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I have a question regarding the SOA graph and a low Vds voltage (below 100mV), I will get to the questions below but needed to layout some information first.

I am looking at a International Rectifier part IRLMS6802PbF, RDS(on) data from datasheet shown below RDS(on) data for Vgs

So assuming I have a 1A current flowing and the Vgs voltage is a stable -4.5VDC, as I understand Vds should be in the region of 50mV, which would result in a point off to the left of the SOA graph shown below enter image description here The red dot provides an approximate position based on assumed Vds.

What I am trying to understand based on the assumed position on the SOA graph is this will also be outside the RDS(on) limit line for the graph.

So is this an impossible situation? or will the Vds be higher? or Does the RD(on) line go horizontal once it exits to the left on the SOA graph?

Suze1990
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  • I would just continue the line past the left side. In essence, adding a 0.01 mark on the X axis. – DKNguyen Aug 20 '20 at 16:28
  • For low current and low Vds the SOA is limited by Rds(on) https://e2e.ti.com/blogs_/b/powerhouse/archive/2015/05/02/understanding-mosfet-data-sheets-part-2-safe-operating-area-soa-graph – G36 Aug 20 '20 at 16:30
  • @DKNguyen yep makes sense so if we continue the RDS(on) line an then plotted a 50mV point on the same logarithmic scale we would fall outside the RDS(on) limit? – Suze1990 Aug 20 '20 at 16:32
  • Looks like it would barely be above the line. – DKNguyen Aug 20 '20 at 16:36
  • @G36 Thanks for the link I have seen these and the video very informative – Suze1990 Aug 20 '20 at 16:36
  • Continue the RdsOn line down and leftwards. Your red dot will lie on it if accurately plotted. Power dissipation is so low at that point there's to point in a SOA graph covering it. –  Aug 20 '20 at 17:32

2 Answers2

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What I am trying to understand based on the assumed position on the SOA graph is this will also be outside the RDS(on) limit line for the graph.

This line just shows you where RDSON limits current, ie. operation in the area 'limited by RDSON' is impossible because RDSON is too high to support that current at that voltage (even though it is inside the Safe Operating Area). Your operating point is at RDSON and well inside the SOA, so you have nothing to worry about.

So is this an impossible situation? or will the Vds be higher? or Does the RD(on) line go horizontal once it exits to the left on the SOA graph?

If you are wondering why extrapolating the line doesn't precisely line up with your dot, it's because RDSON is not a fixed value (it varies with temperature, threshold voltage etc.) so the numeric specifications and graphs may have slightly different values. Nothing should be inferred from this. Characteristics vary from unit to unit so the graphs only show 'typical' or guaranteed minimum/maximum values. They are not intended to be precise.

You should base your power handling calculations on the specified maximum RDSON and maximum acceptable operating temperature (which for reliability should be much less than 150 °C) at your operating point. That gives you the worst case, so just make sure it is well inside the maximum ratings and everything should be fine.

Bruce Abbott
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Above to the left of the apex , the FET is limited by Ron which is minimum near Vgs/Vt >2.5 to 3 and up.

The upward slope is a 0.05 Ohm e.g. = slope 0.1V/2A or 0.05V/1A

If 0.01 Ohm it is steeper.

The power dissipation depends on heat sink and thermal resistance stack Rja=Rjc+...Rca including insulator.

When the heat velocity is too slow, it is limited by internal junction temps defined by downward slope for each component size. SMT tend to be better only if you read the fine print like 1W / sqin or =≈6sqcm for a 60deg temp rise on 1 side or less if tied to 2 external ground planes to radiate heat, if exposed to convection air. Many variables after the heat temp rise especially if you have high air velocity at surface.

Then the SOA for each size is W-s product. With a W limit or in this case for 0.05 ohm a current limit of 40A.

$Pd=I^2Ron= 40^2 0.05= 8W so for a 40’C rise at 25’C or 40’C ambient, you decide the most reliable max temp and design the heatsink accordingly.

Tony Stewart EE75
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