Mosfet IRF540 Transient(?) Switching Power Help

[Cicero]

Hi guys,

I'm trying to use an IRF540 (**broken link removed**) as a sort of "soft-start" switch for particular piece of equipment with high inrush and high load capacitance properties.

So essentially I'm just ramping the voltage up 'slowly', the time scale for this I haven't bedded down. But I'm thinking around 600ms to 1500ms from 'off' to fully 'on'.

My specs are:
Vin = 55V
Inominal = 2A

With the ramp up in voltage, the current slowly ramps as well, this has potential for a lot of heat and I could blow my FET if I'm not careful and keep the junction below the max temperature.

With that timescale of ramping, how do you work out the power and heat dissipated to make sure I keep my junction at a decent temperature? Is 600ms+ ramp time even considered a transient response?

 

[alec_t]

Here's a sim which might serve as a guide. The FET is modelled as a simple changing resistance, and dissipates ~10 Joules during the 1 sec ramp-up.

 

[MikeMl]

Suppose you configured your NFET as a 2A current source. A current of 2A will charge a capacitive load of 22000uF by a ΔV of 55V in 600ms.

I set up the following simple sim to show this. The upper pane shows the current into and the voltage across the capacitor. The lower pane shows the required power in the current source and the power delivered to the capacitive load during the 600ms period.

The power dissipation in the current source (which will be the same as the power dissipated in the NFET current source) is the yellow trace. It starts out at 110W, and linearly decreases to zero in 600ms.

In order to keep the junction temperature of the NFET within allowed limit, I believe that its tab will have to bolted to a thermal mass substantially bigger than itself.

That will be the problem for a future calculation...

 

[Misterbenn]

considered a transient response?

I'd consider it as pulsed power rather than ramped.

I can't quite envision how your circuit is going to work, do you have a schematic? In your discription it sound more like you'll try and use the MOSFETs linear region right? I've always used a resistor in parallel with a thyristor, so the resistor limits the current until the capacitance has charged and the thyristor is fired.

The simulations people have posted above are a good way to estimate the power in the FET. Remember P=I^2R so longer charge times with lower current will result in lower FET power dissipation.

Also it's more instructive to talk about energy instead of instantaneous power when talking at these smallish time scales. os J instead of W.

 

[alec_t]

I see Mike added a hefty cap in his sim. My bad, I forgot the cap in mine; so here's a revised one :-

 

[MikeMl]

Suppose we drive the NFET in a different way: Lets come up with a control circuit that constrains the power dissipation in the NFET to be a constant, like 20W. Now you can put a heatsink on the NFET that even in the steady-state would keep the tab temperature within limits at 20W of dissipation...

I simulated this case by controlling the behavior of the current source to keep the power in it constant at 20W. The current changes as the voltage across the NFET changes in such a way such that the product always = 20W.

Obviously, it is going to take longer to charge the capacitive load if the peak power is reduced from 110W to a steady 20W, keeping everything else the same.

I added the voltage across the simulated NFET (B1) as the blue trace. The product of the blue trace (Volts) * the current though B1, (red trace, Amps) gives the constant power in B1 (violet trace, W).

Note the resulting slew rate of the voltage across the load (green trace) is now non-linear. It now takes 1.7s to charge the load capacitance vs 0.7s before...

 

[Cicero]

Wow, thanks for all the responses, seriously appreciate it.

I thought I would keep things general, because I am trying to just understand how to deal with pulse situations like this and its effect on the FET, but perhaps its better to just give the full picture to give you guys some background here.

I'm using a hot-swap controller https://www.ti.com/product/tps2490 . I find its features absolutely great and the concept simple, plus Ti has a whole bunch of app notes and calculation guides. However this transient/pulse response is still not clicking in my brain. I've limited the power in the FET to about 24W, and it'll cut the startup if it at any point in the startup it gets there. So essentially I can't turn on at my rated voltage (55V) and final current of 2.1A...which isn't ideal. And I dont understand what I can push that Plimit to in my FET further than 24W. I dont even know how its dealing with that kind of power and surviving actually?

 

[Cicero]

Turns out I had the controller in the wrong mode so I am starting up fine now...but I'm still curious on how to deal with these kind of transient/pulse responses.

This app note https://www.ti.com/lit/an/slva158/slva158.pdf talks about treating the case temperature as constant for small enough transients and devices with a low thermal j-c impedance. I'm still not 100% convinced though.

 

[MrAl]

Hi,

The physics of static heating isnt that hard to understand really, for simple cases. Heat flow is a little more involved.

Basically the material being heated has a property called the specific heat capacity. This along with how much material the object contains tells you how fast the temperature will rise with a given drive wattage. For example, with 500 watts you can heat a cup of water from room temperature to 100 degrees C in about 2.5 minutes if there are not too many heat losses.
For a smaller quantity, it will be faster. for half a cup 1.25 minutes, a quarter cup 0.625 minutes, an eighth of a cup 0.3125 minutes, etc.

I went down to an eighth of a cup for a reason, to show the minutes would be 0.3125 (19 seconds), and metal like copper has a specific heat capacity roughly 1/8 that of water.
A cup of water weighs in at about 236 grams, so 1/8 of that would heat up in about
19 seconds, so 240 grams of copper would heat up to 100 C in about 19 seconds.
Assuming we have 1/10 of that (24 grams of copper) it would take about 2 seconds
to heat up to 100 C.

That's at 500 watts, however. At 100 watts, 24 grams of copper in 10 seconds.
That is very roughly about the size of the transistor metal, or maybe half of that,
so say around 5 seconds. We could look closer at the actual volume to see how it fairs.


You can also add a heatsink but then we get into heat flow which is a little more involved:
Ut=k*D^2*U

where Ut is the partial of U with respect to t, and D^2 is the Laplacian, and U is the function, or simplified in one dimension:
Ut=k*Uxx

where Uxx is the second partial of the function U.

So you see how simple this is without the heat flow part, and with the heat flow part we'd have to know all the dimensions, but one thing is clear, that the heat flow does not get from the junction to the tips of the heat sink in zero time, which means the static case does not hold (junction gets much hotter than the tips of the heat sink over a short time period).

What i would suggest is some experimentation. Heat up the device without a heat sink to learn it's characteristics, then use a heat sink and take more measurements and compare.

 

[MikeMl]

All of this discussion about heat capacity presumes that you only have one start-up transient, and then don't have another one within the thermal time constant of how long it takes to cool off after the first one... It still requires a finned heat sink; not just a block of copper...

 

[MrAl]

Hi Mike,

Yes, good point. If the device is going through start up several times then we also have to calculate the cool down time and figure out if it's fast enough for the intended duty cycle. Fins and a fan would help. But we still have to make sure the thermal gradient stays small enough or the junction burns up before we get time to cool it down.

 

[Cicero]

I am able to vary the startup times, to allow for adequate cooling, so thats not an immediate problem but definitely something I will consider.

I am however now running into a few problems. The controller is working in a power limitting mode, pretty much like Mike suggested above where it'll only allow the FET to work within a set power limit. I made a capacitive load to test with, and increased the allowable power for the FET and promptly blew two. So obviously with the capacitive load now its taking strain and my calculations were off. With a large capacitive load 10000-15000uF and the times it takes to charge at 50V, I can't use the traditional methods of junction temp calculations for pulse tests, turn 'on' just takes too long. A different method for slightly longer times but staying within the pulse SOA while allowing the case temp to rise slightly also doesn't work out because of the long charge time.

My thoughts now, is to perhaps use two FETs in parallel, and heat sink them both adequately enough to allow it to turn on over a couple of seconds. What are the limitations to this, considering the long(er) switch on times, will they share the load quite evenly? Or does that only really happen under saturation (fully on) conditions?

 

[MrAl]

Hi,

Each one probably has to be individually controlled. A small series resistance to measure the current in each device, with feedback to regulate each one.

 

[Cicero]

Hi,

Each one probably has to be individually controlled. A small series resistance to measure the current in each device, with feedback to regulate each one.

I was hoping not to, because that would make it completely unfeasible

From what I've read, each one needs to have its own gate resistor, combined with decent PCB layout.

I've tried it now with two in parallel and it seems fine, nothing has blown yet, but I really need to stress test this thing before I'll be confident.

 

[MrAl]

Hello,

Series resistors also help to equalize currents, so some small values may help a lot.

Another idea is to use more than one MOSFET but have different value series resistors, and turn one MOSFET on at a time. Just for a rough example, if one MOSFET had 100 ohms, the next 10 ohms, and the next 1 ohm,
you would turn on the 100 ohm stage fully and wait until the voltage reached a certain point and then turn on the second MOSFET and then wait again, then finally the third MOSFET. The resistor values could be adjusted as needed, and because they are resistors you can get them in whatever power you need.
This is guaranteed to work because the MOSFETs will always be turned on fully while the resistors are the only things that heat up. You could even add a fourth stage if you like and set the timing after some trials or set the timing on the fly with some feedback.

 

[ChrisP58]

Have you considered using a current limited buck converter? I've done a charger for a bank of supercapacitors that way.

Even if you only use the switcher to bring the caps up to ~80% of peak, then finish it with a resistor, you would generate a lot less heat.

 

[Cicero]

Have you considered using a current limited buck converter? I've done a charger for a bank of supercapacitors that way.

Even if you only use the switcher to bring the caps up to ~80% of peak, then finish it with a resistor, you would generate a lot less heat.

Yeah Chris, thought about something similar like a PWM switch to turn on slowly, but nowadays there's just so much on the market which does it all in one ("oh, there's probably an app for that" mentality), so I thought I'd try this hot-swap method and try get rid of as many external components as possible. I also need to monitor the steady state current, make sure I deal with any faults on the power.

Looking at it from the outside it seemed too good to be true, and so far it hasn't been too bad - barring the couple of FET's I've blown. The features are great; current measurement, automatic overvoltage/overcurrent fault detection, super controlled startup, madness. Its mainly been my limited FET switching knowledge that's been the big problem for me.