control circuit fuses

[gar]

101117-1948 EST

mull982:

In the AC excited application if the value of R is made zero, then there is no decay of the transient component and the AC current is displaced, has a DC component that is constant, by the instantaneous value of the steady state current at the time of turn on. Thus, the maximum peak current would be 2 times the normal steady state peak value.

For most applications in the real world R is not zero or even close in an inductive circuit. Thus, in the near term, 1/2 cycle, the peak current won't be greater than 2 times the peak steady state value and in most cases much less. At t=0 the current is still always 0.

This kind of magnitude within the first cycle after turn on is not something one would describe as inrush. Inrush implies something large in comparison with normal.

.

 

[kwired]

Most of us will get lost in all the technical details and will not necessarily remember them for long as we don't encounter the need to know that information very often.

The question is:

Doesn't a solenoid type of load have an increase of current at some point after energization that is relatively high as compared to the normal running current after the plunger has pulled in? And if applying overcurrent protection one must account for this current or else you will have undesired tripping for a normal operation of the circuit?

 

[gar]

101118-0911 EST

kwired:

Yes.

You can get a good estimate of the current by measuring the current while preventing the plunger from moving from its rest position.

From my post #22 at
http://forums.mikeholt.com/showthread.php?t=113227&highlight=motor+starter+inductance

Here is some data on a 50 year old AB #2 709COD:
Dc resistance 41 ohms.
Following are 60 Hz measurements:
Voltage Current
..Volts......Amps
50 ........... 0.68
60 ........... 0.10 just pulled in
90 ........... 0.16
120 ........... 0.24
Note at 50 V the power dissipation in the coil is about 8 times that at 120 V. It will almost certainly burn out.

The coil inductance is 100 MH when the coil is de-energized, and 580 MH when I fully advance the solenoid plunger. These inductance measurements are at low voltage and 1 kHz. The corresponding calculated coil reactances are 37.7 and 218. If I calculate impedances from these values and the 41 ohms DC resistance and then use those values to calculate current I get higher current values than the measured values. May be a result of the 1 kHz measurement and the effect of the shading coil.

Using this information and calculating the impedance for the held open state I get Z about 55 ohms. It would be better to actually measure this at 120 V, but 55 is sufficient for the purpose here. At 120 V this calculates to a current is about 2.2 A with the plunger held open. Power dissipation in the coil would be about 41*2.2^2 = 41*4.6 = 204 W. Compare this with 41*0.24^2 = 41*0.058 = 2.4 W. Very large difference. This change of inductance with plunger position is why sticking AC solenoid valves have coil burn out problems, and DC units do not.

When choosing a protective fuse for a solenoid you have to consider its trip time characteristic with the current vs time characteristic of the solenoid. There is only a very short time that the solenoid is drawing a large current. For most probably much less than 100 milliseconds. In many cases possibly in the 8 to 24 millisecond range. These are guesses because I do not want run an experiment at the moment.

.

 

[mull982]

101117-1948 EST

mull982:

In the AC excited application if the value of R is made zero, then there is no decay of the transient component and the AC current is displaced, has a DC component that is constant, by the instantaneous value of the steady state current at the time of turn on. Thus, the maximum peak current would be 2 times the normal steady state peak value.

For most applications in the real world R is not zero or even close in an inductive circuit. Thus, in the near term, 1/2 cycle, the peak current won't be greater than 2 times the peak steady state value and in most cases much less. At t=0 the current is still always 0.

This kind of magnitude within the first cycle after turn on is not something one would describe as inrush. Inrush implies something large in comparison with normal.

.

I agree that at time 0 there will be no current but within the first 1/2 cycle there can be a DC offset decaying current which decays down to steady state. I have seen this with multiple motors ciruicts and other circuits. I always understood that the typicaly value used for the aysmmetrical peak of the offset waveform was 1.6 times the symmetrical value. Of course I am referring to a circuit that has both R and L components.

I think I have a plot of this somewhere for a motor which I'll try to find.

 

[GeorgeB]

For most probably much less than 100 milliseconds. In many cases possibly in the 8 to 24 millisecond range. These are guesses because I do not want run an experiment at the moment.

.

I work in industrial hydraulics. Your 8-24 matches manufacturer literature pretty closely. The lowest I SEE is 10ms, the highest 35ms. It is important to recognize that these are timed by electronically closing the circuit at the same (and maximum effectiveness, I believe) point on the waveform. There is an 8ms variation in the real world depending on when the "contacts" close. That is one reason we prefer DC where accurate timing is important ... slower, but more consistent.

Just looking at few products, "inrush":sealed (actually measured in our products by blocking all movement to get a "steady state" number for perhaps a half-second or so) varies from 4:1 to 10:1.

Added in edit: And that 4:1 to 10:1 is handled by using time delay fusing by some, slight oversizing by others; there are advantages both ways. NOTE that NEC maxima are never exceeded ... to protect PLC outputs, I may use a 1A time delay or 2A fuse, for example, usually with 16 or 14 AWG wiring.

 

[mull982]

101117-1948 EST

mull982:

In the AC excited application if the value of R is made zero, then there is no decay of the transient component and the AC current is displaced, has a DC component that is constant, by the instantaneous value of the steady state current at the time of turn on. Thus, the maximum peak current would be 2 times the normal steady state peak value.

For most applications in the real world R is not zero or even close in an inductive circuit. Thus, in the near term, 1/2 cycle, the peak current won't be greater than 2 times the peak steady state value and in most cases much less. At t=0 the current is still always 0.

This kind of magnitude within the first cycle after turn on is not something one would describe as inrush. Inrush implies something large in comparison with normal.

.

I went ahead and played around with a circuit simulator to see how different RL values effected current after a switch was closed.

That attached plots show the corrosponding current waveform after a switch was closed at t=0.

What I noticed is there is an DC offset that exists with the decay rate of this offset determined by the value of the inductance reactance in comparison with the resistance. As the inductive reactance value was increased the decay time from the maximum assymettrical DC offset down to steady state symettrical current was increased. For values of resistance much smaller than the inductive reactance the decaying offset was very small and appeared to be non-existant.

Even with the resistor removed from the circuit there appeard to be a decaying offset with just the pure inductor. I'm not sure however if the source modeled had an internal resistance that was being included. The source was a 120V 60Hz sinewave source, and the R and L values used are shown on the attachemet.

What I gather from this is that there appears to always be a decaying DC offset present for inductive circuits where inductive reactance is much larger than circui impedance.

My tests only show however the changing duration of the DC offset and do not do much to show how the peak current is effected at the time of switching since I didn't really pat attention to current magnitude but rather was looking at the decaying DC offset. I may try to run these tests again to show exactly how much the waveform is offset, or how much the peak magnitude is effected for different circuit values.

Since I did not plot voltage these tests dont take into account where on the voltage waveform the switch was closed. They also do not show the current at time t=0 since we have no voltage reference to compare the currents against. I may try to run these tests again showing voltages for these exact reasons.

I'd love to hear comments.