gyraf

Re: Why is there inrush current in a capacitor?
« Reply #20 on: March 30, 2016, 03:59:16 AM »
..don't get too hung up on job interview questions or claims - it's common  practice to try spooking applicants to see how they react to cognitive dissonance/stress. And/or to try pushing into creative thinking..

Jakob E.
(who took the class)
..note to self: don't let Harman run your company..


walter

Re: Why is there inrush current in a capacitor?
« Reply #21 on: April 01, 2016, 11:39:49 AM »
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.
 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.”
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.

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.

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.
Blown like a fuse

JohnRoberts

Re: Why is there inrush current in a capacitor?
« Reply #22 on: April 01, 2016, 01:24:06 PM »
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.   
Quote

 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.
Quote
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.
Quote
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.
Quote
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
Visit https://circularscience.com to hear what properly "cleared" drums sound like.

DaveP

Re: Why is there inrush current in a capacitor?
« Reply #23 on: April 01, 2016, 01:38:59 PM »
Quote
Sorry no mention of capacitor inrush current, its not a thing.

Weeeeeell........

I recently changed the design of my tube power supply, going from 47uF to 470uF.  I decided to check the spec of the standard rectifier diode to see if it could cope.  The 1N4007 has a maximum surge current of 30A (Inrush???) whereas the 1N5400 series is 200A, so I changed to that type for reliability.  Just sayin...............

DaveP
Soundcloud: Delayed Action.

JohnRoberts

Re: Why is there inrush current in a capacitor?
« Reply #24 on: April 01, 2016, 02:02:25 PM »
Quote
Sorry no mention of capacitor inrush current, its not a thing.

Weeeeeell........

I recently changed the design of my tube power supply, going from 47uF to 470uF.  I decided to check the spec of the standard rectifier diode to see if it could cope.  The 1N4007 has a maximum surge current of 30A (Inrush???) whereas the 1N5400 series is 200A, so I changed to that type for reliability.  Just sayin...............

DaveP
"Surge current" is a real thing for diodes...  "Inrush current" is a real thing in transformers and solenoids.

While wasting too much time looking at cap data sheets, I noticed some (pulse) film caps specify a max rate of change in volts per uSec. Which no doubt maps out to a peak current for that rate of change.

Sorry I am doing what I accused others of, arguing about unimportant details.

I'd never get hired by that interviewer.  ;D

JR
Visit https://circularscience.com to hear what properly "cleared" drums sound like.

joaquins

Re: Why is there inrush current in a capacitor?
« Reply #25 on: April 10, 2016, 09:56:20 PM »
  So many equations about caps, currents, voltages and charges and nobody even mentioned an integral or a differential of nothing, my eyes burns!!!!

  You take the differential version of the cap equation  i =C*∂v/∂t and you change the voltage very fast you get a big spike. Voila math explains it. Any math approach without the differential (or integral) will be wonky. The only admisible math without them is the solution (the exponential) but it doesn't explain why it happens, just what.

Q=∫i*dt
v=Q/C

-->v=1/C*∫i*dt

∂/∂t--> ∂v/∂t=i/C

That's the math, I've checked it with wikipedia (no I didn't).

Note that if current is constant, Q=i*t, but making Q=V*C at the same time would make V to rise constantly and would not explain inrush. Making it around v*C=i*t and doing anything with the shape of v makes very little sense, because the equation assumed current was a constant.
(and you can NOT divide by zero, it isn't infinity) if confused ask youtube about atmel µC dividing by zero

Nothing against you @DaveP, talking in general, your post was the one with more math but there are other errors (more like horrors maybe) around.

You could give the physical explanation without any math and it can be right, nothing wrong with that, some math could be thrown around to complete it. If you go for the math approach it needs to be complete, and correct. Otherwise there is no reason for inrush current if there are no integrals or derivates, as if it is a resistor.

JS
« Last Edit: April 10, 2016, 10:15:37 PM by joaquins »
If I don't know how it works, I prefer don't turn it on.

ruffrecords

Re: Why is there inrush current in a capacitor?
« Reply #26 on: April 11, 2016, 04:40:08 AM »
@joaquins

I applaud anyone who uses maths to help explain what happens in the real world. Unfortunately you forgot to include the series resistor!

Here is one explanation:

http://www.intmath.com/differential-equations/6-rc-circuits.php

Cheers

Ian
www.customtubeconsoles.com
https://mark3vtm.blogspot.co.uk/
www.eztubemixer.blogspot.co.uk


'The only people not making mistakes are the people doing nothing'

DaveP

Re: Why is there inrush current in a capacitor?
« Reply #27 on: April 11, 2016, 05:07:48 AM »
Well I can't take the credit or the criticism because I just lifted it from Morgan Jones ;D

DaveP
Soundcloud: Delayed Action.

JohnRoberts

Re: Why is there inrush current in a capacitor?
« Reply #28 on: April 11, 2016, 11:27:02 AM »


Differential equations do not add much clarity for me, but perhaps that is a personal problem.  :-[

I memorized the e^-t/rc equation because I found it extremely useful to describe the voltage in RC circuits versus time.

 Current flow is well defined by the R between the voltage and the cap.

JR
Visit https://circularscience.com to hear what properly "cleared" drums sound like.

ruffrecords

Re: Why is there inrush current in a capacitor?
« Reply #29 on: April 11, 2016, 02:33:05 PM »


Differential equations do not add much clarity for me, but perhaps that is a personal problem.  :-[

I memorized the e^-t/rc equation because I found it extremely useful to describe the voltage in RC circuits versus time.

 Current flow is well defined by the R between the voltage and the cap.

JR

Sometimes math is a help, others it's a hindrance.

I remember in my second year of university, one of the courses was microwaves (taught by one of the guys who invented micro-strip). He started off with Maxwell's equations and then derived everything else, like evanescent waves, directly from them. To pass the exam, all I had to remember was Maxwell's equations and the derivations - about 4 pages of A4 to revise.

Cheers

Ian
www.customtubeconsoles.com
https://mark3vtm.blogspot.co.uk/
www.eztubemixer.blogspot.co.uk


'The only people not making mistakes are the people doing nothing'


Matt Nolan

Re: Why is there inrush current in a capacitor?
« Reply #30 on: April 11, 2016, 03:04:45 PM »
I remember in my second year of university, one of the courses was microwaves (taught by one of the guys who invented micro-strip). He started off with Maxwell's equations and then derived everything else, like evanescent waves, directly from them. To pass the exam, all I had to remember was Maxwell's equations and the derivations - about 4 pages of A4 to revise.
Wow! That's impressive. I only ever got A-levels down to 4 sides of A4. Degree modules were somewhat harder to compress.

I don't remember the microwave course so well but I remember the Opto-electronics one, laser diodes, etc. We had 2 lecturers and in the exam you had to choose 3 questions out of 5 to answer. One lecturer's subjects would take up 2 of the questions and the other lecturer's would take up 3 of them. The first lecturer was utterly incomprehensible. The only thing I remember him saying was how smart Sherlock Holmes (or rather Conan-Doyle) was to know that red light carries further in fog than blue light. So, I risked everything on only revising the second lecturer's subjects and would therefore have no choice on the paper, I would have to do whatever he asked.

There were 106 different equations and proofs, derivations, etc. for just over half of that one module. In my revision, I managed to boil it down to 65 or so, from which the others could be derived. I think it's the most stuff I've had in my head at any one time. I did remarkably well on that exam. The one the following day - really easy subject - digital communication networks - I barely scraped a pass. Head full. No more room.

ruffrecords

Re: Why is there inrush current in a capacitor?
« Reply #31 on: April 11, 2016, 04:24:18 PM »
That reminds me of another 'cramming' technique I used. I would read all my notes for the subject aloud and record them to tape (reel to reel of course). Then sit back and listen to them over and over.

Another technique a colleague and I used was to play chess. Each player had 5 minutes to make his move during which time the other would revise. We had some really great chess games and could honestly say we had spent half the day revising.

Cheers

Ian
www.customtubeconsoles.com
https://mark3vtm.blogspot.co.uk/
www.eztubemixer.blogspot.co.uk


'The only people not making mistakes are the people doing nothing'

JohnRoberts

Re: Why is there inrush current in a capacitor?
« Reply #32 on: April 12, 2016, 12:10:10 PM »


Differential equations do not add much clarity for me, but perhaps that is a personal problem.  :-[

I memorized the e^-t/rc equation because I found it extremely useful to describe the voltage in RC circuits versus time.

 Current flow is well defined by the R between the voltage and the cap.

JR

Sometimes math is a help, others it's a hindrance.

I remember in my second year of university, one of the courses was microwaves (taught by one of the guys who invented micro-strip). He started off with Maxwell's equations and then derived everything else, like evanescent waves, directly from them. To pass the exam, all I had to remember was Maxwell's equations and the derivations - about 4 pages of A4 to revise.

Cheers

Ian
I am math challenged but even I used derivatives and integrals to reduce the number of equations I had to memorize for physics tests...  position/velocity/acceleration equations can be derived from each other.

JR

Visit https://circularscience.com to hear what properly "cleared" drums sound like.

CJ

Re: Why is there inrush current in a capacitor?
« Reply #33 on: April 13, 2016, 02:40:35 PM »
it is not about the plates, but what is in between,

the plates serve to rotate the moment of the molecules in between,

these molecules fight back as voltage goes up,

energy is stored in the rotated molecules

just like magnetic domains in transformer cores, they can not get rotated past 180,  180 is saturation,

180 is probably not achieved in a dialectic, maybe 90 degrees will mean a full charge

the bigger the plates, the more molecules in between to rotate,

pictures are helpful also,

The time constant is the time taken (in seconds) by the capacitor C that is fed from a resistor R to charge to a certain level. The capacitor will charge to 63% of the final voltage in one time constant, 85% in two time constants, and 100% in five time constants. If you graphed the % charge against time constant, the result is exponential.
« Last Edit: April 13, 2016, 02:48:15 PM by CJ »
If I can't fix it, I can fix it so nobody else can!
Frank's Tube Page: www.mif.pg.gda.pl/homepages/frank/vs.html
Guitar Amps: http://bmamps.com/Tech_sch.html

JohnRoberts

Re: Why is there inrush current in a capacitor?
« Reply #34 on: April 13, 2016, 03:32:26 PM »
it is not about the plates, but what is in between,

the plates serve to rotate the moment of the molecules in between,

these molecules fight back as voltage goes up,

energy is stored in the rotated molecules

just like magnetic domains in transformer cores, they can not get rotated past 180,  180 is saturation,

180 is probably not achieved in a dialectic, maybe 90 degrees will mean a full charge

the bigger the plates, the more molecules in between to rotate,

pictures are helpful also,

The time constant is the time taken (in seconds) by the capacitor C that is fed from a resistor R to charge to a certain level. The capacitor will charge to 63% of the final voltage in one time constant, 85% in two time constants, and 100% in five time constants. If you graphed the % charge against time constant, the result is exponential.
Actually 5 time constants gets you to 99.3% but close enough for government work, and a good rule of thumb to remember.   (sorry for TMI).

JR
Visit https://circularscience.com to hear what properly "cleared" drums sound like.

CJ

Re: Why is there inrush current in a capacitor?
« Reply #35 on: April 13, 2016, 05:59:53 PM »
those guys probably worked on the Hubble, you are right, always round up not down

see that in theory the cap never gets fully charged, at least in our lifetime,

http://web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide05.pdf
« Last Edit: April 13, 2016, 08:22:31 PM by CJ »
If I can't fix it, I can fix it so nobody else can!
Frank's Tube Page: www.mif.pg.gda.pl/homepages/frank/vs.html
Guitar Amps: http://bmamps.com/Tech_sch.html

JohnRoberts

Re: Why is there inrush current in a capacitor?
« Reply #36 on: April 13, 2016, 11:25:49 PM »
those guys probably worked on the Hubble, you are right, always round up not down

see that in theory the cap never gets fully charged, at least in our lifetime,

http://web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide05.pdf

"Asymptote",,,, There is good "engineer" joke about asymptotes. 
======
"A mathematician and an engineer are sitting at a table drinking when a very beautiful woman walks in and sits down at the bar.

The mathematician sighs. "I'd like to talk to her, but first I have to cover half the distance between where we are and where she is, then half of the distance that remains, then half of that distance, and so on. The series is infinite. There'll always be some finite distance between us."

The engineer gets up and starts walking. "Ah, well, I figure I can get close enough for all practical purposes."
======
 ;D

JR

Visit https://circularscience.com to hear what properly "cleared" drums sound like.

ruffrecords

Re: Why is there inrush current in a capacitor?
« Reply #37 on: April 14, 2016, 02:40:28 AM »
it is not about the plates, but what is in between,

the plates serve to rotate the moment of the molecules in between,

Does that mean there is no capacitance between a pair of plates in a vacuum?

Cheers

Ian
www.customtubeconsoles.com
https://mark3vtm.blogspot.co.uk/
www.eztubemixer.blogspot.co.uk


'The only people not making mistakes are the people doing nothing'

Matt Nolan

Re: Why is there inrush current in a capacitor?
« Reply #38 on: April 14, 2016, 04:31:14 AM »
it is not about the plates, but what is in between,

the plates serve to rotate the moment of the molecules in between,

Does that mean there is no capacitance between a pair of plates in a vacuum?

Cheers

Ian
I can only assume CJ was making an assumption based on the K of the dielectric being so large as to make the permittivity of the vacuum negligible in comparison.

CJ

Re: Why is there inrush current in a capacitor?
« Reply #39 on: April 14, 2016, 06:49:20 AM »
what was i thinking of course you will have capacitance between two plates, look at a vacuum tube,

interesting to test C in a 6L6 and then let the air out and check it again,

but no, it looks like the constant for air is close to that of a vacuum,

look at this, vacuum capacitors>

https://en.wikipedia.org/wiki/Vacuum_variable_capacitor

note that there will be a mechanical attraction between the plates of a capacitor, so you can use the work-energy theorem to get the work done to hold the plates apart and relate that to energy,

"The physical constant ε0, commonly called the vacuum permittivity, permittivity of free space or electric constant, is an ideal, (baseline) physical constant, which is the value of the absolute dielectric permittivity of classical vacuum. Its value is

    ε0 = 8.854 187 817... × 10−12 F·m−1 (farads per meter).

The relative permittivity of a material is its dielectric permittivity expressed as a ratio relative to the permittivity of vacuum.

Permittivity is a material property that affects the Coulomb force between two point charges in the material. Relative permittivity is the factor by which the electric field between the charges is decreased relative to vacuum.

Likewise, relative permittivity is the ratio of the capacitance of a capacitor using that material as a dielectric, compared to a similar capacitor that has vacuum as its dielectric. Relative permittivity is also commonly known as dielectric constant, a term deprecated in physics and engineering, but one which is still commonly used in chemistry."

the whole idea of charge is very interesting, when you charge a cap, where do the electrons go? you can not force them into the steel atoms, right? nor can you strip electrons off steel atoms to create negative charge? kind of like holes and electrons in transistor theory, borders on the conceptual side,

another interesting thing is the stored energy in the dialectic, when you torque those molecules you store energy, but when the molecules relax, how does that transfer energy back into the system?

glad i don't teach physics,  ;D
« Last Edit: April 14, 2016, 07:09:08 AM by CJ »
If I can't fix it, I can fix it so nobody else can!
Frank's Tube Page: www.mif.pg.gda.pl/homepages/frank/vs.html
Guitar Amps: http://bmamps.com/Tech_sch.html