<div dir="ltr">Pete,<div><br></div><div>In this case the body diode will not conduct unless the inductive kick generates a voltage high enough to cause a breakdown failure of the FET or body diode. A freewheel/snubber diode's job is to provide a current path between the inductor terminals to allow the current to recirculate (can be called a recirculation diode) until the losses in the circuit dissipate the energy stored in the magnetic field. The FET body diode does not provide a path between the inductor terminals. If the body diode could conduct (say you instantly reverse the body diode polarity) then the coil is simply energized through a forward biased diode and says on.</div><div><br></div><div>Interestingly you can kill a MOSFET at relatively low currents if using low gate voltages. A MOSFET is actually not a single switching element but a large array of parallel devices. Under typical use conditions (full gate drive) each element in the FET has a resistance with a positive temperature coefficient meaning that the elements carrying the most current get hotter than their neighbors and therefore develop a higher resistance and this reduces the current through them, allowing the large arrange of elements to effectively share their load. This is the big benefit that MOSFETs have over BJTs. BJTs have a negative temperature coefficient and require external bias resistors to prevent one transistor from getting hot, reducing its resistance, taking more current and getting hotter until the smoke gets out. </div><div><br></div><div>But at low gate voltages, MOSFETs have a negative temperature coefficient like BJTs so you can get localized heating of a cell, lowering it's resistance, stealing current from neighbors cells until the cell has a meltdown. This is why using MOSFET's in linear applications must be done with great care (ensuring you stay above the Zero Temperature Coefficient point. Google Spirito Effect for more details that you could ever want. </div><div><br></div><div>Shane</div></div><br><div class="gmail_quote"><div dir="ltr">On Fri, Mar 11, 2016 at 6:31 PM Pete Soper via TriEmbed <<a href="mailto:triembed@triembed.org">triembed@triembed.org</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div bgcolor="#FFFFFF" text="#000000">
Double oops. Yes, BVDss is exactly what I thought: Vdrain - Vsource.
Beyond this there is conduction, right? For this particular part
this happens in an "all at once" fashion and the heat is spread
evenly so it takes a relatively large, sustained current to risk
damaging the chip, right?<br>
So if the body diode has avalanched it's close to a short circuit
and so the current goes like this:<br>
<br>
<pre> +V <<----|</pre>
<pre> coil |</pre>
<pre> source |</pre>
<pre> drain |</pre>
<pre> ground >>|
</pre>
Now the body diode is a snubber/freewheeler/<insert five other
synonyms>. If the supply voltage, inductance, turn off time ,
transistor selection are all OK then an external diode is a waste of
money.<br>
<br>
What am I missing here?</div><div bgcolor="#FFFFFF" text="#000000"><br>
<br>
-Pete</div><div bgcolor="#FFFFFF" text="#000000"><br>
<br>
<div>On 03/11/2016 04:31 PM,
<a href="mailto:kschilf@yahoo.com" target="_blank">kschilf@yahoo.com</a> wrote:<br>
</div>
<blockquote type="cite">
<div style="color:#000;background-color:#fff;font-family:HelveticaNeue,Helvetica Neue,Helvetica,Arial,Lucida Grande,sans-serif;font-size:16px">
<div>
<div>
<div style="color:#000;background-color:#fff;font-family:HelveticaNeue,Helvetica Neue,Helvetica,Arial,Lucida Grande,sans-serif;font-size:16px">
<div>
<div>
<div style="color:#000;background-color:#fff;font-family:HelveticaNeue,Helvetica Neue,Helvetica,Arial,Lucida Grande,sans-serif;font-size:16px">
<div>
<div>
<div style="color:#000;background-color:#fff;font-family:HelveticaNeue,Helvetica Neue,Helvetica,Arial,Lucida Grande,sans-serif;font-size:16px">
<div>
<div>
<div style="color:#000;background-color:#fff;font-family:HelveticaNeue,Helvetica Neue,Helvetica,Arial,Lucida Grande,sans-serif;font-size:16px">
<div>Hi
Pete,</div>
<div><br clear="none">
</div>
<div>
<div>The
voltage at each of the three
terminals of the transistor (gate,
drain, source) is a function of the
rest of the circuit. You can bias
(set voltages, and draw currents)
the transistor anyway you want, once
you understand its behavior (and
limits) at whatever operating point
you set.</div>
<div><br clear="none">
</div>
<div>
<div>It
is possible to bias the source
such that Vsource > Vdrain (Vds
< 0).</div>
<div>BVDss
the maximum voltage difference
(Vdrain - Vsource) exerted before
you possibly damage the part.
This value is temperature
dependent.<br clear="none">
</div>
</div>
<div><br clear="none">
</div>
<div dir="ltr">
<div dir="ltr">Born
before Wikipedia, I still believe
in books. :-)</div>
<div dir="ltr"><br clear="none">
</div>
<div dir="ltr">Since
textbooks ain't cheap, borrow a
sophomore level circuits text
(NCSU library, etc.). Peruse the
chapter on BJT's and MOSFET's.
That should clear up some of the
mystery. :-)</div>
<div dir="ltr"><br clear="none">
</div>
<div dir="ltr">
<div>Don't
let the smoke out (at least
while anybody is looking!) :-)</div>
<div><br>
</div>
<div>Sincerely,<br>
</div>
</div>
</div>
<div>Kevin
Schilf</div>
</div>
<div><br clear="none">
<br clear="none">
</div>
</div>
</div>
</div>
<div>
<div style="font-family:HelveticaNeue,Helvetica Neue,Helvetica,Arial,Lucida Grande,sans-serif;font-size:16px">
<div style="font-family:HelveticaNeue,Helvetica Neue,Helvetica,Arial,Lucida Grande,sans-serif;font-size:16px">
<div><br clear="none">
<br clear="none">
</div>
<div>
<div>
<div>
<div>
<div dir="ltr">
<font face="Arial" size="2"> </font>
<hr size="1"> <b><span style="font-weight:bold">From:</span></b>
Pete Soper via TriEmbed
<a href="mailto:triembed@triembed.org" target="_blank"><triembed@triembed.org></a><br clear="none">
<b><span style="font-weight:bold">To:</span></b>
Shane Trent
<a href="mailto:shanedtrent@gmail.com" target="_blank"><shanedtrent@gmail.com></a>;
<a href="mailto:triembed@triembed.org" target="_blank">"triembed@triembed.org"</a>
<a href="mailto:triembed@triembed.org" target="_blank"><triembed@triembed.org></a>
<br clear="none">
<b><span style="font-weight:bold">Sent:</span></b>
Friday, March 11, 2016 12:38
PM<br clear="none">
<b><span style="font-weight:bold">Subject:</span></b>
Re: [TriEmbed] N-MOSFET Symbol<br clear="none">
</div>
<div><br clear="none">
<div>
<div>
If the transistor shorts
out at 60 volts it's hard
to get the source above 60
volts, right?<br clear="none">
-Pete<br clear="none">
<br clear="none">
<div>
<div>On 03/11/2016 12:36
PM, Shane Trent wrote:<br clear="none">
</div>
<blockquote type="cite">
<div dir="ltr">Pete,
<div><br clear="none">
</div>
<div>Sorry,I do not
understand the
question. </div>
<div><br clear="none">
</div>
<div>Shane</div>
</div>
<br clear="none">
<div>
<div dir="ltr">On
Fri, Mar 11, 2016
at 11:50 AM Pete
Soper <<a rel="nofollow" shape="rect" href="mailto:pete@soper.us" target="_blank"><a href="mailto:pete@soper.us" target="_blank">pete@soper.us</a></a>> wrote:<br clear="none">
</div>
<blockquote style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div>
That was very
clear about the
other transistor
forward
conducting. One
last question.
Here's the
datasheet for
the transistor
Brian Grawburg
started us with:<br clear="none">
<br clear="none">
<a rel="nofollow" shape="rect" href="http://datasheet.octopart.com/FQP30N06L-Fairchild-datasheet-82531.pdf" target="_blank"><a href="http://datasheet.octopart.com/FQP30N06L-Fairchild-datasheet-82531.pdf" target="_blank">http://datasheet.octopart.com/FQP30N06L-Fairchild-datasheet-82531.pdf</a></a><br clear="none">
<br clear="none">
In the context
of the simple
case of one of
these
transistors
driving a motor
what does it
mean for the
drain-source
breakdown
voltage BVdss to
be the same as
the max
drain-source
voltage Vdss
together with
the avalanche
current and
diode recovery
specs?</div>
<div><br clear="none">
<br clear="none">
<br clear="none">
-Pete</div>
<div><br clear="none">
<br clear="none">
<br clear="none">
<div>On
03/11/2016
10:40 AM,
Shane Trent
wrote:<br clear="none">
</div>
<blockquote type="cite">
<div dir="ltr">Pete,
<div><br clear="none">
</div>
<div>I think
it is easier
if you look at
a half-bridge
using just two
transistors
with a
bi-polar power
supply. </div>
<div><br clear="none">
</div>
<div>Let's
assume we have
+/12V on the
power rails
with one
terminal of
the motor
grounded and
the other
connected to
your
half-bridge
output. We run
the motor
forward by
turning on the
top FET and
applying +12V
to the motor
terminal and
run it
backward by
turning on the
bottom FET and
applying -12V
to the motor
output. In
this case when
you cut the
power to the
motor the body
diode of the
FET that was
NOT conducting
acts as the
catch diode
for the motor
(the body
diode of the
FET that was
used to apply
power does not
conduct any
current). So
if you decide
to drive the
motor in only
one direction
and remove one
of the FETs,
you will have
to add a catch
diode since
you removed
the body diode
of the 2nd FET
which was
acting as your
catch diode. <span style="line-height:1.5">This is why h-bridge and half-bridge circuits
with BJTs
include catch
diodes and
ones with
MOSFET
typically do
not.</span><span style="line-height:1.5"> </span><span style="line-height:1.5">I like
to imaging my
explanations
makes sense
but I am never
sure. So, did
that make
sense to you?</span></div>
<div><br clear="none">
</div>
<div>You can
use external
catch diodes
with a MOSFET
full or
half-bridge
but you need
to ensure the
external
diodes have a
lower Vf than
the FET body
diodes to
ensure the
external
diodes conduct
before the
body diodes.
You may also
see fast
external
diodes used
with a FET to
clamp
inductive
current spikes
faster than
the FET body
diode can
conduct,
clamping the
current spikes
a lower
voltage.</div>
<div><br clear="none">
</div>
<div>Shane</div>
</div>
<br clear="none">
<div>
<div dir="ltr">On
Thu, Mar 10,
2016 at 11:51
PM Pete Soper
<<a rel="nofollow" shape="rect" href="mailto:pete@soper.us" target="_blank"><a href="mailto:pete@soper.us" target="_blank">pete@soper.us</a></a>> wrote:<br clear="none">
</div>
<blockquote style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div> Out in
the world
there are
droves of H
bridge motor
control
circuits with
beefy MOSFETS
and no diodes
in sight
except the
body diodes.
How is that
possible?</div>
<div><br clear="none">
-Pete</div>
<div><br clear="none">
<div>On
03/10/2016
05:59 PM,
Shane Trent
wrote:<br clear="none">
</div>
<blockquote type="cite">
<div dir="ltr">Pete,
<div><br clear="none">
</div>
<div>I believe
you still need
the snubber
even with the
body diode. A
snubber is
typically
placed across
the inductor
(motor or
solenoid or
relay coil)
and not across
the switching
element. </div>
<div><br clear="none">
</div>
<div>For
example, if
you turn off
an N-FET
supplying
several amps
to a large
solenoid, when
you turn the
FET off the
collapsing
magnetic field
of the coil
will cause the
voltage across
the solenoid
terminals to
increase. The
N-FET will
neither
forward
conduct or
reverse
conduct via
the body diode
until the
transistors
breakdown
voltage (Vds
max) is
exceeded and
the FET
fails. </div>
<div><br clear="none">
</div>
<div>The
tradeoff with
using a diode
snubber (it
seems to be
more of a
voltage clamp)
across the
coil is that
it will act as
a catch diode
or
recirculation
diode and
cause the
solenoid to
turn off more
slowly. You
can strike a
balance
between
voltage and
turn-off speed
by combining a
regular diode
and Zener
diode to allow
the voltage to
increase
across the
solenoid
without
exceeding the
FET's maximum
voltage
rating. But
there are MANY
ways to design
inductive
clamps. </div>
<div><br clear="none">
</div>
<div>Shane<br clear="none">
<br clear="none">
<div>
<div dir="ltr">On
Thu, Mar 10,
2016 at 4:24
PM Pete Soper
via TriEmbed
<<a rel="nofollow" shape="rect" href="mailto:triembed@triembed.org" target="_blank"><a href="mailto:triembed@triembed.org" target="_blank">triembed@triembed.org</a></a>>
wrote:<br clear="none">
</div>
<blockquote style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div> This may
come across as
high-minded,
but really I
just want to
pass it along
as something
that's
hopefully on
target. This
topic forced
me to go study
and read and
I'm looking
for
confirmation
I'm not
misleading
anybody.<br clear="none">
<br clear="none">
The specific
motor control
application
that I think
might be
relevant to
Brian's kids
is treated
with the
"freewheeling
diode"s link
on this page:<br clear="none">
<br clear="none">
<a rel="nofollow" shape="rect" href="https://en.wikipedia.org/wiki/Power_MOSFET#Body_diode" target="_blank"><a href="https://en.wikipedia.org/wiki/Power_MOSFET#Body_diode" target="_blank">https://en.wikipedia.org/wiki/Power_MOSFET#Body_diode</a></a><br clear="none">
<br clear="none">
Here is the
transistor
Brian's kids
are going to
use:<br clear="none">
<br clear="none">
<a rel="nofollow" shape="rect" href="https://www.fairchildsemi.com/datasheets/FQ/FQP30N06L.pdf" target="_blank"><a href="https://www.fairchildsemi.com/datasheets/FQ/FQP30N06L.pdf" target="_blank">https://www.fairchildsemi.com/datasheets/FQ/FQP30N06L.pdf</a></a><br clear="none">
<br clear="none">
This
transistor can
handle 32 amps
of avalanche
current and is
specifically
designed for
inductive
loads. The
body diode in
this
transistor
qualifies as a
snubber when a
motor is
turned off and
is
"freewheeling".
The energy
will go
straight to
ground without
incident.
Searching for
this part
number and
"motor" gives
a number of
hits where
hobby folks
are putting
rectifiers
across the
motor
windings. This
strikes me as
redundant. (At
this point one
might think
"but wait,
this
transistor is
only rated at
60 volts
source to
drain". But
when the coil
field
collapses and
the source
voltage shoots
up the
transistor
junction
"avalanches"
and begins to
conduct
current very
quickly,
yanking the
voltage right
down close to
ground. The
"avalanche
feature" of
the transistor
is
manufacturing
technique that
avoids "hot
spots" that
might ruin the
part.)<br clear="none">
<br clear="none">
Sorry for
assuming we
more or less
knew the
application:
wimpy little
low power
motors with
massive
overkill
components.
And I'm
probably
running the
risk of
causing folks
to blow up
their parts by
not simply
recommending a
separate
snubber. It
may be going
too far to
suggest that
the body diode
should be
included in
the schematic
when it can be
considered a
snubber, but I
confess this
the frame of
mind I'd
developed
before the
discussion
woke me up.
I'll be
reading
datasheets
more carefully
in the future!<br clear="none">
<br clear="none">
Ah, but we
haven't
mentioned
improperly
switching the
transistor and
having it sit
in its linear
zone. I claim
the local
record for how
fast a MOSFET
can desolder
itself when
this happens
at six amperes
to a small
SMD. :-)</div>
<div><br clear="none">
<br clear="none">
-Pete</div>
<div><br clear="none">
<br clear="none">
<br clear="none">
<div>On
03/09/2016
06:44 PM, <a rel="nofollow" shape="rect" href="mailto:kschilf@yahoo.com" target="_blank"><a href="mailto:kschilf@yahoo.com" target="_blank">kschilf@yahoo.com</a></a>
wrote:<br clear="none">
</div>
<blockquote type="cite">
<div style="color:#000;background-color:#fff;font-family:HelveticaNeue,Helvetica Neue,Helvetica,Arial,Lucida Grande,sans-serif;font-size:16px">
<div>Hi Pete,</div>
<div><br clear="none">
</div>
<div>Good note
about warning
flags.</div>
<div><br clear="none">
</div>
<div dir="ltr">I
have no idea
about the
application.
Current in an
inductor can
not change
instantaneously.
If you are
going to
interrupt the
circuit, you
should provide
a path to
allow the
inductor
current to
continue
(catch diode
in a switching
power supply)
or diminish
(diode across
a relay
winding),
etc. If not,
you let Mr.
Murphy
determine
where the
energy will
go, sometimes
with exciting
consequences.
:-)</div>
<div dir="ltr"><br clear="none">
</div>
<div dir="ltr">Sincerely,</div>
<div dir="ltr">Kevin
Schilf<br clear="none">
</div>
<div><span></span></div>
<div><br clear="none">
<br clear="none">
</div>
<div style="display:block">
<div style="font-family:HelveticaNeue,Helvetica Neue,Helvetica,Arial,Lucida Grande,sans-serif;font-size:16px">
<div style="font-family:HelveticaNeue,Helvetica Neue,Helvetica,Arial,Lucida Grande,sans-serif;font-size:16px">
<div dir="ltr">
<font face="Arial" size="2"> </font>
<hr size="1">
<b><span style="font-weight:bold">From:</span></b>
Pete Soper via
TriEmbed <a rel="nofollow" shape="rect" href="mailto:triembed@triembed.org" target="_blank"><a href="mailto:triembed@triembed.org" target="_blank"><triembed@triembed.org></a></a><br clear="none">
<b><span style="font-weight:bold">To:</span></b>
<a rel="nofollow" shape="rect" href="mailto:triembed@triembed.org" target="_blank"><a href="mailto:triembed@triembed.org" target="_blank">triembed@triembed.org</a></a>
<br clear="none">
<b><span style="font-weight:bold">Sent:</span></b>
Wednesday,
March 9, 2016
5:25 PM<br clear="none">
<b><span style="font-weight:bold">Subject:</span></b>
Re: [TriEmbed]
N-MOSFET
Symbol<br clear="none">
</div>
<div><br clear="none">
I'm pretty
sure about 70%
of Brian's
interest in
this subject
involves <br clear="none">
dealing with
inductive
loads. The
body diode in
the schematic
symbol is <br clear="none">
a merciful
hint. If his
kids can
remember that
the lack of a
body diode <br clear="none">
is a red flag
they might
avoid blowing
up their BJTs
or adding
redundant <br clear="none">
components.<br clear="none">
<br clear="none">
-Pete
<div><br clear="none">
<br clear="none">
<br clear="none">
_______________________________________________<br clear="none">
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