Considering Inrush Current in Voltage Drop Calculations

[richwaskowitz]

In order to size conductors properly for a 24 VDC distributed control system where some field devices can have a considerable inrush current (often 10 times the steady-state current), is it not important to use the inrush current when calculating voltage drop? For example, the control side of a particular Allen-Bradley Armorstart has a 24 VDC current consumption of only 0.75 amps whereas the inrush is listed as almost 5.0 amps. It is my contention that the 5.0 amp value (regardless of its relatively short duration) must be used when calculating voltage drop, otherwise the relays or contactors in these drives which are contributing to the high inrush current may not pull-in during power up.

Can someone please confirm my thinking or else straighten me out?

 

[richwaskowitz]

Update

Update

After some investigating, I got A-B to reconsider their previous published value of 5 Amps. Now they say it is only 0.75 Amps on inrush (capacitors charging, they say) and around 0.65 Amps steady-state. My question remains unchanged however.

 

[gar]

170206-1305 EST

richwaskowitz:

Let steady state current be Iss, peak inrush at the time of application of power to the device be Iir, and peak current at various times when the device is in operation be Ipo.

If a device is the only load at the end of a long wire, then:

1. At initial application of power to the device the voltage drop on the long wire from Iir is of no importance if the device can get up to a normal operating state within a reasonable time.

2. The voltage drop on the long wire from Iss is of no concern so long as there is sufficient margin that for temperature variations, or nominal variations in source voltage the device will work.

3. Voltage drop from Ipo is likely to be time dependent upon energy storage in the device (capacitors for example). So here you need to know the time dependent characteristics of the device.

If other devices are on this long wire, then the voltage drop at these devices can be important. Much more complex problem.

Two very simple devices to consider.

1. A DC relay. There is no inrush current. The total loop resistance has to be low enough that it exceeds the pull-in current of the relay. There is essentially no change in current with time other than the slow increase in current from this being a series LR circuit. Typically in the millisecond range.

2. An AC relay or solenoid. There is a large inrush current. This results from the change of inductance when the relay or solenoid is unenergized to when the armature or plunger is fully actuated. Total loop impedance at the time of application of source power (voltage) has to be low enough to insure that the relay or solenoid will actuate. As the armature or plunger is moving the inductance is changing and so is the coil impedance.

Thru this time period of energization whatever combination of impedance and required current is necessary at any instant to close the device that criteria must be exceeded.


For electronic devices you need to know their characteristics.

.

 

[richwaskowitz]

Good information. Thanks. And I agree with what you are saying as to the device eventually getting enough voltage to "wake-up". I believe that the higher inrush current in these devices is due to caps in these products and not coils as I might have thought of earlier. As well, once I held A-B's feet to the fire, they backed way down on the published inrush current of <5 Amps down to 0.75 Amps. The inrush is not such a big concern now, but i still want to understand the impact of high inrush in general.

I do want to question your statement "A DC relay. There is no inrush current.'. How can you make that general statement when some larger contactors (i.e A-B bulletin 100 60 Amp ratings and above) list a pick-up load which is as much as 60 times the holding current?

 

[winnie]

IMHO you need to calculate voltage drop during inrush, and compare this to the operational requirements of the device. As long as the device can function with the (momentarily) reduced supply voltage, then I believe that any voltage drop is acceptable.

Many devices will function perfectly well even with huge voltage drop. Consider the inrush on a typical incandescent lamp; voltage drop would actually reduce the stress on the filament during inrush, and would not harm operation.

But there are loads of situations where voltage drop during startup could prevent proper operation, in which case you may need to either reduce voltage drop or use some local energy storage to reduce the inrush current seen by the wires.

-Jon

 

[Ingenieur]

In order to size conductors properly for a 24 VDC distributed control system where some field devices can have a considerable inrush current (often 10 times the steady-state current), is it not important to use the inrush current when calculating voltage drop? For example, the control side of a particular Allen-Bradley Armorstart has a 24 VDC current consumption of only 0.75 amps whereas the inrush is listed as almost 5.0 amps. It is my contention that the 5.0 amp value (regardless of its relatively short duration) must be used when calculating voltage drop, otherwise the relays or contactors in these drives which are contributing to the high inrush current may not pull-in during power up.

Can someone please confirm my thinking or else straighten me out?

no, use steady state
the current doesn't do a 'step' change, it is an inductive load so it climbs exponentially to peak, then trails off
so initially the voltage is 24, then may sag, then rises again once the system is 'charged'
so it may, or may not pick up until the surge is over, but it will pick up
we are talking mSec's

 

[gar]

170206-1428 EST

Ingenieur:

so it may, or may not pick up until the surge is over, but it will pick up

Not so with an AC relay. However, most of the surge will never disappear until the coil burns out. Then, it clearly won't pickup.

 

[Ingenieur]

170206-1428 EST

Ingenieur:

Not so with an AC relay. However, most of the surge will never disappear until the coil burns out. Then, it clearly won't pickup.

???

it's a coil, an inductor
it will always have an inrush, and the resultant increased v drop
you are assuming it 'stalls' and Z is low and the coil burns out, but V is also low so I is limited until charged, impedance increases, as does v, so i decreases and the device 'strokes'
similar to locked rotor
why would the coil burn out?

 

[Phil Corso]

In order to size conductors properly for a 24 VDC distributed control system where some field devices can have a considerable inrush current (often 10 times the steady-state current), is it not important to use the inrush current when calculating voltage drop? For example, the control side of a particular Allen-Bradley Armorstart has a 24 VDC current consumption of only 0.75 amps whereas the inrush is listed as almost 5.0 amps. It is my contention that the 5.0 amp value (regardless of its relatively short duration) must be used when calculating voltage drop, otherwise the relays or contactors in these drives which are contributing to the high inrush current may not pull-in during power up.
Can someone please confirm my thinking or else straighten me out?

Rich...

There is no "inrush", per se, as is found with AC coils. Instead, the current changes in several distinct, but observable, increments:

a) upon initial application of voltage, there is an exponential increase of current to the point when the armature starts to move (pick-up).
b) as the armature begins moving it generates a back-emf opposing the applied voltage.
c) as the armature accelerates the back-emf increases causing the current to drop somewhat below the level reached in point b), above.
d) When the armature is fully seated the coil becomes more inductive than when it was open.
e) this results in a slower growth of current until it reaches an upper limit.
f) finally, as the coil heats up, its resistance increases, thereby lowering the current to a final value.
g) for most coils the crest reached in e), above, should never be much higher than 1-3% of the coil's nominal current-rating.

Regards, Phil Corso

 

[gar]

Transient current to LR circuit

Transient current to LR circuit

170206-2121 EST

Too many people have the misconception that there is an inrish current to an inductor.

A basic characteristic of an inductor is that you can not instantaneously change the current thru an inductor.

For a simple series LR ciccuit with a constant voltage applied to the input the series current and therefore the inductor current vs time after the application of the voltage starts at 0 current (assuming 0 initial current in the inductor) and then rises to a constant value defined by the series resistance and source voltage. The current curve vs time follows an exponential curve.

Following is a plot of applied voltage to a series RL circuit (red), and the associated current curve (blue). The actual values are not important to this discussion, just the shape of the curve.
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In a following post I will show a DC relay.

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[gar]

170206-2210 EST

Now the experiment is with a Potter & Brumfield 24 V DC relay.

The first plot is with the lowest voltage to which I could set the particular power supply. It was sufficient to start to move the armature, but not enough to close the relay within the shown time frame. Thus, you see some slight distortion of the current waveform from an ideal exponential curve.



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The second plot is with a 28 V source. Now the armature moves and you see in the current a change as the inductance changes from a lower initial value to a higher value. A new constant value of inductance exists after the armature fully moves.

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Compare the slope of the current curve fom 0 to 2 mS with 11 to 18 mS. It is a little over a 2 to 1 ratio. Thus, inductance appears to be about twice as great when the relay is energized fully.



The next plot is essentially the same as the last except that the time base has been changed, and I performed a circuit trick to display when the SPDT contact was changing from the normally closed contact to the normally open contact.

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The normally closed contact opens at about 7.2 mS with very slight contact bounce. Then closes on the normally open contact at about 10.8 mS with much more bounce. The transition time is about 3.6 mS, but the armature is not fully seated for another 0.4 mS.

This site has real problems on photo attachments and editing, and time out of developing a post.

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[Phil Corso]

Gar...

Proof that 2 pictures are worth 2,000 words!

Phil

 

[richwaskowitz]

Correct about the inductor

Correct about the inductor

I see what you are saying about no inrush for an inductor. I had to remember ELI the ICE man. Voltage leads current with an inductor and current leads voltage with a capacitor. It is capacitors which causes the inrush of current. I believe it is the capacitors in the controller boards in the ArmorStarts which upon power-up draw the large charging current until they reach equilibrium.

However, wouldn't the "spike" in current until the AC contactor's moving armature seats against the pole pieces be considered inrush?

I loved the scope traces, thank you. Pictures ARE worth a thousand words.

 

[gar]

170207-0926 EST

richwaskowitz:

On an AC contactor or relay "inrush" is a reasonable word to use. But not because it is an inductor, but because of the physical characteristics of it being a relay.

Later I will get back on AC relays, and also to your original question.

.

 

[richwaskowitz]

Right. The inrush is not because the coil is an inductor, the inrush is a result the moving core (in the case of a solenoid) or armature (in the case of a contactor) and the time it takes to complete its motion, lowering the inductance in the circuit.

 

[Ingenieur]

nice work

plot 1
V ~18 vdc?
did the relay eventually close?
if so, how long?

170206-2210 EST

Now the experiment is with a Potter & Brumfield 24 V DC relay.

The first plot is with the lowest voltage to which I could set the particular power supply. It was sufficient to start to move the armature, but not enough to close the relay within the shown time frame. Thus, you see some slight distortion of the current waveform from an ideal exponential curve.

View attachment 16728

 

[gar]

170107-1225 EST

Ingenieur:

Yes at the about 18 V the relay closes. About 48 mS from voltage application to opening of the normally closed contact, and 48 mS of transit time before the normally open contact closes. The velocity of the armature and contact blades is slower ( F = M*A ). It is an interesting plot. Later I will present it.

Lowering the voltage further will cause the relay to never cloase.

.

 

[gar]

170207-1517 EST

richwaskowitz:

About your original question.

Whenever you have a device that requires power, then you may need to study the characteristics of the device during application of power as an independent consideration from how the device responds after power is initialized.

This probably means you take an extremely conservative approach, or if that approach is too wasteful, then in a complex device (your Allen-Bradley Armorstart) you my need to run your own tests to determine how poorly defined or undefined specificications relate to your question.

You have guessed that the AB specified high inrush is probably just charging a capacitor. It is quite possible this can be totally ignored in any voltage drop consideration.

Probably the important information is:
1. What is the minimum voltage at the device for proper operation?
2. What is the peak current required at the device when it is performing its intended functionled? This may be application dependent.
3. How much voltage margin do you want at the device under worst primary source voltage conditions?

I took a quick look at one AB manual and it is massive amount of material to look thru.

A question I have is why not use a local 24 V DC supply?

Independent of what AB may tell you I suggest that you at least run some limited tests, and probably ask more questions of AB.

So i clearly have not helped much.

In some applications rate of rise of initial voltage can be important. For example the reset pulse to a microprocessor, and this may be a problem at the time of powering up the device.

.

 

[gar]

DC relay close to pull-in voltage

DC relay close to pull-in voltage

170208-1636 EST

With my setup to display when the relay contacts are in transition from normally closed to normally open and operating close to the pull-in voltage the results for the P&B KUP 24 V DC relay are:

Below 17 V the relay never pulls-in.


Source voltage 18 V:
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48 mS to open, 48 mS transit time.



About 17.1 V DC for source, at 17 V the relay won't pull-in.
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About 60 mS to open, and about 500 mS for transition.

The difference in RL slope between the armature far from the coil core and in contact with the coil core is very obvious.


Once pulled-in drop-out is about 5.3 V

Coil trsistance is 480 ohms.

Clearly the force balance between the spring and the electromagnet is quite small to get this long tranition time.

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[Ingenieur]

Gar
the pb kup series pull in spec for dc is -25% or 18 vdc, -15% for ac
your results are spot on
it's not often someone will go out of their way to teach/illustrate for others
kudos

if you feel like it set this up
i = relay current, say 0.05 A (obviously use actual value)
size an R for 25% drop or 6 vdc (simulate wire R)
6/0.05= 120 Ohm

put this in series with the relay coil
then run the same tests with a 24 vdc source/supply
plot v, i and contact position

v should start at 24 and decay to 18
curious about relay response

thanks