Inductor current can't be suddenly cutoff if no freewheel d

[Patrick Chung]

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

 

[RobertMacy]

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.

 

[Helmut Wabnig]

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.

 

[Jim Thompson]

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 |
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