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To begin, I've thoroughly searched for answers to my question/s and have yet to find any clear & concise data.

I'm in the progress of determining an appropriate N-Channel MOSFET to be used as a switch for an array of LED's. I need to remain in the SOA (safe operating area) of the MOSFET which is generally determined by a chart in the datasheet based on current and drain source voltage (\$V_{DS}\$).

The LED array will be composed of 2 strings of 10 LEDs (20 total) and use 1 A of current at 36 V.

The anode of the first LED in the strings will be connected to the 36 V power rail and the cathode of the last will be connected to the drain on the MOSFET. The source will be connected to ground.

The question I pose is this... How do I determine the \$V_{DS}\$ to be used in the SOA chart in my application?

Initially I assumed that I should be looking for a MOSFET that has a \$V_{DS}\$ of 36 V and current of 1 A that falls within the Safe Operating Area. But as I thought about it further, the MOSFET will never be exposed to 36 V because each LED diode has a forward voltage drop and the MOSFET itself will be at the end of the chain. Am I being led astray with this assumption?

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    \$\begingroup\$ How are you controlling current? You have a feedback circuit to control the FET gate? Or you have a series limiting resistor? \$\endgroup\$
    – The Photon
    Mar 27, 2014 at 18:21
  • \$\begingroup\$ to keep things simple, this particular project will utilize a regulated 36v constant voltage source instead of a constant current source. The circuit will be fuse protected against over current. There will be a small series resistor to keep voltage within safe specs. \$\endgroup\$
    – GabeNix
    Mar 27, 2014 at 18:45
  • \$\begingroup\$ you will still need to include a current limiting resistor in each leg of your LED trail. 10 in series seems a bit high ... \$\endgroup\$
    – smashtastic
    Mar 27, 2014 at 22:30

2 Answers 2

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How do I determine the VDS of the MOSFET in my application?

For power dissipation, given you are using an external part (series resistor) to limit the current, you just need to know the Rds(on) of the FET and the load current. Then Vds in the "on" state is just \$I \times{} R_{ds}(on)\$.

the MOSFET will never be exposed to 36v because each LED diode has a forward voltage drop and the MOSFET itself will be at the end of the chain. Am I being led astray with this assumption?

As horta points out, you also have to consider the case when the MOSFET is "off". Then there is no current through the LEDs and the Vds on the MOSFET is the whole 36 V.

About Safe Operating Area (SOA)

NXP app note AN11158 explains the reason for the limits on different parts of the SOA curve:

enter image description here

Because you are using an external current limiting device, you should be in part (1) of the curve, limited by resistive self-heating.

When the LEDS are off, you will be in part (4) of the curve, limited by breakdown of the body diode or other junctions in the device.

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  • \$\begingroup\$ I still do not understand what basis the MOSFET will see the entire 36v. When I measure the Vds across a MOSFET in a similar 24v LED string, I only read 6v Vds in the "off" state. I also checked with an oscilloscope for a transient voltage spike, there is none, just 6v Vds. This seems to account for an 18v voltage drop across the LEDs. \$\endgroup\$
    – GabeNix
    Mar 27, 2014 at 19:20
  • \$\begingroup\$ When there is no current flowing, the voltage across each LED is 0. Therefore the drain of the FET will be at 36 V. I'm assuming a very simple circuit with just power supply, string of series LEDs, and FET d-s path in one loop. The other LED string you measured might deliberately allow some leakage in the "off" state to allow using a cheaper MOSFET, or there might be something else going on. \$\endgroup\$
    – The Photon
    Mar 27, 2014 at 19:24
  • \$\begingroup\$ But... Even if the MOSFET does see 36v in the "off" state, there will be no current, so the voltage at this particular time will not be a factor in determining the SOA, although it would still obviously need to be accounted for in the maximum Vds rating. \$\endgroup\$
    – GabeNix
    Mar 27, 2014 at 19:24
  • \$\begingroup\$ You could maybe put a 5-10 V zener (or even a well-chosen resistor) in parallel with the FET's d-s path. This would allow a very small leakage through the LEDs when "off", hopefully nearly invisible except in a very dark room, but enough to limit the Vds seen by the MOSFET. I have no idea if this is or isn't cheaper than just finding a FET with the required Vds rating. \$\endgroup\$
    – The Photon
    Mar 27, 2014 at 19:27
  • \$\begingroup\$ I will elude back to the main question which is about the SOA chart on the datasheet. The determination factor is not about the maximum Vds, it's about the Vds when under the full load in the "on" position. Are you saying that I should consider a Vds of 36v at the 1A draw when referring to the SOA chart? \$\endgroup\$
    – GabeNix
    Mar 27, 2014 at 19:46
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The VDS is important to your application when you want the LEDs shut off. When they shutoff, there will be the entire rail voltage over the drain and source of the mosfet. The mosfet must be able to withstand your entire rail voltage. If it can't it will cause catastrophic breakdown and smoke will rise when you try and shut the mosfet off.

Photon brings up a good point that you should be controlling the current in some way if your power supply is capable of sourcing more current than your LED's or MOSFET can handle when on.

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