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Inductor current can't be suddenly cutoff if no freewheel diode is attached

Started by Patrick Chung September 2, 2013
Inductor current can't be suddenly cutoff  if no freewheel diode is attached. New tutorial is ready

http://www.cirvirlab.com/index.php/tutorials/105-r-l-circuit-with-mosfet-as-switch-and-freewheel-diode.html 

and online simulation

http://www.cirvirlab.com/simulation/r-l_circuit_with_mosfet_as_switch_and_freewheel_diode.php
On Mon, 02 Sep 2013 05:49:52 -0700, Patrick Chung <pchung705@gmail.com>  
wrote:

> Inductor current can't be suddenly cutoff if no freewheel diode is > attached. New tutorial is ready > > http://www.cirvirlab.com/index.php/tutorials/105-r-l-circuit-with-mosfet-as-switch-and-freewheel-diode.html > > and online simulation > > http://www.cirvirlab.com/simulation/r-l_circuit_with_mosfet_as_switch_and_freewheel_diode.php
You may have noticed that a single diode takes quite a while to get the current back down to zero. Try two, or even three, in series and check the time.
On Mon, 2 Sep 2013 05:49:52 -0700 (PDT), Patrick Chung
<pchung705@gmail.com> wrote:

>Inductor current can't be suddenly cutoff if no freewheel diode is attached. New tutorial is ready > >http://www.cirvirlab.com/index.php/tutorials/105-r-l-circuit-with-mosfet-as-switch-and-freewheel-diode.html > >and online simulation > >http://www.cirvirlab.com/simulation/r-l_circuit_with_mosfet_as_switch_and_freewheel_diode.php
You say:
>It is important to notice that without freewheel diode in parallel with the inductor, > sudden cutoff would damage the mosfet or other switch.
This is not generally true. Look at the circuit upper left corner. http://img69.imageshack.us/img69/2623/6vsv.jpg When the current through the FET is cut off, a large voltage spike will build up and load the HV multiplier. You have to take care that the voltage does not rise faster and higher than the FET can handle. So the turn off peak is used to generate a high voltage. A freewheel diode would supress the turn off peak. w.
On Mon, 02 Sep 2013 16:34:17 +0200, Helmut Wabnig <hwabnig@.- ---
-.dotat> wrote:

>On Mon, 2 Sep 2013 05:49:52 -0700 (PDT), Patrick Chung ><pchung705@gmail.com> wrote: > >>Inductor current can't be suddenly cutoff if no freewheel diode is attached. New tutorial is ready >> >>http://www.cirvirlab.com/index.php/tutorials/105-r-l-circuit-with-mosfet-as-switch-and-freewheel-diode.html >> >>and online simulation >> >>http://www.cirvirlab.com/simulation/r-l_circuit_with_mosfet_as_switch_and_freewheel_diode.php > > > >You say: >>It is important to notice that without freewheel diode in parallel with the inductor, >> sudden cutoff would damage the mosfet or other switch. > > >This is not generally true. > > >Look at the circuit upper left corner. >http://img69.imageshack.us/img69/2623/6vsv.jpg > >When the current through the FET is cut off, >a large voltage spike will build up and load the HV multiplier. >You have to take care that the voltage does not rise faster and higher >than the FET can handle. So the turn off peak is used to generate a >high voltage. A freewheel diode would supress the turn off peak. > >w.
Your example is a boost converter. C2 limits the peak voltage on the transistors (MFT1 and T1 sure is a weird configuration :-) On the first "pop"... 0.5*C2*(Vc2)^2 = 0.5*L1*(Il1)^2 I see no method, in that drawing, of controlling the current level in L1 where the transistors turn off... just a flaky oscillator :-( Everything controlled, a gezillion years ago... http://www.analog-innovations.com/SED/FlybackDC-DC-Converter.pdf ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | San Tan Valley, AZ 85142 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.