Author Topic: Why I frequently see MOSFETs with a reversed diode in parallel with drain source  (Read 19648 times)

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Offline king.osloTopic starter

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Hello there,

Will somebody in simple terms explain to me why I frequently see designs where MOSFETs have a reversed diode in parallel with drain and source?

I have googled the subject. I understand the very lowest basics of P- and N-FETS (and diodes, including Schottky), but that's about it.

Thanks,
Marius
« Last Edit: January 20, 2012, 12:42:05 am by king.oslo »
 

Offline amspire

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Are you talking about the internal reversed biased diode in a mosfet?

In a N channel MOSFET, you have a P doped silicon substrate, and the source and drain are N doped regions within the substrate. So in normal operation, both the source and the drain are in fact diode junctions to the substrate. The source is also connected to the substrate as for the gate to work properly as the substrate has to be at the same potential as the source. There are actually some small signal MOSFETs that have the substrate connected instead to a separate fourth pin, but these are now increasingly rare, and I have never seen it done in a power MOSFET.

So in normal operation, the drain region is a reverse biased diode junction to the substrate which is connected to the source. If the drain voltage is taken below the source voltage, this diode junction is now forward biased.

Richard
 

Offline king.osloTopic starter

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Thanks.

1. In the datasheet for the MAX8792 page 11, the low side fet has a diode, whilst the other doesn't. What does this do? http://datasheets.maxim-ic.com/en/ds/MAX8792.pdf

2. So this is true: Are you saying that NFETs always work as a reversed diode between drain and source? When voltage is lower at source drain relative to source, voltage flows from d-s?

3. What about PFETs? Are they forward biased diodes?  :o  ;D

Thank you for your time.M
 

Offline amspire

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N and P mosfets have a reversed biased diode gate to source. P channel MOSFETS have a N substrate with P source and gate. So it is the same story, just that everything is opposite.

Now the diode on Page 11 is a different story.  On most IC's, including the MAX8792, you cannot take any pin to more then 0.3V below the negative supply pin or the IC can latch up.  The inbuilt reverse biased diode in a MOSFET will only limit the reverse voltage to 0.7 to 1V below the negative supply when the inductor causes pin 4 to go negative, so you need to add an extra Schottky diode across the low side MOSFET with a turn on voltage of under 0.3 volts.

That IC will be turning ON the low side MOSFET when pin 4 is being dragged down, so it wouldn't surprise me if that circuit would work fine without the Schottky, but if just once the MOSFET does not turn on in time, the MAX IC will latch up. Adding the Schottky diode is an extra level of protection.

It is all about protecting the control IC.

Dave did a great blog about this:

« Last Edit: January 20, 2012, 01:48:43 am by amspire »
 


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