Short Circuit Protection; addendum
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Partial quote of my former reply:
"Now we know that in order to not exceed 0.025 amps at Vcc of 5v, we'll need
a total Rl of 200 ohms. I've assumed that your Rl is 50 ohms; therefore
you'll need to add 150 ohms to the circuit somewhere. Adding the extra
resistance will mean that your voltage across the load will be very
significantly decreased. This might be OK if your load is inductive (eg: a
relay's coil). If not, I suggest examining a different type of device that
is capable of handling more output current."
Note the underlined sentence; I omitted a reference to the use of a
"clamping diode" across an inductive load (relay coils, stepper motor
windings, etc). This is very important when you consider the
characteristics of an inductive load.
You can think of an inductive load as "inertia" in the real world. Hop in
your car, and head out on the freeway. You don't just hit the "go 65"
button and instantly proceed at 65MPH - your car has mass, which must be
accelerated from 0mph to 65mph by the engine.
Inductive loads are very similar, when you consider current flow through the
device. In the first few nanoseconds of the application of power, there is
practically zero current flow. It takes time for the magnetic field to
build up around the inductor, during which the current flow becomes
increasingly strong. Ok, now we're humming along - the magnetic field has
built up, and we have current flow through the inductive load. Suddenly,
the power is cut off (your transistor is driven into cutoff). What happens
now?
Back to the car analogy. You're humming along at 65mph, and someone yells
"STOP!" You notice a huge boulder by the side of the road, and slam into it
head-on, transferring thousands of foot-pounds of energy to the boulder,
moving it several feet. Effective, but undesireable, as the car is much
worse the wear after this stunt.
Back at the inductive load - it was humming along happily with current
flowing through it. Suddenly, there is no path for the current to flow
through, so the inverse voltage across the inductor builds quite high in an
attempt to maintain the current flow - it's bound & determined to keep those
'trons moving. All of a sudden, you have 40v to 100v (guesstimate) in
reverse polarity across the inductor, and your transistor won't be happy
about this high voltage. What to do about this? Place a "clamping" diode
across the inductive load. The diode should be wired in so that when Vcc is
applied to the coil, no current flows through the diode. It's simply there
to provide a path for that reverse polarity current generated by the
collapsing magnetic field of the coil. Your drivers will be subjected to
much less "stress" afterwards.
It doesn't even require a very high-rated diode - just a general purpose
diode with a PIV rating of your Vcc + 50% or so. If your Vcc is 5v, then
choose a diode with a PIV of 7.5v or greater.
Hope this helps...
Steve
"Now we know that in order to not exceed 0.025 amps at Vcc of 5v, we'll need
a total Rl of 200 ohms. I've assumed that your Rl is 50 ohms; therefore
you'll need to add 150 ohms to the circuit somewhere. Adding the extra
resistance will mean that your voltage across the load will be very
significantly decreased. This might be OK if your load is inductive (eg: a
relay's coil). If not, I suggest examining a different type of device that
is capable of handling more output current."
Note the underlined sentence; I omitted a reference to the use of a
"clamping diode" across an inductive load (relay coils, stepper motor
windings, etc). This is very important when you consider the
characteristics of an inductive load.
You can think of an inductive load as "inertia" in the real world. Hop in
your car, and head out on the freeway. You don't just hit the "go 65"
button and instantly proceed at 65MPH - your car has mass, which must be
accelerated from 0mph to 65mph by the engine.
Inductive loads are very similar, when you consider current flow through the
device. In the first few nanoseconds of the application of power, there is
practically zero current flow. It takes time for the magnetic field to
build up around the inductor, during which the current flow becomes
increasingly strong. Ok, now we're humming along - the magnetic field has
built up, and we have current flow through the inductive load. Suddenly,
the power is cut off (your transistor is driven into cutoff). What happens
now?
Back to the car analogy. You're humming along at 65mph, and someone yells
"STOP!" You notice a huge boulder by the side of the road, and slam into it
head-on, transferring thousands of foot-pounds of energy to the boulder,
moving it several feet. Effective, but undesireable, as the car is much
worse the wear after this stunt.
Back at the inductive load - it was humming along happily with current
flowing through it. Suddenly, there is no path for the current to flow
through, so the inverse voltage across the inductor builds quite high in an
attempt to maintain the current flow - it's bound & determined to keep those
'trons moving. All of a sudden, you have 40v to 100v (guesstimate) in
reverse polarity across the inductor, and your transistor won't be happy
about this high voltage. What to do about this? Place a "clamping" diode
across the inductive load. The diode should be wired in so that when Vcc is
applied to the coil, no current flows through the diode. It's simply there
to provide a path for that reverse polarity current generated by the
collapsing magnetic field of the coil. Your drivers will be subjected to
much less "stress" afterwards.
It doesn't even require a very high-rated diode - just a general purpose
diode with a PIV rating of your Vcc + 50% or so. If your Vcc is 5v, then
choose a diode with a PIV of 7.5v or greater.
Hope this helps...
Steve
Comments
"inverse voltage" build up when the current path is removed? Maybe I need to
better understand the part about "humming along - the magnetic field has built
up, and we have current flow through the inductive load".
Sorry that this is (I'm guessing) more a physics question that purely a Basic
Stamp question.
Thanks,
Kevin
Steve Wilke wrote:
> Back at the inductive load - it was humming along happily with current
> flowing through it. Suddenly, there is no path for the current to flow
> through, so the inverse voltage across the inductor builds quite high in an
> attempt to maintain the current flow - it's bound & determined to keep those
> 'trons moving. All of a sudden, you have 40v to 100v (guesstimate) in
> reverse polarity across the inductor, and your transistor won't be happy
> about this high voltage.
to pass a wire through a magnetic field to observe the electric current
being generated as the wire is in motion.
A relay or solenoid contains a length of wire wound in a coil -- this is an
electromagnet. Power is applied to the coil and the electromagnet pulls on a
metal plate or plunger and closes the electric contacts. As long as the
power is applied, a constant magnetic field surrounds the coil.
When power to the device is turned off, the magnetic field surrounding the
coil of a relay or solenoid collapses and it generates a reverse voltage
which may damage the device used to control the relay. A diode is hooked up
backwards across the coil to short-circuit this reverse voltage.
Think about making a wave in the bathtub -- it goes away from you until it
hits the side of the tub, then its reflected back.
************
Original Message
> So I know I should have studied this in school. But why does such a large
> "inverse voltage" build up when the current path is removed? Maybe I need
to
> better understand the part about "humming along - the magnetic field has
built
> up, and we have current flow through the inductive load".
>
> Sorry that this is (I'm guessing) more a physics question that purely a
Basic
> Stamp question.
>
steady magnetic field and a wire doesn't do anything. That's why generators
have to rotate (or at least move) and transformers don't work at DC.
So when the relay is on, the current flowing through it is steady and that's
OK. When the transistor breaks the circuit, the difference in voltage
between the two coil terminals drops to 0. This makes the current drop to 0
also. Changing magnetic field + wire = electricity! The faster the drop, the
more voltage is generated.
Years ago, I worked for a large semi company as a failure analyst. Our tech
assesment group was building a big transistor current source circuit with
power transistors. The problem with working on things that fail is you tend
to think everything fails (especially if you have no experience). At lunch
these guys told me how they had bought 5 or 6 transistors for this circuit
"and they were all bad!" I told them that didn't sound likely, and asked if
I could see the schematics. They showed me and it looked fine on paper.
Finally, I asked to see the real circuit. The emitter resistor was wire
wound. These resistors have significant inductance. Base width modulation
caused small but rapid fluctuations in the emitter voltage. These changes
were causing inductive voltage spikes and that would kill the BE junction in
the transistor. Solution: non-inductive wire wound resistors (these are
wound 1/2 in one direction and then 1/2 in the other so the inductance
cancels).
Hope that helps.
Al Williams
AWC
* Floating point A/D
http://www.al-williams.com/awce/pak9.htm
>
Original Message
> From: Kevin Olalde [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=GPG1FxDAcWn4D5zuE8IPE7NxijoP5DpP8CoX_k1voeOei3WRT1Yu5Q26a78ZSpzlLhiJyjwcm0WbirA]kolalde@h...[/url
> Sent: Thursday, October 04, 2001 7:59 AM
> To: basicstamps@yahoogroups.com
> Subject: Re: [noparse][[/noparse]basicstamps] re: Short Circuit Protection; addendum
>
>
> So I know I should have studied this in school. But why does such a large
> "inverse voltage" build up when the current path is removed?
> Maybe I need to
> better understand the part about "humming along - the magnetic
> field has built
> up, and we have current flow through the inductive load".
>
> Sorry that this is (I'm guessing) more a physics question that
> purely a Basic
> Stamp question.
>
> Thanks,
> Kevin
>
> Steve Wilke wrote:
>
> > Back at the inductive load - it was humming along happily with current
> > flowing through it. Suddenly, there is no path for the current to flow
> > through, so the inverse voltage across the inductor builds
> quite high in an
> > attempt to maintain the current flow - it's bound & determined
> to keep those
> > 'trons moving. All of a sudden, you have 40v to 100v (guesstimate) in
> > reverse polarity across the inductor, and your transistor won't be happy
> > about this high voltage.
>
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>
We were looking at some BOSCH relays and they listed an option that
included a 'shunt resistor' and the ones we currently purchase with an
included diode. Now, the question is why is the shunt resistor there? It
is my impression that the shunt is a LOW resistance value for other
purposes than to protect a relay coil. Am I getting wrapped up in the
wording and missing the point?
TIA,
Mark
Original Message
From: Rodent [noparse]/noparse]SMTP:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=UBRBwmPbpJcRxBb1oJ71xxnU7SpxrAmn0Ajb25SHQ_KlFWI4cw2EgIepkIiFcjxOuWih4tJiqYtdYJPC0g]daweasel@s...[/url
Sent: Thursday, October 04, 2001 8:09 AM
To: basicstamps@yahoogroups.com
Subject: Re: [noparse][[/noparse]basicstamps] re: Short Circuit Protection; addendum
One of the science experiments you typically do in middle or high school is
to pass a wire through a magnetic field to observe the electric current
being generated as the wire is in motion.
A relay or solenoid contains a length of wire wound in a coil -- this is an
electromagnet. Power is applied to the coil and the electromagnet pulls on
a
metal plate or plunger and closes the electric contacts. As long as the
power is applied, a constant magnetic field surrounds the coil.
When power to the device is turned off, the magnetic field surrounding the
coil of a relay or solenoid collapses and it generates a reverse voltage
which may damage the device used to control the relay. A diode is hooked up
backwards across the coil to short-circuit this reverse voltage.
Think about making a wave in the bathtub -- it goes away from you until it
hits the side of the tub, then its reflected back.
************
Original Message
> So I know I should have studied this in school. But why does such a
large
> "inverse voltage" build up when the current path is removed? Maybe I
need
to
> better understand the part about "humming along - the magnetic field has
built
> up, and we have current flow through the inductive load".
>
> Sorry that this is (I'm guessing) more a physics question that purely a
Basic
> Stamp question.
>
To UNSUBSCRIBE, just send mail to:
basicstamps-unsubscribe@yahoogroups.com
from the same email address that you subscribed. Text in the Subject and
Body of the message will be ignored.
Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
similar function as the clamping diode would. However, a simple shunt
resistor would, of course, increase the current dissipation of the circuit,
requiring a more robust power supply. This practice, IMHO, is not
environmentally friendly, as increased power consumption demands more from
your local power company (or more gasoline, if an automotive application)
Re-visiting the clamping diode theory: let's say the inductive load (relay
coil, in your case) is energized, and we open the circuit. The collapsing
magnetic field attempts to continue the current flow; however the only path
is through the clamping diode. The current flow may be quite high, which
may exceed the rating of the clamping diode. To fix this problem, we could
add a current limiting resistor in series with the clamping diode; a
properly selected resistor value would prevent voltage "spikes" across the
inductive load, while limiting the amount of current through the clamping
diode. Variables to consider here are the impedance of the inductor
(Henries), the voltage applied across the inductive load, and the max
current rating of the clamping diode.
Hope this helps...
Steve
Mark Adams [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=VCz4thFLjM2NmGAGfQWgD5MQITCcNIbOrokzOd6iozb7azl7voHPUqAiACLkboUBQcGvixX_U8D8TA]madams@a...[/url wrote:
Here is another basic question...
We were looking at some BOSCH relays and they listed an option that
included a 'shunt resistor' and the ones we currently purchase with an
included diode. Now, the question is why is the shunt resistor there? It
is my impression that the shunt is a LOW resistance value for other
purposes than to protect a relay coil. Am I getting wrapped up in the
wording and missing the point?
<snip>