Parasitic effect of Capacitor question

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In summary, the frequency of a sin wave is 0 when the current through the parasitic resistance is equal to the current through the capacitor.
  • #1
Learnphysics
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Homework Statement



eW4Q1hS.png


A 68 nF capacitor has a parasitic parallel resistance Rp = 1 MΩ.
If a voltage V = Vmsin(ωt) is applied, as shown in the diagram, find the frequencies at which:
1. The current amplitude through the parasitic resistance is 1% of the current amplitude
through the capacitor.
2. The amplitudes of the currents through the resistor and the capacitor are equal.

Homework Equations



I=C(dv/dt) Through capacitor

I=V/R Through Resistance.

The Attempt at a Solution



I want to try to solve the second part of the question first because it would provide be a more fundamental solution.

In any case, I've equated the current through the capacitor with the current through the parasitic resistor:

Cdv/dt = V/Rp

Simple enough.

Now V=Vmsin(wt)
so
dv/dt = (w)(Vm)(Cos)(wt)Substituting in we get:

(C)(w)(Vm)(Cos)(wt) = Vmsin(wt)/(Rp).

so:
(Rp)(C)(w)(Vm)(Cos)(wt) = Vmsin(wt) <-- just moving the resistance to the other side

At t=0. We know that sin waves have 0 amplitude, and cos waves have 1 amplitude. So setting t=0.

(Rp)(C)(w) = 0

Or in another words w=0, and frequency also =0.

Clearly that isn't correct though. What am I missing.

---------------------------------------
At the stage
Cdv/dt = V/Rp

I've tried integrating both sides with respect to dv and dt. which produces:

(C)(Rp)V = -(1/W)(V)(Cos(wt))
When t=0.

-(C)(Rp)(W) = 1
And we can solve for W
and thus frequency.

that produces the right answer, but I can't see why the first approach produces the answer f=0. I think the problem is to do with how I've used/understood calculus in the first attempt. Can anyone shed any light?
 
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  • #2
Learnphysics said:

Homework Statement



eW4Q1hS.png


A 68 nF capacitor has a parasitic parallel resistance Rp = 1 MΩ.
If a voltage V = Vmsin(ωt) is applied, as shown in the diagram, find the frequencies at which:
1. The current amplitude through the parasitic resistance is 1% of the current amplitude
through the capacitor.
2. The amplitudes of the currents through the resistor and the capacitor are equal.

Homework Equations



I=C(dv/dt) Through capacitor

I=V/R Through Resistance.

The Attempt at a Solution



I want to try to solve the second part of the question first because it would provide be a more fundamental solution.

In any case, I've equated the current through the capacitor with the current through the parasitic resistor:

Cdv/dt = V/Rp

Simple enough.

Now V=Vmsin(wt)
so
dv/dt = (w)(Vm)(Cos)(wt)Substituting in we get:

(C)(w)(Vm)(Cos)(wt) = Vmsin(wt)/(Rp).

so:
(Rp)(C)(w)(Vm)(Cos)(wt) = Vmsin(wt) <-- just moving the resistance to the other side

At t=0. We know that sin waves have 0 amplitude, and cos waves have 1 amplitude. So setting t=0.

(Rp)(C)(w) = 0

Or in another words w=0, and frequency also =0.

Clearly that isn't correct though. What am I missing.

---------------------------------------
At the stage
Cdv/dt = V/Rp

I've tried integrating both sides with respect to dv and dt. which produces:

(C)(Rp)V = -(1/W)(V)(Cos(wt))
When t=0.

-(C)(Rp)(W) = 1
And we can solve for W
and thus frequency.

that produces the right answer, but I can't see why the first approach produces the answer f=0. I think the problem is to do with how I've used/understood calculus in the first attempt. Can anyone shed any light?

The problem is that the currents they are referring to are not instantaneous currents at a particular time t, they are average currents or RMS (root mean square) currents. The instantaneous current ratio has no particular value. You need to square the currents and then average over a cycle then take the square root. The average of cos^2(wt) is equal to the average of sin^2(wt).
 
Last edited:
  • #3
You wrote
Ic = CdV/dt
Ir = V/R
but V = V0sin(wt)
and dV/dt = wV0cos(wt)

So just use the amplitudes of Ic and Ir:
wCV0 = 0.01V0/R or wC = 0.01/R for Ic = 0.01Ir
and wCV0 = V0/R or wC = 1/R for Ic = Ir.
Solve for w in both cases.
 

FAQ: Parasitic effect of Capacitor question

1. What is a parasitic effect of a capacitor?

A parasitic effect of a capacitor refers to the unintended and undesirable behaviors that can arise in a capacitor due to its physical construction and other external factors. These effects can impact the performance and reliability of the capacitor and the circuit it is a part of.

2. What are some examples of parasitic effects in capacitors?

Some common examples of parasitic effects in capacitors include leakage current, dielectric absorption, self-resonance, and parasitic capacitance. These effects can cause issues such as loss of charge, distortion of signals, and interference with neighboring components.

3. How do parasitic effects affect the performance of a capacitor?

Parasitic effects can significantly impact the performance of a capacitor. They can lead to changes in capacitance, resistance, and inductance, which can affect the behavior of the capacitor in a circuit. This can result in errors in circuit calculations, reduced efficiency, and even malfunction of the circuit.

4. How can parasitic effects be minimized?

Parasitic effects can be minimized by carefully selecting the type of capacitor, its physical placement in the circuit, and using appropriate decoupling techniques. Using high-quality components and minimizing the length of the connecting leads can also help reduce parasitic effects.

5. Can parasitic effects be completely eliminated?

No, it is not possible to completely eliminate parasitic effects in capacitors. However, their impact can be minimized through proper design and component selection. It is important to consider parasitic effects during the design stage to ensure the overall performance and reliability of the circuit.

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