In a MOSFET, the obvious cause of asymmetry is the fact that the source is tied to the body. But in a JFET, no such obvious answer exists, as the body, gate, and drain are all the same piece of silicon with no material differences between them (as far as I'm aware). And indeed, most JFETs seem to be symmetric.

But there are some JFETs on the market have specifications that would imply asymmetry. See for example the UJ3N120070K3S that I just used in a design; it's rated for a gate-to-source voltage of no more than -20 volts, but a drain-to-source blocking voltage of 1200 volts. But if \$V_{ds} = 1200\ \mathrm V\$, and \$V_{gs} = -20\ \mathrm V\$, that leaves \$V_{gd} = -1220\ \mathrm V\$! Even if \$V_{gs} = 0\ \mathrm V\$, \$V_{gd}\$ is still vastly higher than the rated \$V_{gs}\$.

If the FET were symmetric, that would imply that the gate-source and gate-drain ratings should be the same, but they clearly can't be here, as if they were, the drain-source rating couldn't possibly be higher than the gate-source rating, and applying rated \$V_{ds}\$ would fry the gate junction.

What about its physical construction makes this JFET, and others like it, asymmetric? Why can \$V_{gd}\$ be so much higher than \$V_{gs}\$?

  • 1
    \$\begingroup\$ @Andyaka While troubleshooting the circuit I used it in, I was able to measure that the gate was indeed a diode junction. The forward voltage didn't seem much like a SiC pn junction though, now that you mention it... Perhaps a SiC Schottky junction, though. I didn't think to test if it had the characteristic body diode of a MOSFET, but I can check that on Monday. \$\endgroup\$
    – Hearth
    Jan 9, 2021 at 18:38
  • 1
    \$\begingroup\$ @DKNguyen I did. He points at this patent. \$\endgroup\$ Jan 10, 2021 at 17:12
  • 1
    \$\begingroup\$ @MarcusMüller Funny thing: when you wrote asking me, it was entirely by coincidence that I'd read the patent a few years back. I had no real need at the time. I just enjoy reading about new ideas and there was a news item or something that triggered my interest. Since I had a bookmark handy, I was able to respond instantly when you mentioned this question. I don't feel as though I know enough to do a credible job of debating/discussing the details. So no answer from me. I've no problem writing the inventors. But Marcus, I think you may be in a better position for that. Why not? We'd all learn! \$\endgroup\$
    – jonk
    Jan 12, 2021 at 8:10
  • 1
    \$\begingroup\$ @Theodore I can't see how that asymmetry would affect breakdown voltages, though. It would certainly affect high-frequency performance with capacitance differences and such, but breakdown voltages depend on the characteristics of the junction itself. \$\endgroup\$
    – Hearth
    Mar 23, 2021 at 14:20
  • 1
    \$\begingroup\$ @Theodore The semiconductor breakdown occurs far before package dielectric breakdown, I do know that much. In fact, you'd see the air around the package breaking down before dielectric breakdown in the package happens, in most cases! (MOSFETs and IGBTs being a notable exception, with their very thin gate oxide) \$\endgroup\$
    – Hearth
    Mar 23, 2021 at 15:37

1 Answer 1


Many JFETs are symmetric and this is sometimes indicated in the datasheet such as the

enter image description here Source: https://shop.micross.com/pdf/LSM_2N4117A_TO-71.pdf

If it is indicated then you know its symmetric otherwise you would have to assume either.

I think this is question is backwards:

What about its physical construction makes this JFET, and others like it, asymmetric? Why can Vgd be so much higher than Vgs?

If there is asymmetry you can infer this from measurements or from the datasheet because if the physical construction were symmetrical there would be no difference.

A better question would be: Can I infer that a JFET is asymmetric if there is a difference between Vgd than Vgs. I believe the answer is yes, but you could never be sure unless you actually check the physical construction or a patent. Also if there were Cgd and Cgs listed and they were different I also think you could assume asymmetry.

This J308 "N-Channel JFET" has a different Cgd and Cgs.

enter image description here Source: https://www.digchip.com/datasheets/parts/datasheet/078/J308-pdf.php

  • 2
    \$\begingroup\$ Yes, but what I'm asking is specifically what makes some JFETs asymmetric--I'm well aware that most of them are symmetric, but the ratings of specifically the U3JN120070K3S and others like it indicate asymmetry (and considering I've actually blown one by exceeding its gate-source voltage, measurements also indicate asymmetry). \$\endgroup\$
    – Hearth
    Mar 5, 2021 at 1:40
  • 1
    \$\begingroup\$ In short, I'm asking what in the physical construction could possibly make a JFET asymmetric, because I can't seem to understand how a JFET could possibly have that different Vgs and Vgd ratings. \$\endgroup\$
    – Hearth
    Mar 5, 2021 at 1:41
  • \$\begingroup\$ One is the body might be tied to one of the terminals (might account for differences in capacitance like in the below example), the other is that the gate is not symmetric and not equal amounts of material to the gate from the 'source' or 'drain'. Now that I think about it, you could check with some kind of impedance measurement for asymmetry, if it wasn't physically symmetrical the impedance measurement could not be the same \$\endgroup\$
    – Voltage Spike
    Mar 5, 2021 at 3:46
  • \$\begingroup\$ The old Siliconix data books were brilliant. They included metallization patterns. Have a look at p182 of this link and you will see that source and drain geometry are not symmetric: archive.org/details/bitsavers_siliconixdixFETDatabook_24133064/… \$\endgroup\$
    – 10ppb
    Apr 10, 2021 at 1:28
  • \$\begingroup\$ @10ppb p182 is a mosfet though... the JFET on page 186 is symmetric \$\endgroup\$
    – BeB00
    May 6, 2023 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.