# In-rush current and line loss

1. May 14, 2013

### sgtpepper777

I have an Altronix AL600ULX power supply which will output 24VDC and is rated for a maximum of 6A. I want to have this power supply activate a locking mechanism which is located several hundred feet away. I have already purchased 18AWG cable to connect the locking mechanism to the power supply. I am trying to determine the allowable distance between the lock and power supply before the lock will simply not activate.

The documentation on the lock indicates that it is rated for 24VDC +/- 10%, so I'm assuming my lock will still operate at 21.6VDC. The lock draws 0.3A at 24VDC when engaged, but it has an in-rush current of 1.3A. I'm not sure whether I should use the in-rush current or the continuous duty current rating to calculate my line loss.

I'm thinking I need to use the in-rush current for my calculation - the logic being that if my cable run is too long, the voltage will not be high enough on the lock side to give it that initial kick to activate the solenoid. The in-rush is only a fraction of a second, though, so I'm wondering if there's something special to the circumstances of the in-rush situation that would make it not apply to the line loss calculation.

Maybe it's just wishful thinking, because I'd really like it if the cable and device locations that I've picked out would work :-)

Thanks for any insight you can offer!

2. May 14, 2013

### Staff: Mentor

Welcome to the PF.

I would be inclined to use the continuous current, but it would help if you could post links to the datasheet and any application notes.

3. May 15, 2013

### sgtpepper777

Altronix Power Supply: http://www.altronix.com/products/installation_instructions/AL600ULXseries.pdf
Lock: http://www.vonduprin.com/pdf/9899_catalog.pdf#page=46

The documentation on the lock doesn't specifically mention the in-rush current, but I've been told that the in-rush is in fact 1.3A. I'm thinking that because of the resistance in the length of cable I want to run, there will be a drop in voltage equal to I*R. At the instant that I=1.3A, the voltage will not be high enough at the lock to activate it. Maybe I'm wrong, though?

4. May 15, 2013

### sophiecentaur

Unless you need this lock to operate within a few milliseconds, the 'inrush current' will soon drop and you will end up with the required operating voltage. I imagine the lock works with a motor, which takes a short while to get going. The initial 1.3A will be the 'stall current', I imagine. As soon as the motor starts to move at all, the 1.3A will reduce, giving you more torque and a 'win-win' situation. If there is any backlash in the system at all, there will be a very low mechanical load on the motor when it starts to rotate so it should speed up very quickly (before the full load bites).
Why not contact the manufacturer abut this perfectly valid concern you have? I reckon they are bound to have figures for acceptable supply cable resistance, if their lock system is to be used for remote gates (?) or similar. According to my ancient tables the resistance of 1000ft of 18AWG wire is about 6 Ohms (i.e. 500ft each way?). That would involve a big voltage drop (8V ish) when passing 1.3A. If you do have a problem, you can always use two lengths in parallel and reduce the voltage drop to half that value, so it's not a disaster .

5. May 16, 2013

### skeptic2

I agree with you. When the plunger of a solenoid is out of the coil, the coil has the weakest pull on the it so you need a stronger field to start it moving. Once the plunger has stopped within the coil the magnetic field is much, much stronger and it takes much less current to hold it in place.

6. May 16, 2013

### sophiecentaur

If the lock is worked by a solenoid then there will be no 'in-rush' current. A solenoid is basically a high value Inductance which will delay the current build-up, unlike a motor starting from rest, which will take a high initial current until its speed builds up.
Has this in-rush current actually been measured (by whom?)? Are we sure what the lock mechanism actually consists of? Is it a motor or a solenoid?

Also, the armature of the solenoid could be arranged to be anywhere when at rest. It would not have to be out of the end. In fact, if a high initial force were needed, the armature would start right in the middle by design.

7. May 16, 2013

### jim hardy

That Vonduprin latch datasheet you linked to gives on page 43 a value of 16 amperes for current inrush to the DC 24 volt latch. I'd guess it is a two coil design with a big pull-in coil and a small holding coil like on 1940's Ford overdrives.
The datasheet also mentions on page 47 Vonduprin's power supply which they say is capable of 16 amp surge.

The Altronix installation instruction brochure that you linked for the supply does not say whether the supply is capable of 16 amps, it only says six amps continuous.

I think you'd better try it out before installing it.

8. May 18, 2013

### sophiecentaur

Jim
You seem to know these things.
The spec says it is a solenoid and a solenoid is only an inductor. An inductor does not have an 'inrush current' but rather, it acts as a 'choke', slowing any change / increase in current. I can't quite square this with what's being said, here unless there is some added complication in the device. Is it possible that the unit has two windings, with two windings in circuit initially and one winding disconnected once the armature / plunger has made physical contact with a pole, requiring fewer Ampere Turns to hold in?

9. May 18, 2013

### milesyoung

I don't think it's an inrush current in the sense of AC solenoids where you'll have that initial low-reactance surge. They probably just use it as a term for when a device draws a lot of current as an initial transient.

It would make sense to have one coil to do the mechanical work of moving the plunger and another to hold it.

10. May 18, 2013

### sophiecentaur

What "reactance"? The only reactance which would give a surge would be a massive parallel C. OR you can get a surge with a tungstan filament at switch on. I'd go for the two coil option.

11. May 18, 2013

### milesyoung

I wasn't talking about DC excitation of the solenoid(s) in the lock when I mentioned reactance, maybe that wasn't clear.

12. May 18, 2013

### jim hardy

I've never seen the door actuator, but reading their brochure that's exactly what I believe it has.

Just yesterday I helped a friend with his Ford dumptruck - the fuel shutoff solenoid on its big diesel engine is just such a device.
When starter engages,
A high current "actuate coil" pulls the fuel valve open, against a return spring,
a small "hold coil" holds fuel valve open thereafter so long as ingnition is "on",
and a contact stops current to actuate coil so it won't burn up.
.. when power to hold coil is removed (by switching ignition "OFF") the stout return spring pushes fuel valve closed to stop engine..
My 1949 Ford overdrive transmission had a similar scheme on the solenoid that engages the planetary gears .

Of course it takes fewer amp-turns to hold than it does to pull-in because of the air gap which starts out large but becomes infinitesimal as the solenoid plunger moves to its 'actuated' position.
So for DC solenoids that have to move anything substantial, one either uses a two coil device or switches in series resistance to reduce coil current to holding value.
That's why AC solenoids pull high inrush, their inductance is low because of the large airgap. Once they actuate and close the airgap their inductance goes up so coil current goes down to holding value with no outside intervention.

Things of beauty, both !
(if you can't tell - I really enjoy old Fords and old Evinrudes)

So the door latch manufacturer used, in my opinion, a poor terminology.
He should have called it "Solenoid Pull In" or "Un-Latch" current, and given a duration..
Term "Inrush" in this case confuses us guys who deal with authentic inrush current.

old jim

Last edited: May 18, 2013
13. May 19, 2013

### sophiecentaur

@Jim
That seems to have sorted out my problem about the practicalities of 'real' solenoids', thanks. The two coil thing makes sense. Hold in current is soooo much lower (gas appliances all use the output of a thermocouple to keep the valve open).
I still have a bit of a problem with what, precisely, this "inrush" current is. Where there's any purely inductive component, I can't see how the current can be any higher at switch on than at any other change of polarity. Even when switch on occurs at an AC voltage maximum, inductance will still limit the current during that initial spike. Am I still missing something?

14. May 19, 2013

### milesyoung

There's no inrush current if you're thinking in terms of the instantaneous current through the coil during a couple of AC cycles or so.

There's a large "inrush" current, though, if you're thinking in terms of the RMS current through the coil as the plunger/piston closes the airgap and the reluctance of the magnetic circuit drops abruptly.

15. May 19, 2013

### sophiecentaur

Right. So the Inductance suddenly goes up. Under steady state, though, the current is only limited by the winding resistance. Can there ever be more current than that?

16. May 19, 2013

### milesyoung

If the circuit consists of nothing more than a constant voltage source and a solenoid then, regardless of how the inductance of the solenoid changes, there can be no inrush current and I don't think anyone is arguing otherwise.

I think the manufacturer just uses "inrush current" to tell you that the device draws a transient current when engaged that's large in magnitude compared to the current it settles at.

17. May 19, 2013

### jim hardy

I think we're all saying the same thing here. Let me just ramble to clarify , if only for me.
If my anecdotes bother you just skip the post.

Two cases here, the AC solenoid and the DC one.

AC solenoid current might look like this:

where the larger cycles are while the plunger is closing the air gap.
The Westinghouose AC relays in my plant were very fast and pulled in in ~ one cycle so you saw only one large peak, somewhat faster than the right hand trace.
We called it "inrush" though "Latching" or "Pull-in" might have been a better adjective.
Pull in current was an amp, holding about 1/10 amp.

Because of its air gap, an AC solenoid core can't saturate on energization like a transformer does.
When airgap closes both inductance and XL increase so current goes down to holding value.
Curent will be V/(R+jXL) and XL is initially small because of the air gap.
If a piece of debris gets into the airgap and prevents full travel the relay cannot close the airgap to reduce its current and its coil burns up. That used to happen with our early relays. They had a ventilation hole right above the plunger , and when the construction electricians skinned wires it was not unusual for a small piece of plastic insulation to fall into that hole and block armature travel. Maunfacturer quickly redesigned to close the top vent hole.

Wish I could find a trace for DC solenoid current.
A DC solenoid's current will rise toward V/R on its L/R time constant.
Note theres a glitch when airgap closes and L becomes larger, but it keeps on increasing at the longer time constant.
Small DC relays can be designed with pull-in current that won't overheat the coil so are just a simple solenoid.
Bigger solenoids employ something external to reduce current once the plunger has completed its travel.

Aha ! here's a document with a trace for DC solenoid, see page 114 of this huge PDF:
(i'd paste it here if I knew how, maybe a mentor would help?)
That is a closed loop current controller applying rectified 3 phase to an electromagnetic latch that holds one control rod in a plant.
Look at filtered signal in middle:
At d, voltage is applied and current starts to rise
at c, the electromagnet latches and a current glitch occurs because energy = half LI^2and L went up. One can hear the 'clank' on the loose parts monitor.
After latching , the current controller holds current at desired value of 8 amps for that coil. It could as easily be driven lower and indeed for some other coils in same system it is halved to 4 amps.
When regulator fails current can rise to V/R and blow the fuse and rod drops.

sorry for ramble. Getting these practical basics 'down pat' can help out young folks, and judging by some of the google search results on 'solenoid inrush' there's a lot of confusion on this particular detail. I hope this helps someone.

old jim

Last edited: May 19, 2013
18. May 19, 2013

### sophiecentaur

That's useful stuff Jim. Thanks again. Jeez, that document was big but I did see the figure of current vs time. It's interesting that you can see the change in slope (increasing L) after the magnetic circuit is completed. Something weird is going on where the current seems to drop, momentarily - some brief back emf effect perhaps as the reluctance suddenly changes to a lower value.

I think I can stoip worrying now. I guess the thing that was bugging me at the start was that the term "inrush" is not a well defined standard one and is interpreted by different people in different ways and it covers a range of different circumstances.

19. May 19, 2013

### jim hardy

I have just always accepted that it is in accordance with Lenz's law, induced voltage will oppose the change in flux.
So when reluctance disappears and flux starts to increase, an induced EMF appears that lowers current in a futile attempt to keep flux constant. Counter EMF is as good a name as any I suppose, though we typically think of counter emf as result of changing the current or moving the conductor, not changing the reluctance.

Doubtless there's a more elegant derivation.

Learning this rod drive system was really good for my basics.

I cant access the IEEE standards, no longer a member. Apparently there's one called "The IEEE Standard Dictionary of Electrical and Electronics Terms ", IEEE 100 , that gives a definition. Could be it's less ambiguous and the term is just mis-used a lot. If there's a IEEE member reading perhaps he will enlighten us.

old jim

Last edited: May 19, 2013