Inrush current filament Electrical

[techstu]

Filament of incand bulb are coiled in geomtry. cold resistance is 1/15 (aprox) of hot resistance. The coiled filament have inductance ''L". question Does the equation Ldi/dt is applicable?
if so is the inrush current is limited?

 

[Pumblechook]

The inductance will be small. It will have no effect at 50 or 60 Hz or slow down the rise of a DC current.

 

[engr]

strictly spkng , there is an effect.

L is small but dt is very small and di=15 , so di/dt will be considerably high. Ldi/dt will have a value which opposes the applied voltage. The impact do not have any relation with supply frequency.

 

[Pumblechook]

Doubt it.

Gut feeling is that the inductance is pretty small. Self resonance would be several MHz. The coiling is a very small diameter and there is no high permeability core obviously.

Can you measure the inductance or the resonant frequency with a 50pF capacitor in parallel?

 

[Pumblechook]

Come to think of it ordinary bulbs are used as dummy loads or even a means of estimating RF power well into VHF and even UHF.. Would suggest quite a low inductance.

Randall and Boot used a bulb to estimate the ouput from the first cavity magnetron working at 3000 MHz.

The only figure I can find is 20 microhenries but no mention of the power or voltage of the bulb.

You have to consider the rise time of the whole circuit from alternator to bulb and how this compares to thermal rise time of the bulb.

I know that quite high value fuses can blow. I was fooled by it till I found out that there is such a big difference between the cold and hot resistance of a bulb.

Only other thoughts would be to measure the peak current and even plot the current v time if that was possible.

 

[berkeman]

strictly spkng , there is an effect.

L is small but dt is very small and di=15 , so di/dt will be considerably high. Ldi/dt will have a value which opposes the applied voltage. The impact do not have any relation with supply frequency.

Pumblechook is 100% correct. The inductance of the filament coil is totally negligible.

 

[engr]

Pumblechook,what is the resistance vs temperature graph looks like? either use 'strict' or use 'normal'

 

[The Electrician]

Pumblechook,what is the resistance vs temperature graph looks like? either use 'strict' or use 'normal'

You can find resistance vs temperature numbers for tungsten here:

http://it.stlawu.edu/~koon/classes/221.222/222L/Stefan.html

but you'll have to plot your own graph.

 

[engr]

Thanks ,it is for filament temperature vs resistance.if it is replaced by equal resistor with another shape (not coil shape) will these values same?

 

[The Electrician]

If the other resistor is made of tungsten, then of course the resistance vs temperature characteristic will be the same. The resistance vs temperature characteristic of a resistor made of metal is independent of the geometry of the resistor.

I found a burned out 60W incandescent bulb in the basement and broke it open.

The filament is a coiled coil. Treating the smaller coil as a single wire, the filament would be a solenoid of .025" diameter, wound with a wire .007" in diameter. The filament is 1" long, with about 45 turns in the solenoid. A calculation of the inductance gives a value of about 36 nanohenries. The fact that the outer coil is not wound with a solid wire, but with another coil will increase the inductance. I don't have a simple formula for that, so I measured the inductance. The measured inductance is about 190 nanohenries.

 

[engr]

consider tungston of 'L' meter length.
voltage 110V/60Hz
case1:

the resistor in stright line , as a long wire ,measure voltage vs current for 0-10 milli sec , .5ms band.

now make the wire a coil / coiled coil and do the same experiment , Will the graph in the both cases be same?

 

[The Electrician]

consider tungston of 'L' meter length.
voltage 110V/60Hz
case1:

the resistor in stright line , as a long wire ,measure voltage vs current for 0-10 milli sec , .5ms band.

now make the wire a coil / coiled coil and do the same experiment , Will the graph in the both cases be same?

Probably not.

This is not, of course, the same question you asked in posts #7 and #9. Now you're including time, which you didn't in those previous posts.

Do you want a plot of resistance of the filament vs time, when the light bulb is suddenly energized? Maybe when you say, "...measure voltage vs current for 0-10 milli sec", you're really asking for voltage/current vs time; is that the case?

 

[ScottXe]

I am studing a failure mode of tungsten filament of a low power incandescent light bulb. There are two issues for this studing. One is why the tungsten broke after use for 1 to 2 months. Two is why it drew over 10A current resulting in a big bang noise when the filament broke.

In my reseach, the early failure may be due to poor quality of filament and/or filament winding. The second one might be due to the arcing when the filament broke with insert gas. However, the bulb does not contain any gas and just vacuumed.

Has anyone done similar reserch and can share with me any result or direction to go for.

Thanks!

 

[engr]

Electrician sorry If I made you confuse :(

I will make the question clear.

A bulb filament is normally coild coil.

Coduct an experiment for 0-15ms (milli seconds)

case1:

plot a graph for current vs time.

case 2:

make the filament stright ( like a wire )

conduct the same experiment.

my question is will the both graph same?
(what will be the position of the graph , upper or lower )

I hope the question is clear now thanks to all those responded to this post

 

[The Electrician]

As I said in post #12, the curves will probably not be the same.

The straight wire will get rid of heat faster, and so will conduct high currents longer.

 

[The Electrician]

Here are a couple of scope photos showing the current drawn by a 60 watt incandescent bulb with suddenly applied voltage.

The first photo shows the current with normal 120 VAC (U.S.) applied at the peak of the sine wave.

The second photo shows the current with 120 volts DC applied.

http://www.freeimagehosting.net/uploads/adbe0bd5b1.png

http://www.freeimagehosting.net/uploads/4463acf782.png

This filament is a coiled coil. Sorry I can't get a curve for a filament stretched out as a straight wire. :-(

 

[engr]

Electrician, thanks.

"The straight wire will get rid of heat faster, and so will conduct high currents longer"

Agree to you at this point, but what about the "inductive" effect due to the inductance of coilied/coil?

( Pls do not confuse:- due to coiled coil geometry of the filament there should be some inductance right?

what will be the effect of this inducatance to the inrush current? which hapens in very short time.

 

[The Electrician]

The controlling relationship is e(t) = L*di/dt, ignoring the resistance in the circuit.

The second scope photo referred to in post #16 is for a 120 volt DC voltage source suddenly applied to the light bulb.

In post #10 I gave the measured inductance of the coiled coil filament as 190 nanohenries. Plugging the 120 volts and 190 nanohenries into the expression above, the initial rate of rise of the current is 632 amps/microsecond. Since the scope photo shows the peak initial current to be 6 amps, then the time taken for the current to rise to 6 amps should be 6/632E6 = 9.5 nanoseconds.

But in reality, the 190 nanohenries of the coiled coil will be negligible compared to the other inductances in the circuit unless extreme care is taken to minimize those stray inductances.

I haven't shown the scope photo, but I did capture the leading edge of the current spike at turn on, and it measures about 200 nanoseconds, which suggests that the stray inductances of my setup were substantially larger than 190 nanohenries.

If the coiled coil is straightened out, then the resulting single strand of wire, which will be over a meter long, will have an inductance which is larger than the coiled coil if it is formed into a circular loop, or smaller if formed into a narrow hairpin shape. Whatever the inductance is, the ideal rate of rise can be calculated from the formula above (neglecting stray inductances).

If the light bulb is energized from the power grid, the inductances associated with the pole transformer and the cable into the house will completely swamp out the 190 nanohenries of the filament.

 

[engr]

Electrician Thanks , I am sorry I could not follow you fully.

"If the coiled coil is straightened out, then the resulting single strand of wire, which will be over a meter long, will have an inductance which is larger than the coiled coil "

this looks very strange , can you check this statement please?

**********
"inductances in the circuit " this parameter is existed always , so I think the total effect of this parameter will be inductive reactance ( together with the circuit resistance it will be impedance , as the resistance is very high when cmpare with resistance and so the effect is also negligible,

so the wawe form ( current) will be ' inphase ' with the voltage

I agree these points
***

but why there is a spike ( not inphase with voltage waveform during starting?

if you say " due to circuit inductance , I will have to say that circuit inductance are always available in the circuit

****

please do not just say ínrush current!

 

[The Electrician]

I said:
"If the coiled coil is straightened out, then the resulting single strand of wire, which will be over a meter long, will have an inductance which is larger than the coiled coil IF IT IS FORMED INTO A CIRCULAR LOOP, OR SMALLER IF FORMED INTO A NARROW HAIRPIN SHAPE"

The filament is a tungsten wire about .0018 inches in diameter. If you straighten it out and form it into a circular loop, the loop will be about 39.37/Pi = 12.5 inches in diameter. The inductance of such a loop is about 1.8 uH. The coiled coil has a measured inductance of about 190 nanohenries.

But, then you ask:

"...why there is a spike ( not inphase with voltage waveform during starting?"

The spike is due to the fact that when the voltage is first applied, the temperature of the filament is at room temperature (about 20 degrees centigrade). But within a short time (milliseconds), the temperature rises several thousand degrees. This causes the resistance to increase by a factor of about 15.

The initial current is about 6 amps, but after several tens of milliseconds, it decreases to about .4 amps.

The inductance hardly changes at all due to the very great increase of temperature.

This very large change in resistance is the dominant effect. The inductance of the filament has a negligible effect on the initial rise time of the current pulse compared with the other circuit inductances.

You say:
"...circuit inductance are always available in the circuit"

This is quite correct, and the circuit inductances don't change as the filament becomes white hot. It is the change in filament resistance that determines the shape of the current pulse vs time.

In one of your posts, you suggest that strictly speaking, the inductance of the filament will have an effect on the current. This is true, but the effect is completely negligible compared to other effects, especially the change in resistance of the filament vs time as it heats up.

The effect of circuit inductances will be seen mainly in the initial rise time of the current pulse, which I measured as about 200 nanoseconds. If the sweep speed of the oscilloscope is set to display 10 milliseconds/centimeter, as in this scope photo:

http://www.freeimagehosting.net/uploads/4463acf782.png

then the rise time of the current pulse appears almost instantaneous, but it is actually about 200 nanoseconds. The circuit inductances have essentially no noticeable effect on the current pulse after the initial rise time. The large decrease in the current occurring in a few milliseconds is due to the increase in the resistance of the filament as its temperature rises to several thousand degrees.

 

[Redbelly98]

... The measured inductance is about 190 nanohenries.

Thanks for measuring that. So the impedance is

|Z| = | j ω L |

= 2 π 60s^-1 190x10^-9H

= 72 μΩ

Compare that to the filament resistance of order 100Ω hot or 10Ω cold.

So yes, totally negligible as has been said.

 

[engr]

Redbelly98,

The subject here is not impedance of the filament ,(As the inductance is very low when coparing with resistance the inductive reactance will also is very very low , please refer the the earlier posts,)

the point is the effect of rate of change of current on the inductance of coiled coil filament which is Nothing to do with impedance of the filament

 

[engr]

Electrecian thanks,

thanks for the link :)

insted of tungston , if a copper wire with same geometry of the filament coil do you think 'absolute effect' comparing the two situation will the same?

 

[The Electrician]

If copper wire of the same diameter (.0018", 45 gauge), were used to form the filament, its cold resistance would be much lower than the tungsten wire. When the voltage was applied, there would be a much larger initial current, the power dissipated in the wire would be much larger, and the wire would promptly melt; it might even explode. You wouldn't see much of the sort of pulse of current you saw with the tungsten wire.

 

[engr]

Thanks Electrician,

before conclude I have one more question,

1.If a solid resistor of same value but length equal to the length of filament ( not in the stright postion , coiled coil geometry length) , same material do you think the wave form will be the same?