<html><head></head><body><div style="color:#000; background-color:#fff; font-family:HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, sans-serif;font-size:16px"><div id="yiv0647688466"><div id="yui_3_16_0_1_1457732229592_2595"><div id="yui_3_16_0_1_1457732229592_2594" style="color:#000;background-color:#fff;font-family:HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, sans-serif;font-size:16px;"><div id="yiv0647688466"><div id="yiv0647688466yui_3_16_0_1_1457731473767_2572"><div id="yiv0647688466yui_3_16_0_1_1457731473767_2571" style="color:#000;background-color:#fff;font-family:HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, sans-serif;font-size:16px;"><div id="yiv0647688466"><div id="yiv0647688466yui_3_16_0_1_1457730728150_2607"><div id="yiv0647688466yui_3_16_0_1_1457730728150_2606" style="color:#000;background-color:#fff;font-family:HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, sans-serif;font-size:16px;"><div id="yiv0647688466"><div id="yiv0647688466yui_3_16_0_1_1457730136768_3125"><div id="yiv0647688466yui_3_16_0_1_1457730136768_3124" style="color:#000;background-color:#fff;font-family:HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, sans-serif;font-size:16px;"><div id="yiv0647688466yui_3_16_0_1_1457729256661_8666">Hi Pete,</div><div id="yiv0647688466yui_3_16_0_1_1457730136768_3126"><br clear="none"></div><div id="yiv0647688466yui_3_16_0_1_1457730136768_3127"><div id="yiv0647688466yui_3_16_0_1_1457730136768_3904">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 id="yiv0647688466yui_3_16_0_1_1457730728150_2705"><br clear="none"></div><div id="yiv0647688466yui_3_16_0_1_1457730728150_2701"><div id="yiv0647688466yui_3_16_0_1_1457731473767_2610">It is possible to bias the source such that Vsource > Vdrain (Vds < 0).</div><div id="yiv0647688466yui_3_16_0_1_1457731473767_2611">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 id="yiv0647688466yui_3_16_0_1_1457730728150_2605"><br clear="none"></div><div dir="ltr" id="yiv0647688466yui_3_16_0_1_1457731473767_2612"><div dir="ltr" id="yiv0647688466yui_3_16_0_1_1457731473767_4229">Born before Wikipedia, I still believe in books. :-)</div><div dir="ltr" id="yiv0647688466yui_3_16_0_1_1457731473767_7101"><br clear="none"></div><div dir="ltr" id="yiv0647688466yui_3_16_0_1_1457731473767_7100">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" id="yiv0647688466yui_3_16_0_1_1457731473767_12052"><br clear="none"></div><div dir="ltr" id="yiv0647688466yui_3_16_0_1_1457731473767_12054"><div id="yui_3_16_0_1_1457732229592_2683">Don't let the smoke out (at least while anybody is looking!) :-)</div><div id="yui_3_16_0_1_1457732229592_3044"><br></div><div id="yui_3_16_0_1_1457732229592_3045">Sincerely,<br></div></div></div><div id="yiv0647688466yui_3_16_0_1_1457731473767_12072">Kevin Schilf</div></div><div class="yiv0647688466qtdSeparateBR" id="yiv0647688466yui_3_16_0_1_1457729256661_8404"><br clear="none"><br clear="none"></div></div></div></div><div id="yiv0647688466yui_3_16_0_1_1457730136768_3130"> <div id="yiv0647688466yui_3_16_0_1_1457729256661_8327" style="font-family:HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, sans-serif;font-size:16px;"> <div id="yiv0647688466yui_3_16_0_1_1457729256661_8326" style="font-family:HelveticaNeue, Helvetica Neue, Helvetica, Arial, Lucida Grande, sans-serif;font-size:16px;"> <div class="yiv0647688466qtdSeparateBR" id="yiv0647688466yui_3_16_0_1_1457730136768_3129"><br clear="none"><br clear="none"></div><div class="yiv0647688466yqt3064303358" id="yiv0647688466yqt16561"><div class="yiv0647688466yqt6465671602" id="yiv0647688466yqt54586"><div class="yiv0647688466yqt0867570876" id="yiv0647688466yqt19683"><div class="yiv0647688466yqt9061048935" id="yiv0647688466yqt51373"><div dir="ltr" id="yiv0647688466yui_3_16_0_1_1457729256661_8325"> <font size="2" face="Arial"> </font><hr size="1"> <b><span style="font-weight:bold;">From:</span></b> Pete Soper via TriEmbed <triembed@triembed.org><br clear="none"> <b><span style="font-weight:bold;">To:</span></b> Shane Trent <shanedtrent@gmail.com>; "triembed@triembed.org" <triembed@triembed.org> <br clear="none"> <b><span style="font-weight:bold;">Sent:</span></b> Friday, March 11, 2016 12:38 PM<br clear="none"> <b id="yui_3_16_0_1_1457732229592_3084"><span id="yui_3_16_0_1_1457732229592_3083" style="font-weight:bold;">Subject:</span></b> Re: [TriEmbed] N-MOSFET Symbol<br clear="none"> </div> <div class="yiv0647688466y_msg_container" id="yiv0647688466yui_3_16_0_1_1457729256661_8607"><br clear="none"><div id="yiv0647688466"><div id="yiv0647688466yui_3_16_0_1_1457729256661_8606">
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 class="yiv0647688466yqt6102815973" id="yiv0647688466yqt89537"><div class="yiv0647688466moz-cite-prefix" id="yiv0647688466yui_3_16_0_1_1457729256661_8608">On 03/11/2016 12:36 PM, Shane Trent
wrote:<br clear="none">
</div>
<blockquote id="yiv0647688466yui_3_16_0_1_1457729256661_8611" type="cite">
<div dir="ltr" id="yiv0647688466yui_3_16_0_1_1457729256661_8610">Pete,
<div id="yiv0647688466yui_3_16_0_1_1457729256661_8609"><br clear="none">
</div>
<div>Sorry,I do not understand the question. </div>
<div><br clear="none">
</div>
<div id="yiv0647688466yui_3_16_0_1_1457729256661_8612">Shane</div>
</div>
<br clear="none">
<div class="yiv0647688466gmail_quote" id="yiv0647688466yui_3_16_0_1_1457729256661_8615">
<div dir="ltr">On Fri, Mar 11, 2016 at 11:50 AM Pete Soper <<a rel="nofollow" shape="rect" ymailto="mailto:pete@soper.us" target="_blank" href="mailto:pete@soper.us">pete@soper.us</a>>
wrote:<br clear="none">
</div>
<blockquote class="yiv0647688466gmail_quote" id="yiv0647688466yui_3_16_0_1_1457729256661_8614" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">
<div id="yiv0647688466yui_3_16_0_1_1457729256661_8613"> 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" target="_blank" href="http://datasheet.octopart.com/FQP30N06L-Fairchild-datasheet-82531.pdf">http://datasheet.octopart.com/FQP30N06L-Fairchild-datasheet-82531.pdf</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 class="yiv0647688466gmail_quote">
<div dir="ltr">On Thu, Mar 10, 2016 at 11:51 PM Pete
Soper <<a rel="nofollow" shape="rect" ymailto="mailto:pete@soper.us" target="_blank" href="mailto:pete@soper.us">pete@soper.us</a>>
wrote:<br clear="none">
</div>
<blockquote class="yiv0647688466gmail_quote" 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 class="yiv0647688466gmail_quote">
<div dir="ltr">On Thu, Mar 10, 2016 at 4:24
PM Pete Soper via TriEmbed <<a rel="nofollow" shape="rect" ymailto="mailto:triembed@triembed.org" target="_blank" href="mailto:triembed@triembed.org">triembed@triembed.org</a>>
wrote:<br clear="none">
</div>
<blockquote class="yiv0647688466gmail_quote" 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" target="_blank" href="https://en.wikipedia.org/wiki/Power_MOSFET#Body_diode">https://en.wikipedia.org/wiki/Power_MOSFET#Body_diode</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" target="_blank" href="https://www.fairchildsemi.com/datasheets/FQ/FQP30N06L.pdf">https://www.fairchildsemi.com/datasheets/FQ/FQP30N06L.pdf</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" ymailto="mailto:kschilf@yahoo.com" target="_blank" href="mailto:kschilf@yahoo.com">kschilf@yahoo.com</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 size="2" face="Arial">
</font><hr size="1"> <b><span style="font-weight:bold;">From:</span></b>
Pete Soper via TriEmbed <a rel="nofollow" shape="rect" ymailto="mailto:triembed@triembed.org" target="_blank" href="mailto:triembed@triembed.org"><triembed@triembed.org></a><br clear="none">
<b><span style="font-weight:bold;">To:</span></b>
<a rel="nofollow" shape="rect" ymailto="mailto:triembed@triembed.org" target="_blank" href="mailto:triembed@triembed.org">triembed@triembed.org</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">
Triangle, NC Embedded
Computing mailing list<br clear="none">
<a rel="nofollow" shape="rect" ymailto="mailto:TriEmbed@triembed.org" target="_blank" href="mailto:TriEmbed@triembed.org">TriEmbed@triembed.org</a><br clear="none">
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TriEmbed web site: <a rel="nofollow" shape="rect" target="_blank" href="http://triembed.org/">http://TriEmbed.org</a><br clear="none">
</div>
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<br clear="none">
</div>
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</div>
</div>
</div>
</blockquote>
<br clear="none">
</div>
_______________________________________________<br clear="none">
Triangle, NC Embedded Computing mailing
list<br clear="none">
<a rel="nofollow" shape="rect" ymailto="mailto:TriEmbed@triembed.org" target="_blank" href="mailto:TriEmbed@triembed.org">TriEmbed@triembed.org</a><br clear="none">
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TriEmbed web site: <a rel="nofollow" shape="rect" target="_blank" href="http://triembed.org/">http://TriEmbed.org</a><br clear="none">
</blockquote>
</div>
</div>
</div>
</blockquote>
<br clear="none">
</div>
</blockquote>
</div>
</blockquote>
<br clear="none">
</div>
</blockquote>
</div>
</blockquote></div>
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