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While reading schematics diagrams from several switch mode power supplies of LCD TVs, I noticed that the pin which delivers the PWM pulse to the gate of a MOSFET has a diode and resistor in parallel.

Some diagrams don't have it. But there are lot of which they have it. I guess it's some protection to the driver to the IC controller.

Although I'm not sure. In first diagram there are diode and resistor in parallel, and in second there is not.

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Transistor
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NIN
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2 Answers2

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The idea is to get the MOSFET to turn off more quickly than it turns on. When the MOSFET is driven "on" the gate charge is supplied through (say) R915 + R917 = 51.7 ohms.

When it turns off, the gate charge is sucked out through the diode in series with the 4.7 ohm resistor.

You can think of the gate as looking a bit like a large capacitor (gate-source capacitance plus a typically much larger component from the drain-gate capacitance, the latter has a bigger influence due to the Miller effect- the drain typically changes in potential by a much larger amount, multiplying the effect of the drain-gate capacitance.

In the case of the FMV111N60ES, the gate charge can be as much as 73nC.

This can be used to help prevent two MOSFETs from being "on" at the same time, causing shoot through (which wastes power and can damage the MOSFETs) or just to control the waveforms a bit better.

Spehro Pefhany
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    Spehro On second diagram there're resistor 10 ohms from driver of 1533 to gate of mosfet. Why no just put directly the driver pinout to gate of mosfet ? – NIN Feb 16 '19 at 13:01
  • To make the MOSFET switch more slowly. If the MOSFET switches too fast it can cause problems such as bouncing the source below ground enough to damage the driver (due to inductance in the source power circuit) and will cause more EMI than necessary. Of course slower switching means more switching losses, but engineering involves trade-offs. – Spehro Pefhany Feb 16 '19 at 16:29
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    Spehro. Your help has been extremly useful. I don't have words to thank you . Because this question so specific it's almost impossible find it in internet. – NIN Feb 16 '19 at 23:55
  • One question : when you say " to make MOSFET switch more slowly" , you mean that resistor make the MOSFET's slope (change between on and off ) longer, for example 2 nS to 20nS ? – NIN Feb 17 '19 at 00:49
  • Yes, that's correct. See references such as this: http://www.ti.com/lit/an/slla385/slla385.pdf and this http://www.ti.com/lit/an/slyt664/slyt664.pdf. – Spehro Pefhany Feb 17 '19 at 01:57
  • Just one word : Awesome – NIN Feb 17 '19 at 03:14
  • OMG . . Waooooo. . . . Those articles from TI are awesome. . . OMG. . Now i understand. . . Man you are really my hero man – NIN Feb 17 '19 at 04:50
  • Spehro and I think resistor of 47k is because while time MOSFET turn off that resistor prevent MOSFET's gate load with electrostatic volts right ? – NIN Feb 17 '19 at 13:05
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In addition to Spehro's excellent answer, there are a few other considerations.

RF emissions from circuits increase with fast-switching devices, but there's also the gate driver limits to account for. As the transistors drive inductive loads, faster switching will not actually increase performance for a given circuit. The circuit is tuned to operate at a certain frequency, so faster switching can lead to greater driver cost with no benefit.

The context changes dramatically when you replace the MOSFET with a GAN-HEMT transistor, as they can handle higher loads and switch at far higher speeds, 500kHz switching of KW range supplies are not unheard of. This is when ground bounce and RF emissions can become a serious design headache.

SamGibson
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Peter
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  • Wow! That's impressive! Can you recommend any application note for reading more on ground bounce and RF with heavy loads? – Pranav Feb 17 '19 at 04:27