The real function of the resistor connected in parallel between the gate and source of a MOSFET!
In MOSFET driver circuits, this resistor is sometimes seen and sometimes not. Common values for this resistor are 5kΩ and 10kΩ. But what is its purpose?
Before analyzing this problem, a simple experiment can be conducted:
Find a MOSFET, leave its gate (G) floating, and then apply a voltage across its drain (DS). What happens? The MOSFET burns out when the input voltage is only a few tens of volts because it becomes conductive.
The reason why a MOSFET can conduct without a driving signal (e.g., when the driver chip is not powered on or is damaged and the chip's driver pin is in a high-impedance state) is because there are junction capacitances Cdg and Cgs between the drain (DG) and gate (GS) terminals of the MOSFET. Therefore, the voltage applied between the drain and source terminals will charge Cgs through Cdg, thus raising the voltage at the gate terminal until the MOSFET turns on.
Therefore, when the drive circuit is not working and there is no discharge circuit, the MOSFET is easily damaged. If a transformer is used for drive, the transformer winding can play a discharge role, so even without a gate resistor, the MOSFET will not turn on by itself when there is no drive.
Furthermore, the book "Diagnosis and Troubleshooting of Switching Power Supplies" also mentions practical problems encountered.
Summarize
1. Preventing ESD damage to MOSFETs (one reason I saw was: because the junction capacitance is relatively small, according to the formula U=Q/C, a small Q will also lead to a large voltage, causing the MOSFET to fail).
2. Provides a fixed bias. When the preceding circuit is open, this small resistor can ensure that the MOS is effectively turned off (reason: when the gate is open, when a voltage is applied to the DS terminal, it will charge Cgd, causing the gate voltage to rise and preventing effective turn-off).
3. Below is an explanation of the resistance value. If it is too small, the drive current will be large, and the drive power will increase; if it is too large, the turn-off time of the MOS will increase.
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