# Understanding purpose of components in a half bridge driver circuit

I'm currently trying to learn and understand about creating an H-Bridge using two half bridge drivers. The half bridge drivers I have are IR2184.

I found a circuit diagram someone else has designed using the same driver ICs but I have some questions about it.

I've seen several similar circuits where resistors are placed in line with the gates of the mosfets - What are those resistors for?

Also on the circuit above the designer has placed some diodes parallel to those resistors. Is there a reason for doing that?

You should always provide a gate resistor.

At the very least it actually helps define the current that will flow in and out of the GATE during switching transients (to charge and discharge the gate capacitance).

Without any gate resistor you could have very large (relatively speaking) peak currents flowing. Best-case... its an EMC concern, worst-case... you potentially burn out the gate of the FET.

You also run the risk of creating a Pierce Oscillator.

In this instance you can see a series gate resistance with a diode in parallel. This limits the current and thus the switching time for TURN-ON.

The diode then "shorts" the resistor out permitting a higher charge transfer for a TURN-OFF.

This is a cheap/simple means to mitigate shoot-through's of a H-bridge leg during PWM transitions as you now have fast TURN-OFF and slow TURN-ON

There's another important reason to the presence of the diode antiparalled to the gate resistor. In a bridge configuration, when a transitor goes on the switching, the voltage acroos the other transitor has a increased dV/dt.

In a MOSFET, that's the case of your H-brigde, there's three intrinsic capacitances composing its body, like in the image below (that shows other stray components of the Mosfet body):

When the Vds voltage increases, all the three capacitances need to be charged, like in the schematic below:

simulate this circuit – Schematic created using CircuitLab

Where the red arrows are current flowing through the circuit (when Vds are increasing).

So, if this resistance of the discharge isn't low, the Cgs capacitange could had a peak of charge, creating a short-time short-circuit, named punch through.