I'm building a Royer Oscillator, and I'm not sure why my initial MOSFET (U2) is shorting? My LT Spice simulation works somewhat, however I get a lot of current coming out of the drain (6A+), and likewise the 2 inductors drawing massive currents. Why would this be the case?

Is it because of the MOSFETs M1 and M2? Are they what's causing this massive surge in current? And hence the short in the initial MOSFET.

The power supply I'm using is rated to 6A, and it actually goes up to 6A when I turn it on, but the voltage reduces drastically.

Anyone have any ideas? (Ground is the bottom rail). Royer Oscillator

  • 7
    \$\begingroup\$ U2 is directly across the power rails. Each time a pulse turns it on it will short out. \$\endgroup\$
    – HandyHowie
    Jul 29, 2016 at 7:24
  • \$\begingroup\$ over-heat and/or over-voltage \$\endgroup\$
    – mkeith
    Jul 29, 2016 at 7:25
  • \$\begingroup\$ +mkeith , its not that. Its because the 'Fet is directly across the power rails. \$\endgroup\$
    – Alex
    Jul 29, 2016 at 9:31
  • \$\begingroup\$ @Alex, use the character '@' to tag people, not +. FET's fail when they get too hot or when they have too high a voltage across channel or at gate. If you can put one across a power supply and turn it on and off without doing either of those things, it will not fail. \$\endgroup\$
    – mkeith
    Jul 29, 2016 at 15:20
  • 1
    \$\begingroup\$ @Alex, actually it is due to over-heating. Resistive losses during high pulse current cause excessive heating which causes failure. \$\endgroup\$
    – mkeith
    Jul 30, 2016 at 18:02

1 Answer 1


As HandyHowie told you, you put the Mosfet across the battery.

Look on Fig 1 and 2. http://www.irf.com/product-info/datasheets/data/irfr4105.pdf

11 V as Vds and 10V as Vgs and the Mosfet will allow ~ 100A, assuming you battery is capable, a nice 1kW.

The devices will blow up in ~ 100 us. (Zth ~ 0.2 C/W)


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