Hello.
First sorry for poor english.
I have a question.
What factors cause that a capacitor burst in a Circuit, Specially Electrolytic capacitors?
How about inductor,Does it burst too?
tnx.
Electrolytic capacitors explode usually from overvoltage , reverse voltage or hi ripple current , although the last usually causes the capacitor to get hot and it will bulge first before exploding.
Inductors normally dont explode, although if the inductor is a ferrite type , the ferrite can get hot and crack.
This is usually a sign of the wrong grade of ferrite for the frequency in use, or the inductor is being overloaded by too much current causing saturation.
leoncorleone:
Hello.
First sorry for poor english.
I have a question.
What factors cause that a capacitor burst in a Circuit, Specially Electrolytic capacitors?
How about inductor,Does it burst too?
tnx.
Electrolytic capacitors, being filled with an oily electrolyte (i.e. a liquid) can explode if they get too hot (causing the liquid to boil).
Things that can make them too hot are:
- Reverse polarity
- Higher voltage than rated for
- Excessive ripple current
Inductors are simply a length of wire wound around a form (air core) or wound around a core of ferrite, iron or even aluminum (!).
An inductor will not explode, but it can overheat and burn. Things that can cause an inductor to overheat and burn:
- Too many volts per turn (saturated core)
- Too much DC current flowing through it
The first one (too many volts per turn) is a function of the frequency of the AC signal across the inductor. For example, a certain inductor with 50 volts AC at 200 khz across it may run happy and cool, but 50 volts AC at 100 hz will fry it.
Lookup Faraday's Law to learn more about induction.
Typically its not just age, its leaving them unpowered for long periods of time (years), then powering up
to full voltage. The insulating oxide layer degrades, becoming conducting enough to cause heating
and bursting. A cap held at its working voltage will automatically maintain the oxide layer through
electrolytic action (clue is in the name!), and have minimal leakage current.
Especially for high voltage caps its recommended to re-condition them if been in storage, charging
to nominal working voltage through a fairly high value resistor for some hours, so that full voltage
is only reached after the oxide has been re-grown and the leakage current drops away to safe levels.
You can monitor the voltage across the resistor to measure the leakage current (make sure the resistor
can handle the power if a high voltage cap!)
Krupski:
The first one (too many volts per turn) is a function of the frequency of the AC signal across the inductor. For example, a certain inductor with 50 volts AC at 200 khz across it may run happy and cool, but 50 volts AC at 100 hz will fry it.
Lookup Faraday's Law to learn more about induction.
Does this depend on practical resistance in the winding?
The resistance is what generates the heat that causes the damage, but both the resistance and inductance
limit the current that flows, but inductive effect is proportional to frequency. Or put in maths
Z = R + wL
where R is the resistance, L the inductance, w is frequency in rad/s (w should be omega really, but my keyboard doesn't have that)
Z is the total impedance (being the phase-aware version of resistance). V = IZ is the phasor (ie complex)
version of V = IR
Southpark:
Does this depend on practical resistance in the winding?
Well, yes and no.
It's a tad complex, but I'll try.
An inductor is typically wire with "N" number of turns wrapped around a core. The wire has, of course, DC resistance. But, being an inductor, it also has AC resistance (properly called "impedance").
The symbol for impedance is "Z", and the formula to calculate impedance is [b]Z[sub]L[/sub] = 2 * PI * F * L [/b]where F is frequency in Hertz and L is inductance in Henrys.
So, you can see that if F is zero, then the AC "resistance" (impedance) is also zero and therefore the only "resistance" in the system is the wire.
So you can see that too much current through whatever the R of the inductor is the complete cause of heating.
Now, let's look at the AC side of things. The magnetic flux density in the core depends on the current flowing through the winding. So, looking again at the formula above, you can see that as F goes up, ZL also goes up and therefore the current in the winding goes down. But of course, if the frequency if too low, the current (and therefore flux density) goes up.
At F=0 (DC) the flux density would try to be infinity (only limited by the DC resistance, R, of the winding).
BUT!! Every inductor core has a flux density limit, above which it cannot increase. When you hit that point, the core is said to be "saturated" and no more current can go into making the flux density higher (the excess is burned off as resistive loss heat).
Now, if you have a transformer (two inductors coupled by one core), you need a certain number of turns in order to keep the core from saturating. You can also accomplish this by using a larger (area) core.
So, do you see that if you build a transformer, it can be smaller and use less turns if the AC frequency is higher?
That's the key to switching power supplies. For example, a 1000 watt PC power supply using an iron core transformer would probably be a cubic foot in size and weigh 30 pounds. By using a high frequency AC (250 khz or so), the supply works fine with much smaller transformer cores and less turns of wire.
Another thing to point out is that core MATERIAL differs at different frequencies. For low frequencies (like 1 50/60 hz transformer), you use IRON core. For high frequencies (50 khz or more... or RF) you use FERRITE cores.
Indeed, an iron core would not work at 50 khz, nor would a ferrite core wotk at 60 hz.
Hope this all makes sense....
Krupski and markT...... thanks very much for the nice explanations! Definitely helped in the understanding of inductor failure if a source AC voltage is applied across it..... at relatively low source frequency.
Yeah.... the impedance of the inductor is the j.omega.L
The reactance is the portion without the imaginary 'i' value.
Thanks for explaining again!