Battery Contactor Diode?

"Scott" wrote in message
.. .
Nope, not really a concerning factor. Any of the "garden variety"
diodes in the 1N400X series should be fine. Just be sure to put the
cathode to the +12 or +24 volt side of the contactor coil and the anode
to the ground side (assuming a negative ground system).

I am not very good with the electron understanding, but I would be grateful
to understand this whole discussion.

Is the cathode normally the downstream side of the diode? What does the
installation like this, do, to help with the current spike?
--
Jim in NC

"Morgans" wrote

I am not very good with the electron understanding, but I would be
grateful
to understand this whole discussion.

Is the cathode normally the downstream side of the diode? What does the
installation like this, do, to help with the current spike?

Oh, I also want to add, that if Jim Weir responds, will someone reply to the
post with his entire post quoted?

For some reason, my ISP seems to have "blacklisted" his ISP, so the only
way I see his posts is in a response.

Also, to say I don't understand electrons, is a huge understatement! g
--
Jim in NC

"Morgans" wrote in message
news

"Scott" wrote in message
.. .

Nope, not really a concerning factor. Any of the "garden variety"
diodes in the 1N400X series should be fine. Just be sure to put the
cathode to the +12 or +24 volt side of the contactor coil and the anode
to the ground side (assuming a negative ground system).

Scott, at and above the 1N4004, the chip inside the case is physically more
robust and will take a spike of current a little better than, say, a 1N4001.





I am not very good with the electron understanding, but I would be
grateful
to understand this whole discussion.

Is the cathode normally the downstream side of the diode? What does the
installation like this, do, to help with the current spike?


Jim, as you know, a diode conducts current in one direction and blocks it
from conducting in the other direction. A positive voltage on the anode
forward biases the diode and the voltage will appear as a positive voltage
(less some small conduction drop of a volt or so) at the cathode. A
negative voltage on the anode will reverse bias the diode and it will not
appear as a negative voltage at the cathode.

Likewise, a negative voltage at the cathode will appear as a negative
voltage at the anode.

When a coil has a collapsing field, it produces a spike of energy as the
field attempts to keep the current through the coil constant. THis large
negative spike can and will cause some of the aircraft electronics to fail
catastrophically. One way of making sure that negative spike doesn't kill
the avionics is to shunt it to ground through a forward biased diode. That
way the most that can sneak through is a volt or so as opposed to several
hundred volt spikes without the diode. How do you forward bias a diode with
a negative voltage? You connect the CATHODE to the "hot" +12 volt terminal
of the coil and the ANODE to ground. THe negative spike is effectively
clipped at a volt or thereabouts.

Jim

Bummer. ANd I just switched ISPs also. I'm now using wildblue and if
you've got wildblue blocked, you won't get any of us using satellite ISPs.

Jim


For some reason, my ISP seems to have "blacklisted" his ISP, so the only
way I see his posts is in a response.
--
Jim in NC

Hi Jim,
Schematically, the cathode is the bar and the anode is the arrow part.

With the diode connected with its cathode to the positive voltage and
its anode connected to ground, it is reverse biased and looks like an
open circuit during normal operation. When the contactor relay (master
solenoid) is turned off, the collapsing magnetic field will generate a
field of opposite polarity (the voltage spike we are talking about) such
that the side of the coil that is connected to the +12V bus will now be
at a large NEGATIVE voltage, putting the cathode of the diode more
negative than its anode and it will now conduct. The voltage across a
conducting diode is about 0.7 Volts, so the spike will be limited to
about a -0.7V along the +12V bus in the aircraft. Without the diode,
hundreds of negative volts could be placed on the bus (for a short
period of time, in the milliseconds), but solid state devices don't get
along well with these high voltages, ESPECIALLY since it is opposite
polarity of what they are designed to operate at.

Scott


Morgans wrote:

"Scott" wrote in message
.. .

Nope, not really a concerning factor. Any of the "garden variety"
diodes in the 1N400X series should be fine. Just be sure to put the
cathode to the +12 or +24 volt side of the contactor coil and the anode
to the ground side (assuming a negative ground system).


I am not very good with the electron understanding, but I would be grateful
to understand this whole discussion.

Is the cathode normally the downstream side of the diode? What does the
installation like this, do, to help with the current spike?

Morgans wrote:
"Morgans" wrote

I am not very good with the electron understanding, but I would be
grateful
to understand this whole discussion.

Is the cathode normally the downstream side of the diode? What does the
installation like this, do, to help with the current spike?

Oh, I also want to add, that if Jim Weir responds, will someone reply to the
post with his entire post quoted?

For some reason, my ISP seems to have "blacklisted" his ISP, so the only
way I see his posts is in a response.

Also, to say I don't understand electrons, is a huge understatement! g
--
Jim in NC


Try reading RAH here

http://groups.google.com/group/rec.a...nk=gschg&hl=en

"Scott" wrote in message
news
Hi Jim,
Schematically, the cathode is the bar and the anode is the arrow part.

With the diode connected with its cathode to the positive voltage and
its anode connected to ground, it is reverse biased and looks like an
open circuit during normal operation. When the contactor relay (master
solenoid) is turned off, the collapsing magnetic field will generate a
field of opposite polarity (the voltage spike we are talking about) such
that the side of the coil that is connected to the +12V bus will now be
at a large NEGATIVE voltage, putting the cathode of the diode more
negative than its anode and it will now conduct. The voltage across a
conducting diode is about 0.7 Volts, so the spike will be limited to
about a -0.7V along the +12V bus in the aircraft. Without the diode,
hundreds of negative volts could be placed on the bus (for a short
period of time, in the milliseconds), but solid state devices don't get
along well with these high voltages, ESPECIALLY since it is opposite
polarity of what they are designed to operate at.

Thanks, that was a clear explanation (I think g) of the type I was
looking for.

So the .7 volts in buss at the wrong polarity is not too much for all of the
electronics to deal with?
--
Jim in NC

Correct!

Scott

Morgans wrote:

"Scott" wrote in message
news
Hi Jim,
Schematically, the cathode is the bar and the anode is the arrow part.

With the diode connected with its cathode to the positive voltage and
its anode connected to ground, it is reverse biased and looks like an
open circuit during normal operation. When the contactor relay (master
solenoid) is turned off, the collapsing magnetic field will generate a
field of opposite polarity (the voltage spike we are talking about) such
that the side of the coil that is connected to the +12V bus will now be
at a large NEGATIVE voltage, putting the cathode of the diode more
negative than its anode and it will now conduct. The voltage across a
conducting diode is about 0.7 Volts, so the spike will be limited to
about a -0.7V along the +12V bus in the aircraft. Without the diode,
hundreds of negative volts could be placed on the bus (for a short
period of time, in the milliseconds), but solid state devices don't get
along well with these high voltages, ESPECIALLY since it is opposite
polarity of what they are designed to operate at.


Thanks, that was a clear explanation (I think g) of the type I was
looking for.

So the .7 volts in buss at the wrong polarity is not too much for all of the
electronics to deal with?

If a switch is used to apply the 12 volts to the solenoid coil, and then the switch is opened, how does the voltage
generated during the collapse of the magnetic field get to the positive buss?


"Scott" wrote in message news : Hi Jim,
: Schematically, the cathode is the bar and the anode is the arrow part.
:
: With the diode connected with its cathode to the positive voltage and
: its anode connected to ground, it is reverse biased and looks like an
: open circuit during normal operation. When the contactor relay (master
: solenoid) is turned off, the collapsing magnetic field will generate a
: field of opposite polarity (the voltage spike we are talking about) such
: that the side of the coil that is connected to the +12V bus will now be
: at a large NEGATIVE voltage, putting the cathode of the diode more
: negative than its anode and it will now conduct. The voltage across a
: conducting diode is about 0.7 Volts, so the spike will be limited to
: about a -0.7V along the +12V bus in the aircraft. Without the diode,
: hundreds of negative volts could be placed on the bus (for a short
: period of time, in the milliseconds), but solid state devices don't get
: along well with these high voltages, ESPECIALLY since it is opposite
: polarity of what they are designed to operate at.
:
: Scott
:
:
: Morgans wrote:
:
: "Scott" wrote in message
: .. .
:
: Nope, not really a concerning factor. Any of the "garden variety"
: diodes in the 1N400X series should be fine. Just be sure to put the
: cathode to the +12 or +24 volt side of the contactor coil and the anode
: to the ground side (assuming a negative ground system).
:
:
: I am not very good with the electron understanding, but I would be grateful
: to understand this whole discussion.
:
: Is the cathode normally the downstream side of the diode? What does the
: installation like this, do, to help with the current spike?

..Blueskies. wrote:
If a switch is used to apply the 12 volts to the solenoid coil, and then the switch is opened, how does the voltage
generated during the collapse of the magnetic field get to the positive buss?


The master solenoid coil has one terminal connected to the hot
(battery +) switching terminal, and the other goes to the master
switch, which grounds it when the switch is on. In this way, there's no
direct connection to the battery's positive terminal and no unloaded
feed line (with its risk of short-circuit and fire) leading into the
cabin.
When the master switch is turned off, the abrupt arresting of
the current causes a collapse of the magnetic field within the
solenoid's coil. This collapsing field tries to keep the current
flowing, and it is violent enough that the electrons will jump the gap
in the switch and complete the circuit to some degree. The spike is
extremely sudden and can be arcing across the switch's contact gap
before those contacts are very far apart. The electron flow is from
ground, through the partially open master switch contacts, through the
coil and into the solenoid's hot terminal, then through the solenoid's
big contacts (which are still closed during this almost instantaneous
event) and into the aircraft's electrical system. Any sensitive
electronic device (like an expensive radio) without its own internal
protection goes poof.

It's the same sort of process in the magneto. When the points
open, coil current is arrested and the collapse of the field causes a
big spike in the secondary winding, making spark plugs do their thing.
The points have a capacitor across them to absorb the initial surge
that tries to jump them as they open; an arc here would both burn the
points and prevent the really sudden arresting of the current that we
want to get the biggest jolt out of the magneto. The capacitor will
fill up, and by that time the points are too far apart for the curent
to jump them, so the arresting is more complete. That magneto won't
work worth a hoot without the capacitor (sometimes called condenser).
A diode would definitely not work in the magneto; we *want* a
big spike.
The diode across the solenoid's coil provides an easy path
from the solenoid's positive terminal back to it's negative, almost
completely draining off the spike and keeping it out of the system.

A long answer to a simple question, huh?

Dan