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.