# safe operating area of power transistor

Discussion in 'Datasheets, Manuals and Component Identification' started by adeel abid, Sep 4, 2012.

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Sep 4, 2012
Dear All,

i am a final student and working to design an inverter. for this purpose i have selected a power MOSFET IRF740. but i am very much confused about the safe operating area figure and saturation characteristics. Considering the FIARCHILD datasheet SOA curve (fig 4) says that Id=0.8A for VDS=1v but fig 6 says that Id= 2.5A for Vds=1V.

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2. ### Harald KappModeratorModerator

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Nov 17, 2011
Your datasheet lacks the figure numbers. I took this one from the Internet.

A note in advance: the datasheet specifies the data for a transistor without additional cooling by a heatsink.

With respect to figure 4 this datasheet shows the area left of the SOA curve as "limited by Rdson". This means that you can calculate the power across the source-drain region by P=Uds²/Rds or P=Ids²*Rds. Depending on how much thermal power you can dissipate using a heatsink, you can operate the transistor within this region, therefore you can operate it at Vds=1V, Ids=2A, too, provided the heatsink can spread the thermal energy.

Did you note the remark "pulse duration = 80µs" in figure 6? Such a short pulse will develop not enough heat to destroy the transistor immediately. This way this characteristic can be evaluated. If you apply the load for longer than 80µs, you have to decrease the load to stay within the SOA or you have to add a heatsink.
Compare also to figure 2. Without a heatsink the case temperature will rapidly rise to >100° and therefore your drain current would be limited.

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Sep 4, 2012

it means that it is equivalent to increase in Rds(on) with case temperature thus limiting the drain current for given Vds and automatically brings the transistor to SOA.
I came across IRF740 datsheet from Intrernational Rectifiers. its characteristics are similar to one in FAIRCHILD but its SOA gives Rds(on) LIMIT region to have Rds(on)=0.5 ohm as well as its Id-Vds curve. Can you please explain what causes it to be like this?

Last edited: Sep 5, 2012
4. ### Harald KappModeratorModerator

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Nov 17, 2011
Sorry, I cannot explain this. Each manufacturer is free to characterize his component at his own. If a component is available from different manufacturers under the same type/name, this doesn't necessarily mean the components are 100% compatible. It only means that the manufacturers hold this component as reasonable similar to be substituted against each other (unlsee, of course, a manufacturer explicitly specifies his product as a second source for another product).

This is not uncommon and for special purposses (e.g. certain certifications) the manufacturer of an electronic product will specify not only the type of acomponent but also the manufacturer from whom to buy this component. Components from other manufacturers are then not allowed, even if the specifications are (nearly) identical.

Last edited: Sep 5, 2012

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Sep 4, 2012

I was asking about SOA because i have designed a boost converter. It is based on IRF740. Driving gate signal is generated by microcontroller which is then pulled up with !k resistor to 12V. So Vgs is either 12V or 0V with ton=190us and min. toff=10us giving about 93.3% duty cycle.The gate driving signal is not continuous but actually based upon monostable multivibrator that is only trigerred if load voltage is less than desired value & simultanoeusly indcutor current is reduced to zero (this is sensed by sensing voltage across 1ohm resistor).

Problem i am facing is that when MOSFET is on for 190us as per ideal calculation id should be ramp with ID(max)=8A. During this phase only the battery, inductor and IRF740 is in the circuit. But practically after implementation i am getting ID(max)=3.5 to 4A (measured by sensing peak voltage acorss 1 ohm resistor).

There is not so much heating of IRF740 to increase RDs(on) to limit the drain current to this value because i have tested it with no load i.e. without 500 ohm resistance (and hence multivibrator pulse repetition period is quite large to allow cooling between generation of pulses). This is limiting the input power of boost converter and hence the output power delivered to the load.

please provide reason for this behaviour.

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6. ### Harald KappModeratorModerator

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Nov 17, 2011
I don't know what exactly you are measuring.
During the time when the MOSFET is ON, you cannot measure Ids with the 1 Ohm resistor because this resistor is not in the path of Ids.
The 1 Ohm resistor measures current when the MOSFET is OFF. The current is then supplied by the energy stored in the 300µH inductor.

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Sep 4, 2012
To let you know how i am measuring the current please conider the attached inductor current waveform.
MODE 1: during this mode transistor is ON during interval ton and inductor current rises from 0 to TD(max).
MODE 2: during this mode transistor is OFF during interval toff after and inductor current starts decreasing from ID(max) to zero. This inductor current passes through 47uF capacitor, load resistance(if exists) and 1ohm resistor. Thus the voltage across 1 ohm resistor is actually the waveform of inductor during mode 2. Its peak value is inductor current peak value.

Also i have connected the DC ammeter in series with inductor to measure average DC current. this average is then multiplied by 2 to get its peak value.(Iavg=0.5*Idmax)

please comment if am wrong in calculation or else.

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8. ### Harald KappModeratorModerator

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Nov 17, 2011
The figure is surely from a datasheet and does not represent what you have in your circuit. Have you looked at the waveforms using an oscilloscope?
At your duty cycle the current through the inductor will not go to zero in mode 2. Try to reverse the duty cycle: 10µs on, 190 µs off. What happens?

Your calculation Iavg=0.5*Imax is valid only if the time for mode 3 = 0 and if the current starts at I=0 for mode 1 and ends at I=0 for mode 2. At your ducty cycle it will not go to zero at the end of mode 2 and thus your calculation is wrong.

Is the inductor rated for 8A? Did you include the resistance of the inductor in your considerations? 8A peak current is for an ideal coil (0 Ohm series ressitance) only. If you include a series resistance of 1Ohm into your calculations, the current will approximately be halfed (I did a short simulation - not representative) which comes near to your measured value.

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Sep 4, 2012
you are very right. the figure i have posted is for purpose of understanding how i am measuring the peak current & not the actual representation of inductor current in my circuit.

Secondly i have mentioned that min. toff = 10us next pulse of monostable multivibrator of 190us pulse is triggered only if inductor current (i.e. voltage across 1ohm resistor falling edge) is reduced to zero as well as output voltage is less than desired value (logical and operator between vout and I).

Thirdly in this case mode 3 cannot exist in my circuit. so Iavg=0.5*Imax is valid in this case.

for inductor i have used 22 gauge wire and measured its value to be 300uH from a LCR meter.

regarding 1 ohm resistance is it possible that Rds(on) has increased to 1ohm from 0.5 ohm but due what reason. if not what else can cause this error?

Thanks

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Sep 4, 2012
please find the attached real oscilloscope picture.

upper trace is of gate drive signal and lower trace is of voltage across 1ohm resistance.

may be this can help you to find the real source of problem

thanks

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11. ### Harald KappModeratorModerator

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Nov 17, 2011
O.K., so it looks like my assumption of the current not reachin 0 during the OFF time of the transistor is wrong. I apologize. I still think the resistance of the coil reduces the pass current during the ON phase of the MOSFET. Have you measured the DC resistance of the coil?

12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010

What current is it capable of?

What does the voltage at the input side of the inductor look like during operation (using your scope)?

Can you check with your oscilloscope to see what the actual voltage across the inductor looks like during operation (careful if your power supply is earthed)

Have you checked the inductance of the inductor. It is possible that the inductance is slightly higher at low currents (if the inductance is specified at (say) 8A) and the initial rise in current may be slightly slower than you expect.

How fast is your mosfet switching off? If it is switching slower than you expect, it may provide time for the current to ramp down a little.

I'm pretty sure the voltage across the inductor is likely to tell a story though.

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Sep 4, 2012
Dear Steve,

voltage at input side of inductor fig1 with 1v/div and 0.1ms/div.

voltage across inductor fig 2 with 10X probe.5v/div and 0.1ms/div. please note that lower levels are not exactly parallel to reference but has some +ve non zero slope indicating large Rds(on).

Inductance is checked with LCR meter and found to 300uH. also inductor slightly heated up during operation. inductor is wound on ferrite core with 18 gauge wire.

switching off should be quite faster because of gate being pulled down by npn BJT fig 3 upper trace.

inductor current measured with 0.1ohm resistance in series fig 4 0.1v/div and 0.1ms/div with 0 line 1 div below reference line.

total current measured with 0.1ohm resistance in series with battery fig 5

Also i have checked Rds(on) with ohmmeter and found it to be 0.8ohm

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• ###### total current.jpg
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Last edited: Sep 9, 2012
14. ### Harald KappModeratorModerator

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Nov 17, 2011
You still haven't told us the DC resistance of the coil. It's not hard to measure...

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Sep 4, 2012
i have measured it and it fluctuating between 0 and 0.1 ohm

16. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Firstly, thanks for reading my questions and providing *exactly* what I asked for.

I presume you're using a 10x probe here? So we're seeing about a 3V drop in the battery voltage as the current increases.

if this is a 1x probe, then you have very little to worry about.

I'm guessing it's a 1x probe because other measurements taken later would not make sense if they were done with a 10x probe (so I'll assume 10x only when you say so).

Was it AC coupled here?, if not it looks like your battery was supplying only about 11 to 11.5V at minimum inductor current.

OK, I'm assuming that it's probably AC coupled again. I can see the voltage across the inductor falling slightly as the current increases, and the sharp spike in voltage during the off period.

"slight heating" is OK.

How is the mosfet and the schottky diode for temperature?

Yeah, switch off is fast and that will contribute to lower power dissipation in the mosfet. The switch on time is slower, but that is not of great consequence.

OK, so we see a number of things here. Firstly the inductor current doesn't fall to zero.

By my calculation, the voltage across the 0R1 resistor varies between 0.06V and 0.5V (I think you must be using a 1x probe).

The second (and important) important thing here is that we don't see the current increasing at an increasing rate at any point, so the inductor isn't nearing saturation.

However it does seem (based on your measurement of Rds(on) or 0R8) like the voltage across the mosfet rises to about 4V during this period, and that is going to be a little wasteful.

It will see a peak current of 500mA, and an average current of about 250 mA during the ON cycle, and let's also say it's an 80% duty cycle. So this means about 0.08W on average being dissipated by the mosfet.

However, since you're only driving a 500 ohm load (that should draw no more than about 50mA), I'm concerned about why the current is so high. I would be expecting an average current closer to 130mA or thereabouts.

Let's assume 8V across the 300uH inductor for 2ms. The current should rise by about 53 amps. The current is rising by about 1% of that, so I would guess that your frequency is actually about 40,000 Hz

OK, that's a classic shape, but I need to know the scale and where zero is. In any case, it seems fine. It mimics the current through the inductor, which is that I'd be expecting (I presume there is other circuitry that you've not shown us )

Everything seems to be going along quite swimmingly, except the current you're seeing is way higher than I'd expect.

Several things come to mind:

1) are you *really* using a 1N5404 diode? to rectify the output?

2) do you have some other (unshown) feedback path to limit the voltage across the load?

If you are really using a 1N5404 diode, then pull it out and get a suitable rated schottky diode. I would expect that if you placed a current sense resistor in series with the1N5404, you would see significant reverse current flowing. This may both increase the input current required, AND increase the current flowing through your mosfet.

If I were to suggest anything, it would be to replace this diode.

You show no feedback to maintain the output voltage at 24V I assume there must be something, otherwise the output voltage will rise significantly, often until something (mosfet, diode, or capacitor) decides to start conducting current when it's not supposed to.

Last edited: Sep 9, 2012

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Sep 4, 2012
Actually the load used for these readings is 220V, 100W tungsten filament bulb.

voltage at input side of inductor is measured with 1x probe & AC coupling.

inductor voltage is measured with DC coupling and frequency is 4KHz.

MOSFET is also heated and i have not used shottkey diode but only body-drain diode of IRF740 is used.

please explain how can u say that inductor current is not reducing to 0. on time is 190us or approx 0.2ms (not 2ms).

0v is always the reference line except for fig 3 & fig 4.

i have never mentioned that desired vout = 24V. it is actually 210V and there is a separate mechanism to maintain it.

BUT MY PROBLEM IS STILL UNSOLVED "WHY I AM NOT GETTING THE PEAK CURRENT OF APPROX 8A DURING TURN ON PERIOD (190us FIXED IN ANY CASE) OF MOSFET? BECAUSE DURING THIS PERIOD THERE ARE NO OTHER COMPONENTS IN CIRCUIT EXCEPT BATTERY, INDUCTOR & MOSFET"

18. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
No, in that circuit diagram you show 2 diodes. Neither is the body diode (which is a very poor diode typically)

You say "inductor current measured with 0.1ohm resistance in series fig 4 0.1v/div and 0.1ms/div with 0 line 1 div below reference line." for one of the graphs. Clearly you're telling me that the inductor current doesn't fall to zero. (i.e. you said it, not me)

But which line is that?

Yeah, not sure where I go that from.

Some of my calculations earlier were based on that. Had you specified the output voltage previously?

Well, if it really only 300uH, and 12V, and 0.8 ohm, in 190uS...

Oh, is that all.

Well theory suggests it should climb to 15A. Clearly something is preventing that from happening. The additional 0.1 ohm resistor should limit it to 13.3A

If the reason is some other resistance (and there is almost certainly some) then the value of that resistance is about 0.7 ohms (to limit it to 8A) or an additional 1.3 ohms to limit it to 6A.

Use the formula i(t) = Vs(1-e^(R/tL)) (that assumes the current falls to zero though.)

At 8A, the mosfet is dissipating about 23W by your firures, and I thus expect it to get warm. Without knowing the size of the heatsink, I can't comment on how warm that might be. Without a heatsink it would be *very* warm (and short lived)

At 6A, you're looking at around 15W, with similar observations on temperature. (actually, I can calculate it. Without a heatsink, the junction temperature will rise to 930 degrees C above ambient -- which is hot in anyone's language).

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Sep 4, 2012
Please find the attached pictures of cores i am using to design inductor for boost converter. Please advice about its specs, material. whether it can give incorrect reading with my LCR meter its inductance increases as current through it increases? and whether it is suitable for boost converter application with Imax=8A, ton=190us and toff>=10us. if not suitable then which core should be used, advice about its physical appearence and other details. i am in doubt whether i am using the correct core for this application. color of core is green/blue ans yellow/white