EveryCircuit
Contact
Reviews
Home
PrathikP
modified 5 years ago

How to Drive a MOSFET - Type 1 Gate Driver

11
3
649
08:29:52
A MOSFET, when used as a switch, must be driven by a proper circuit, especially at high frequencies. The longer a MOSFET spends transitioning between is OFF state and ON state, the higher it's power dissipation becomes. This power, which is dissipated during the transition, is called SWITCHING LOSS. The faster the MOSFET changes states, the lower is the switching loss and the more efficient it becomes. Before talking about a good driver, let us first see what an improperly driven MOSFET looks like: http://everycircuit.com/circuit/6628118843097088 (link opens on phone). As you can see, the gate voltage is not a perfect square wave. It is due to the input capacitance. Also, if you closely observe, you will see two periods during which the gate voltage stays "flat". This is called "gate plateau voltage" and is due to the reverse transfer capacitance or "Miller" capacitance. Let us try to understand these two capacitances. The first, the input capacitance, is the parasitic capacitance formed between the gate and the source of the MOSFET. We are required to charge this capacitance up in order to raise the gate-source voltage above and beyond the threshold voltage of the MOSFET, in order to turn it on. The second, the reverse transfer capacitance, is the parasitic capacitance formed between the gate and the drain. This is a little harder to understand. Due to it's location, it works against the driver. It sucks current which the driver tries to supply to the input capacitance, thus hindering it's charging and discharging and causing the plateau or a "pause" in the gate voltage, as you saw in the waveform. In order to raise the gate voltage quickly to ensure that the MOSFET spends little time transitioning between the OFF and the ON state, we must overcome the effects of these two capacitances. By supplying a high current to the gate, the input capacitance charges up quickly and the "Miller" capacitance (which steals current and hinders the charging process) has a reduced effect on the gate voltage. This can be achieved by charging and discharging the MOSFET quickly by supplying a high gate current using a GATE DRIVE. In the circuit, you can see a diode, a PNP transistor and a series resistor (in order to limit the peak current) are used to drive the MOSFET properly. When the signal source goes high, the diode forward biases and charges the gate of the MOSFET through the 4.7ohm resistor quickly. During this time, the PNP transistor is OFF and has no effect. When the signal goes low, the diode reverse biases, but the transistor turns ON with the help of the 100ohm resistor. Upon turning on, it connects the gate through through the 4.7ohm resistor to the source, thus discharging is quickly. Such a gate driver (single transistor type) is used in isolated gate drives, such as when a MOSFET is driven using a pulse transformer, like http://everycircuit.com/circuit/6347832565694464 .
published 5 years ago
schernya
5 years ago
Great explaination! Would it switch faster without 4.7 Ohm resistor?
PrathikP
5 years ago
Yes but the charging current would be too high. Apart from stressing the diode/transistor, it would stress the signal source or cause a dip in the power supply.
yankozz
4 years ago
In the circiuit is n-chanel mosfet, what about p-chanel?

EveryCircuit is an easy to use, highly interactive circuit simulator and schematic capture tool. Real-time circuit simulation, interactivity, and dynamic visualization make it a must have application for professionals and academia. EveryCircuit user community has collaboratively created the largest searchable library of circuit designs. EveryCircuit app runs online in popular browsers and on mobile phones and tablets, enabling you to capture design ideas and learn electronics on the go.

Copyright © 2025 by MuseMaze, Inc.     Terms of use     Privacy policy