S
Steve Evans
Guest
Anyone know?
tnx,
steve
--
Fat, sugar, salt, beer: the four essentials for a healthy diet.
tnx,
steve
--
Fat, sugar, salt, beer: the four essentials for a healthy diet.
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J
Jamie
Guest
Steve Evans wrote:
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.
coil when energized source is removed quickly.Anyone know?
tnx,
steve
used to absorb the reverse (flyback) voltage release from a
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
Guest
Jamie <jamie_5_not_valid_after_5_Please@charter.net> wrote
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?)
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
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
Guest
On 28 Nov 2004 04:51:28 -0800, dmb06851@yahoo.com (dB) wrote:
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.
That makes sense. But the diverted engery has to be dissipatedJamie <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.
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.
S
Steve Evans
Guest
On Sun, 28 Nov 2004 21:52:57 -0000, "john jardine"
<john@jjdesigns.fsnet.co.uk> wrote:
[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!
--
Fat, sugar, salt, beer: the four essentials for a healthy diet.
<john@jjdesigns.fsnet.co.uk> wrote:
[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!
--
Fat, sugar, salt, beer: the four essentials for a healthy diet.
T
Tom Del Rosso
Guest
"Active8" <reply2group@ndbbm.net> wrote in message
news:xrvyax8bmvgv$.dlg@news.individual.net...
it's a current and not a voltage that is proportional to the rate of change
in a magnetic field? I don't see how it can be both.
--
Reply in group, but if emailing add
2 more zeros and remove the obvious.
news:xrvyax8bmvgv$.dlg@news.individual.net...
Mike, can you reconcile that please with the numerous references that sayOn Tue, 30 Nov 2004 02:44:47 GMT, Tom Del Rosso wrote:
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
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.
--
Reply in group, but if emailing add
2 more zeros and remove the obvious.
J
John Popelish
Guest
Steve Evans wrote:
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.
--
John Popelish
I think this term is often used to describe the application where aAnyone know?
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.
--
John Popelish
J
John Popelish
Guest
Steve Evans wrote:
but in the more practical copper wound coils, the wire resistance
consumes most of the energy.
http://focus.ti.com/lit/an/slup100/slup100.pdf
http://powerelec.ece.utk.edu/pubs/pesc2002_sssi.pdf
--
John Popelish
If the coil were made of superconductor, this might well be the case,On 28 Nov 2004 04:51:28 -0800, dmb06851@yahoo.com (dB) wrote:
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?
but in the more practical copper wound coils, the wire resistance
consumes most of the energy.
http://www.maxim-ic.com/appnotes.cfm/appnote_number/848Someonne 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://focus.ti.com/lit/an/slup100/slup100.pdf
http://powerelec.ece.utk.edu/pubs/pesc2002_sssi.pdf
--
John Popelish
C
CBarn24050
Guest
There seems to be some confusion here, a flywheel diode is a diode connectedSubject: Re: Whats a "snubber diode"?
From: John Popelish jpopelish@rica.net
Date: 28/11/2004 17:06 GMT Standard Time
Message-id: <41AA05A7.485845A7@rica.net
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?
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,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?
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
Guest
"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:q9rjq0t62e2eojj7tr0m3jv9pophoi2qtj@4ax.com...
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
news:q9rjq0t62e2eojj7tr0m3jv9pophoi2qtj@4ax.com...
If you short circuit a charged up inductor then absolutely nothing happens!.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.
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
T
Tom Del Rosso
Guest
"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:fpokq0hsdqgth4qgprkbvrrbi2g6ectqrh@4ax.com...
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.
--
Reply in group, but if emailing add
2 more zeros and remove the obvious.
news:fpokq0hsdqgth4qgprkbvrrbi2g6ectqrh@4ax.com...
Somebody check me on this.On Sun, 28 Nov 2004 21:52:57 -0000, "john jardine"
john@jjdesigns.fsnet.co.uk> wrote:
[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!
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.
--
Reply in group, but if emailing add
2 more zeros and remove the obvious.
S
Steve Evans
Guest
On Tue, 30 Nov 2004 00:34:21 GMT, "Tom Del Rosso"
<ng01@att.net.invalid> wrote:
--
Fat, sugar, salt, beer: the four essentials for a healthy diet.
<ng01@att.net.invalid> wrote:
IIrc,. they both store *charge*."Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:fpokq0hsdqgth4qgprkbvrrbi2g6ectqrh@4ax.com...
On Sun, 28 Nov 2004 21:52:57 -0000, "john jardine"
john@jjdesigns.fsnet.co.uk> wrote:
[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,
--
Fat, sugar, salt, beer: the four essentials for a healthy diet.
T
Tom Del Rosso
Guest
"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:nginq09n4575q6fm2k8mbc8v7pknrfss5e@4ax.com...
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.
--
Reply in group, but if emailing add
2 more zeros and remove the obvious.
news:nginq09n4575q6fm2k8mbc8v7pknrfss5e@4ax.com...
Actually the inductor stores a magnetic field, which is converted back to aOn Tue, 30 Nov 2004 00:34:21 GMT, "Tom Del Rosso"
ng01@att.net.invalid> wrote:
AIUI a cap stores voltage and an inductor stores current,
IIrc,. they both store *charge*.
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.
--
Reply in group, but if emailing add
2 more zeros and remove the obvious.
A
Active8
Guest
On Tue, 30 Nov 2004 02:44:47 GMT, Tom Del Rosso wrote:
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.
--
Best Regards,
Mike
Nope. Faraday's Law states that a time varying flux (which is caused"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:nginq09n4575q6fm2k8mbc8v7pknrfss5e@4ax.com...
On Tue, 30 Nov 2004 00:34:21 GMT, "Tom Del Rosso"
ng01@att.net.invalid> wrote:
AIUI a cap stores voltage and an inductor stores current,
IIrc,. they both store *charge*.
Actually the inductor stores a magnetic field, which is converted back to a
current.
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.
it is in a cap because it takes work to get that charge in there.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
Yeah, like C and V.either device, and measurement of columbs only takes you farther away from
the parameters that matter.
--
Best Regards,
Mike
A
Active8
Guest
On Tue, 30 Nov 2004 06:43:51 GMT, Tom Del Rosso wrote:
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.
--
Best Regards,
Mike
Show me a law of physics that states such a proportion"Active8" <reply2group@ndbbm.net> wrote in message
news:xrvyax8bmvgv$.dlg@news.individual.net...
On Tue, 30 Nov 2004 02:44:47 GMT, Tom Del Rosso wrote:
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
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.
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.
--
Best Regards,
Mike
T
Tom Del Rosso
Guest
"Active8" <reply2group@ndbbm.net> wrote in message
news:1xbc54nvqk8k6.dlg@news.individual.net...
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?
--
Reply in group, but if emailing add
2 more zeros and remove the obvious.
news:1xbc54nvqk8k6.dlg@news.individual.net...
Obvious. Should have seen that. My physics is very rusty.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.
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?
--
Reply in group, but if emailing add
2 more zeros and remove the obvious.
A
Active8
Guest
On Tue, 30 Nov 2004 15:48:47 GMT, Tom Del Rosso wrote:
no, I don't see how you could, but I might be brain farting here.
--
Best Regards,
Mike
Ok. That would have been in reply to someone else, but lemme see..."Active8" <reply2group@ndbbm.net> wrote in message
news:1xbc54nvqk8k6.dlg@news.individual.net...
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?
no, I don't see how you could, but I might be brain farting here.
--
Best Regards,
Mike
J
John Popelish
Guest
Tom Del Rosso wrote:
squared, and inductors store energy proportional to current squared.
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.
--
John Popelish
More accurately, capacitors store energy proportional to voltage"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:fpokq0hsdqgth4qgprkbvrrbi2g6ectqrh@4ax.com...
On Sun, 28 Nov 2004 21:52:57 -0000, "john jardine"
john@jjdesigns.fsnet.co.uk> wrote:
[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,
squared, and inductors store energy proportional to current squared.
For fixed inductances and capacitances, I think these are goodso 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.
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.
--
John Popelish