walter said:
I had a re-do interview and when asked this question again, why is there inrush current in a capacitor? A capacitor reacts to an instantaneous change in voltage with high current, charging or discharging. Was well received with the reply- a capacitor cannot react instantaneously to a change in voltage, and an inductor cannot react instantaneously to a change in current.
Practical real world capacitors and inductors have non-ideal characteristics that come into play for these theoretical hypotheticals. I would also add there is no such thing as an instantaneous voltage step (in the real world it will have a rise time).
A capacitor can begin to react arbitrarily quickly but will have additional electrical parameters like ESR (equivalent series resistance) and ESL (equivalent series inductance) that just like they sound will behave like a series resistance and a series inductance that will limit peak current.
My response was based on this statement I found on-line “The current through a capacitor is given as the capacitance times the change of the voltage versus time. So, the faster the voltage changes, the greater the current. Also, the greater the current, the greater the voltage change. If the voltage were to change instantly, that would mean that the current would be extremely large - infinitely large.”
Yes, the theoretical relationships are useful to predict terminal voltage behavior for modest rates of change. In the extremes the non-ideal characteristics (ESL, ESR, etc) can become significant.
I had thought about the situation of a completely discharged capacitor behaving like a wire at t=0, but suspected that a partially charged capacitor also has inrush current so I did some math. My circuit has a 10v supply, 10 Ohm resistor, and 1uF capacitor. 10uS tau, or time constant. I looked at five time constants with the fully discharged cap, and then five time constants with a cap that has a 2v charge at t=0. The fully discharged cap after one time constant: (10v)(1-(1/e^-10uS/10uS)) = 6.32,second time(….(20uS/10uS)) =8.65, Third = 9.5, 4th =9.82, 5th = 9.93.
The initial or starting voltage of a capacitor does not affect the capacitor's impedance, just how much charge it contains.
When starting with a cap with 2v, there is a difference of 8v so: (8v)(1-(1/e^-10uS/10uS)) = 5.06 so 5.06v + 2v = 7.06v on the cap after one time constant, 2nd = 6.92 + 2 = 8.92, 3rd = 7.60 + 2 = 9.60, 4th = 7.85 + 2 = 9.85, 5th = 7.94 + 2 = 9.94. The inrush voltage is similar with a completely discharged cap, and a cap that has a charge.
I then looked at the math for inrush current. Q/t =I for both the discharged cap, and the cap with 2v at t=0. For each time constant, I used the voltage value from above to determine the value of Q from the equation Q=CV. So the equation I used is CV/t = I, changing only V for each time.
I am still uncomfortable calling that "in rush" current. I checked my edition of S-100 (IEEE reference standard dictionary of terms) and they list a "inrush current (1) (electronic power transformer)"- talking about transformer start up current, and "inrush current (2) (packaging machinery) " - talking about current of a solenoid or coil with the armature blocked ..
Sorry no mention of capacitor inrush current, its not a thing.
The discharged cap initially draws 1A, and after each 10uS: 368mA, 135mA, 50mA, 18mA, 7mA.
The cap with 2v initially draws 800mA, and after each 10uS: 294mA, 108mA, 40mA, 15mA, 6mA.
The inrush curve is similar with a completely discharged cap, and a cap that has a charge. This leads me to believe that a fully discharged cap behaving like a wire is insignificant- as long as the cap is not bad.
A cap will always behave like a cap... perhaps with a small wire (ESR) in series.
For your circuit a RC with 10 ohms R and 1uF C, I didn't check your math.
Just for chuckles I looked at typical ESR for 1uF caps. According to the mouser look up parts selection 1uF aluminum caps have ESR over a range of 3 ohms to more than 300 ohms.
http://www.mouser.com/Passive-Components/Capacitors/Aluminum-Electrolytic-Capacitors/_/N-75hqt?P=1z0wrj5
A 1 uF film cap might have lower ESR.
Back to the real world, PS reservoir caps can see the kind of charging currents you describe and some modern low ESR caps (designed for switching PS use) can be quite good. For them the useful spec is ripple current or how much charging current they are happy with. Note: in that case the R for the RC is the transformer winding resistance, while the diode in the middle makes it more interesting.
To explore real RCs more practical currents are mA or uA. I hope I didn't add to the confusion.
I apologize for being so pedantic about the terminology, use of terms vaguely or incorrectly can impact how an interviewer will evaluate your skill level. Maybe tell them they are using "inrush" wrong. 8) 8) 8) :
JR