MOSFET Switching Circuit

Here is simple a circuit for a MOSFET used as a switch.

 

 

 

The diode is needed if the load is inductive. The diode protects the MOSFET from high voltage back emf pulses by allowing the load current to die away gradually.

Drive a MOSFET from a PIC Chip

If 5V is applied to the input above, some MOSFETS would not turn fully on (saturate) so they would not act correctly as a switch. Here is a circuit that fixes this problem. If the bipolar transistor is OFF, the MOSFET gate potential rises to +12V. This will completely turn on the MOSFET. If the bipolar transistor is turned on, the MOSFET gate potential will drop to about 0.2V. This will completely turn off the MOSFET. Note that logic zero at the input turns on the device (light bulb in this example). By using this circuit, you lose the advantages of fast switching and high input impedance of the MOSFET.

 

Prevent Electrostatic Damage

Because MOSFETS can be damaged by electrostatic discharge, the inputs can be protected as shown below. The zener diodes prevent input voltages greater than about +/- 15.7 V.

 R1 and R2 form a potential divider. The input voltage is reduced by about 1% by these components.

• D1 and D2 are 15 volt zener diodes. This means the voltage at the FET gate can never be more than +/- 15.7V. This makes electrostatic damage to the FET very unlikely. If a very large voltage was applied to the input, R1 would fail. This component is not expensive and is easy to replace.
• The FET will turn on if a positive potential is applied to the input. (+ 4 to 6 volts needed).
• If an inductive load is being used, D3 provides a path for the inductor current to die away slowly when the FET is turned off suddenly. D3 can be omitted if the load is resistive.
• R3 is optional. It acts as a dummy load.