Why do eGaN FETs conduct in the reverse (S-D) direction with about 1.5V?

If the eGaN FETs have no reverse recovery charge and no reverse diode, why do they conduct in the reverse (S-D) direction with about 1.5V? Do you have any technical paper on this or can you refer me to where I can find additional information on the reverse characteristics of your eGaN FETs?

The FET itself performs the diode function automatically. We have a virtual diode. When the gate become positively biased relative to the source (Vgs) or, surprise, the drain (Vgd), the FET will turn on. The Gate does not care which terminal, Source or Drain, it becomes positively biased against. So there is a Vgs transconductance curve and a Vgd transconductance curve. There is very little difference between the Vgs curve and the Vgd curve. The virtual diode forward voltage follows the Vgs transconductance curve voltage needed to support the given drain current flow which is why it is much higher than a traditional diode Vf. This same property applies to MOSFET, However the MOSFETs parasitic drain-source PN junction diode conducts and clamps the Vds voltage PREVENTING a Vgd voltage from being developed that is high enough to turn on the MOSFET. Hence this property is never talked about with a MOSET because it is not an electrically accessible function of the MOSFET. If one made a MOSFET with the parasitic diode out of the way, you would see it work. We do not show a transconductance curve labeled Vgd in the datasheet. Instead we show Vsd curve (Figure 8 in the datasheet). Notice how it looks compared to the Vgs transconductance curve (Figure 2). The diode or Vsd curve assume the gate is shorted to the source which mean Vgs=0 at all times. Any none zero Vgs voltage will add or subtract the virtual diode voltage. For more information view the webinar Understanding the Impact of Dead-time, QRR, and COSS