2. The source is grounded and the voltage is 0. When the voltage of the gate (that is, the right-eye driver 1 in the figure) is greater than the conduction voltage Vth (generally 0.7V), a conductive channel can be formed between the source and the drain to generate current.
3. When the grid has no voltage (or the voltage is less than Vth), the electron tube is closed and the resistance is very high. The voltage of 12V is basically applied to the tube, the voltage of R 12 is very small, and the circuit current is almost zero. When the gate voltage is greater than the turn-on voltage, the electron tube is turned on and the resistance drops rapidly. R 12 shares part of the voltage and generates current in the circuit. This current satisfies both Ohm's law (resistance current-voltage equation) and Chih-TangSah's equation (tube current-voltage equation), that is, the upper and lower currents are the same, which satisfies the current consistency.
4. The diode plays a protective role. When the drain-source voltage is high, the diode will reverse breakdown before the source-drain punch-through, thus protecting the lamp.
5. If the diode is a voltage regulator, it is used to keep the drain-source voltage constant, so the current can be set by selecting the voltage regulator. At this time, the current flowing through R 12 is equal to the sum of the current flowing through the diode and the current flowing through the tube, which satisfies Kirchhoff's current law, that is, the current flowing into the node is equal to the current flowing out of the node.