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Inrush Current for Capacitive Load

electronicsLearner77

Jul 2, 2015
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I read that for the capacitive loads there will be inrush current. But if you see the circuit below or in general and if a step input of 12V given,
upload_2022-6-30_13-29-30.png

As per the impedance formula the Z=1/j*w*C, the capacitor shall offer infinite impedance and the capacitor will act as open circuit and the initial current will be 0, then why will there be high inrush current? It is applicable for inductors only?
 

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Harald Kapp

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The formula to use here is I = C × dV/dt.
With a step input dV/dt is infinite and the inrush current is mathematically also infinite.
In practice no real capacitor is ideal. There is always a series resistance, called ESR. In LTSPICE you can enter this value directly via a right mouse click:
upload_2022-6-30_10-38-30.png
Unfortunately I'm unable to find the default ESR in LTSPICE, so you'll have to enter a reasonable value, preferably from the capacitor's datasheet.
 

electronicsLearner77

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The standard ESR resistance is between 0.1 - 0.01 Ohm. If i do the calculations,
Inrush current = 12/0.1 = 120 Amps.
The current formula is = 12/0.1*e^{-t/RC} = 120*e^{10^6}*t; Is this correct? The LTspice also shows as 120A. But i am confused of the formula dv/dt, i am not calculating dV/dt anywhere.
upload_2022-7-1_8-50-17.png
 

Harald Kapp

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i am not calculating dV/dt anywhere.
Maybe you don't, but LTSPICE does it for you internally. Every SPICE variant uses differential calculus (or mathematical approximations like the Runge-Kutta method) to obtain the results.

Using the impedance XC = 1/(2 × Pi × f × C) is applicable only to stationary, sinusoidal signals.
 

electronicsLearner77

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Maybe you don't, but LTSPICE does it for you internally. Every SPICE variant uses differential calculus (or mathematical approximations like the Runge-Kutta method) to obtain the results.
What I am trying to say is the inrush current is simply V/R. Why is it required to use dv/dt may be in this particular scenario.
 

Harald Kapp

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Why is it required to use dv/dt may be in this particular scenario.
Because V changes over time in a non-stationary, non-sinusouidal manner. It is explained e.g. here or find another explanatio that suits you better?

Or let me trry to explain it without differential equations:
At t= 0 the capacitor is discharges, Vc = 0 V.
Thus when the input voltage jumps from 0 V to 12 V, this voltage is fully across the resistor and I = V/r applies with the resulting 120 A for a 0.1 Ω resistor, as you observed.
Now take a step in time forward: The current has charged the capacitor, thus the voltage across the capacitor starts to rise. The voltage across the resistor on the contrary is reduced since this voltage is the difference between input voltage and the voltage across the capacitor: Vr = Vin-Vc. The reduced Vr in turn leads to a reduced current: I = Vr/R = (Vin-Vc)/R.
Take another step forward in time: Again the capacitor's voltage has increased, but more slowly since the current is reduced compared to the initial condition. Apply the same reasoning as above: Rising capacitor voltage reduces the voltage across the resistor which in turn reduces the current.
This is why the current slowly but steadily decreases (exponentially) while the voltage across the capacitor increases in the same way.
When you make the time steps infinitesimally small you arrive at the differential equations.


In a real circuit there are, apart from the capacitor's ESR, other factors limiting the current:
  • wiring resistance
  • limited current drive capability of the voltage source (inner resistance or enforced current limiting)
These all can easily be modeled in LTSPICE either as parameters of the component (as shown in my post #4 - this is the preferred method!) or as separate components (not preferred due to the way these parasitics can be handled more effectively when entered as component parameters.
 

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hevans1944

Hop - AC8NS
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I read that for the capacitive loads there will be inrush current. But if you see the circuit below or in general and if a step input of 12V given,
View attachment 55494

As per the impedance formula the Z=1/j*w*C, the capacitor shall offer infinite impedance and the capacitor will act as open circuit and the initial current will be 0, then why will there be high inrush current? It is applicable for inductors only?
The very first thing wrong with this statement is your "impedance formula" where you state the impedance is Z = 1/j*w*C. Last time I looked, the capacitive reactance "formula" is XC = 1/jωC, where ω is the radian frequency, equal to 2πf, and C is the capacatnce. A somewhat similar "formula" applies to inductive reactance: XL = jωL, where L is the inductance. Note that both of these are vector quantities. Also note that these "formulas" apply ONLY to circuits with steady sinusoidal currents. Transient currents, as when a sinusoidal voltage, or a step-voltage is applied, require analysis by differential equation calculus, as @Harald Kapp described in his post #9 above.

Your initial example of a 12 volt step voltage applied to a 10 μF capacitor has no real world solution. The step voltage would have to be provided by a voltage source of infinite current capacity to instantaneously develop 12 volts across the uncharged capacitor terminals. This would require that this step-voltage source an infinite amount of current instantaneously and charge the capacitor to 12 volts instantaneously. None of these conditions exist in the real world.

As for "formulas" I suggest you dig a little deeper to discover where and how those "formulas" are derived, and under what circumstances a "formula" applies.
 

CircutScoper

Mar 29, 2022
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I read that for the capacitive loads there will be inrush current. But if you see the circuit below or in general and if a step input of 12V given,
View attachment 55494

As per the impedance formula the Z=1/j*w*C, the capacitor shall offer infinite impedance and the capacitor will act as open circuit and the initial current will be 0, then why will there be high inrush current? It is applicable for inductors only?

Try looking at it this way.

The charge Q (Coulombs) stored by a capacitor C (Farads) with V (Volts) applied Q = CV = 120uCb in your example of a 10uF cap when it's charged to 12V. That 120uCb of charge has to be transferred from the 12V supply to the cap when the supply is switched on. The current that transfers this initial charge is the "inrush" we're talking about.

1 Ampere is defined as 1A = 1Cb/s. Therefore, in this example, the inrush in Amps will be 120uCb divided by the time T (seconds) that the cap takes to charge. I = Q/T = 120uCb/T.

Eg: if T = 1us (as when R1 = 0.1Ohm in Harald's diagram), then I = 120uCb/1us = 120A and heaven help your switch!
 
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electronicsLearner77

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Using the impedance XC = 1/(2 × Pi × f × C) is applicable only to stationary, sinusoidal signals.
In most of the books i have come across there will be a statement which considers a case where the frequency of the sinusoidal signal is "0", which is a pure DC. That is the reason i have put the formula, one thing i forgot is, this is applicable when the 12V has become stabilized or linear, but frequency is not 0 when the signal goes from 0 to 12V an ideal scenario which is not possible as pointed above.
 

CircutScoper

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In most of the books i have come across there will be a statement which considers a case where the frequency of the sinusoidal signal is "0", which is a pure DC. That is the reason i have put the formula, one thing i forgot is, this is applicable when the 12V has become stabilized or linear, but frequency is not 0 when the signal goes from 0 to 12V an ideal scenario which is not possible as pointed above.

That's the reason 1/(2piFC) is of no help in quantifying and designing for the inrush. One way that is useful, however, is to consider the associated energy the inrush will dissipate, which is easily calculated thusly.

The energy drawn from the power supply during the inrush = QV = (CV)V = CV^2 = 10E-6 x 12^2 = 0.00144J.
Meanwhile, the energy that will be delivered and stored in the capacitor is = CV^2/2.
Therefore the energy dissipated by the inrush = CV^2 - CV^2/2 = CV^2/2 = 0.00072J
 

electronicsLearner77

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Eg: if T = 1us (as when R1 = 0.1Ohm in Harald's diagram), then I = 120uCb/1us = 120A and heaven help your switch!
I am not sure why you have considered 1us, in case if i consider 1ns it will be multiplied by a factor of 10^3 or 120*10^3 A, which is wrong i think.
 

CircutScoper

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I am not sure why you have considered 1us, in case if i consider 1ns it will be multiplied by a factor of 10^3 or 120*10^3 A, which is wrong i think.

As explained, a timeconstant (T=RC) of 1us and 120A inrush result from Harald's simulation when R1 = 0.1 Ohms, which is a fairly plausible value. 0.0001 Ohms, as required for RC = 1ns, is (much) less so.

Meanwhile, the inrush energy dissipation calculation (1/2CV^2) is totally independent of R1.

upload_2022-7-1_7-17-57-png.55497
 
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