diode on relay coil for back emf protection???
Christopher Zoeller
I've noticed that designs using relays typically have a
diode reverse-biased across the relay coil. I think the
diode is placed across the coil to "protect?" from back
emf? Is this assumption correct? -- and either way:
what's back emf?
Thank you -- any help is greatly appreciated!!!!!!
(My question pertains to small relays and small-signal diodes--
however, I don't think this distinction really matters in theory.)
Don Yuniskis
In article <51mtpn$4...@jetty.nsg.telerate.com>,
Matt Daughtrey <mdaug...@ccgate.sysdev.telerate.com> wrote:
>In article <51l8gs$m...@news-e2b.gnn.com>,
>when the current is removed from the relay coil and the magnetic field around
>the coil collapses, inducing a current in the coil of opposite polarity to the
>current that energized the relay. The diode simply shorts it out, preventing
>damage to the rest of the circuitry.
Note that this technique is also employed when the "drive electronics"
are NOT semiconductors! (i.e. switches and other relays driving
large contactors, etc.)
--don
Douglas W. Jones,201H MLH,3193350740,3193382879
From article <51l8gs$m...@news-e2b.gnn.com>,
by CZoe...@gnn.com (Christopher Zoeller):
> I've noticed that designs using relays typically have a
> diode reverse-biased across the relay coil. I think the
> diode is placed across the coil to "protect?" from back
> emf? Is this assumption correct? -- and either way:
> what's back emf?
Absolutely right. Consider this circuit:
+ Supply _____________
| __|__
The subject _|_ | |
of this /|\ | coil|
discussion | |_____|
|________|
|
Control ___________|/ any kind of
|\ switching device
Ground _____________|
If the diode is missing, turning on the coil produces no excitement.
Because the coil is an inductor, the current through the coil rises
gradually to the full current when the switch is turned on, and then
stays there. When the switch is turned off, though, there is a
problem. The coil is, after all, an indutor! The current wants to
keep flowing, and with the switch turned off, the inductor puts a
theoretically infinite voltage across the switch trying to keep the
current flowing.
In fact, of course, the voltage isn't infinite. Real switches have
some capacitance, and real switches do not turn off instantly. Both
of these considerations change the infinity to merely a few thousands
or tens of thousands of volts.
This phenomonon is used deliberately in your car to make the spark
that ignites the gasoline in your cylinders, using a supply voltage
of only 12 volts. It is also used deliberately in some transformerless
DC to DC converters.
In electronically driven relays, this phenomonon will destroy almost
any drive circuit, so we put the diode in, shorting the potentially
infinite voltage to the positive supply, and allowing the current in the
coil to coast down from the operating current to zero when the switch
is turned off.
If you want the coil to deenergise more quickly, that is, if you want
the current recirculating through the diode to go to zero more quickly,
put a resistor or a zener diode in series with the diode. Size it
carefully in terms of the breakdown voltage of your switch!
Doug Jones
jo...@cs.uiowa.edu
James P. Meyer
On Tue, 17 Sep 1996, Matt Daughtrey wrote:
> This assumption is correct. The back EMF (Electromotive Force )is generated
> when the current is removed from the relay coil and the magnetic field around
> the coil collapses, inducing a current in the coil of opposite polarity to the
> current that energized the relay. The diode simply shorts it out, preventing
> damage to the rest of the circuitry.
>
The voltage across the coil changes polarity, but the current
through the coil continues to flow in it's original direction.
Jim
Kevin AstirCS 1U KO0B
CZoe...@gnn.com (Christopher Zoeller) wrote:
>I've noticed that designs using relays typically have a
>diode reverse-biased across the relay coil. I think the
>diode is placed across the coil to "protect?" from back
>emf? Is this assumption correct? -- and either way:
No.
>what's back emf?
Back EMF is minor effect in relay...as armature moves, the relay
"generates" a small voltage which effectivly subtracts from the
appliied voltage, causing the coil current to ramp up a little slower
than it would otherwise. back emf is a major effect in DC motors,
however, where it serves to establish the speed/voltage characteristic
of the motor. You could consider a relay to be a motor with very
limited angular travel....but the armature never has a chance to get
up to "cuise speed" due to mechanical stops.
In no case does back emf require protective measures...it is self
limiting, and smaller than voltage (and same sign) as voltage which
should be there anyway.
The diodes provide a path for the current that is flowing through the
coil when the switch is opened, allowing the current to slowly ramp to
zero.
If the current were suddenly interupted (no diode) then the collapsing
magnetic field would generate a large voltage spike which would arc
the switch terminals, or avilanche a solid state switch.
This is not the only way to rech this end however. Z1ener diodes, or
RC snubber networks serve a similar purpose, and can result in faster
de-energization of the relay.
If you substitute the phrase "inductive spike" for "back emf" then
your posting had it about right...but they are different things.
-KF-
JBlessing
In article <51l8gs$m...@news-e2b.gnn.com>, CZoe...@gnn.com says...
>
>I've noticed that designs using relays typically have a
>
>diode reverse-biased across the relay coil. I think the
>
>diode is placed across the coil to "protect?" from back
>
>emf? Is this assumption correct? -- and either way:
>
>
>Thank you -- any help is greatly appreciated!!!!!!
>
>(My question pertains to small relays and small-signal diodes--
>
>however, I don't think this distinction really matters in theory.)
>
>
>
>
--
As the magnetic field from the relay coil colappses it can generate a
spike of fairly high voltage that will feed back into the circuit and
possibly harm other devices. The diode simply shorts the spike out before
it gets anywhere,
Jordan Blessing L1 Master Tech
*For Information on my new booklet:
*What You Should Know Before the Tow: A Consumers Guide to Auto Repair
*Reply to this message via Email with your request.
Matt Daughtrey
Gary Morris
Christopher Zoeller wrote:
>
> I've noticed that designs using relays typically have a
>
> diode reverse-biased across the relay coil. I think the
>
> diode is placed across the coil to "protect?" from back
>
> emf? Is this assumption correct? -- and either way:
>
> what's back emf?
>
> Thank you -- any help is greatly appreciated!!!!!!
>
> (My question pertains to small relays and small-signal diodes--
>
> however, I don't think this distinction really matters in theory.)
Yes, you are correct. The diode is placed across the coil to protect the
switching device from back EMF.
The distinctive characteristic of an inductor, such as a relay coil, is
that inductors try to maintain a constant current through themselves.
When the switching device is turned on, current begins to flow through
the inductor causing a magnetic field to begin building up. As the field
builds, it cuts across the windings of inductor, generating a voltage of
polarity that opposes the source voltage causing the current. This back
EMF never quite equals the source voltage, so rather than maintaining
zero current, the current builds up over time to a value determined by
the the source voltage and the DC resitance of the circuit (Ohm's law).
Then when the switching device is turned OFF, the current tries to stop.
But this causes the magnetic field to begin collapsing. The inductor now
becomes a voltage source that tries to keep the current constant. When
the switching device is OFF, it is a VERY high resistance. The inductor
must generate a VERY high voltage to try to keep the current going
through the VERY high resistance. This high voltage is high enough to
cause arcing across the contacts of mechanical switch contacts, or to
cause breakdown of transistor switches.
The diode is forward biased by the inductor's back EMF. Since the
forward biased diode is a low resistance, only a low voltage is required
to maintain the current. Thus the back EMF does not become high enough
to cause arcing or breakdown.
--
Gary Morris
gmo...@gv.net
Steve Weiler
Kevin AstirCS 1U KO0B wrote:
>
> CZoe...@gnn.com (Christopher Zoeller) wrote:
>
> >I've noticed that designs using relays typically have a
>
> >diode reverse-biased across the relay coil. I think the
>
> >diode is placed across the coil to "protect?" from back
>
> >emf? Is this assumption correct? -- and either way:
>
>
> >what's back emf?
>
Becuase a relay coil can be quite inductive (it is inducing
a magnetic field, remember :-) ) any sharp change in current
(i.e. de-energising the coil quickly) will cause an EMF
proportional to the change in current and also proportional
to the inductance of the coil. The diode clamps the coil's
"Back EMF" and gives a path for the current to discharge
slowly and keep potentially high transient voltages out of
the circuit.......
--
+---------------------------------------------------------------------+
|Steve Weiler ste...@vikingmem.com |
|Design Engineer |
|Viking Components, Inc. http://www.vikingmem.com |
+---------------------------------------------------------------------+
| "Have you heard of the wonderful one-hoss shay..." |
| O.W. Holmes |
| |
| The opinions expressed.. bla bla bla... mine and no-one elses.... |
| bla bla bla... reflect only on me... bl bla bla |
----------------------------------------------------------------------+
Thomas A Maier
I hate to dip into the calculus, because many people don't understand it
but this is a case where the true meaning of what goes on in the relay
coil very nicely drops out of the equation.
The voltage across an inductor is defined by: e = L * di/dt,
that is, "the inductance times the rate of change of the current with
respect to time".
In english this means that when there is a rapid change in the current
flow, e.g., when you turn off the transistor that is pulling the bottom
of the relay coil to ground, the di/dt part of the equation becomes VERY
large--a relatively small current but and even smaller period of time.
So the product can be a very large number....if the diode isn't there to
give the induced current someplace to go.
An interesting sidelight is that since the diode conducts, current
continues to flow thru the coil for *quite a while*--electronically,
speaking--after the switch opens. So the relay is slow to drop out.
Normally, the few extra milliseconds doesn't mean much, but if the relay
is carrying information it was a different matter. You youngster
probably don't remember teletype machines, but having the teletype
electromagnets *stick on* would mean that single bits of data were often
lost. Even data running at the phenomenal rate of 45.45 baud (!)--a
standard speed of the pre-PC, *hollow-state era*--could be garbled. One
had to place a resistor in series with the diode to limit the current
flow to permit the relay to drop quicker. This, of course, meant that
the voltage across the coil would rise, so you had to trade speed off
against spike amplitude.
Tom, Uni of Akron, BSEE 1970. Any Zips out there?
Steve White
On 17 Sep 1996, Douglas W. Jones,201H MLH,3193350740,3193382879 wrote:
...[clip]...
> This phenomonon is used deliberately in your car to make the spark
> that ignites the gasoline in your cylinders, using a supply voltage
> of only 12 volts. It is also used deliberately in some transformerless
> DC to DC converters.
...[clip]...
Aha! Maybe this is why they are called "arc supression diodes"?
--
.............................................................
Steve White, Lab Technician @ Human Interface Technology Lab
e-mail: ste...@hitl.washington.edu voice: 206.616.1411
http://www.hitl.washington.edu/ (My page under construction)
.............................................................
James P. Meyer
On Tue, 17 Sep 1996, Kevin AstirCS 1U KO0B wrote:
> >diode reverse-biased across the relay coil. I think the
> >diode is placed across the coil to "protect?" from back
> >emf? Is this assumption correct? -- and either way:
>
> Back EMF is minor effect in relay...as armature moves, the relay
> "generates" a small voltage which effectivly subtracts from the
> appliied voltage, causing the coil current to ramp up a little slower
You are defining the term "back emf" too narrowly. Back emf is
generated *whenever* the current changes in an inductor *or* when the
inductance changes with a constant current flowing.
> If you substitute the phrase "inductive spike" for "back emf" then
> your posting had it about right...but they are different things.
The term "inductive spike" and "back emf" are equally "right" in
this case.
Jim
Kevin AstirCS 1U KO0B
OK, looks like "back EMF" wins a popular usage contest in spite of my
protests. Guess I've worked with motors too much where "back EMF" is
an effect due to motor acting as a generator, and needs to be
considered in a much different vein from the inductive nature of the
motor windings.
Gary Morris <gmo...@gv.net> wrote:
> This back
>EMF never quite equals the source voltage, so rather than maintaining
>zero current, the current builds up over time to a value determined by
>the the source voltage and the DC resitance of the circuit (Ohm's law).
OK, gotta niggle with this othewise good discription of current
ramping up in an inductor, due to an applied voltage step.. The
current builds over time because if it didn't the magnetic field would
be static, and couldn't generate any EMF a'tall.
If current initially equals zero, then NO voltage difference is
required to force a change in current. As current builds, the
required voltage difference is established by the parasitic
resistance of the circuit....no voltage difference is required to
allow the the current to increase, however. It is the increasing
current which produces the voltage in opposition to the applied
voltage, not a shortage of opposing voltage which permits the current
increase.
.
Mechanical analogy is appropriate, consider pushing a car. The car
must exert same force on pusher, as pusher exerts on car, (or else
pusher would pass through car) yet car accelerates (at an initial rate
determined by mass) to a speed determined by point at which forces due
to friction equal force supplied by pusher..
It is the acceleration which produces the initial back force against
the pusher, NOT a difference in forces which permits the acceleration.
If you want to argue the force on two sides of interface is not equal,
fine, instead, pull the thing with a rope....think the tension is
different at two ends of a (let it be massless) rope?...no way!
Anyhow, my argument boils down to the distinction between a
differential, and a difference...which I _finally_ appreciate.
-KF-
Roy McCammon
"James P. Meyer" <jim...@acpub.duke.edu> wrote:
>On Tue, 17 Sep 1996, Matt Daughtrey wrote:
>
>> This assumption is correct. The back EMF (Electromotive Force )is generated
>> when the current is removed from the relay coil and the magnetic field around
>> the coil collapses, inducing a current in the coil of opposite polarity to the
>> current that energized the relay. The diode simply shorts it out, preventing
>> damage to the rest of the circuitry.
>>
>
>through the coil continues to flow in it's original direction.
Exactlty right. And it makes sense if you remember Lenz's law which says
more or less that the magnetic field will change in a way to oppose
change. That means, the magnetic field in the inductor will try to
keep the current flowing in the same direction, even if it has to
generate enough voltage to break down the transister that is trying
to block the current.
Opinions expressed herein are my own and may not represent those of my employer.
Mr. Brooke Clarke
Christopher Zoeller wrote:
>
> I've noticed that designs using relays typically have a
>
>
> diode is placed across the coil to "protect?" from back
>
> emf?
Christopher:
In a car with old fashioned spark ignition, a current is flowing
through the "coil" and when the current is stopped by the "points"
opening, the coil tries to keep the current going so it generates
a high voltage (could be over a hundred volts on the low voltage
side of the coil).
Ths same thing happens in a relay, when the current stops, the coil
tries to keep the current going and generates a voltage that can be
much higher than the voltage that was driving the coil. By placing a
diode across the coil so that during normal operation the diode is
reverse biased, when the coil tries to generate a high voltage, the
diode now is fordward biased and clamps the high voltage to a safe
level.
Hopes this helps,
Brooke, N6GCE
Roy McCammon
kfer...@aquilagroup.com (Kevin AstirCS "1U" KO0B) wrote:
>OK, looks like "back EMF" wins a popular usage contest in spite of my
I would think that just EMF would be the right term.
>If current initially equals zero, then NO voltage difference is
>required to force a change in current. As current builds, the
>required voltage difference is established by the parasitic
>resistance of the circuit....no voltage difference is required to
>allow the the current to increase, however. It is the increasing
>current which produces the voltage in opposition to the applied
>voltage, not a shortage of opposing voltage which permits the current
>increase.
an inductor is a circuit element that enforces the relationship V = L (di/dt)
It is equally valid to view the current rate of change as causing the voltage
or the voltage as causing the current rate of change. Either one could
be the independent variable.
>Mechanical analogy is appropriate, consider pushing a car. The car
>must exert same force on pusher, as pusher exerts on car, (or else
>pusher would pass through car) yet car accelerates (at an initial rate
>determined by mass) to a speed determined by point at which forces due
>to friction equal force supplied by pusher..
>
>It is the acceleration which produces the initial back force against
>the pusher, NOT a difference in forces which permits the acceleration.
I think the conventional understanding is that acceleration of the car arrises
from the difference between the the force supplied to the car by the pusher
and the frictional forces opposing the motion.
brian whatcott
In article <51l8gs$m...@news-e2b.gnn.com>, CZoe...@gnn.com says...
>
>I've noticed that designs using relays typically have a
>diode reverse-biased across the relay coil. I think the
>diode is placed across the coil to "protect?" from back
>emf?
It's true regular designers put a diode across a relay coil.
But *good* designs have a diode and resistor: that magnetic energy
has to go somewhere - either a slowish decay through a diode's R
or a faster decay through an extra (small) R
brian whatcott <in...@intellisys.net>
Altus OK
Alan
As I remember that a diode across a relay coil will eliminate
ringing of the relay caused by the back EMF in the coil. The
potential of the back EMF will bias the diode and the current
will flow through the easiest path, in this case the diode
instead of the coil. This will greatly enhance the life of the
relay contacts.
Alan
Ayman El-Khashab
Alan <10426...@CompuServe.COM> wrote in article
<527ls2$4m0$1...@mhafc.production.compuserve.com>...
Actually, the relay contacts are only a minor concern. True
the contacts will get oxidized from the sparks, but that is
not the big problem. If you remember, whenever you generate
an EM field, you have a coil with a current going through it,
when you suddenly break the circuit, the em field continues
generating lots of em feedback which will go backwards through
your circuit blowing up semiconductors on its way to ground.
di/dt!=0 instantly for an inductive load.
You use the diodes to short the em feedback and prevent it
from going backwards through your circuit. Not many semiconductors
like 500 volts to show up at their outputs.
- Ayman
Thomas A Maier
> The em field continues generating lots of em feedback which will go
to ground.
An interesting manifestation of this same effect--only magnified 1000s
of times--can be seen in an aluminum smelting plant.
A typical plant has about 200 smelting "pots" in series across a supply
of about 1000 VDC . To give you some perspective--the pots are each
typically as big or bigger than a railroad car and the current flowing
thru them is about 250,000 AMPS !!! (Yep, that's right, a Quarter of a
Million Amps.) The pots are arranged side-by-side to form a long "U"
with the ends of the "U" connecting to the power supply. The
conductors from the bottom of one pot to the top of the next are large!
(4"x12" cross-section), paralleled aluminum bars.
The magnetic field generated on a potline is strong enough that the 6 ft
steel bars that are used to break the crust that forms on the pot can be
*stored* by simply releasing your grasp on them. The magnetic field
takes over and slams the bar against the side of the pot, where it stays
until the next time that you need it.
Generally the power is slowly applied and removed to avoid the inductive
kick sorts of problems. The inductance isn't all that high--we only
have a 1 *turn* coil, after all--, but in the event of a power failure,
di/dt becomes...well, LARGE and *lightning bolts* MANY feet in length
can be generated. Instrumention on smelting plants is, obviously, quite
a problem.
Richard Torrens
In fact a lot of semiconductors, when driving a relay, will survive the
inductive kick-back: they go into avalanche or zener breakdown. Whether they
survive this depends on the energy stored in the coil (1/2 L I squared) and
whether this energy is within the acceptable repetitive breakdown rating of
the semiconductor.
However - it's a lot easier to put a 'flywheel' diode in than to calculate
the effects of the inductive energy discharge. Also the high voltage spike
can cause unforseen interactions elsewhere in the ciruit.
There is another point: energise a relay (via a transistor) from 12v and the
relay will take, perhaps, 3 milliseconds to make (due to the rate of chage
of the current). De-energise it without a flywheel diode and it tries to
de-energise as quick as it can - maybe 1/2 mSec (depending on where the
energy goes). Put a flywheel diode in and the current keeps circulating so
the release time could be, perhaps 10 milliseconds.
To get a release time about the same as the energise time, stick a resistor
in series with the diode: resistor's value should be about the same as the
coil's d.c. resistance.
--
/| Richard Torrens - 4...@argonet.co.uk
/ |
/ | 4 Q D
/ | | We manufacture
/ /| | MOSFET controllers for battery operated motors
/ / | | See us on http://www.argonet.co.uk/users/4qd
/_/__| |____ our www site contains FAQ sheet on motors & controllers
/_____ ____\ and a selection of interesting circuit diagrams
/ _ \| | _ \
| | | | | | | | Phone/fax +44 1638 741 930
| |_| | | |_| |
\__\_\ |____/ We use an Acorn RISC-PC 32 bit RISC computer
Tony Williams
In article <na.7c559647...@argonet.co.uk>, Richard Torrens
<URL:mailto:4...@argonet.co.uk> wrote:
>
> In fact a lot of semiconductors, when driving a relay, [snip]
>
> To get a release time about the same as the energise time, stick a resistor
> in series with the diode: resistor's value should be about the same as the
> coil's d.c. resistance.
>
I prefer to put the Flyback Clamp directly across the driving device, in this
way the driver is exactly protected and you get fastest discharge of the
stored energy. Zeners or Avalanche Surge Suppressors work best.
> We use an Acorn RISC-PC 32 bit RISC computer
So do we... Let the rest of the world be PC-masochists.
--
[Tony Williams, Ledbury, Herefordshire, UK.------------Pagewidth=78----------]
James P. Meyer
On Thu, 26 Sep 1996, Tony Williams wrote:
> I prefer to put the Flyback Clamp directly across the driving device, in this
> way the driver is exactly protected and you get fastest discharge of the
> stored energy. Zeners or Avalanche Surge Suppressors work best.
Topologically, and if the power supply is well bypassed and low
impedance, the connection you describe is equivalent to putting the device
across the coil. Additionally, since the two connections are equivalent,
there is *no* speed advantage that is the result of your connection. The
speed advantage comes from your choice of devices only.
Jim
ROTHKICKS
Why does the magnetic field collaspe after the driving force is removed
from the coil? Why doesnt the magnetic field just stay "out there" ?? If
it has to collaspe, could one very quickly move the coil physically out of
the field and minimize its impact on the circuit ? Is it possible to move
this field somewhere and save it for later?
Bill Pendleton
Whatever you call it, the reverse biased diode prevents the reverse voltage
spike from leaving the coil when it de-energizes. A second forward diode in
series with the positive voltage supply further protects sensitive circuits.
James P. Meyer (jim...@acpub.duke.edu) wrote:
: On Tue, 17 Sep 1996, Kevin AstirCS 1U KO0B wrote:
: > >diode reverse-biased across the relay coil. I think the
: > >diode is placed across the coil to "protect?" from back
: > >emf? Is this assumption correct? -- and either way:
: >
: > Back EMF is minor effect in relay...as armature moves, the relay
: > "generates" a small voltage which effectivly subtracts from the
: > appliied voltage, causing the coil current to ramp up a little slower
: You are defining the term "back emf" too narrowly. Back emf is
: generated *whenever* the current changes in an inductor *or* when the
: inductance changes with a constant current flowing.
: > If you substitute the phrase "inductive spike" for "back emf" then
: > your posting had it about right...but they are different things.
: The term "inductive spike" and "back emf" are equally "right" in
: this case.
: Jim
--
Bill
Thomas A Maier
> Whatever you call it, the reverse biased diode prevents the reverse voltage
> spike from leaving the coil when it de-energizes. A second forward diode in
> series with the positive voltage supply further protects sensitive circuits.
I don't completely agree with this statement. If you put to diodes is
series and place the two diodes across the coil, the voltage across the
coil during discharge will be twice the forward drop of a single diode.
The result will be that the field will collapse more quickly.
The same is true if you put a resistor or a Zener (watch the direction)
in series with the standard diode. Both will allow the voltage to rise
higher and cause the field to decay quicker. HOWEVER, there is an
increase spike to consider. It still won't be as high as the "no diode"
case. The spike voltage in the zener case will be the zener voltage + 1
diode forward volt drop.
Tom
Thomas A Maier
This is a joke...right?
Tony Williams
In article <Pine.SOL.3.91.96092...@godzilla3.acpub.duke.edu>,
identical. However you do have to make some drastic and perhaps
impractical assumptions about the Supply Impedance that, if believed,
can/will result in early failure of the relay driving device.
The original proposal, diode directly across the coil, is the most widely
used configuration and reasonably trouble-free. If you want faster turnoff
of the relay you have something in series with the diode that allows a higher
reverse swing and hence quicker discharge of the stored energy. This of
course increases the voltage across the driving device, to Vsupply plus the
Vbackswing.
In practice, relay supplies tend to be rough and unregulated with easy
feedthrough of Mains spikes and, in a control panel, the relay-bank may be
a yard or so from the final smoothing cap. As the relay is switched off
Vsupply may tend to kick upwards slighly. The nice, tidy, paper-view does
not really apply here and the point I was trying to make was that the
real purpose of the Flyback Clamp is to protect the driving device and
therefore best placed directly across the driving device.
This configuration allows you to specify a Clamp Voltage as near to the
driver Breakdown Voltage as possible and be confident that the driver is
safe, even during mains spikes coincident with switchoff, and you still have
as high a backswing voltage as can be afforded. The Safety is inherent.
Tony.
Ray Heasman
Thomas A Maier (tma...@andrew.cmu.edu) wrote:
: > Why does the magnetic field collaspe after the driving force is removed
: This is a joke...right?
Hmm, it may be a joke, but its still a good question. We have absolutely no
idea what a 'field' is. Okay, we have a definition for a magnetic or
electric field, but we still dont know what they _really_ are.
Cheerio,
Ray
--
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Bill Pendleton
: Whatever you call it, the reverse biased diode prevents the reverse voltage
: spike from leaving the coil when it de-energizes. A second forward diode in
: series with the positive voltage supply further protects sensitive circuits.
Sorry for the confusion, always believe everybody knows what I am thinking...
An attempt to draw a schematic
V+
----------------|>-|-----.---.
D1 | C
- O
D2 ^ I
V0 _ L
-------------------------.----.
The idea is D1 conducts to energize the coil. On cutting power to the coil, D2
conducts to absorb the reverse polarity inductive spike. D1 now functions to
further isolate sensitive control circuitry. If needed a similiar diode can
be added to the ground line for extremely low noise appplications.
Regards,
Bill Pendleton
: James P. Meyer (jim...@acpub.duke.edu) wrote:
: : On Tue, 17 Sep 1996, Kevin AstirCS 1U KO0B wrote:
: : > >diode reverse-biased across the relay coil. I think the
: : > >diode is placed across the coil to "protect?" from back
: : > >emf? Is this assumption correct? -- and either way:
: : >
: : > Back EMF is minor effect in relay...as armature moves, the relay
: : > "generates" a small voltage which effectivly subtracts from the
: : > appliied voltage, causing the coil current to ramp up a little slower
: : You are defining the term "back emf" too narrowly. Back emf is
: : generated *whenever* the current changes in an inductor *or* when the
: : inductance changes with a constant current flowing.
: : > If you substitute the phrase "inductive spike" for "back emf" then
: : > your posting had it about right...but they are different things.
: : The term "inductive spike" and "back emf" are equally "right" in
: : this case.
: : Jim
: --
: Bill
--
Bill
Richard Torrens
In article <52eogl$g...@newsbf02.news.aol.com>, roth...@aol.com (ROTHKICKS)
wrote:
>
> Why does the magnetic field collaspe after the driving force is removed
> from the coil? Why doesnt the magnetic field just stay "out there" ?? If
> it has to collaspe, could one very quickly move the coil physically out of
> the field and minimize its impact on the circuit ? Is it possible to move
> this field somewhere and save it for later?
>
Yes - there are static magnetic fields. This is what a permanent magnet
causes. I have never seen an explanation of the mechanism.
We are talking about the magnetic field associated with an electrical
current: remove the curent and the field collapses.
The field is associated with the current and the inductance and has energy
stored in it. This energy tries to keep the field going by keeping the
current flowing.
I can't explain why - but I'm not small enough to see an electron. Maybe
that would help. All I can do is to notice the results. Even a formula
expressing the relationship would not explain why!
--
/| Richard Torrens - 4...@argonet.co.uk
/ |
/ | 4 Q D
/ | | We manufacture
/ /| | MOSFET controllers for battery operated motors
/ / | | See us on http://www.argonet.co.uk/users/4qd
/_/__| |____ our www site contains FAQ sheet on motors & controllers
/_____ ____\ and a selection of interesting circuit diagrams
/ _ \| | _ \
| | | | | | | | Phone/fax +44 1638 741 930
| |_| | | |_| |
\__\_\ |____/ We use an Acorn RISC-PC 32 bit RISC computer
Douglas W. Jones,201H MLH,3193350740,3193382879
Someone asked:
> Why does the magnetic field collaspe after the driving force is removed
> from the coil? Why doesnt the magnetic field just stay "out there" ?? If
> it has to collaspe, could one very quickly move the coil physically out of
> the field and minimize its impact on the circuit ? Is it possible to move
> this field somewhere and save it for later?
The field collapse is a necessary consequence of the termination of the
current through the coil, and in fact, the field collapse is usually
necessary to the application. For example, if the coil is part of a relay,
if the field didn't collapse, the relay wouldn't open!
But, yes, you can stop the current flowing through one coil and get the
stored energy to go somewhere else!
For example, if you wind two coils around a relay armature, the primary
winding being N turns of wire used for the normal drive of the relay, and
the secondary coil being one turn of very heavy copper wire, with the two
ends solidly connected to each other, when you turn off a current of 1 amp
in primary coil, what will happen is that you'll quite suddenly have a
current of N amps through the secondary. In this case, you avoided the
inductive spike, but you did so by providing a second current path so that
the field won't collapse. The field will slowly collapse in this case,
because the secondary winding has some resistance. Relays with up to a
1 second time delay have been built using this technology (the relay only
closes a second or so after it is energised, and releases about a half
second after it is deenergised).
A second trick to save the energy that was stored in the coil is to store
the energy in a capacitor. In this case, the voltage spike is used to
pump electrons "uphill" into a capacitor, storing the energy there until
it is needed elsewhere. Note that you're not storing the magnetic
field itself, you're storing the energy that was in the magnetic field,
but you're storing it as an electric field.
Doug Jones
jo...@cs.uiowa.edu
Claude Frantz
Tony Williams <to...@ledelec.demon.co.uk> writes:
>In article <Pine.SOL.3.91.96092...@godzilla3.acpub.duke.edu>,
> The original proposal, diode directly across the coil, is the most widely
> used configuration and reasonably trouble-free. If you want faster turnoff
> of the relay you have something in series with the diode that allows a higher
> reverse swing and hence quicker discharge of the stored energy.
A good idea is to use a varistor as unique protection device.
If you need short on and off switching time, consider the US patent 3,896,367.
Have a look on ftp://bauv106.bauv.unibw-muenchen.de/pub/claude/p3896367/
> This of
> course increases the voltage across the driving device, to Vsupply plus the
> Vbackswing.
Of course.
--
Claude
(cla...@bauv106.bauv.unibw-muenchen.de)
The opinions expressed above represent those of the writer
and not necessarily those of her employer.
Jeremy Miller
: > from the coil? Why doesnt the magnetic field just stay "out there" ?? If
: > it has to collaspe, could one very quickly move the coil physically out of
: > the field and minimize its impact on the circuit ? Is it possible to move
: > this field somewhere and save it for later?
: This is a joke...right?
I hope its a joke 'cause I think he might actually be smart enough to be
dangerous.
Roy McCammon
brian whatcott wrote:
>
> In article <51l8gs$m...@news-e2b.gnn.com>, CZoe...@gnn.com says...
> >
> >I've noticed that designs using relays typically have a
> >diode is placed across the coil to "protect?" from back
> >emf?
>
> It's true regular designers put a diode across a relay coil.
> But *good* designs have a diode and resistor: that magnetic energy
> has to go somewhere - either a slowish decay through a diode's R
> or a faster decay through an extra (small) R
Actualy, when you use just a diode, the energy is mainly disapated in the
resistence of the relay coil. So, assuming low resistence drivers and
low
resistence diodes, the electrical time constant of enerization is about
the same as de-energization.
Good designs meet the system requirements without extra parts or cost. A
simple clamp diode will usually meet the requirements. Sometimes you
really
do want to get as much speed out of the relay as possible and then a
series
resister will help.
Sir Spamalot
On the contrary, this is seemingly subtle effect is analogous to
the scientific reason that the back-emf that is blocked by the
uni-directional speaker cables provides the reset energy to back
bias the output transistors of a class-a or class AB1 audio/RF
power amplifier, when it's operating as a current pump, thus
removing any hint of NPN/PNP interactivity crossover distortion
production.
Simple, basic electronics.
Roy McCammon
Thomas A Maier (tma...@andrew.cmu.edu) wrote:
> Why does the magnetic field collaspe after the driving force is removed
> from the coil? Why doesnt the magnetic field just stay "out there" ?? If
> it has to collaspe, could one very quickly move the coil physically out of
> the field and minimize its impact on the circuit ?
The standard model that we use to predict these things is that the magnetic
field exits as an effect of a source. Take away or eliminate the source, and
the field collapses.
There are two sources. The current in the coil is one source. When the current
stops, this source is effectively eliminated. The other source is the elementry
dipole magnets in the relay material. The field caused by the current causes these
elementry magnets to align, thereby adding to the total field. In fact, the
net contribution of these magnets is dominent, so, obviusly yanking the relay
around yanks the sources right along with it and the fields.
When the magnetizing current stops, the elementry diples do stay aligned
and support the field for a little while, but thermal agitation causes them
to become unaligned fairly quickly. It wouldn't be a very good relay if it
became permanently magnetized and wouldn't drop.
terry moreau
RELAY COIL DIODES...the bottom line!
Here is the straight poop from The Relay Handbook...
Use the diode across the coil when concerned about back emf in the driving
circuit and when the relay contacts are NOT driving and inductive load. IE. a
transistor switch like 2N2222 could get damaged upon fast turn-off of a big
inductive load like a relay coil. IE. a low power relay driving a high power
relay could get shortened contact life from arcing as it breaks open the
inductive load of the high power relay's coil.
However, the diode is actually a bad choice when the relay drives an un-clamped
and un-snubbed inductive load (or eneven a resistive load near maximum rating).
This is because the the diode 'slows' the decay of magnetic field and thus can
slow the opening of the contact. The whole idea with inductive load turn-off
is to open the contacts quickly to break the arc as fast as possible and
minimize contact damage. In some cases the addition of the diode across the
relay coil will indeed shorten the relay contact life!
A compromise is to use a bilateral transorb, or a zener and regular diode is
series. The idea is to permit a back emf which allows the relay coil's field to
collapse quickly but to prevent high voltage sikes which might damage a delicate
transistor driver or produce some annoying EMI.
The other thing is to avoid having relay contacts switch inductive loads
without the addition of a clamp or snubber network. When you do this, the speed
of contact opening is not as big an issue.
Some relays/contactors are inherently fast opening and the diode across the coil
ruins the high speed mechanical operation. Other relays are so slow at opening
that the diode across the coil makes little difference. BEWARE! and TAKE CARE!
Like most things there are pro/cons....you need to know the application to make
the best choice.
~~~~~___~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~/0 0\~~~~~~~~~~Terry Moreau, Electronics Designer~~~~~
ooo~\\_//~ooo~~~~~~PGP available on request~~~~~~~~~~~~~~~
THE CURIOUS SEEK~~~tmo...@direct.ca~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Some fish get eaten by sharks!~~Others get caught in a NET
Life's a beach, the whales get stuck & sand crabs clean up
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
terry moreau
Richard Torrens <4...@argonet.co.uk> wrote:
>In fact a lot of semiconductors, when driving a relay, will survive the
>inductive kick-back: they go into avalanche or zener breakdown. Whether they
>survive this depends on the energy stored in the coil (1/2 L I squared) and
>whether this energy is within the acceptable repetitive breakdown rating of
>the semiconductor.
True..but in practice the avalanche rating is HIGHLY variable and one of the
most widest varying and most difficult to control in manufacture and in
operation of semiconductors. You are putting an enormous peak stress on the
semiconductor...and breif as it may be there is still a failure point which is
highly unpredictable. If you've designed equipment that comes close to
avalanche ratings...I would never buy it! You are only asking for reduced
reliability needlessly! Data sheets rarely explain that it is " typical" data
and that in in reality the peak energy rating in avalanche can vary 4:1. Some
devices are 100% tested at avalanche conditions however cummulative effects are
not considered. After years of this repetivitive punishment these published
ratings may no longer be valid! NEVER NEVER design to the limit on avalanche if
you can avoid it!
Isaac Wingfield
> Richard Torrens <4...@argonet.co.uk> wrote:
>
> >In fact a lot of semiconductors, when driving a relay, will survive the
> >inductive kick-back: they go into avalanche or zener breakdown. Whether they
> >survive this depends on the energy stored in the coil (1/2 L I squared) and
> >whether this energy is within the acceptable repetitive breakdown rating of
> >the semiconductor.
>
Another neat trick, if you need a faster turn-off, is to use no diode, and
put a Miller cap on the driving transistor (you *do* use a discrete driver,
don't you?). Use a high beta device for the driver, and select a high
breakdown voltage (80-100 V) for faster turnoff.
By "tuning" the cap, you can control how fast the relay drops, by allowing
a higher voltage to show up across the driver. Done properly, this is
perfectly safe, as there is no uncontrolled collapse of the coil's field.
Dropout time can be significantly reduced.
Isaac
Richard Torrens
In article <5360he$2...@orb.direct.ca>, tmo...@direct.ca (terry moreau)
wrote:
>
>
>
> Here is the straight poop from The Relay Handbook...
>
>
> Use the diode across the coil when concerned about back emf in the driving
> circuit and when the relay contacts are NOT driving and inductive load.
[rest snipped]
Would someone like to explain this! Some authors write cr*p!
How can the load on the contacts of a relay have ANY effect on the back emf
of the coil!
Just because it's in a book does NOT mean it is corrrect!
Richard Torrens
In article <5361vb$6...@orb.direct.ca>, tmo...@direct.ca (terry moreau)
wrote:
>
>
> >In fact a lot of semiconductors, when driving a relay, will survive the
> >inductive kick-back: they go into avalanche or zener breakdown. Whether
> they
> >survive this depends on the energy stored in the coil (1/2 L I squared)
> and
> >whether this energy is within the acceptable repetitive breakdown rating
> of
> >the semiconductor.
>
> the
> most widest varying and most difficult to control in manufacture and in
> operation of semiconductors. You are putting an enormous peak stress on
> the semiconductor...and breif as it may be there is still a failure point
> which is highly unpredictable. If you've designed equipment that come
> for reduced reliability needlessly! Data sheets rarely explain that it
> is " typical" data and that in in reality the peak energy rating in
> avalanche can vary 4:1.
> Some devices are 100% tested at avalanche conditions however cummulative
> effects are not considered. After years of this repetivitive punishment
> these published ratings may no longer be valid! NEVER NEVER design to the
> limit on avalanche if you can avoid it!
'Semiconductors' as Terry uses the description is FAR too sweeping a
statement.
Older semiconductors did not have specified avalanche ratings for the
reasons Tery states.
However Avalanche effect devices are available and are designed to be used
in Avalanche mode.
Also modern power MOSFETs do have Avalanche energy specified. It there is a
peak, single occurrence rating - I would not design for this. However if
there is a guaranteed safe repetitive rating this can safely be used (or is
Terry accusing all MOSFET manufactureres of lying).
As with all specs - it is generally best to UNDERSTAND what they mean and to
use them as the manufacturer intended. I.e. don't approach too close to
maximum ratings.
Yes - most bipolar devices are very unpredictable, though I have noticed
more predictability with improving technology. There are also voltage
breakdown mechanisms other than Avalanche (e.g. zener breakdown).
Zener breakdown is always positive slope: avalanche may have a negative
slope and devices can go into oscillation because of this negative slope.
This is a good reason for not using avalanche.
However - if the manufacturer gives specifications for the avalanche mode
then he is predicting his own product's performance. These specifications
are intended to be used.
Ralph R. Boudreaux
In article <32528D...@mmm.com>, rbmcc...@mmm.com (Roy McCammon) writes:
>Actualy, when you use just a diode, the energy is mainly disapated in the
>resistence of the relay coil. So, assuming low resistence drivers and
>low
>resistence diodes, the electrical time constant of enerization is about
>the same as de-energization.
Wrong. Energizing the coil places the supply voltage across the terminals.
A simple diode clamps the discharge voltage to the forward voltage of the
diode. For a 12V supply, the turn off time will be roughly 12 times longer
than the turn on time. The quickest (and simplest) way to turn one off is
to use an avalanche rated MOSFET with the highest practical breakdown
voltage and *no* diode. The turn off voltage will be the breakdown voltage
of the FET minus the supply voltage.
-Rusty Boudreaux
>
>Good designs meet the system requirements without extra parts or cost. A
>simple clamp diode will usually meet the requirements. Sometimes you
>really
>do want to get as much speed out of the relay as possible and then a
>series
>resister will help.
>
Winfield Hill
Isaac Wingfield, <i...@witzend.com> said...
>
>Another neat trick, if you need a faster turn-off, is to use no diode, and
>put a Miller cap on the driving transistor (you *do* use a discrete driver,
>don't you?). Use a high beta device for the driver, and select a high
>breakdown voltage (80-100 V) for faster turnoff.
A second key ingedient for sucess with this scheme: controlled drive current.
The miller integrator speed is be determined not only by the feedback
capacitor, but by the circuit's drive current. Therefore a current-limiting
resistor, or some similar scheme, must be employed to keep this parameter
predictable. Then you can sleep at night, knowing that a change in h_FE or
some other volatile parameter won't cause a failure!
Sleep. That's next for me!
--
Winfield Hill hi...@rowland.org
The Rowland Institute for Science _/_/_/ _/_/_/_/
100 Edwin Land Blvd. _/ _/ _/_/ _/
Cambridge, MA USA 02142-1297 _/_/_/_/ _/ _/ _/_/_/
_/ _/ _/ _/ _/
http://www.artofelectronics.com/ _/ _/ _/_/ _/_/_/_/
Pin 2 Hot
Richard Torrens wrote:
>
> In article <5360he$2...@orb.direct.ca>, tmo...@direct.ca (terry moreau)
> wrote:
> >
> > RELAY COIL DIODES...the bottom line!
> >
> >
> > Here is the straight poop from The Relay Handbook...
> >
> >
> > Use the diode across the coil when concerned about back emf in the driving
> > circuit and when the relay contacts are NOT driving and inductive load.
>
> [rest snipped]
>
> Would someone like to explain this! Some authors write cr*p!
>
> How can the load on the contacts of a relay have ANY effect on the back emf
> of the coil!
I think what he meant was that a diode on the coil will slow down the
contacts opening, and with an inductive load, will promote arcing of the
contacts due to EMF from the load.
Gene
Roy McCammon
Ralph R. Boudreaux wrote:
>
> In article <32528D...@mmm.com>, rbmcc...@mmm.com (Roy McCammon) writes:
> >Actualy, when you use just a diode, the energy is mainly disapated in the
> >resistence of the relay coil. So, assuming low resistence drivers and
> >low
> >resistence diodes, the electrical time constant of enerization is about
> >the same as de-energization.
>
> Wrong. Energizing the coil places the supply voltage across the terminals.
right
> A simple diode clamps the discharge voltage to the forward voltage of the
> diode.
right, but remember that the inductance of the relay coil doesn't see the
terminal voltage, rather it sees the terminal voltage in series with the voltage
drop across the copper resistence of its winding.
For a 12V supply, the turn off time will be roughly 12 times longer
> than the turn on time.
Wrong. Example. Assume 12 volt supply. Vsat of driver = .4V,
Vf of diode = .7V, winding resitence = 100 ohms. Energize the relay,
the inductance sees 12-.4 = 11.6V initially. After everything settles,
the current is 11.6V/100 ohms = 116 ma. The winding resistence is absorbing
the 11.6V of terminal voltage. The voltage across the inductance is the
terminal voltage (11.6V) minus the drop accross the copper (11.6V) or zero.
Now, de-energize the relay. The 116ma is still flowing and producing 11.6V
but the terminal voltage is now -0.7V, so the inductance sees
-0.7V - 11.6V = -12.3V, so you see the voltage that is initially opposing the
relay current is actually a little higher upon discharge than the voltage
that tended to establish that current upon enerization.
Perhaps you remember that the electrical time constant is just L/R. Assuming
that both the diode and the driver are both zero resistence when "on" then
R is just the winding resistence of the relay and is the same is both cases.
Of course I assume that L is the same in both cases which isn't necessarily true,
since the magnetic circuit in the relay does change.
> The quickest (and simplest) way to turn one off is
> to use an avalanche rated MOSFET with the highest practical breakdown
> voltage and *no* diode. The turn off voltage will be the breakdown voltage
> of the FET minus the supply voltage.
I'd have to agree that would be simple and fast.
John van Veen
Richard Torrens (4...@argonet.co.uk) wrote:
: In article <5360he$2...@orb.direct.ca>, tmo...@direct.ca (terry moreau)
: wrote:
: >
: > RELAY COIL DIODES...the bottom line!
: >
: >
: > Here is the straight poop from The Relay Handbook...
: >
: >
: > Use the diode across the coil when concerned about back emf in the driving
: > circuit and when the relay contacts are NOT driving and inductive load.
:
: [rest snipped]
:
: Would someone like to explain this! Some authors write cr*p!
:
: How can the load on the contacts of a relay have ANY effect on the back emf
: of the coil!
:
I think what they are talking about is doing something with the energy
stored in the inductive circuit/load. For instance, a 3-phase variable
speed ac drive I used to service would switch the load to a large
grounding resistor whenever it was turned off.
have a better day
it annoys those won't
John
Matthew Williams
I hope you guys also use a resistor/capacitor snubber across the coil
also. The pair will snub the initial voltage spike for a few
microseconds which is necessary for the mosfet diode to turn on. You
could use a siemens BTS117, SGS vnp07n04, or similiar mosfet with
Overvoltage, Current limit, and Thermal limit protection built in. When
you let the mosfet do all the snubbing, you need to make sure you do not
exceed the power rating of the device. ie: a 72v breakdown voltage @ 2
amps of inductor current causes the mosfet to heat up quickly. As
switching frequency increases, so does the power dissipated by the
mosfet. At work, we have yet to destroy the siemens or sgs part with
overvoltage, overcurrent, and heatgun thermal limiting.
Matthew
Pin 2 Hot
I think that resistor was for dynamic braking of the motor.
Roy McCammon
John van Veen wrote:
>
> Richard Torrens (4...@argonet.co.uk) wrote:
> : In article <5360he$2...@orb.direct.ca>, tmo...@direct.ca (terry moreau)
> : wrote:
> : > RELAY COIL DIODES...the bottom line!
> : > Here is the straight poop from The Relay Handbook...
> : > Use the diode across the coil when concerned about back emf in the driving
> : > circuit and when the relay contacts are NOT driving and inductive load.
> :
> : [rest snipped]
> :
> : Would someone like to explain this! Some authors write cr*p!
> :
> : How can the load on the contacts of a relay have ANY effect on the back emf
> : of the coil!
The conern is that an inductive load causes possible destructive arcing on the
relay contacts, hence a desire to operate the contacts as quickly as possible.
There are methods which collapse the magnetic field of the relay faster than
the simple diode suppressor and would thus be expected to allow the relay
to open the contacts more quickly.
mcn...@snet.net
rbmcc...@mmm.com (Roy McCammon) wrote:
Inductors (relay coil) don't like current to be interrupted suddenly.
Diodes across the coil, give it a path to flow while it decays
gradually.