Transient Response of Capacitance

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bsrivastava

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Hi,
I want to know the Lenz's law equivalent in electrostatics.
To elaborate the same:
The inductor resist any instantaneous change of current across it and can easily be modeled by replacing the inductor with voltage source to continue the current to flow in same direction. This is in line with Lenz's law.

Now, the capacitance also resist any instantaneous change of voltage across its plates. How can it be better understood or modeled.

Thanks
-Bharat
 

You may use the same principle as for inductor -- Current source with a resistor and measure the voltage across the resistor.

But you rarely use such models in transient analysis. Consult any standard textbook
 

in transient capacitor acts like voltage, and blocks current, and develop voltage.




Naveed
 

For linear capacitor, the capacitor current and voltage are related by

\[ i_C(t) = C \frac{d v_C(t)}{dt}\]

So, voltage on the capacitor is the proportional to the integral of the
current injected into it. In the more general case, where the capacitor is not
necessarily linear, we have

\[ i_C(t) = \frac{d (C v_C(t))}{dt} = \frac{d q(t)}{dt}\]

where \[q(t)=C v_C(t)\] is the charge on the capacitor.
This might seem like a trivial definition but when the capacitance is
not a constant, for example, if it is a function of the capacitor voltage itself,
we get \[C=C(v_C)\]. In this case, to derive the current we have to
take the derivative of the product \[C(v_C) v_C\] which will
involve the use of the chain rule from calculus. We cannot just write the
current in the form of the first equation above which only applies if the
capacitance is a constant.

Best regards,
v_c
 

The transient response for capacitance is it opposes change of voltage accross it and acts like a short circuit.
 

pmonon said:
The transient response for capacitance is it opposes change of voltage accross it and acts like a short circuit.
Click to expand...

You are right, it does oppose the change of voltage across it. In fact you cannot change the voltage across the capacitance instantaneously. That would require infinite current. You can see this from the i=C dv/dt expression. If you want to change the voltage instantaneously the derivative of the voltage (slope) would be infinite, which would make the current infinite.

Capacitor does not act like a short circuit in general.
It will only act like a short circuit (low impedance) at higher frequencies.
Capacitor acts like an open circuit at DC when the capacitor voltage is not changing (it is either at DC excitation or has reached equilibrium). This is also apparent from the i=C dv/dt expression. At dc equilibrium, the dv/dt term is zero since the capacitor voltage is not changing with time. Since dv/dt equals zero, so does the current in the capacitor. Zero current in the capacitor means open circuit.

Best regards,
v_c
 


You are missing the point. The question is about transient response not the steady state. In transient case, the capacitor is a short circuit even if the input is DC. This is because the change of voltage is abrupt and dv/dt is infinity.
 

pmonon,

I think we might be talking about different things as we interpret the question
asked by the original poster. I am talking about the behavior of a capacitor in a circuit where there is a time dependent source of some kind.

I agree that dv/dt would be infinity only if there was a step voltage source right across the capacitor. At that time, the capacitor would be a short as infinite current would flow across it as you indicate. However, this is only a special case. In general, you cannot say that the capacitor acts like a short circuit for transient response. Take an RC circuit with a step voltage source excitation for example. Can you say that the capacitor acts like a short circuit during the transient response? Maybe initially, but once the transient has started, even if we are not in steady state, the capacitor is building up its voltage as it is getting charged and is behaving more like a time dependent voltage source -- which is not really a short circuit.

It is up to the original poster to clarify what his intent was as we (you and I) seem to be more interested in this topic than him

Best regards,
v_c
 

 

v_c and pmonon,

I am very much involved in the ongoing discussion being the original poster.

I am rephrasing my doubt " Capacitor resists any instantaneous change across its plates"

In the light of above; Think of resistor divider circuit and replace the resistances with equal capacitances. If the input of this circuit is a unit step function of voltage V, what will be the voltage in between in transient and in steady state.
 

bsrivastava -- good question!

If you apply a voltage of \[u(t)\] to the two capacitors, each capacitor will instantaneously have voltages equal to \[u(t)/2\] and therefore the voltage across the bottom capacitor from the node that you marked \[V_{ref}\] to ground will be equal to \[u(t)/2\] as well.

The reason that the capacitor voltage will change instantaneously in this case is because infinite current will flow through the capacitors. This current is equal to the voltage across the times the derivative of the voltage applied across them. To make it easier, you can imagine that the voltage \[u(t)\] is being applied across a single equivalent capacitance of value \[C/2\], since this is the series combination of two capacitors in series. Based on this current through the capacitors is \[(C/2)dv/dt=(C/2) du(t)/dt=(C/2) \delta (t)\], where the derivative of the step is the familiar Dirac delta function.

In a more realistic circuit, the input voltage source would have a finite risetime to go from 0 to 1V. With the parasitic series and parallel resistance in the circuit would limit, this would limit current and therefore the current would be finite. The duration of the current pulse would be some small value but not zero as in the case of the impulse. However, the each of the capacitor voltages would still eventually settle to 0.5V.

Below are some simulation results, as you can see the input is a unit step \[u(t)\], the voltage across the capacitor is \[0.5u(t)\] and the current is impulsive in nature. The zoomed in picture shows more detail. The magnitude of the current is not infinite. Its magnitude is just \[C/2\] times the derivative of the voltage input. Since the input voltage has a slope of \[1V/1ns\], the current is \[(C/2) dv/dt = (0.5 \mu F) (1V/1ns)= 500A\].

Does this help?

Best regards,
v_c
 

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