Confused by FET "safe operating area"

[elt]

I want to make a PWM style dump controller but, PWM or not, I think there's a time I have to expect the switch to be "on."

My plan is a 1.2KW dump load to handle 40 amps and 30 volts. (I'm guessing 29 or 30 volts isn't unreasonable for a 24 volt system.)

I'm looking at spec sheet and am not totally sure what the graphs mean but it looks to me like even the big honking devices (100-200 amps)are rated at some low voltage... when I look on the "safe operating area" chart and look for amps were DC and 30 volts intersect, it appears to be only 1/8 to 1/16 or the devices rating. I've found some $15-$20 devices that look like they'll do 10 or 12 amps DC at 30 volts but ganging four of those seems expensive... Is ganging eight 5 amp units reasonable? Ten 4 amp devices? Am I interpreting "safe operating area" correctly? ...

I really need some help with device selection; thank you very much!

- Ed.

[edy252]

Hi...

I'm not sure if i understand you right.....but i can tell you that a FET lets lots of amps go through it and that doesnt require that you look at volts charts....(volts are applied to your load ... and not the FET).... i.e. when there's a high voltage applied between the drain and source of your FET, there's very little (or no) current passing through it....when the current is allowed to pass (action on the gate - positive voltage for N-channel FET for example) the current will pass from drain to source and the voltage between those will be very low (or zero)

                 +V

                 |

               |  |

               |  |   Load

               |  |

               |  |

                |

           D----    

           |      

------ G - |

           |

           S---

               |

               |

             -----

              ---

               -

the maximum voltage you can see in the chart is the maximum you can apply between D and S, ie. when there's no current passing through it...(there's another maximum which is gate to source, but that's definately not as high as 100V)

Hope that helps....if it doesnt let me know and ill try to explain more and btw...experimenting with this will let you understand it better i guess

[edy252]

actually there's gonna be a relatively low voltage between D and S  and it will be equal to Id x Rds

Id = drain current

Rds = resistance between D and S

but it still very low

for example with the IRF520, Rds(max) = 0.4 ohms and Id (max) = 8A ...

--> Vds(on-state) = 3.2 V

It also has a maximum of 100V for V(DS) (off state)

[elt]

Here's the graph of a part listed at 30 amp and 60 volts

I don't know if this graph means something else that I don't have to be concern about but I read it to say for "DC on" you can have 2 volts at 30 amps or 60 volts at 2 amps (or 30 volts at three amps.)

I was thinking that I could use two or three of these guys...

Also, with my dump load rated at 40 amps and an rdson of .047 ohm; I'll need a heat sink for about 75 watts and I'm not sure how to size that...

Thank you!

- Ed.

[mtbandy]

Yes, it is completely normal to parallel MOSFETs like that. Afterall, they are only tiny devices - and there's no way you are going to want to push 40 amps through a single device continuously, even if the ratings seem to allow it. This is not only due to the lead thickness and physical size of the device, but also the Rds(on) and the problems it causes with heating and wasted energy. Parallelling 4 of those devices will reduce your dissipation to around 20 watts.

I've seen commercial inverters that parallel 2 sets of 3 or even 4 60 amp devices for this exact reason, getting the Rds(on) down to about .0045 ohm. The higher you go with MOSFET D-S voltage, the higher the resistance tends to be, so choose a device with a suitable voltage (like you already have done) and then look at specs for current capability, alongside Rds(on) and cost. Try and find the cheapest combination which isn't going to blow up or overheat. If you get it right, you should be able to get away with minimal or no heatsinking at all. Solving the problem with a honking great device isn't the way to go! Make sure all MOSFETs are tied together closely, and driven hard with a decent MOSFET driver. Hope that helps a bit!

[Nando]

ED:

Please send your email to me to communicate extensively about it.

SOA is the transit time of the FET that limits the power dissipation in the SOA curve, the FET needs to transit fast and not to stay close to the SOA curve.

The dump controller you need for 1.2 KW needs to designed fro about 1.5 to 1.8 KW to protect you against unknown eventualities in your system.

You may have two types of controllers, the bang-bang or the proportional controller.

the best one is the proportional and just loads the generator to keep the battery at its charged voltage or at certain defined generator voltage.

Since you have a 24 volts bank, the Fet breakdown should be at least 50 volts.

The Fet current needs to be defined accordingly to the Rdson of the FET for minimum FET dissipation

Assuming 30 volts and 1,5 KW the current is 50 amps, then we need to assume the FET dissipation let's say 2 % of 1500 watts or 30 watts , which gives 30 w / 50 a = 0.6 ohms total, what about instead we say let's reduce the fet dissipation to 5 watts, then the Rdson = 5 w / 50 a = 100 milliohms = 0,1 ohms -- this require a minimum heat sink.

Now we need to find the FET, 100 to 50 volts, RDson 0,1 -- what about if 4 fets are used and each has 25 millioms RDson or even less ?

Going to www.onsemi.com I found several but one is good NTP52N10, 100 Volts, 30 milliohms, 60 amps capable, TO220 and $1.5 price.

If we used 3 in parallel the RDson would be 10 millohms at 25 Celsius and about 20 millohms at much higher temperature.

I have not tried to see how the MosFet availability is, which may require a change in the type with more commonly available.

This is the basic initial design.

Nando

[elt]

I don't see were the the "2%" comes from; perhaps "experience?"

Just to double check, W = I*I*R ... I think you used W/I = R in your example.

But if I follow what you've written, 4x of the rfp30n06 devices in parallel with rdson of .047 ohms would have a parallel resistance of .012 ohms. watts = 50 amps * 50 amps * .012 ohms = 30 watts total or about 7.4 watts per device? ... I've got three in my hand left over from my MPPT controller so that'd be about 13 watts per device? (Data sheet says max dissipation in 96 watts.) Should I look for devices with a lower rdson, use more of them or is 13 watts per device still a small heat sink?

Thanks again,

- Ed.

[SamoaPower]

elt,

I'll try to shed some light on your original question about the "safe Operating Area" chart.

It's easy to think that FETs (or bipolars) are used mostly as switches, where they are either fully on or fully off. Such is not the only case. There are many applications where they are used in their so-called linear region where they are partially on most/all of the time. An audio amplifier is one example.

The Vds axis of the chart does not refer to the supply voltage but to the instantaneous voltage from drain to source. You will notice that the scale doesn't go below 1V where you would normally expect to find Vds when it is fully on as in  switch mode.

When the gate of a FET is biased so that so that it is only partially enhanced (on), it is behaving more like a resistor from drain to source with a resistance much greater than Rds. When current is allowed to flow from drain to source there is a voltage drop across this resistance and THAT is the Vds on the chart. This, of course, results in more power being dissipated in the device compared to switch mode.

As you might imagine, there are numerous combinations of Id and Vds that result in a given power dissipation. The chart relates these parameters to show an operating area where the power dissipation won't allow the internal device temperatures to exceed ratings.

The area bounded by "DC" is for steady state conditions. The other areas, labeled with time, are for when those operating conditions are turned off and on and the time refers to the on time pulse width.

The chart is not applicable to switch mode operation.

 

[Nando]

let me start saying that I made a big mistake

I divided watts/amps and made it ohms, it should have been volts

The equations is

R = watts / Amps^2

and in this case if CURRENT = 50 AMPS and WATTS = 5 THEN

R = WATTS / AMPS^2 = 2 MILLIOHMS

I was going at full force without paying attention, what a mistake ( I was trying to finish fast because I needed to go out).

Nando

[Nando]

The decision to have certain dissipation is based in experience, needs, parts availability plus PWM frequency, Fet switching dissipation, besides the RDson dissipation, this due to the current.

With let's say 1500 watts load, one needs to define how much can be dissipated by the switching FET, in this case one can say, since it is a protection circuit that the dissipation of the FET could be one just enough to keep it happy with the heat sink one may have available or can use.

The device you have RFP30N06P means that it is a 60 volts device with 30 amps capability, which really should not go above 15 or so amps, the RDson increases with temperature to almost twice at around 100 degrees die temperature(silicon inside). Though I can not find who produce it

Once a dissipation is assumed at 100 celsius then the device find is needed.

Your device 0.047 ohms at 25 Celsius, may have around 0.09 ohms at 100 Celsius

This is the value you should use for optimum design.

Your 3 devices would have the equivalent of 0.03 ohms ( at 100 Celsius) then with 50 amps.

watts = 2500 * 0.03 = 75 watts at 100 c. and around 38 watts at 25 C.

Heat sink in this case would be large, finned with 1 inch long fin and 3 fins per inch may be 6 inches wide by 12 inch long, for maximum dissipation and reduce temperature rise since is going to be in a rough environment

My software is in another crashed hard disk that I am trying to save, so I have to postpone the calculation since it is a bit long done by hand.

Nando

[ghurd]

Nando,

My current understanding of parallel fets says they should be connected thermally to avoid run away of a single fet.  

Won't that greatly affect the thickness of the heat sink, if not the surface area?

G-

[boB]

>>Nando,
>>My current understanding of parallel fets says they should be connected thermally to >>avoid run away of a single fet.

FET sharing is helped in that if one FET heats up, its resistance goes up, so the
FET that is cooler should conduct more current and itself heat up. And so on..

Also, the SOA curve is sort of misleading.  Take for instance the SOA curve in this thread...
At 10 Amps and 10 Volts, it dissipates 100 Watts (10 X 10) but that assumes that the
FET Case is held at 25 degrees C (Kind of like PV and STC)...  How likely is it going
to be that you can actually dissipate 100 Watts in the FET AND keep its case at 25C ??
Usually, Not very.

boB

[elt]

I'm not married to those FETs, they're just "in hand". But I think getting a better understanding of one applies to others...

The rdson modifier for 100C is 1.5 so with three devices, .047/3*1.5 = .0235 ohms * 50 * 50 = 60 watts total.

Likewise, I have solid copper CPU heatsink that'll handle 60 watts and should keep the devices at 50C or less (just guessing cause that's what it does with a cpu) so I think rdson will be a little lower. I don't want to depend on a fan on the heatsink in the long run (unless I have to!) but I think that'll get me going.

Heatsinks are still guesswork for me. I went to a manufacturer's web site (a while ago, don't recall which) and the math didn't scare me, it just seemed like I couldn't figure out the numbers to put into the formulas. I got frustrated so more or less I've just looked at heatsinks used in similar applications. But since this my first 1.5KW application, I'm not sure what's similar... maybe my inverters, they're both 1.8KW?

[scottsAI]

Hello elt,

What generator are you using?

Since you need load dump I assume PMA? (and looking at your other posts)

PMA rated for 24 v will put out many times more than that in high winds.

The output voltage is directly proportional to wind speed. The voltage is 2x for wind speed above cut in etc. With that said if the load is placed on the battery it will limit the voltage seen by the load dump.

What if your working on the battery?

Do you want the load dump to work or shut down the wind gen?

Remember the higher winds will produce higher volts, with a higher output voltage, when connected to the battery the voltage is absorbed as heat in the stator. I have seen many reports of fried stators. Allowing the output voltage to rise will place the heat in the load...

Things to consider.

FET voltage rating must be 1.5x higher than the highest voltage you expect to see. Why? Well what if you think 100v is max, then something you did not plan for occurs, and 150v happens, Device may fail, I say may fail due to the manufacturing testing only confirms rating, may be better then may not.

FET resistance has a spec of +-30%, so a 100mohm may be 70 to130mohm, make sure you use all the values in your calculations for worst case conditions.

Same thing for current, do not plan on using long term the devices max ratings, leave some room for error. If you have great understanding of what is going on then design to max. I design to max for cost reasons due to millions of units in production = $$.

Food for thought, what is the purpose of a load dump? Limiting RPM.

With low winds load dump is not needed, it will spin faster without load, should be within RPM limit.  Base the load dump on RPM. Many ways to do this.

Have fun,

Scott.

[Nando]

TRYING to add to boB statement the insertion software has kept me for the last 45 minutes in an endless loop.

SOA exists for every temperature you may be keeping the Fet on.

I am attaching two charts, one is saturation voltages at 25 and 175 Celsius and the second is the SOA at 25 Celsius.

Due to RDson, the saturation voltage varies with temperature and current.

One can observe this difference by reading the Drain-Source voltages of both curves at known current and the minimum voltage is observed at the slope of the GATE voltage.

The MosFet DC operating curve is defined by the SOA at the working temperature and the Curve shown is good for all temperatures -- one needs to set the RDson curve at the RDson variations and adjust the SOA accordingly -- the operating SOA area is always below the RDson and the transient time of the FET ( Toff & Ton in microseconds) so if the design switches the FET slow the SOA is below that slow switching time, if fast, then the SOA looks like a rectangle with one corner cut by the RDson limitations.

The SOA area = V(drain-source) * I (drain) under the switching time of the Fet.

Nando

[Nando]

Do not use 1.5 factor go for 2 for safety,so double the RDson.

You are not informing about your copper heat sink, I think that you mean to say that is one used to cool CPU and it it is then it is not enough for the power you are trying to keep coll.

Also, if it has a blower then if it fails the heat will raise greatly maybe damaging the Fets.

Unless you add a Fet saturation detector for fast report (alarm).

Nando

[Flux]

Not a lot I can add to all this. As others have said the makers data is somewhat misleading in that it is based on a temperature of 25 deg C. Unless you have infinite heat sinks much of the ratings can't be achieved. There is also a package limit that determines how far you can push things. The short time pulse ratings are perfectly acceptable but under steady dc conditions you are limited to the dc SOA line at normal currents.At the high current end there is a package limitation that is rarely mentioned.

The very high currents that can be tolerated for short pulses can not be used at dc. the package limitations especially for those miserable TO220 things more or less dictate that under dc conditions there is little real difference between any of the devices with very low on resistance, despite very large differences in claimed rating.

If you devices with low on resistance and heatsink them really well then you may manage about 30A per device but aiming for 20 is likely better. There is no real problem paralleling lots of devices as long as you have drivers capable of driving the gate capacitance. Short connections with equal circuit resistance is needed and individual gate resistors of a few ohms are desirable to prevent spurious oscillation.

Flux

[ghurd]

Now you know what you should do and why, I'll throw in what I do and why.

I use more fets to keep them about 1W each, don't run them too fast, and switch them hard.  The `Bigger Hammer is Better' approach.

I have an affinity for the IRFZ44. It doesn't get warm with 6A continuous at 14V without a heat sink, so I try to keep them under 5A just in case, and heat sink them. They are certainly cheap enough to use a handful.

That's 8 fets at 5A for a 40A dump load.  Only a couple dollars extra to have a bit of piece of mind.  Not needing a fan or extra related controls may totally offset the cost.

G-

[elt]

That's a fine looking device but I'm using logic-level FETs... not totally sure if I'm stuck with that with my simple FET driver or not.  I'm using Samoa's suggestion with 5 volts going to the 2n4401 to drive the logic FETs.

Can I raise that to 12 or 15 volts to use devices like the irfz44 and still switch the 2N devices with the 5 volt output from my microprocessor? (I realize that that's a very basic question but I am analog-ignorant.)

Here's my prototype -

It's hard to tell what's going on - that's a copper CPU heatsink (it has about a square foot of copper in it), I added a second fan to make that 24 volts, the FETs are screwed onto the bottom. Some solid copper wire buses their pins and is wrapped around some #10 machine screws on each side to make the high-current terminals.

Thank you,

- Ed.

[elt]

> What if your working on the battery?

yes, I've made the driver board separate from the FET's so I can beef them up later.

(The socket will get a "Tiny45" micro processor.)

 - Ed.

[Nothing40]

Maybe this will help also..just a bit more info.

http://sound.westhost.com/soa.htm

;-)

[ghurd]

It looks like it would work at 14V to me.  Not 100% sure about it.

[elt]

I guess I'm asking whether I'll needed a level shifter to drive the driver pair... something like this?

[SamoaPower]

elt,

The original circuit will drive the gate to about 4.3V with 5V drive, a little shy for hard switching of logic level FETs although they will switch. You really want about 6V or more.

Adding the extra stage will take care of the level but will invert the PWM. This could probably be handled in the software. I'd also eliminate the 1k from the base of the first stage to ground.

In any case make sure the FETs have a Vgs rating of at least 15V and your 12V supply isn't going above 12V.