebook, kind of describes the function....
    http://www.allaboutcircuits.com/vol_3/chpt_5/1.html

    As far as determining what side is source, I think it's an eny-meny-miny-mo, kinda deal:)
     
    tindel likes this.
  3. #12

    Expert

    Nov 30, 2010
    16,355
    6,852
    Does that apply to ALL jfets?
     
  4. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    Yes, unless you know of something different. The only thing I can think of where this wouldn't hold would be if the gate was planted at the source or drain, that way the polarity would matter, otherwise, as long as there are two ends, with oppositely doped well(s) at a point(s) in the bulk, the source and drain are interchangeable.....since the reversed biased PN junction, and it's corresponding applied voltage is responsible for depleting the bulk silicon...
     
  5. #12

    Expert

    Nov 30, 2010
    16,355
    6,852
    I don't know, I just assumed the manufacturers labeled the source and drain because it matters.
     
  6. JDT

    Well-Known Member

    Feb 12, 2009
    658
    85
    Although the basic JFET is symmetrical, the substrate (the body of the chip) is often connected to the source. This pin is then the reference for the input on the gate. This is true for a 3-pin device.

    If your JFET has 4 pins, then the substrate is connected to this 4th pin and the source and drain are truly symmetrical. The input voltage is then between the gate and substrate.
     
    #12 likes this.
  7. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,395
    1,607
    Maybe it does and maybe it don't. One job I once worked on used a jfet in die (raw chip) form, and the "geometry" drawing from the manufacturer showed the pads to be "G" "D/S" and "S/D" meaning they could be swaped, but same manufacturer for the same device in packaged form had no such indication.
     
    #12 likes this.
  8. tindel

    Thread Starter Active Member

    Sep 16, 2012
    568
    193
    This is a great explanation of how the part works! As I said, though, I understand the operation of the device quite well, assuming I know which is the source.

    But how do I know which is the source? For example, say I have a p-channel JFET. If I apply 5Vds and I apply 5Vgs, then the Vgd is then 0V. How do I know that the jfet won't decided the drain is really the source and it's now conducting current when I expect it to not be conducting current?

    What decides the source?
     
  9. w2aew

    Member

    Jan 3, 2012
    219
    64
    As far as I recall, the drain and source are often interchangable. However, for some applications and packages, they may not be. The leadframe in the package may have more capacitance to the Gate on one of the two remaining lead, and the one labeled Drain would be the one with lower capacitance to the gate.
     
  10. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    I couldn't find anything before in the AAC book, but, another look yielded this.

    It would seem that the device is not always symmetrical along the length of the body, as seen here:
    [​IMG]

    Either can be used as a source, but capacitance and ensuring the PN junction is properly biased may differ, depending on orientation...largely, it wouldn't matter, but small discrepancies in the manufacturing of the device makes one source/drain combination more appropriate.
     
    tindel likes this.
Loading...