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I am checking out a schematic for a full bridge and I noticed these diodes at the gate of every mosfet and I do not understand their purpose there. Can someone elaborate please.

The IC to the left is a half bridge driver... the output to the motor is at the far right. enter image description here

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    \$\begingroup\$ They bypass the resistors when turning the MOSFETs off. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 7 '18 at 4:29
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    \$\begingroup\$ Did you realize each gate resistor is in parallel with a diode? \$\endgroup\$ – Andy aka Jun 7 '18 at 7:57
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    \$\begingroup\$ @Andyaka A case where the schematic adds confusion. That print could certainly have been done better. \$\endgroup\$ – Chris Knudsen Jun 7 '18 at 12:31
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    \$\begingroup\$ Including in the answers, depending on the control IC, they will keep the gate emitter impedance low. This will help maintain the gate emitter voltage low during switching. Because of Miller Effect, a current goes to the gate, and the it can go into conduction. Good PCB design and reducing parasitic inductance is another factor. \$\endgroup\$ – Diego C Nascimento Jun 7 '18 at 13:23
  • \$\begingroup\$ @ChrisKnudsen I agree drawing them visually in parallel with the resistor would have provided more insight and aided my intuition on their purpose. \$\endgroup\$ – Edwin Fairchild Jun 9 '18 at 19:54
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'Shoot through' is a rather descriptive term for what happens in a bridge when one set of devices turns on before the other set of devices have turned off. A large spike of current flows through the bridge. At best it reduces efficiency and generates EMI, at worst it overheats or destroys the FETs.

The simplest scheme to avoid this is to use a resistor in series with the gate, to slow down the FET turn on (R1, R5, R8, R9 etc), but then bypass it with a diode, to speed up turn off (D3, D5 etc).

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