Off-topic, but the learned denizens here are pretty good at asking bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note that
this is a one time test circuit, and I don't want to break the bank. My
initial thought was to parallel a couple of 28 VDC relay coils to get the 12
A current. What I'm finding is that most of the relay coils draw less than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero luck
finding out how much an automotive starter solenoid draws, if I could find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.

Why does the load need to be inductive?

Bob Swinney
"Ken Finney" wrote in message
...
Off-topic, but the learned denizens here are pretty good at asking bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note that
this is a one time test circuit, and I don't want to break the bank. My
initial thought was to parallel a couple of 28 VDC relay coils to get the
12
A current. What I'm finding is that most of the relay coils draw less
than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero
luck
finding out how much an automotive starter solenoid draws, if I could find
a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.

On Wed, 30 Mar 2005 18:34:59 GMT, "Ken Finney"
wrote:

Off-topic, but the learned denizens here are pretty good at asking bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note that
this is a one time test circuit, and I don't want to break the bank. My
initial thought was to parallel a couple of 28 VDC relay coils to get the 12
A current. What I'm finding is that most of the relay coils draw less than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero luck
finding out how much an automotive starter solenoid draws, if I could find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.

Use the run winding of an ordinary induction motor. I just measured
the inductance of a 1/3 HP 110-volt Dayton motor to be about 5.4 mH.
You might need a bit of series resistance to keep the DC current down
to 12 amps. The equivalent circuit for any inductive load is a pure
inductor in series with it's own resistance, so adding series
resistance to get the current you want is a valid approach.

In article ,
Don Foreman wrote:

On Wed, 30 Mar 2005 18:34:59 GMT, "Ken Finney"
wrote:

Off-topic, but the learned denizens here are pretty good at asking bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note that
this is a one time test circuit, and I don't want to break the bank. My
initial thought was to parallel a couple of 28 VDC relay coils to get the 12
A current. What I'm finding is that most of the relay coils draw less than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero luck
finding out how much an automotive starter solenoid draws, if I could find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.

Use the run winding of an ordinary induction motor. I just measured
the inductance of a 1/3 HP 110-volt Dayton motor to be about 5.4 mH.
You might need a bit of series resistance to keep the DC current down
to 12 amps. The equivalent circuit for any inductive load is a pure
inductor in series with it's own resistance, so adding series
resistance to get the current you want is a valid approach.

?????

But when working with DC, other than "moving" stuff (Motors, speakers,
relays - something that "taps" energy from the magnetic field in some
manner, rather than leaving it stand there idle) an inductor is
effectively nothing but a dead short. That's one of the reasons
transformers are worthless for changing voltage up or down in DC
systems. Or so I've been led to believe my for entire "electronics
life"... ?

If enough energy is being pumped into it, heat starts building, and in
surprisingly (to those who haven't seen it happen before...) short
order, the whole shebang has transformed itself into a (puddle of
melted)/(cloud of vaporized) conductor, an open circuit, and an
optional, variably sized, cloud of smoke.

You've succeeded in confusing me...

Should I be reading that last quoted sentence of yours as saying that an
inductive load is the same thing as a coil and resistor (of arbitrary
value) wired in series, or are you saying that an inductive load is the
same thing as a coil wired to a resistor matching the resistance of the
coil? I'm definitely hazy as to which was your intent. Your (seeming)
misuse of "it's" doesn't help clarify things any...

(Hint: "its" = "belonging to it" - "it's" = "it is")

--
Don Bruder - - New Email policy in effect as of Feb. 21, 2004.
Short form: I'm trashing EVERY E-mail that doesn't contain a password in the
subject unless it comes from a "whitelisted" (pre-approved by me) address.
See http://www.sonic.net/~dakidd/main/contact.html for full details.

Ken Finney wrote:
Off-topic, but the learned denizens here are pretty good at asking bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note that
this is a one time test circuit, and I don't want to break the bank. My
initial thought was to parallel a couple of 28 VDC relay coils to get the 12
A current. What I'm finding is that most of the relay coils draw less than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero luck
finding out how much an automotive starter solenoid draws, if I could find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.

Just a quick back-of-an-envelop calculation
shows you'd need about 500 feet of 16 gauge
wire for a dc resistance of 2.3 ohms in your
coil. I'd not want to use anything less than
16 gauge and even then for just a short time.
It's going to have to dissipate 336 watts once
the core is saturated.

That's a big coil, either to wind or to find
in an existing device.

Could you tell us why you need to do this.
We might be able to come up with a better
approach.

In article ,
"Ken Finney" wrote:

Off-topic, but the learned denizens here are pretty good at asking bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note that
this is a one time test circuit, and I don't want to break the bank. My
initial thought was to parallel a couple of 28 VDC relay coils to get the 12
A current. What I'm finding is that most of the relay coils draw less than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero luck
finding out how much an automotive starter solenoid draws, if I could find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.




I might try a microwave transformer from a junk oven, the secondary
might be close enough or you could series/parallel primaries/secondaries
from several to get what you want. Very inductive and will stand a lot
of watts for a short time, and cheap.

--
Free men own guns, slaves don't
www.geocities.com/CapitolHill/5357/

On Wed, 30 Mar 2005 20:55:29 GMT, Don Bruder wrote:

In article ,
Don Foreman wrote:

On Wed, 30 Mar 2005 18:34:59 GMT, "Ken Finney"
wrote:

Off-topic, but the learned denizens here are pretty good at asking bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note that
this is a one time test circuit, and I don't want to break the bank. My
initial thought was to parallel a couple of 28 VDC relay coils to get the 12
A current. What I'm finding is that most of the relay coils draw less than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero luck
finding out how much an automotive starter solenoid draws, if I could find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.

Use the run winding of an ordinary induction motor. I just measured
the inductance of a 1/3 HP 110-volt Dayton motor to be about 5.4 mH.
You might need a bit of series resistance to keep the DC current down
to 12 amps. The equivalent circuit for any inductive load is a pure
inductor in series with it's own resistance, so adding series
resistance to get the current you want is a valid approach.

?????

But when working with DC, other than "moving" stuff (Motors, speakers,
relays - something that "taps" energy from the magnetic field in some
manner, rather than leaving it stand there idle) an inductor is
effectively nothing but a dead short. That's one of the reasons
transformers are worthless for changing voltage up or down in DC
systems. Or so I've been led to believe my for entire "electronics
life"... ?

Real coils have real resistance, some more than others. Aside from
superconductivity, there's no such thing as a "dead short". Coils
wound with many turns of fine wire, as used in DC solenoid valves and
relays, have considerable resistance.

If enough energy is being pumped into it, heat starts building, and in
surprisingly (to those who haven't seen it happen before...) short
order, the whole shebang has transformed itself into a (puddle of
melted)/(cloud of vaporized) conductor, an open circuit, and an
optional, variably sized, cloud of smoke.

That's why I suggested adding series resistance as necessary to make
the steady-state current be 12 amps when excited with 28 VDC. A 1/3
HP 110-volt induction motor can certainly take 12 amps of DC with
none of the dramatic effects you describe.

You've succeeded in confusing me...

Should I be reading that last quoted sentence of yours as saying that an
inductive load is the same thing as a coil and resistor (of arbitrary
value) wired in series, or are you saying that an inductive load is the
same thing as a coil wired to a resistor matching the resistance of the
coil? I'm definitely hazy as to which was your intent. Your (seeming)
misuse of "it's" doesn't help clarify things any...

You're right, Don's error. I shouldn't have typed an apostrophe in
"it's". I intended posessive case, not contraction of "it is".

In your paragraph above, the second statement is right. An inductive
load may be viewed as being comprised of a pure inductance (no
resistance) in series with a resistor whose value is the DC
resistance of the coil, admittedly neglecting AC effects like
eddycurrent loss and hysteresis loss.

On Wed, 30 Mar 2005 13:21:49 -0800, Jim Stewart
wrote:

Ken Finney wrote:
Off-topic, but the learned denizens here are pretty good at asking bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note that
this is a one time test circuit, and I don't want to break the bank. My
initial thought was to parallel a couple of 28 VDC relay coils to get the 12
A current. What I'm finding is that most of the relay coils draw less than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero luck
finding out how much an automotive starter solenoid draws, if I could find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.

Just a quick back-of-an-envelop calculation
shows you'd need about 500 feet of 16 gauge
wire for a dc resistance of 2.3 ohms in your
coil. I'd not want to use anything less than
16 gauge and even then for just a short time.
It's going to have to dissipate 336 watts once
the core is saturated.

That's a big coil, either to wind or to find
in an existing device.

The induction motor winding will have some resistance. The rest can
be made up with 1/16" dia stainless TIG rod. I measured one such 36"
piece to be about 0.313 ohms. I don't think it'd overheat at 12
amps. It's probably settle out at about 315C. .

"Jim Stewart" wrote in message
...
Ken Finney wrote:
Off-topic, but the learned denizens here are pretty good at asking
bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note
that
this is a one time test circuit, and I don't want to break the bank. My
initial thought was to parallel a couple of 28 VDC relay coils to get
the 12
A current. What I'm finding is that most of the relay coils draw less
than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero
luck
finding out how much an automotive starter solenoid draws, if I could
find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.

Just a quick back-of-an-envelop calculation
shows you'd need about 500 feet of 16 gauge
wire for a dc resistance of 2.3 ohms in your
coil. I'd not want to use anything less than
16 gauge and even then for just a short time.
It's going to have to dissipate 336 watts once
the core is saturated.

That's a big coil, either to wind or to find
in an existing device.

Could you tell us why you need to do this.
We might be able to come up with a better
approach.



I knew I was too close to the problem, and that there had to be a simple
solution. Your solution is likely what we will use.

As for the why, we are testing a switch to ensure it won't set off an
explosive atmosphere. The test method isn't too clear other than we need to
test the switch at maximum current into an inductive load. Rather than try
to justify why this really means "a partially inductive load", we'd rather
worst-case the test conditions.



#! rnews 1609
Xref: xyzzy rec.crafts.metalworking:615226
Newsgroups: rec.crafts.metalworking
Path: xyzzy!nntp
From: "Ken Finney"
Subject: 17th Century Biodiesel?
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References: y9F2e.87890$SF.48851@lakeread08
Date: Wed, 30 Mar 2005 23:34:21 GMT


"Glenn Ashmore" wrote in message
news:y9F2e.87890$SF.48851@lakeread08...
#2 daughter is working on a story for her creative writing class. It is
about a modern recreational sailboat in route between Charleston and the
Caribbean that gets caught in the infamous Bermuda time warp and is
transported back to the 17th century. It is turning into a pretty
interesting story but being an economics major much of it is about
surviving
by trading ordinary things from the modern world. I am technical advisor
for things nautical. :-)

One thing has me stumped. When the boat starts running low on diesel she
has them trading an old aluminum jelly glass to a passing New England
Whaler
for a barrel of whale oil. I have no idea what whale oil is like. Is it
conceivable you could run a diesel engine on it?

What else could you use for diesel fuel that existed in the 17th century?


I believe Rudolph Diesel used peanut oil when he developed his engine.

Ken Finney wrote:
Off-topic, but the learned denizens here are pretty good at asking bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note that
this is a one time test circuit, and I don't want to break the bank. My
initial thought was to parallel a couple of 28 VDC relay coils to get the 12
A current. What I'm finding is that most of the relay coils draw less than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero luck
finding out how much an automotive starter solenoid draws, if I could find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.





Unless I missed something, Ken.....

I'm sitting here scratching my head wondering how the thread got this
far along without anyone asking you if you've got any idea what value of
inductance you need to create the right test setup and how much
saturation of that inductor would be acceptable at 12 amps dc.

Without that information, I can't imagine how you can expect to run a
very meaningful test.

If you'd tell us more about what the thing you are testing has to do,
and what kind of inductive device it normally controls, there's lot's of
guys here who could give you a better handle on what to use.

Jeff

--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

"As long as there are final exams, there will be prayer in public
schools"

Ken Finney wrote:

snipped


I knew I was too close to the problem, and that there had to be a simple
solution. Your solution is likely what we will use.

As for the why, we are testing a switch to ensure it won't set off an
explosive atmosphere. The test method isn't too clear other than we need to
test the switch at maximum current into an inductive load. Rather than try
to justify why this really means "a partially inductive load", we'd rather
worst-case the test conditions.



Now, even more than I posted to this thread an hour ago, I think you
really need to find out just what inductance the authority governing the
certification on your "switch" requires for that test, and also the
composition and temperature/pressure conditions of the "explosive
atmosphere".

It sounds like a job for UL, Factory Mutual or CSA to me.

Years ago I used to design "intrinsically safe" monitoring circuits for
use in hazardous locations like the inside of gasoline storage tanks and
had to get my stuff certified by actual tests at those three agencies.

They had pretty involved standards and procedures for testing such
things and used carefully controlled mixtures of gasses and air they
tried to get our stuff to set off. The same kind of gas test cells along
with standardized loads were used for testing "explosion proof"
equipment such as switches.

So forgive me if I'm misunderstanding your requirements, but the answer
to your question may not be as simple as you hope it is, particularly in
today's litigeous world.

Jeff

--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

"As long as there are final exams, there will be prayer in public
schools"

In article , Jeff Wisnia says...

I'm sitting here scratching my head wondering how the thread got this
far along without anyone asking you if you've got any idea what value of
inductance you need to create the right test setup and how much
saturation of that inductor would be acceptable at 12 amps dc.

Inductors as a load *is* sort of a non-sequiter. They don't
absorb energy, at least in the steady state.

Switching a large current into an inductor won't absorb any
energy, to start. The current flow is zero at the instant
the switch is thrown.

I'm also at a bit of a loss as to the application.

Jim


--
==================================================
please reply to:
JRR(zero) at pkmfgvm4 (dot) vnet (dot) ibm (dot) com
==================================================

"Don Foreman" wrote in message
...
On Wed, 30 Mar 2005 13:21:49 -0800, Jim Stewart
wrote:

Ken Finney wrote:
Off-topic, but the learned denizens here are pretty good at asking
bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note
that
this is a one time test circuit, and I don't want to break the bank.
My
initial thought was to parallel a couple of 28 VDC relay coils to get
the 12
A current. What I'm finding is that most of the relay coils draw less
than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero
luck
finding out how much an automotive starter solenoid draws, if I could
find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.

Just a quick back-of-an-envelop calculation
shows you'd need about 500 feet of 16 gauge
wire for a dc resistance of 2.3 ohms in your
coil. I'd not want to use anything less than
16 gauge and even then for just a short time.
It's going to have to dissipate 336 watts once
the core is saturated.

That's a big coil, either to wind or to find
in an existing device.

The induction motor winding will have some resistance. The rest can
be made up with 1/16" dia stainless TIG rod. I measured one such 36"
piece to be about 0.313 ohms. I don't think it'd overheat at 12
amps. It's probably settle out at about 315C. .

The flaw with that theory is that while a "real inductor" is indeed
inductance
in series with the resistance of the winding, putting a resistor external to
the
winding takes away from the amount of inductance you would have if that
was all in the coil -- taken to the extreme, you could say 1 turn of wire
and
a 2.3 ohm resistor is the same as the 500 feet of 16 gauge wire -- from
a DC viewpoint, that would be true, but for the purposes of an inductive
load
with a changing field, they are definitely NOT the same circuit. Of
course it also depends if it is wound on an iron core or air core or ...
Your best bet to meet your requirements as stated would probably be to
find a large transformer with 2-2.5 ohms dc resistance and use that.
There is still the part about liability that you need to make sure you are
covered there -- as another poster said, probably UL, or others.

mikey

I'm wondering whether you may have it BACKWARDS. Could it be that they want
you to run 12 amps through an inductive load, and then OPEN the circuit?
That condition will tend to make the switch arc, which could set off an
explosive atmosphere. If that's the case, the amount of inductance needs to
be specified, since this will determine the amount of arcing.

On Wed, 30 Mar 2005 19:47:47 -0500, Jeff Wisnia
wrote:

Ken Finney wrote:

snipped


I knew I was too close to the problem, and that there had to be a simple
solution. Your solution is likely what we will use.

As for the why, we are testing a switch to ensure it won't set off an
explosive atmosphere. The test method isn't too clear other than we need to
test the switch at maximum current into an inductive load. Rather than try
to justify why this really means "a partially inductive load", we'd rather
worst-case the test conditions.



snip


Your original post didn't turn up on my server so I
may have missed some of the earlier comments.

For specifying switching conditions an inductive
load can be accurately defined as a pure inductance in
series with a resistance.

On switching "break" or contact bounce the inductive
stored energy release is 1/2 L x I squared. This means that
energy storage is directly proportional to inductance so
that worst case testing requires an infinite inductance !!

Testing is meaningless unless the value of the
inductance of the 12A load is specified or switching is
tested with a worst case sample of the device(s) that are to
be switched.

If you need a larger design margin it's better to
test at increased current (energy is I squared) rather than
mess about with non representative inductive components.

Jim

In article ,
Ignoramus19234 wrote:

I am curious if a capacitor run in series with a resistor, would
constitute an inductive load.

No.

Caps and coils are two *COMPLETELY* different animals that *CAN* achieve
a similar result *UNDER SOME CIRCUMSTANCES*, but are a long, long, LONG
way from being interchangable with each other.

--
Don Bruder - - New Email policy in effect as of Feb. 21, 2004.
Short form: I'm trashing EVERY E-mail that doesn't contain a password in the
subject unless it comes from a "whitelisted" (pre-approved by me) address.
See http://www.sonic.net/~dakidd/main/contact.html for full details.

On Wed, 30 Mar 2005 17:35:23 -0800, "Mike Fields"
wrote:


"Don Foreman" wrote in message
.. .
On Wed, 30 Mar 2005 13:21:49 -0800, Jim Stewart
wrote:

Ken Finney wrote:
Off-topic, but the learned denizens here are pretty good at asking
bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note
that
this is a one time test circuit, and I don't want to break the bank.
My
initial thought was to parallel a couple of 28 VDC relay coils to get
the 12
A current. What I'm finding is that most of the relay coils draw less
than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero
luck
finding out how much an automotive starter solenoid draws, if I could
find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.

Just a quick back-of-an-envelop calculation
shows you'd need about 500 feet of 16 gauge
wire for a dc resistance of 2.3 ohms in your
coil. I'd not want to use anything less than
16 gauge and even then for just a short time.
It's going to have to dissipate 336 watts once
the core is saturated.

That's a big coil, either to wind or to find
in an existing device.

The induction motor winding will have some resistance. The rest can
be made up with 1/16" dia stainless TIG rod. I measured one such 36"
piece to be about 0.313 ohms. I don't think it'd overheat at 12
amps. It's probably settle out at about 315C. .

The flaw with that theory is that while a "real inductor" is indeed
inductance
in series with the resistance of the winding, putting a resistor external to
the
winding takes away from the amount of inductance you would have if that
was all in the coil -- taken to the extreme, you could say 1 turn of wire
and
a 2.3 ohm resistor is the same as the 500 feet of 16 gauge wire -- from
a DC viewpoint, that would be true, but for the purposes of an inductive
load
with a changing field, they are definitely NOT the same circuit. Of
course it also depends if it is wound on an iron core or air core or ...
Your best bet to meet your requirements as stated would probably be to
find a large transformer with 2-2.5 ohms dc resistance and use that.
There is still the part about liability that you need to make sure you are
covered there -- as another poster said, probably UL, or others.

I specified a lumped inductance able to handle 12 amps without
saturating: an induction motor. Whatever the value of L is, the
circuit here must have 2.3 ohms of resistance for steady-state
current to be the specified 12 amps at 28 volts. It doesn't matter
where this resistance is located in the circuit., and adding series R
does not diminish inductance.

I'm not making this up. You'll find it in any first course in AC
circuits for undergraduate EE candidates.

On Wed, 30 Mar 2005 23:32:56 GMT, "Ken Finney"
wrote:


I knew I was too close to the problem, and that there had to be a simple
solution. Your solution is likely what we will use.

As for the why, we are testing a switch to ensure it won't set off an
explosive atmosphere. The test method isn't too clear other than we need to
test the switch at maximum current into an inductive load. Rather than try
to justify why this really means "a partially inductive load", we'd rather
worst-case the test conditions.

If you're dealing with explosive atmospheres, you'd better hire
someone who knows what they're doing to do the testing and certify
the results -- or perhaps tell you to walk away from this one.
12 amps and 28 volts are way far out of intrinsically safe territory
with any switch or other device.

Please have a look at figure four in the following, which addresses
circuits with no inductance but only resistance:
http://www.crouse-hinds.com/CrouseHi...fm?CompanyID=1

I strongly agree with Jeff Wisnia here. You should definitely get
competent professional counsel rather than rely on opinions from
guys on a newsgroup who are not accountable if/when your switch blows
something up.

Bob

I just read your post below and missed the original post by Ken. If ken
is looking for info on the current draw of a 24 volt starter solenoid, he
has to identify which of the two winding in the solenoid he wants to know
the current draw of.
The "over the road, 18 wheeler" type starter solenoids are *not* difficult
to get alot of data on. The diesel starter (Delco 40 MT type) will draw
alot more current from the start button than a Chevy starter (Delco 10 MT).

I wonder how difficult it would be to locate an old 12 volt alternator
rotor that draws about 6 amps field current with 12 volts applied.. I think
an old (1980) Chevy alternator draws about 6 or 8 amps That would probably
handle 24 volts for a while, depending on how well it gets cooled.

I sure wouldnt use a truck starter solenoid as an inductive load. They
have two windings with the same number of turns but much different wire
size.
If Ken wants to phone Marks One, in Phoenix Arizona, he can get all the
automotive solenoid current draw info he'll ever need. Or Ken can contact
me. I do alot of work with Marks One (solenoid builders)

Jerry


"Robert Swinney" wrote in message
...
Why does the load need to be inductive?

Bob Swinney
"Ken Finney" wrote in message
...
Off-topic, but the learned denizens here are pretty good at asking
bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note that
this is a one time test circuit, and I don't want to break the bank. My
initial thought was to parallel a couple of 28 VDC relay coils to get the
12
A current. What I'm finding is that most of the relay coils draw less
than
an amp, and I don't want have to parallel 16 relays to draw that much
current. Does anyone have a better suggestion for me? I'm having zero
luck
finding out how much an automotive starter solenoid draws, if I could
find a
24 VDC diesel starter solenoid that requires 3 or more amps, that would
probably be doable.

Don/Jeff sez:
I strongly agree with Jeff Wisnia here. You should definitely get
competent professional counsel rather than rely on opinions from
guys on a newsgroup who are not accountable if/when your switch blows
something up.

Well it finally became clear why he wanted an inductive load.

Bob Swinney




"Don Foreman" wrote in message
...
On Wed, 30 Mar 2005 23:32:56 GMT, "Ken Finney"
wrote:


I knew I was too close to the problem, and that there had to be a simple
solution. Your solution is likely what we will use.

As for the why, we are testing a switch to ensure it won't set off an
explosive atmosphere. The test method isn't too clear other than we need
to
test the switch at maximum current into an inductive load. Rather than
try
to justify why this really means "a partially inductive load", we'd rather
worst-case the test conditions.

If you're dealing with explosive atmospheres, you'd better hire
someone who knows what they're doing to do the testing and certify
the results -- or perhaps tell you to walk away from this one.
12 amps and 28 volts are way far out of intrinsically safe territory
with any switch or other device.

Please have a look at figure four in the following, which addresses
circuits with no inductance but only resistance:
http://www.crouse-hinds.com/CrouseHi...fm?CompanyID=1

"Jeff Wisnia" wrote in message
...
Ken Finney wrote:

snipped


I knew I was too close to the problem, and that there had to be a simple
solution. Your solution is likely what we will use.

As for the why, we are testing a switch to ensure it won't set off an
explosive atmosphere. The test method isn't too clear other than we
need to
test the switch at maximum current into an inductive load. Rather than
try
to justify why this really means "a partially inductive load", we'd
rather
worst-case the test conditions.



Now, even more than I posted to this thread an hour ago, I think you
really need to find out just what inductance the authority governing the
certification on your "switch" requires for that test, and also the
composition and temperature/pressure conditions of the "explosive
atmosphere".

It sounds like a job for UL, Factory Mutual or CSA to me.

Years ago I used to design "intrinsically safe" monitoring circuits for
use in hazardous locations like the inside of gasoline storage tanks and
had to get my stuff certified by actual tests at those three agencies.

They had pretty involved standards and procedures for testing such
things and used carefully controlled mixtures of gasses and air they
tried to get our stuff to set off. The same kind of gas test cells along
with standardized loads were used for testing "explosion proof"
equipment such as switches.

So forgive me if I'm misunderstanding your requirements, but the answer
to your question may not be as simple as you hope it is, particularly in
today's litigeous world.


The test method we are testing to (MIL-STD-202, method 109, available at
http://www.dscc.dla.mil/Downloads/Mi...section100.pdf)
isn't the best I've seen. As for the inductance, it states "Proper
precaution shall be taken to duplicate the normal load in respect to torque,
voltage, current, inductive reactance, etc. In all instances it shall be
considered preferable to operate the specimen as it normally functions
during service use." However, when you look at MIL-SPEC switches that are
already rated for use in explosive atmosphere, that are always tested at the
maximum inductive current "in accordance with MIL-STD-202, method 109", so
that is what we are trying to do.

And as far a litigation, we are covered by "The Government Contractor
Defense". ;^)


#! rnews 1379
Xref: xyzzy rec.crafts.metalworking:615347
Newsgroups: rec.crafts.metalworking
Path: xyzzy!nntp
From: "Ken Finney"
Subject: (OT) Need inductive load
X-Nntp-Posting-Host: e244847.nw.nos.boeing.com
Message-ID:
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References:
Date: Thu, 31 Mar 2005 17:28:30 GMT


"Leo Lichtman" wrote in message
news
I'm wondering whether you may have it BACKWARDS. Could it be that they
want
you to run 12 amps through an inductive load, and then OPEN the circuit?
That condition will tend to make the switch arc, which could set off an
explosive atmosphere. If that's the case, the amount of inductance needs
to
be specified, since this will determine the amount of arcing.



Sorry I wasn't clearer, yes, we have to close and open the switch under
test. And no, the inductance isn't specified (even though it should be).

Ken Finney wrote:
"Jeff Wisnia" wrote in message
...

Ken Finney wrote:

snipped

I knew I was too close to the problem, and that there had to be a simple
solution. Your solution is likely what we will use.

As for the why, we are testing a switch to ensure it won't set off an
explosive atmosphere. The test method isn't too clear other than we

need to

test the switch at maximum current into an inductive load. Rather than

try

to justify why this really means "a partially inductive load", we'd

rather

worst-case the test conditions.



Now, even more than I posted to this thread an hour ago, I think you
really need to find out just what inductance the authority governing the
certification on your "switch" requires for that test, and also the
composition and temperature/pressure conditions of the "explosive
atmosphere".

It sounds like a job for UL, Factory Mutual or CSA to me.

Years ago I used to design "intrinsically safe" monitoring circuits for
use in hazardous locations like the inside of gasoline storage tanks and
had to get my stuff certified by actual tests at those three agencies.

They had pretty involved standards and procedures for testing such
things and used carefully controlled mixtures of gasses and air they
tried to get our stuff to set off. The same kind of gas test cells along
with standardized loads were used for testing "explosion proof"
equipment such as switches.

So forgive me if I'm misunderstanding your requirements, but the answer
to your question may not be as simple as you hope it is, particularly in
today's litigeous world.



The test method we are testing to (MIL-STD-202, method 109, available at
http://www.dscc.dla.mil/Downloads/Mi...section100.pdf)
isn't the best I've seen. As for the inductance, it states "Proper
precaution shall be taken to duplicate the normal load in respect to torque,
voltage, current, inductive reactance, etc. In all instances it shall be
considered preferable to operate the specimen as it normally functions
during service use." However, when you look at MIL-SPEC switches that are
already rated for use in explosive atmosphere, that are always tested at the
maximum inductive current "in accordance with MIL-STD-202, method 109", so
that is what we are trying to do.

And as far a litigation, we are covered by "The Government Contractor
Defense". ;^)


#! rnews 1379
Xref: xyzzy rec.crafts.metalworking:615347
Newsgroups: rec.crafts.metalworking
Path: xyzzy!nntp
From: "Ken Finney"
Subject: (OT) Need inductive load
X-Nntp-Posting-Host: e244847.nw.nos.boeing.com
Message-ID:
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X-Msmail-Priority: Normal
Lines: 18
Sender: (Boeing NNTP News Access)
Organization: The Boeing Company
X-Newsreader: Microsoft Outlook Express 6.00.2800.1437
References:
Date: Thu, 31 Mar 2005 17:28:30 GMT


"Leo Lichtman" wrote in message
news
I'm wondering whether you may have it BACKWARDS. Could it be that they

want

you to run 12 amps through an inductive load, and then OPEN the circuit?
That condition will tend to make the switch arc, which could set off an
explosive atmosphere. If that's the case, the amount of inductance needs

to

be specified, since this will determine the amount of arcing.




Sorry I wasn't clearer, yes, we have to close and open the switch under
test. And no, the inductance isn't specified (even though it should be).




Well, it does now sound like you've got a problem on your hands and if
it were me I'd jump up and down screaming that the inductance should be
specified or at the very least, the specific piece of equipment being
controlled by that switch, such as a motor, a big solenoid actuator or
whatever be defined as the expected (and only) load your swich will be
rated to work with.

Is this perhaps an "explosion proof" switch? i.e. one mounted inside a
strong metal housing so that even if a hazardous atmosphere inside the
enclosure gets touched off by a spark from the switch it won't "leak
out" enough hot gas to ignite a hazardous atmosphere outside of the
enclosure. IIRC one operating principle of such housings is that the
path length of the cover flanges on them is long enough so that any hot
gasses from an explosion inside them which squirt through the joints get
cooled by contact with the metal path so that they aren't hot enough to
ignite the atmosphere outside the enclosure.

IMHO breaking 12 amps of dc with even a tiny amount of inductance in the
circuit is quite liable to set off the atmosphere specified in the
document you referenced, I'd run like hell if the switch isn't an
"explosion proof" design.

Jeff

--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

"As long as there are final exams, there will be prayer in public
schools"

In article , Jeff Wisnia says...

Well, it does now sound like you've got a problem on your hands and if
it were me I'd jump up and down screaming that the inductance should be
specified...

Heck no. If I were in his shoes, I'd use the standard
value for that: 2 picoHenries!

Jim


--
==================================================
please reply to:
JRR(zero) at pkmfgvm4 (dot) vnet (dot) ibm (dot) com
==================================================

On Thu, 31 Mar 2005 14:57:41 -0500, Jeff Wisnia
wrote:


Well, it does now sound like you've got a problem on your hands and if
it were me I'd jump up and down screaming that the inductance should be
specified or at the very least, the specific piece of equipment being
controlled by that switch, such as a motor, a big solenoid actuator or
whatever be defined as the expected (and only) load your swich will be
rated to work with.

Is this perhaps an "explosion proof" switch? i.e. one mounted inside a
strong metal housing so that even if a hazardous atmosphere inside the
enclosure gets touched off by a spark from the switch it won't "leak
out" enough hot gas to ignite a hazardous atmosphere outside of the
enclosure. IIRC one operating principle of such housings is that the
path length of the cover flanges on them is long enough so that any hot
gasses from an explosion inside them which squirt through the joints get
cooled by contact with the metal path so that they aren't hot enough to
ignite the atmosphere outside the enclosure.

IMHO breaking 12 amps of dc with even a tiny amount of inductance in the
circuit is quite liable to set off the atmosphere specified in the
document you referenced, I'd run like hell if the switch isn't an
"explosion proof" design.

At 12 amps and 28 volts, the inductance value doesn't matter. Even a
purely resistive load will produce enough spark to ignite an explosive
mixture. Adding inductance increases the duration, length and
energy of the spark, but once ignition is accomplished, more ignition
is irrelevant.

jim rozen wrote:
In article , Jeff Wisnia says...


Well, it does now sound like you've got a problem on your hands and if
it were me I'd jump up and down screaming that the inductance should be
specified...


Heck no. If I were in his shoes, I'd use the standard
value for that: 2 picoHenries!

Jim



Yes, do it that way and he can probably save himself about $21 million
and still get into orbit.G

http://home.btconnect.com/wildwings/orbitalsf.html

Jeff

--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

"As long as there are final exams, there will be prayer in public
schools"

"Don Foreman" wrote in message
...
On Wed, 30 Mar 2005 17:35:23 -0800, "Mike Fields"
wrote:


"Don Foreman" wrote in message
.. .
On Wed, 30 Mar 2005 13:21:49 -0800, Jim Stewart
wrote:

Ken Finney wrote:
Off-topic, but the learned denizens here are pretty good at asking
bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note
that
this is a one time test circuit, and I don't want to break the bank.
My
initial thought was to parallel a couple of 28 VDC relay coils to
get
the 12
A current. What I'm finding is that most of the relay coils draw
less
than
an amp, and I don't want have to parallel 16 relays to draw that
much
current. Does anyone have a better suggestion for me? I'm having
zero
luck
finding out how much an automotive starter solenoid draws, if I
could
find a
24 VDC diesel starter solenoid that requires 3 or more amps, that
would
probably be doable.

Just a quick back-of-an-envelop calculation
shows you'd need about 500 feet of 16 gauge
wire for a dc resistance of 2.3 ohms in your
coil. I'd not want to use anything less than
16 gauge and even then for just a short time.
It's going to have to dissipate 336 watts once
the core is saturated.

That's a big coil, either to wind or to find
in an existing device.

The induction motor winding will have some resistance. The rest can
be made up with 1/16" dia stainless TIG rod. I measured one such 36"
piece to be about 0.313 ohms. I don't think it'd overheat at 12
amps. It's probably settle out at about 315C. .

The flaw with that theory is that while a "real inductor" is indeed
inductance
in series with the resistance of the winding, putting a resistor external
to
the
winding takes away from the amount of inductance you would have if that
was all in the coil -- taken to the extreme, you could say 1 turn of wire
and
a 2.3 ohm resistor is the same as the 500 feet of 16 gauge wire -- from
a DC viewpoint, that would be true, but for the purposes of an inductive
load
with a changing field, they are definitely NOT the same circuit. Of
course it also depends if it is wound on an iron core or air core or ...
Your best bet to meet your requirements as stated would probably be to
find a large transformer with 2-2.5 ohms dc resistance and use that.
There is still the part about liability that you need to make sure you
are
covered there -- as another poster said, probably UL, or others.

I specified a lumped inductance able to handle 12 amps without
saturating: an induction motor. Whatever the value of L is, the
circuit here must have 2.3 ohms of resistance for steady-state
current to be the specified 12 amps at 28 volts. It doesn't matter
where this resistance is located in the circuit., and adding series R
does not diminish inductance.

I'm not making this up. You'll find it in any first course in AC
circuits for undergraduate EE candidates.


That was not my point -- the question had been asked if the available
inductor had too low a resistance was it legit to add series resistance.
My answer stands -- you are correct that series resistance does not
change the value of the inductance, HOWEVER, if you simply pick
some coil and add series resistance to limit the current, you do NOT
have anywhere near the inductance that you would have if you had
enough wire in the coil to limit the current with it's own resistance.
Look at the case I gave -- taking it to the extreme, 1 loop of wire with
a series resistor is NOT the same as a large inductor with 2+ ohms worth
of wire in it. Go back and READ YOUR BOOKS.

On Thu, 31 Mar 2005 18:35:49 -0800, "Mike Fields"
wrote:


That was not my point -- the question had been asked if the available
inductor had too low a resistance was it legit to add series resistance.

That was the question, and yes it is.

My answer stands -- you are correct that series resistance does not
change the value of the inductance, HOWEVER, if you simply pick
some coil and add series resistance to limit the current, you do NOT
have anywhere near the inductance that you would have if you had
enough wire in the coil to limit the current with it's own resistance.
Look at the case I gave -- taking it to the extreme, 1 loop of wire with
a series resistor is NOT the same as a large inductor with 2+ ohms worth
of wire in it. Go back and READ YOUR BOOKS.

OK.