Mosfet SOA chart reading for low Vds - Page 1
Forum section: electronics / projects

author: fedimakni
author_rank: Regular Contributor
timestamp: April 11, 2025, 02:12:43 am
content: Hello,I am designing an ideal diode circuit to protect my circuit and as it requires an n-channel mosfet so i am on the way to choose a suitable one.My system operate at 48V with 8A current draw.There are multiple n channel mosfet i found but i will share 2 of them just to illustrate the issue i am facing.First mosfet: SQJA62EPDatasheet:https://www.vishay.com/docs/75098/sqja62ep.pdfSecond mosfet: BSC034N06NSDatasheet:https://www.infineon.com/dgdl/Infineon-BSC034N06NS-DataSheet-v02_03-EN.pdf?fileId=db3a304342371bb001424c65286a70beI am trying to look for the SOA chart to check if the mosfet will operate in a safe area or notLet's take BSC034N06NS as first example Rds on = 3.4mohm so Vds should be = 0.0272V The graph doesn't provide Vds less than 0.1V so i am worry if the 8A mark will fall within the safe area or notFor SQJA62EP Rds on = 0.0062ohm so Vds =0.04V depending on the chart at Vds = 0.04V it is safe to draw up to 4A which seems a little weird for a mosfet capable to provide 60A of current.Am I missing something, Is there something wrong with the logic?I would like to note that the mosfet is not in a switching configuration so it just turn on or offThank you.

author: Whales
author_rank: Super Contributor
timestamp: April 11, 2025, 02:37:36 am
content: BSC034N06NS:You're perfectly fine.The lines labelled DC, 10ms, 1ms etc are the actual SOA limits. If you cross those the device is not guaranteed to survive or behave correctly.The line on the top isn't an SOA line, it's a practical limit imposed by RDSon. You can't cross it under normal conditions because of physics, but if you did (eg by supercooling the chip) then the device would be fine. You can project this line off the left of the graph (towards lower Vds) using ohm's law if you want to stretch the graph to lower voltages.SQJA62EP:Again you're hitting the RDSon line.> at Vds = 0.04V it is safe to draw up to 4ANope, nothing to do with safety. If 4A is flowing then you will have minimum 0.04VDS. If 8A is flowing then you will have minimum 0.08VDS. At both of those points you're well below the SOA lines under all timing circumstances.

author: Whales
author_rank: Super Contributor
timestamp: April 11, 2025, 02:47:44 am
content: Another way to think about things: if you're operating the mosfet as a switch (either fully on or fully off) then your mosfet will encounter three states:(1) Fully off. Vds = something, I = 0. You're safe as long as you're not exceeding the voltage rating of the mosfet.(2) Turning on/off. In your case you can probably assume this is quick enough and happens seldom enough that it's not a worry, because you're using a mosfet that exceeds your needs by a big margin.(3) Fully on. Vds = I * Rdson. This means you will sit somewhere on the red line of this graph, approximately:ie as long as you stay below 60A or so then you're still within the SOA. You can't operate much above that anyway without hitting other limits of the part anyway.The rest of this graph (the bits not along the red line) are only useful for analysing the turn-on/off transitions (listed as (2) above), or for if you're intentionally operating the mosfet in a state other than fully on or fully off (variable resistor, "linear mode").N.B. the SOA isn't the whole story, the chip still needs appropriate heatsinking and cooling, there are dV/dT limits, Vgs limits, etc.

author: Smokey
author_rank: Super Contributor
timestamp: April 11, 2025, 03:14:18 am
content: The point of SOA is to not melt the die by having it go above its abs max temp. It's an energy thing (power * time). More thermal load than can get dissipated in a given time and the die overheating is what you are trying to avoid. These SOA curves can also be simulated if they provide the RC thermal impedance constants for the thermal junctions (but they don't always include those).

author: PCB.Wiz
author_rank: Super Contributor
timestamp: April 11, 2025, 04:32:14 am
content: Quote from: fedimakni on April 11, 2025, 02:12:43 amHow are you driving the MOSFET ?As above, the +45 degree slope is simply RDS*Id, the SOA info is more focused on the right part of the graphs, which is I-V-Time related.Ideal diode drivers have some turn-off time, and that turnoff under hot-plug conditions, is when the FET load line will need to be checked.The better ones turn off the MOSFET in under 1us, but high currents can flow during that time, depending on cable resistance/inductance/source impedances.

author: fedimakni
author_rank: Regular Contributor
timestamp: April 11, 2025, 04:51:15 am
content: Whales:I didn't quite catch that, So you think i'll be safe with any of these mosfets correct?The limited by Rds on graph simply means that when i draw 8A Vds will be = 80mV correct? So i don't need to worry about it? (just of course temperature rise and power dissipation which are ok depending on my calculation)So other than the right side of the graph which is related to switching ... in the normal operation of a mosfet as a switch what should i take care about in the SOA graph?

author: Whales
author_rank: Super Contributor
timestamp: April 11, 2025, 08:35:25 am
content: Quote from: fedimakni on April 11, 2025, 04:51:15 amWhich parts? If you can point them out then I can try explaining in a different way.QuoteProbably yes, I do not know the rest of your circuit and application so I can't promiseThe numbers you have quoted will not violate the SOA.QuoteYes. Really the RDson line has nothing to do with SOA, but they draw it on there anyway to stop you thinking that you can use the top half of the graph.QuoteNo. "Switching" is a very complicated word that can mean many different things, I think this word is misleading you.Your mosfet is always in one of these three states:(1) Fully off. Vds = something, Ids = almost zero(2) Partially on. Vds = something, Ids = something(3) Fully on. Vds = almost zero, Ids = something (this is the only circumstance where Rdson is achieved)Each of these three states is a different position on the SOA graph. All 3 of them can violate the limits in different ways.Your application requires a mosfet to turn on or to turn off. This is a form of "switching". Depending on how your drive the gate of the mosfet this transition might be fast (microseconds) or slow (seconds). During this time your mosfet is partially on and will be dissipating lots of power.The fully on and fully off states are easy to understand. When the mosfet is fully off it acts like a very high value resistor (eg 1Mohm). When the mosfet is fully on it acts like a very small resistor of size Rdson (eg 0.01ohm). In both cases the mosfet will not be dissipating much heat because either Vds or I will be almost zero (P = V*I)."Partially on" is the complex state that you need the help of the SOA graph for. In this state the mosfet acts like a moderately sized resistor (let's choose 2 ohms as an example) In this state LOTS of heat is generated because both Vds and Ids are reasonably large (P = V*I).Does that make any more sense, or do specific bits not make sense?

author: mtwieg
author_rank: Frequent Contributor
timestamp: April 11, 2025, 11:03:44 am
content: Quote from: fedimakni on April 11, 2025, 04:51:15 amI feel I should make some clarifications to prevent people misinterpret what's being said here...SOA is important when the FET must spend significant time (say >10us) operating in saturation (meaning both Ids and Vds are large at the same time). This is generally an issue for aFET used as load switch, in which the FET may conduct substantial current with a large voltage across it. If you were designing a load switch, then the switching time of of the circuit would be important, and we would need to know the load current during the transient (often dominated by capacitance on the load side).However you say you just want to use theFET as an ideal diode. This means that FET will only turn on/off when the voltage drop is very low (no more than the Vf of the FET's body diode). So ideal diode emulation basically guarantees that SOA isn't going to be a concern for you.However, many "ideal diode controller" ICs actually are capable of more than simple diode emulation, and also add features such as reverse polarity protection, overvoltage protection, external on/off control, etc. If such features are being used, then it is likely you must consider SOA limitations.

author: Smokey
author_rank: Super Contributor
timestamp: April 11, 2025, 08:13:16 pm
content: Quote from: mtwieg on April 11, 2025, 11:03:44 amI've been doing quite a few projects with ideal diode controllers recently. Apart from one condition I can think of, the FETs are either full on or full off. So it's essentially all I^2R losses associated with Rds-on. Depending on what controller you select you can have higher currents for switching faster.The one condition where you would find yourself dissipating real power is if the ideal diode controller (or discrete circuit if you like pain and suffering) implements a turn on ramp for inrush limiting. If that's the case, and you turn it on into a load then it's up to you to set that ramp so you don't exceed your FET's SOA under any possible operating conditions.

author: fedimakni
author_rank: Regular Contributor
timestamp: April 11, 2025, 10:44:15 pm
content: Smokey:Yes as you said, i am implementing the protection features of idea diode and one of them is the inrush current limiter. i chose the turn on time to be around 10ms, What i need to care about in that case?

author: Smokey
author_rank: Super Contributor
timestamp: April 11, 2025, 11:24:44 pm
content: Quote from: fedimakni on April 11, 2025, 10:44:15 pmBe super careful here. If you slow the turn on ramp down that much you are spending significant time in the linear resistive part of the FET curve. It's pretty easy to violate SOA depending on what load you are turning on into (big caps will pull a bigger current spike than you are probably expecting). It's up to you to get this right based on your application requirements and the controller won't protect the FET in this case.Here is a calculation spreadsheet from TI for sizing the FET and inrush time:https://www.ti.com/product/LM7480#allhttps://e2e.ti.com/support/power-management-group/power-management/f/power-management-forum/1256470/lm7480-q1-inrush_fet_soa-margin-calculatorhttps://e2e.ti.com/cfs-file/__key/communityserver-discussions-components-files/196/6052.INRUSH_5F00_FET_5F00_SOA-Margin-Calculator.XLSXIf you have a spice SIM, you can look at power and see how things are effected. Just add a bunch of capacitance after your ideal diode and you will see the turn on spike.Note: I recommend the LM74502 controller. It's pretty inexpensive, and just works. Protections = Ideal Diode/Reverse polarity, reverse current (with second FET), over-voltage, under-voltage, enable, all in one tiny part. There is another variant that also includes over-current protection but it's 2x the cost so if you don't need it then the LM74502 is good. There is also a variant with high gate drive current that you can use to make a high side switch.

author: fedimakni
author_rank: Regular Contributor
timestamp: April 12, 2025, 06:49:21 am
content: Smokey:I am actually going to use the LM7480 as it also has all the features as well as its availability.I am really worry that the one i selected will not be suitable in case of turning off the mosfets when they draw 8A as i am driving them with 48V so the power is high and most of the mosfets will be just on the edge unless i turn them off fast which depends on the capability of the idea diode

author: mtwieg
author_rank: Frequent Contributor
timestamp: April 12, 2025, 11:26:57 am
content: Quote from: fedimakni on April 12, 2025, 06:49:21 amOk so now things are much more complicated than just an ideal diode controller. You need to provide some specific info:1. What sort of arrangement you intend to use the LM7480 in. The datasheet shows two very different arrangements on the first page.2. What is the total capacitance on the output(s).

author: fedimakni
author_rank: Regular Contributor
timestamp: April 14, 2025, 01:12:44 am
content: mtwieg:I use the LM74800 with 2 n channel mosfet back to back as shown in the page 19 without Vout2It's a little hard to provide a precise value of the output cap but it is expected to be about 1000uF

author: mtwieg
author_rank: Frequent Contributor
timestamp: April 14, 2025, 11:34:48 am
content: Ok so let's say you want to keep the inrush from the capacitance below 5A. That means the output dV/dt should be limited to 5000V/s, and Cdvdt should be 11nF. This inrush will last for about 10ms.So now we look at the SOA curves. The SOA curve assumes constant Vds and Ids, but in your case Vds is ramping down linearly, so some approximation is needed. Most conservative approach is to assume Ids/Vds is 5A/48V for the entire 10ms pulse. Looking at the SOA curves for SQJA62EP, the 10ms contour intercepts Vds=48V at around 3.5A. It might work... but I would recommend having more margin. For the BSC034N06NS, its 10ms contour intercepts 48V at around 0.08A, I would forget using this part entirely.Increasing Cdvdt to 22nF will reduce the inrush current to 2.5A and the inrush will last 20ms. The SQJA62EP should handle this fine. But keep in mind this math only accounts for current due to the output capacitance. If anything else on the output side starts drawing significant current during the inrush, it could make these assumptions invalid. Usually UVLO circuits are used on the output side to ensure this doesn't happen until the output is near its final level.