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Whats a "snubber diode"?

J

John Popelish

Jan 1, 1970
0
Steve said:
Anyone know?

I think this term is often used to describe the application where a
diode allows a peak voltage (usually created by an inductor that has
had its current source switched off) to be limited by conducting
current into a capacitor that is precharged to a voltage just below
the desired limit. Often that precharge is just what is left after a
resistor has bled the effect of the previous pulse from the capacitor.
 
J

Jamie

Jan 1, 1970
0
Steve said:
Anyone know?

tnx,

steve
used to absorb the reverse (flyback) voltage release from a
coil when energized source is removed quickly.
the flyback voltage is in reverse of what went in and can get
very high in level which will short out things.
the trick is to place a diode across the coil connections.
the Cathode is connected to the + side and Anode to the - side
when energized, the diode does not conduct. when source is
removed quickly, the release will generate high voltage in reverse
polarity. at this point the diode will conduct and anything above the
cut off voltage of the diode will get absorb in the diode and protect
other voltage sensitive components.
 
D

dB

Jan 1, 1970
0
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.

(What do you mean when you use the word "flyback"? Do you know where
the term came from?)

the flyback voltage is in reverse of what went in and can get
very high in level which will short out things.

the trick is to place a diode across the coil connections.
the Cathode is connected to the + side and Anode to the - side
when energized, the diode does not conduct. when source is
removed quickly, the release will generate high voltage in reverse
polarity. at this point the diode will conduct and anything above the
cut off voltage of the diode will get absorb in the diode and protect
other voltage sensitive components.


I'll give you 3 points out of 10 for English, and 3 out of 10 for the
clarity of your explanation.

The same point is dealt with in this item from a forum:

http://p199.ezboard.com/fbasicelectronicsbasics.showMessage?topicID=1705.topic
 
S

Steve Evans

Jan 1, 1970
0
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?
 
J

John Popelish

Jan 1, 1970
0
Steve said:
On 28 Nov 2004 04:51:28 -0800, [email protected] (dB) wrote:

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?

If the coil were made of superconductor, this might well be the case,
but in the more practical copper wound coils, the wire resistance
consumes most of the energy.
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?

http://www.maxim-ic.com/appnotes.cfm/appnote_number/848
http://focus.ti.com/lit/an/slup100/slup100.pdf
http://powerelec.ece.utk.edu/pubs/pesc2002_sssi.pdf
 
C

CBarn24050

Jan 1, 1970
0
Subject: Re: Whats a "snubber diode"?
From: John Popelish [email protected]
Date: 28/11/2004 17:06 GMT Standard Time
Message-id: <[email protected]>

There seems to be some confusion here, a flywheel diode is a diode connected
across an inductor. It provides a path for the inductor current when the drive
voltage is removed (switch off). If say there was 1 amp flowing and the drive
was removed then the current (1 amp) would circulate through the coil and the
diode, as there is resistance as well as the diodes forward voltage drop, the
current will decay to zero.

Sometimes diodes are used with snubber networks, typicaly accross transistors,
igbt or thyristors. They are in series with the cap and allow the switch off
pulse to pass into the cap,as normal, but prevent the enrgy flowing back into
the transistor when it switches on again. A small resistor accross the cap
dissipates the charge from the cap during the "off" time.
 
J

john jardine

Jan 1, 1970
0
Steve Evans said:
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?

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
 
S

Steve Evans

Jan 1, 1970
0
On Sun, 28 Nov 2004 21:52:57 -0000, "john jardine"

[snip]

Tnx, john (and others). Thats a pretty comprehensive answer, i guess.
I'll ponder on it for a while. I must say your first bold
pronouncement about coils discharging harmlessly and caps desroying
had me confused, but your explanation of these pheonomena is of
considerable help!
 
T

Tom Del Rosso

Jan 1, 1970
0
Steve Evans said:
On Sun, 28 Nov 2004 21:52:57 -0000, "john jardine"

[snip]

Tnx, john (and others). Thats a pretty comprehensive answer, i guess.
I'll ponder on it for a while. I must say your first bold
pronouncement about coils discharging harmlessly and caps desroying
had me confused, but your explanation of these pheonomena is of
considerable help!

Somebody check me on this.

AIUI a cap stores voltage and an inductor stores current, so the result is
that when you discharge a cap you can't see a voltage higher than what you
had put into it, and when you discharge an inductor you can't see a current
higher than what it carried before discharging.

If the cap sees zero resistance then it produces a large current. If the
inductor sees infinite resistance then it produces a large voltage. Both of
these conditions must be avoided.
 
S

Steve Evans

Jan 1, 1970
0
Steve Evans said:
On Sun, 28 Nov 2004 21:52:57 -0000, "john jardine"

[snip]

Tnx, john (and others). Thats a pretty comprehensive answer, i guess.
I'll ponder on it for a while. I must say your first bold
pronouncement about coils discharging harmlessly and caps desroying
had me confused, but your explanation of these pheonomena is of
considerable help!

Somebody check me on this.

AIUI a cap stores voltage and an inductor stores current,

IIrc,. they both store *charge*.
 
T

Tom Del Rosso

Jan 1, 1970
0
Steve Evans said:
IIrc,. they both store *charge*.

Actually the inductor stores a magnetic field, which is converted back to a
current. People say inductors store current because there is a direct
relationship, but there is no direct relationship to charge. Charge is a
function of the number of electrons (6 * 10^18 electrons per coloumb I
think), but that parameter is not characteristic to the stored energy in
either device, and measurement of columbs only takes you farther away from
the parameters that matter.

One big and one small cap/inductor in parallel/series both have the same
voltage/current but not the same charge within them.
 
A

Active8

Jan 1, 1970
0
Actually the inductor stores a magnetic field, which is converted back to a
current.

Nope. Faraday's Law states that a time varying flux (which is caused
by the "field") induces an EMF.

d(Phi)
EMF = - -------
dt

That flux can't drop to zero instantaneously, but you can get a
pretty respectable voltage, anyway.
People say inductors store current because there is a direct
relationship, but there is no direct relationship to charge. Charge is a
function of the number of electrons (6 * 10^18 electrons per coloumb I
think), but that parameter is not characteristic to the stored energy in

it is in a cap because it takes work to get that charge in there.
either device, and measurement of columbs only takes you farther away from
the parameters that matter.

Yeah, like C and V.
 
T

Tom Del Rosso

Jan 1, 1970
0
Active8 said:
Nope. Faraday's Law states that a time varying flux (which is caused
by the "field") induces an EMF.

d(Phi)
EMF = - -------
dt

Mike, can you reconcile that please with the numerous references that say
it's a current and not a voltage that is proportional to the rate of change
in a magnetic field? I don't see how it can be both.
 
A

Active8

Jan 1, 1970
0
Mike, can you reconcile that please with the numerous references that say
it's a current and not a voltage that is proportional to the rate of change
in a magnetic field? I don't see how it can be both.

Show me a law of physics that states such a proportion
mathematically. How are you going to get a current in a transformer
secondary if it's open? Same thing with a generator or motor - the
EMF or back EMF is what we talk about.
 
T

Tom Del Rosso

Jan 1, 1970
0
Active8 said:
Show me a law of physics that states such a proportion
mathematically. How are you going to get a current in a transformer
secondary if it's open? Same thing with a generator or motor - the
EMF or back EMF is what we talk about.

Obvious. Should have seen that. My physics is very rusty.

But I asked (3 of my messages back) if I was correct in stating that the
snubber can't see a current higher than what the coil carried before it was
turned off. What about that?
 
A

Active8

Jan 1, 1970
0
Obvious. Should have seen that. My physics is very rusty.

But I asked (3 of my messages back) if I was correct in stating that the
snubber can't see a current higher than what the coil carried before it was
turned off. What about that?

Ok. That would have been in reply to someone else, but lemme see...
no, I don't see how you could, but I might be brain farting here.
 
J

John Popelish

Jan 1, 1970
0
Tom said:
Steve Evans said:
On Sun, 28 Nov 2004 21:52:57 -0000, "john jardine"

[snip]

Tnx, john (and others). Thats a pretty comprehensive answer, i guess.
I'll ponder on it for a while. I must say your first bold
pronouncement about coils discharging harmlessly and caps desroying
had me confused, but your explanation of these pheonomena is of
considerable help!

Somebody check me on this.

AIUI a cap stores voltage and an inductor stores current,

More accurately, capacitors store energy proportional to voltage
squared, and inductors store energy proportional to current squared.
so the result is
that when you discharge a cap you can't see a voltage higher than what you
had put into it, and when you discharge an inductor you can't see a current
higher than what it carried before discharging.

For fixed inductances and capacitances, I think these are good
generalizations.

If the magnetic path can change (think relay armatures moving,
solenoids with moving plungers, etc.) or if the current changes from
all the turns to some of the turns (or charges through one winding and
discharges through another winding) the current can change.

Likewise, if capacitor plate spacing can change, voltage can change.

There are somewhat unusual conditions but not at all unheard of.
 
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