control circuit fuses

[nizak]

Any thoughts as to what may cause control circuit fuses(.5A) to blow as soon as the 120V coil is energized on a motor starter? 240V primary/120V secondary control Xfmr..FWIW, increasing the fuse size to 3A will not cause the fuses to blow, but instead the coil makes a loud "clattering" sound. I realize this is vague info, but it's a problem that started out of the blue. There does not seem to be any short circuit problem, or overload for that matter. The only reason the 3 A fuses were tried was to see if it would make any difference.Thanks.

 

[chris kennedy]

the coil makes a loud "clattering" sound.

Sounds like an undervoltage situation.

 

[nizak]

Could the under voltage been an issue as well with the .5A fuses blowing? With them there was no chattering of the coil, it just blew the fuses imediatetly. Thanks

 

[jumper]

Dirty contacts maybe a possibility.

 

[dicklaxt]

.5a seems low,do you know the coil burden?



dick

 

[sii]

Is this a new installation or something that has worked for some time but now does not? Does the starter have the proper coil?

 

[kwired]

The coil being an inductive load has an "inrush" of current when first energized that rapidly decreases as the magnetic field increases. If your fuse does not have sufficient time/current characteristics you will blow fuses.

Low voltage will cause chatter as well as increased current levels. Weak contact in a control device could do this as well.

Also make sure the controlled motor is not causing significant voltage drop to reduce the control voltage enough to cause problems with the control circuit.

How does the contactor operate if the motor is disconnected so it will not draw down voltage?

 

[millelec]

are there any other coils of the same rating you can compare it to? if the coil overheats the windings can short together and you can pick it up by taking a resistance reading and comparing it to a new/good coil. w/an overheat condition like that the coil will look swollen & may have small splits in it. another thing to check is the plunger guides for contactor. can wear out and cause misalignment and chattering. like the suggestion for low voltage as well. would definitely cause an overcurrent condition.

 

[gar]

101112-0642 EST

kwired:

The coil being an inductive load has an "inrush" of current when first energized that rapidly decreases as the magnetic field increases.

This is not a correct statement.

One characteristic of an inductor is that its current can not be instantaneously changed.

Thus, in a simple RL series circuit with an open switch, battery, and initial current of zero there is no energy stored in the inductor and the voltage across the inductor is zero.

The instant after the switch closes there is no current flow. The boundary conditions are identical on both sides of t=0. The next instant the current starts to rise. The current rise is exponential with time until at infinite time the steady state current is the battery voltage divided by the series resistance.

At one time constant the current will have risen to about 63% of its steady state value. During the next time constant period the current rises an additional 63% of the remaining current change. After 6 time constants the current is within 0.25% of its steady state value. For a simple series RL series circuit the time constant is L/R.

In contrast for a capacitor in a simple series RC circuit with zero initial charge on the capacitor the initial current after the switch closes is a maximum equal to the source voltage divided by the series resistance.


When a ferromagnetic inductive circuit is considered there are other factors to consider, but the basic characteristics of an inductive circuit are not violated.

Consider a transformer with an iron core, steel if you please. If the iron core is removed, then the air core inductor behaves as described above. When the core is installed, then the inductance is greatly increased. However, the inductance is a function of the flux density in the core. At a sufficiently high flux density the the coil looks like an air core inductor.

Suppose the transformer excitation is turned off such that there is a large residual flux in the core. Next apply a voltage to this coil of a polarity that forces the core further into saturation, then after a short time the input current will be much higher than if the voltage had been of the opposite polarity which would have caused a lowering of the flux density. This relates to the hysteresis curve of the magnetic material.

For a relay or solenoid it is somewhat different. Here there is a mechanical change that causes a change of inductance with time. When the relay or solenoid is de-energized there is a large air gap and the inductance is low. Thus, a high current after a short time. As the armature closes the inductance rises and the current diminishes.

An oscilloscope and some parts will allow a study of some of these characteristics.

You can learn more about these circuits in various books. One such book for circuit calculations is "Analysis of A-C Circuits", by Melville B. Stout, Ulrich's Book Store, 1952. One covering ferromagnetic characteristics is "Electric and Magnetic Fields", by Steven S. Attwood, John Wiley & Sons, 1949.

.

 

[Jraef]

Nizak,
Before we get too far afield on side issues, it sounds as though your Control Power Transformer is too small for the load. With the proper size fuse the fuse blows because the CPT secondary voltage drops and the current spikes. If you increase the fuse, the voltage drop then becomes apparent in the contactor coil, making it chatter.

You can look up coil data on the manufacturer's data sheets for the contactor. it will tell you the Inrush rating f the coil, then look up the inrush capacity of your CPT, the CPT value must exceed the coil inrush value or you will have this problem. If there are multiple devices being powered by the CPT, there are a number of ways to calculate the proper size of CPT for the total load. The most conservative one that I always use is:

Inrush Capacity of CPT = Total Sealed VA (including other loads such as pilot lights etc.) + Largest Inrush VA

 

[billsnuff]

Replace the motor starter and put a .5amp fuse back in.

check mechanicals on motor starter.

 

[kwired]

101112-0642 EST

kwired:

This is not a correct statement.

One characteristic of an inductor is that its current can not be instantaneously changed.

Thus, in a simple RL series circuit with an open switch, battery, and initial current of zero there is no energy stored in the inductor and the voltage across the inductor is zero.

The instant after the switch closes there is no current flow. The boundary conditions are identical on both sides of t=0. The next instant the current starts to rise. The current rise is exponential with time until at infinite time the steady state current is the battery voltage divided by the series resistance.

At one time constant the current will have risen to about 63% of its steady state value. During the next time constant period the current rises an additional 63% of the remaining current change. After 6 time constants the current is within 0.25% of its steady state value. For a simple series RL series circuit the time constant is L/R.

In contrast for a capacitor in a simple series RC circuit with zero initial charge on the capacitor the initial current after the switch closes is a maximum equal to the source voltage divided by the series resistance.


When a ferromagnetic inductive circuit is considered there are other factors to consider, but the basic characteristics of an inductive circuit are not violated.

Consider a transformer with an iron core, steel if you please. If the iron core is removed, then the air core inductor behaves as described above. When the core is installed, then the inductance is greatly increased. However, the inductance is a function of the flux density in the core. At a sufficiently high flux density the the coil looks like an air core inductor.

Suppose the transformer excitation is turned off such that there is a large residual flux in the core. Next apply a voltage to this coil of a polarity that forces the core further into saturation, then after a short time the input current will be much higher than if the voltage had been of the opposite polarity which would have caused a lowering of the flux density. This relates to the hysteresis curve of the magnetic material.

For a relay or solenoid it is somewhat different. Here there is a mechanical change that causes a change of inductance with time. When the relay or solenoid is de-energized there is a large air gap and the inductance is low. Thus, a high current after a short time. As the armature closes the inductance rises and the current diminishes.

An oscilloscope and some parts will allow a study of some of these characteristics.

You can learn more about these circuits in various books. One such book for circuit calculations is "Analysis of A-C Circuits", by Melville B. Stout, Ulrich's Book Store, 1952. One covering ferromagnetic characteristics is "Electric and Magnetic Fields", by Steven S. Attwood, John Wiley & Sons, 1949.

.

I kind of got lost in there, but isn't there a period in there where the current is relatively high compared to normal operating current and eventually drops to normal? That is really what my point was, correct me if I am wrong. How is a common induction motor any different from a high level of current at or during startup?

I have always known (may be a simplified version of what actually goes on but is useful information to most field electricians) that inductive loads are essentially like a short circuit for very short period of time as the inductive reactance increases rapidly as the core becomes magnetized.

I have also always known a similar activity when applying power to a capacitor - one that is not charged will draw high level of current that decreases as the charge on the capacitor increases.

 

[kwired]

Replace the motor starter and put a .5amp fuse back in.

check mechanicals on motor starter.

OP did not mention what size starter he has but this could be a fairly expensive job if it turns out there is nothing wrong with the starter - say the control transformer was too small as has been mentioned.

 

[kc8dxx]

1. The fuse is blowing for a reason. There is a coil in the circuit, and contacts in series tend to take a beating. I would be suspicious of the motor starter. One can overhaul them by replacing the guts, or polishing the contacts, instead of replacing the entire thing.

2. Just because it's in the field doesn't mean it's designed properly to begin with, or maybe the circuits been modified. Maybe what was once just below the edge is now just above the edge. Double check the sizing on the transformer. JRAEF has a simple clear approach that I also like to use.

If it's been in the field for some time, then don't be suprised if there are several contributory causes.

 

[gar]

101113-0830 EST

kwired:

Your sentence

The coil being an inductive load has an "inrush" of current when first energized that rapidly decreases as the magnetic field increases.

logically applies to any and all inductive loads. A basic inductive load with no initial stored energy (meaning that initially there is zero current flow thru the inductor) actually has 0 inrush current at the time of first applying a voltage to the inductor. And if the inductance of the load remains unchanged there is no inrush current following application of voltage. With a constant DC input voltage, invariant resistance and inductance, the current just rises towards a steady state value of V/R. This is a fundamental characteristic of a series RL circuit.

Inrush was not defined but it is reasonable to assume that it meant a somewhat large current after switch closure that lasted for a short time.

Following are two useful references on an RL circuit transient response:
http://www.electronics-tutorials.ws/inductor/LR-circuits.html Voltage and current curves vs time are provided.
http://www.allaboutcircuits.com/vol_1/chpt_16/3.html This only shows voltage vs time. ...tml"]http://http://beta-a2.com/EE-photos.html

For motors much of the inrush problem is a result of the inertia load on the motor.

Inductance does play a part in the possible problem of this thread. The contactor has a coil that has inductance, but that inductance varies as the position of the moving part of the magnetic core changes position.

If you wade thru the following post it may help clarify what happens with a contactor and current:
http://forums.mikeholt.com/showthread.php?t=113227&highlight=motor+starter+inductance
It is important to note that the current thru the contactor coil resistance is the primary cause of coil heating. When the contactor is not fully closed the coil inductance is less and the current is higher than when fully closed. A chattering contactor, not just a humming contactor, would probably have a higher coil current than a correctly closed contactor. Mechanical binding of the contactor could cause chattering, but so could poor circuitry in the coil control path.

You can experiment on your own with a Variac, meters, and an AC contactor or solenoid.

.

 

[weressl]

Dirty contacts maybe a possibility.

More like dirt between the magnet armature surfaces.

 

[mull982]

101112-0642 EST

kwired:

This is not a correct statement.

One characteristic of an inductor is that its current can not be instantaneously changed.

Thus, in a simple RL series circuit with an open switch, battery, and initial current of zero there is no energy stored in the inductor and the voltage across the inductor is zero.

The instant after the switch closes there is no current flow. The boundary conditions are identical on both sides of t=0. The next instant the current starts to rise. The current rise is exponential with time until at infinite time the steady state current is the battery voltage divided by the series resistance.

At one time constant the current will have risen to about 63% of its steady state value. During the next time constant period the current rises an additional 63% of the remaining current change. After 6 time constants the current is within 0.25% of its steady state value. For a simple series RL series circuit the time constant is L/R.

In contrast for a capacitor in a simple series RC circuit with zero initial charge on the capacitor the initial current after the switch closes is a maximum equal to the source voltage divided by the series resistance.


When a ferromagnetic inductive circuit is considered there are other factors to consider, but the basic characteristics of an inductive circuit are not violated.

Consider a transformer with an iron core, steel if you please. If the iron core is removed, then the air core inductor behaves as described above. When the core is installed, then the inductance is greatly increased. However, the inductance is a function of the flux density in the core. At a sufficiently high flux density the the coil looks like an air core inductor.

Suppose the transformer excitation is turned off such that there is a large residual flux in the core. Next apply a voltage to this coil of a polarity that forces the core further into saturation, then after a short time the input current will be much higher than if the voltage had been of the opposite polarity which would have caused a lowering of the flux density. This relates to the hysteresis curve of the magnetic material.

For a relay or solenoid it is somewhat different. Here there is a mechanical change that causes a change of inductance with time. When the relay or solenoid is de-energized there is a large air gap and the inductance is low. Thus, a high current after a short time. As the armature closes the inductance rises and the current diminishes.

An oscilloscope and some parts will allow a study of some of these characteristics.

You can learn more about these circuits in various books. One such book for circuit calculations is "Analysis of A-C Circuits", by Melville B. Stout, Ulrich's Book Store, 1952. One covering ferromagnetic characteristics is "Electric and Magnetic Fields", by Steven S. Attwood, John Wiley & Sons, 1949.

.

Gar

I understand what you are saying about the current after energizing an ideal inductor, but what about the DC offset portion of the current rusulting in the syssmetrical portion of current when a coil is energized?

If I'm not mistaking, when energizing an RL circuit there is an initial current transient which has a DC offset accounting for a decaying ayssmetrical peak? Is this correct?

 

[gar]

101115-2131 EST

mull982:

Go to
http://www.electronics-tutorials.ws/...-circuits.html
look at the series RL, BATTERY, and switch circuit.

Instantaneously you can not change the current thru an inductor. If you assume as an initial condition 0 for the current thru the inductor before the switch is closed, then at the very instant of switch closure the current thru the inductor is still 0. But starting at this point the current gradually increases to a value of V/R at time infinity. This is a monotonic function (never reverses on itself) and a very smooth exponential curve. This smooth current curve is shown in the next figure along with the voltage drop across the inductor.

You can assume some other initial current in the inductor. Whatever that initial inductor current is will be the current thru the switch at the moment just after switch is closed. The RL current curve starts from that point. This will cause a step rise in the input current thru the switch, but not a current higher than any following value on the RL current curve.

The above comments could be misleading if you have an initial inductor current higher than V/R. In this case there will be reverse current thru the switch and into the battery and I suppose you could call this inrush. Not a normal condition you would encounter. For most situations, like turning power on to a device, the initial current is going to be zero.

.

 

[mull982]

101115-2131 EST

mull982:

Go to
http://www.electronics-tutorials.ws/...-circuits.html
look at the series RL, BATTERY, and switch circuit.

Instantaneously you can not change the current thru an inductor. If you assume as an initial condition 0 for the current thru the inductor before the switch is closed, then at the very instant of switch closure the current thru the inductor is still 0. But starting at this point the current gradually increases to a value of V/R at time infinity. This is a monotonic function (never reverses on itself) and a very smooth exponential curve. This smooth current curve is shown in the next figure along with the voltage drop across the inductor.

You can assume some other initial current in the inductor. Whatever that initial inductor current is will be the current thru the switch at the moment just after switch is closed. The RL current curve starts from that point. This will cause a step rise in the input current thru the switch, but not a current higher than any following value on the RL current curve.

The above comments could be misleading if you have an initial inductor current higher than V/R. In this case there will be reverse current thru the switch and into the battery and I suppose you could call this inrush. Not a normal condition you would encounter. For most situations, like turning power on to a device, the initial current is going to be zero.

.

Gar

What I was referring to was energizing an RL circuit with an AC source. I have read and seen that when energizing an RL circuit such as a motor with an AC source the initial current will have a DC offset associated with it depending on how far from the zero crossing the voltage is applied to the circuit. The closer to zero crossing the voltage is applied the higher the current DC ofsset will be with the DC offset decaying to steady state. This is the same used to represent motor inrush as well as fault current which both exhibit an initial DC offset with decaying current towards steady state.

I believe the equation representing AC voltage applied to an RL circuit is:

i = Vmax/Z [sin(wt+alpha-theta)-e^-Rt/L*sin(alpha-theta)]

Where the first half of this equation represents steady state current and the second half represents the DC offset.

Is this different then the simple RL circuit model you are referring to?

 

[gar]

101116-1055 EST

mull982:

Look at your equation and substitute t=0. The result is that i=0 at t=0.

In this AC excitation situation the DC transient component of current is equal and opposite that of the steady state component such that at t=0 i=0.

.