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'[EE] Mosfet question'
2007\12\28@180729 by Enki

picon face

       I'm doing some tests with the FDN337N Fairchild Mosfet.
       It is working switching a 24V @ 500mA relay in series with a 1 ohm
resistor. The relay has the usual 1N4007 diode.
       If I don't use the diode, the back EMF when the Mosfet switches off
goes up to ~40V, the Mosfet breakdown voltage, for 10mS.
       Just out of curiosity, I let the Mosfet switch the relay each second
without the diode, some 100.000 times, 50% duty cycle.
       The Mosfet seems not care about the energy absorved during the
breakdown period. It warms up to some 40C.

       My question is:

       Is there any expected long term problem running the circuit without
the diode?
       
       The design WILL use the diode, my question is just to satisfy my
curiosity.

       Thanks and happy 2008!
       Mark Jordan - PY3SS

2007\12\28@182113 by Sean Breheny

face picon face
Hi Mark,

Power MOSFETs have an avalanche energy rating. This, as I understand
it, is a measure of how much overvoltage they can tolerate. In other
words, if you apply a voltage higher than the breakdown voltage, then
the integral of the voltage times the current over time must not
exceed the avalanche energy or you may damage the FET. This rating is
not so much meant to eliminate the need for diodes as it is to take
care of the fact that even with diodes, there will always be very
brief spikes over the breakdown voltage, even when the diode is
working, due to stray inductance in the diode and FET leads as well as
finite diode speed.

In your case, if you take the inductance of the relay, multiply by
0.5*I^2 where I is the current in the coil at turn-off, this will give
you the max energy available to attack the MOSFET. If this is lower
than the avalanche energy, it should be OK. I believe the mechanism of
damage is simply thermal.

One final complicating factor is that there are usually several
different avalanche ratings, depending on whether it is a single pulse
or repetitive.

Sean


On Dec 28, 2007 6:07 PM, Enki <spam_OUTenkitecTakeThisOuTspamgmail.com> wrote:
{Quote hidden}

> -

2007\12\28@183541 by Bob Blick

face picon face
Hi Mark,

You are looking at avalanche operation, and there has
been a fair amount of study paid to it with regard to
mosfets. Some manufacturers go so far as to have an
entire line of them touted as "avalanche mosfets".

But there's no reason why any mosfet can't be operated
in avalanche mode as long as there are two parameters
that are not violated. One is to make sure that the
drain-gate voltage rating is equal to or higher than
the drain-source rating. As far as I know that is
always the case with modern parts. The other is to
insure that you don't exceed the safe operating area
of the device. I have not looked at the ratings of the
part you mentioned, but that's the curve I'd look at
if I needed to use a part that way. Is 500mA at 40V
within the curve?

And of course EMI is usually another reason to use a
snubber diode, but I know that's not the reason for
the question.

Cheerful regards,

Bob



--- Enki <.....enkitecKILLspamspam@spam@gmail.com> wrote:

{Quote hidden}

2007\12\28@191342 by Bob Axtell

face picon face
Enki wrote:
> I'm doing some tests with the FDN337N Fairchild Mosfet.
>        It is working switching a 24V @ 500mA relay in series with a 1 ohm
> resistor. The relay has the usual 1N4007 diode.
>        If I don't use the diode, the back EMF when the Mosfet switches off
> goes up to ~40V, the Mosfet breakdown voltage, for 10mS.
>        Just out of curiosity, I let the Mosfet switch the relay each second
> without the diode, some 100.000 times, 50% duty cycle.
>        The Mosfet seems not care about the energy absorved during the
> breakdown period. It warms up to some 40C.
>
>        My question is:
>
>        Is there any expected long term problem running the circuit without
> the diode?
>        
>        The design WILL use the diode, my question is just to satisfy my
> curiosity.
>  
Others will poo-poo me, but I have found that 1N4007s don't save MOSFETs
from spikes, because their
turn-on time is too slow. I use SMB varistors or transorbers at 26V to
protect the MOSFETS from relay spiking.
You can see this unsuppressed spike with a very fast analog scope.

Since doing this several years ago, I can't remember a single MOSFET
failure since.

--Bob

>        Thanks and happy 2008!
>        Mark Jordan - PY3SS
>
>  

2007\12\28@194401 by Enki

picon face

       I'm using a 100nF capacitor in parallel with the Mosfet.
       It seems to eliminate the spike.

       Mark Jordan

On 28 Dec 2007 at 17:13, Bob Axtell wrote:

{Quote hidden}

> --

2007\12\28@200035 by Lamebert

flavicon
face
On Fri, 28 Dec 2007 21:07:02 -0200, you wrote:

>        I'm doing some tests with the FDN337N Fairchild Mosfet.
>        It is working switching a 24V @ 500mA relay in series with a 1 ohm
>resistor. The relay has the usual 1N4007 diode.

>        Is there any expected long term problem running the circuit without
>the diode?

You can't say. That MOSFET has no avalanche rating and is presumably not
designed or tested for avalanche operation. You don't know how close to
destroying the device you are.

If you could run it on say 36v for a few minutes without killing it then I
would say it would probably run forever on 24v, it doesn't mean the next
FDN337N would though.
--

2007\12\28@200534 by Chris Smolinski

flavicon
face
{Quote hidden}

You'll want to use a faster diode than a 1N4007.  Also, Google "snubber".

--

---
Chris Smolinski
Black Cat Systems
http://www.blackcatsystems.com

2007\12\28@212242 by Sean Breheny

face picon face
I'm really curious about the idea that slow diodes aren't fast enough
for this task. As I understood it, all silicon diodes are very fast to
turn on (nanoseconds), and all the difference comes in the turn-off
time (Trr - Reverse Recovery Time). That can cause a problem in many
cases, particularly when it may cause high-side to low-side
shoot-thru, but I don't think that should be a problem here.

Of course, it's not 100%, but I just did a spice sim using Diodes,
Inc.'s model of the 1N4001 and the turn-on is faster than 10ns, turn
off is 7 microseconds. 1N914 shows about the same for turn-on, and
turn off is 35 ns.

Sean


On Dec 28, 2007 8:05 PM, Chris Smolinski <csmolinskispamKILLspamblackcatsystems.com> wrote:
> You'll want to use a faster diode than a 1N4007.  Also, Google "snubber".
>
> --
>
> ---
> Chris Smolinski
> Black Cat Systems
> http://www.blackcatsystems.com
>
> -

2007\12\28@234536 by Spehro Pefhany

picon face
At 09:22 PM 12/28/2007, you wrote:
>I'm really curious about the idea that slow diodes aren't fast enough
>for this task. As I understood it, all silicon diodes are very fast to
>turn on (nanoseconds), and all the difference comes in the turn-off
>time (Trr - Reverse Recovery Time). That can cause a problem in many
>cases, particularly when it may cause high-side to low-side
>shoot-thru, but I don't think that should be a problem here.
>
>Of course, it's not 100%, but I just did a spice sim using Diodes,
>Inc.'s model of the 1N4001 and the turn-on is faster than 10ns, turn
>off is 7 microseconds. 1N914 shows about the same for turn-on, and
>turn off is 35 ns.
>
>Sean

An oscilloscope would be better. The standard SPICE model does not accurately
model forward recovery characteristics.

Forward recovery time for a 1N400x is typically in the 100ns range at 100mA
down to 2V. A cap in addition to a diode with stray capacitance and
distributed
capacitance will limit the dv/dt, which is generally a Good Thing.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
.....speffKILLspamspam.....interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com



2007\12\29@045211 by Apptech

face
flavicon
face
> > I'm using a 100nF capacitor in parallel with the Mosfet.
>> It seems to eliminate the spike.

The coils 0.5Li^2 energy will transfer to the capacitor as
0.5CV^2.
As long as C >= L.i^2/V^2 where i is the interrupted current
and V is the maximum acceptable voltage then this will allow
you to then dissipate the energy "at your leisure". You may
(will) get a degree of ringing which may be unacceptable and
the capacitor will "slug" (slow) the magnetic decay and thus
increase release time if it is a solenoid or relay. The
capacitor ESR will change things slightly but probably not
enough to matter here.

A diode will dissipate energy via the coil resistance but
greatly slow the field collapse.

A resistor in series with the diode will allow a V = I.R
increase in voltage to some acceptable limit (here only 30V
all included due to FDN377 rating) and reduce dissipation
time.


       Russell







2007\12\29@151157 by Apptech

face
flavicon
face
> You are looking at avalanche operation, and there has
> been a fair amount of study paid to it with regard to
> mosfets. Some manufacturers go so far as to have an
> entire line of them touted as "avalanche mosfets".

The reason this is done is that special design care is taken
to ensure they work properly in this mode.

> But there's no reason why any mosfet can't be operated
> in avalanche mode as long as there are two parameters
> that are not violated. One is to make sure that the
> drain-gate voltage rating is equal to or higher than
> the drain-source rating. As far as I know that is
> always the case with modern parts. The other is to
> insure that you don't exceed the safe operating area
> of the device.

The trick with avalanche rating a part is to distribute the
dissipated energy evenly enough to prevent localised energy
concentration causing punch through. You know that they have
failed in your application when Rdson is unchanging and is
ohms to a few hundred ohms when there is no power on the
FET. Sometimes there is a small hole in the front of the
case as an added giveaway.

MOSFETs are usually deemed to be immune to the second
breakdown effects that cause problems in bipolar devices
(although this is not universally the case). While there is
some black magic involved, second breakdown is largely held
to be a localised thermal effect where current concentration
cause increased temperatures which cause current
concentration which cause ... oops.

However, when operated in avalanche mode MOSFETs that are
not designed to rigorously distribute the current evenly
over their multiple subdevices tend to concentrate current
at selected sites with results that are not after the event
especially well distinguishable from having been
second-breakdowned. As SOA is not liable to be well defined
or defined at all in avalanche mode in a non avalanche
specified part YMMV and probably will if operating an
unrated part this way.



           Russell

2007\12\29@191051 by Sean Breheny

face picon face
Hi Russell,

That's really interesting - I was not aware that avalanche operation
was limited by second-breakdown type effects.

MOSFETs have a negative temperature coefficient (increased temp gives
less current) in the fully-on state (ohmic) and a positive temperature
coefficient in the linear operating range (partially on - channel not
inverted the whole way across between drain and source). This means
that it is OK to parallel MOSFETs if they are to be operated in switch
applications but not necessarily OK if they are going to spend
significant time in the linear region.

I have been bitten by this numerous times. Once, I made a
constant-current battery discharger and used multiple FETs in parallel
for the very reason that I thought they could always be paralleled.
One got much hotter than the rest.

A more interesting case of this I ran into more recently: I was using
a pair of super low RdsON FETs in parallel in a 20kHz PWM application
at around 80 Amps. I found that one of them would get much hotter than
the other, even though the majority of the time they were purely ON or
OFF. Amazingly, what I discovered (or at least hypothesized based on
the data) is that the turn-on/off transient power dissipation was
comparable to the I2R losses and since the transient power dissipation
happens in linear mode, this means that during turn-on/turn-off, the
hotter FET would take more of the current (effectively turn on faster,
preventing the other from ever seeing a very large voltage and current
at the same time). So, when paralleling FETs, sometimes low RdsON is a
bad thing! I ended up solving the problem by turning the FETs on even
faster, to keep the switching losses below the I2R losses, but that
may not always be possible, in which case one may have to resort to
higher RdsON or slower PWM.

Sean


On Dec 29, 2007 9:49 AM, Apptech <EraseMEapptechspam_OUTspamTakeThisOuTparadise.net.nz> wrote:
{Quote hidden}

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