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I am trying to learn about n-channel JFETs from David. A. Bell's book. In the transfer characteristics graph, there seem to be two curves as shown. Note that VGS(off) = Vp

enter image description here

As you can see, one curve is drawn for the minimum values of VGS(off) and IDSS and the other curve is drawn for the maximum values of VGS(off) and IDSS. The book shows an excerpt from a data sheet and says that the manufacturer has mentioned the values of minimum and maximum of VGS(off) and IDSS and uses this formula to draw the curves:

enter image description here

I can't understand how there can be maximum and minimum values for the cut-off voltage and the drain current. From the input characteristics graph, I understand that cut-off voltage is nothing but the Pinch-off voltage when there is no voltage applied between the gate and the source(ie) when VGS=0. Now, if there is a maximum and minimum cut off voltage then that means for VGS=0 there are two pinch-off voltages which doesn't make any sense. Also what does having a minimum "saturation" current even mean?

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  • \$\begingroup\$ FETs are analogue devices, there is no voltage where it is completely off and 1e-190V later it is completely on, there are gradually increasing electric fields that interact. \$\endgroup\$
    – PlasmaHH
    Oct 12, 2017 at 8:03
  • \$\begingroup\$ @PlasmaHH Are you implying that at the minimum of cut-off voltage the JFET starts to have near zero current and at the maximum of cut-off voltage it actually switches off? \$\endgroup\$
    – VenkiPhy6
    Oct 12, 2017 at 8:28
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    \$\begingroup\$ From Electronic Devices and Circuits, second edition, page 271: One of the problems in using FET's is that each device type does not have a single transfer characteristic. This is because Idss and Vp cannot be specified accurately. Instead, the manufacturer specifies maximum and minimum values for each parameter. Referring to Fig. 12-9, it is seen that, for the 2N5457 FET, Idss(min) = 1 mA, and Idssmax=5mA. Also Vp, which is listed as Vgs,off has a minimum level of 0.5 V and a maximum of 6 V. \$\endgroup\$
    – Bart
    Oct 12, 2017 at 10:27
  • \$\begingroup\$ Electronic Devices and Circuits is the book VenkiPhy6 is referring to. \$\endgroup\$
    – Bart
    Oct 12, 2017 at 10:38
  • \$\begingroup\$ @Bart I actually have the fifth edition(Indian version). I don't see this explanation here. So thanks for going through the trouble of finding the book and letting me know! Your comment also checks out with Neil_UK 's answer... \$\endgroup\$
    – VenkiPhy6
    Oct 12, 2017 at 11:10

1 Answer 1

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Any given FET (at least when at a fixed temperature) will have just one curve, somewhere between those two limits.

Unfortunately, FETs are difficult to make consistently, and if you buy a number of 2Nxyz FETs, you can expect that their curves will not be identical, even if they are from the same batch. If they're from different batches, you should expect a wider spread.

The manufacturer has gone to the trouble of characterising what he expects to be the minimum and maximum possible numbers likely with his manufacturing process, so that you have some idea over what range any FET of that type you buy will have. This is important, as your bias circuit must be able to properly bias any FET within that range. If your bias circuit does not handle that range of numbers, then you need to buy a better specified (probably selected, and so more expensive) FET, or to improve your bias circuit.

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  • \$\begingroup\$ So the two transfer characteristic curves are an artifact of the difficulties in the manufacturing process of JFETs. Got it. I am actually a Physics student who has been forced to take an electronics course to get my degree and I am not taught by an engineering professor. So I don't know these nitty-gritty details. Thanks for the answer. I will wait for a few days to give others a chance to answer too. If none turn up, I will accept yours. \$\endgroup\$
    – VenkiPhy6
    Oct 12, 2017 at 11:15
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    \$\begingroup\$ +1 for the understatement "FETs are difficult to make consistently" \$\endgroup\$
    – Peter Smith
    Oct 12, 2017 at 14:32

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