# What is the purpose of gate resistor in gate bias?

The input driving below circuit has only two states: $$\-10V\$$ for OFF and $$\0V\$$ for ON. Why can't we apply this input directly to the gate? I mean removing the resistor $$\R_G\$$ doesn't seem to change anything?

I understand $$\R_D\$$ is there to ensure the jfet is in hard saturation, operating in ohmic region.

But I don't get the role of $$\R_G\$$. It is not part of voltage divider bias or anything. It is simply shorted to the ground. Why can't I remove it? (Maybe removing $$\R_G\$$ even helps here.. as the the input impedance of jfet gate is very high, so the input signal doesn't have to provide much current.)

• @LarsHankeln no it doesn't. The answer you linked is more about the turn-off or gate floating prevention of power MOSFETs. This question, however, is about biasing the N channel JFETs. Feb 18 at 6:46
• Could you please link the source of the info? Textbook? Website? It seems that this is about self biasing but I reckon there should be a resistor between the JFET's source and the ground to provide proper biasing. Normally RG is there to make the JFET's gate grounded. It's high enough not to load the source. Feb 18 at 6:50
• @RohatKılıç yeah this would be self biasing if there were a resistor at source terminal. I think this is just a direct gate bias meant for switching circuits. Not for active region operation. I'll link the textbook, one sec... Feb 18 at 6:53
• @LarsHankeln I'm going through that link... haven't started MOSFET yet but that link is interesting, it seems to suggest Rg makes the switching faster somehow.. Feb 18 at 6:57