It has two functions:
1. Provide a power supply for the MOSFET, and obtain the voltage division value through the resistor divider, that is, provide a bias voltage;
2. Play the role of anti-ESD static, avoid being in a high resistance state. This resistor can be regarded as a bleeder resistor to prevent the MOS tube from malfunctioning, thereby damaging the gate G-source S of the MOSFET;
First of all, the first role of bias voltage should be easy to understand. I will not explain too much here. Mainly, let's talk about the working principle of the second role, how does it prevent ESD and protect the G pole and S pole of the MOS tube.
Totem pole drive circuit. We know that the impedance between the GS of the MOS tube is very large, so as long as there is a small amount of static electricity on the GS pin and a little current flows, the equivalent capacitance between the GS poles will have a very high value. Voltage, because the large impedance is multiplied by the current, there will be a very large voltage. If the ESD static electricity is not discharged quickly, the high voltage generated at both ends of the junction capacitance Cgs may cause the MOS tube to malfunction, or even It is possible to break down the GS pole of the tube, so adding this resistor between the gate (G) and the source (S) is mainly to discharge the ESD static electricity, so as to protect the MOS tube.
Analysis: This circuit is a totem pole push-pull circuit, which is used to accelerate the rapid turn-on and turn-off of the MOS tube. We know that the MOS tube has a Miller effect. In order to avoid the tube staying on a Miller platform for a long time, we need to accelerate the turn-on and turn-off time of the MOS tube to reduce the loss of the switch. Q3 and Q4 are turned on in turn. The G pole of the MOS tube is constantly charging and discharging. If the power is suddenly turned off at this time, there may be two situations in the gate of the MOS tube:
The first situation is that when the power is off, the gate and source are just in the discharge state, and the junction capacitance Cgs has no charge stored, and it just happened to be discharged;
Another situation is that the gate-to-source (Vgs) is just in the charged state when the power is off. At this time, the junction capacitance Cgs is charged and the electricity is fully charged. The power is disconnected, and Q1 and Q2 are also disconnected at this time, and there is no power. At this time, the charge of Vgs does not have a release loop. This charge has always existed here and can be maintained for a long period of time. The establishment of the MOS tube conduction condition has not disappeared, which is equivalent to that Vgs has always been voltage when it is turned off. In this way, at the moment of turning on again, since the excitation signal has not been established, and the power supply of the drain (D pole) of the MOS tube is randomly provided at the moment of turning on, under the action of the conductive channel, the MOS tube immediately produces an uncontrolled huge drain (D pole). ) The current Id may cause the MOS tube to burn out, which is not what we want.
Therefore, in order to avoid this phenomenon, a bleeder resistor R4 is connected in parallel before the GS of the MOS tube. After the shutdown, the voltage stored in the junction capacitor Cgs is quickly released to GND through this resistor R4. Generally, the resistance of this resistor should not be too large. If it is too large, the discharge will be slower. In order to ensure the rapid release of the charge, it is generally about 5K to several 10K.