Re: Whats a "snubber diode"?

From: john jardine (john_at_jjdesigns.fsnet.co.uk)
Date: 11/28/04 

Date: Sun, 28 Nov 2004 21:52:57 -0000

"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:q9rjq0t62e2eojj7tr0m3jv9pophoi2qtj@4ax.com...
> On 28 Nov 2004 04:51:28 -0800, dmb06851@yahoo.com (dB) wrote:
>
> >Jamie <jamie_5_not_valid_after_5_Please@charter.net> wrote
> >
> >
> >
> >> used to absorb the reverse (flyback) voltage release from a
> >> coil when energized source is removed quickly.
> >
> >
> >The voltage isn't "absorbed" it is prevented from developing.
>
> That makes sense. But the diverted engery has to be dissipated
> somewhere. If the diodes simply in antiparallel with the source, itll
> act as a short circuit on the back emf. Why doesn't that (the energy
> of that reverse pulse) destroy the diode? Someonne else said a cap and
> bleed resitor can be used to store and discharge the pulses
> harmlessly, but no one esle has verfified this. Can we have some
> clarification, please?
> --
>
> Fat, sugar, salt, beer: the four essentials for a healthy diet.

If you short circuit a charged up inductor then absolutely nothing happens!.
On the other hand, a charged capacitor would destroy the shorting wire.
...
Use a mechanical switch and switch off a relay coil without a diode across
the coil. The relay discharges and drops out near instantly and you'll
probably get a spark across the switch contacts as thousands of volts come
back off the coil.
Now put a ('flywheel')diode across the coil and again switch off the relay.
The relay takes ages to discharge and drop out and there is no spark as only
0.7V can come off the coil.
Now put a 33V Zener diode (+ a reverse diode!) across the relay and switch
it off. It drops out quite fast with a peak of 33V across the coil.
Similar effect occurs if you put a diode in series with (say) a 100 ohm
resistor. This time the discharging coil voltage can be quite high and
depends on the inductance value.
Put a capacitor across the relay coil and you've made a nice resonant
circuit that takes many cycles to decay before the relay drops out. (it's
the relay wire resistance that finally dissipates all the stored energy).
Capacitor + resistor across the coil gives a very lossy 'damped' tuned
circuit ' but a number of measurements and calcs then needed to allow a
balancing act on the numbers.

Idea is that with an inductor the stored energy can be removed (discharged)
by allowing it to develop a voltage across some kind of load hence lose
it's stored energy as heat.
Bigger the discharge load resistance, then bigger the voltage, then bigger
the power loss, then quicker the discharge. Put a short circuit across the
inductor and a big current would try to flow but that same current will also
be charging up the inductor so nothing actually happens. Put an open circuit
across the inductor and the voltage screams upwards until something gives.
Tiny current flowing but high voltage and power dissipation hence fast
discharge.

The only inductor formula worth noting are ...

Stored Watt seconds(Joules)=1/2 x Inductance Value x [current through it
^2].

Amps per second though inductor = V across inductor / Inductor value.

[snip 8" of tedious relay coil example calcs]

These simple calcs are useful for test purposes. Generally it's easier and
much more accurate, just to use Spice.

regards
john

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