Capacitor and First-Order Systems

7/30/2019 Capacitor and First-Order Systems

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Capacitors and first order systems

(Note: first order system is a circuit containing only one independent energy storageelement)

Ideal linear capacitors:

Capacitor is an energy storing device. The capacitor stores energy as voltage dropacross its plates.

Memory device: The output of capacitor not only depends on set of inputs butalso on the previous state, that is why it is a memory device.

Capacitance of a capacitor:

The amount of charge stored on the capacitor depends on the capacitance of the

capacitor. The equation for the capacitance is given asC= *A/ d

Where C is capacitance, is permittivity of material between the surfaces(plates in case of parallel plate capacitor); A is area of the surface (plates in case

C

i

Q (charge)

+

v

-


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of parallel plate capacitor) used to store charge. Capacitance is measured inFarads .

(Note: usually permittivity of material other than air is given as relative permittivity i.e.

r ; which is the ratio of permittivity of material to the permittivity of air. So in this case permittivity of material is calculated by multiplying the r with the permittivity of the

free space)

When a potential difference v is applied across the capacitor of capacitance C ,a charge Q is stored on the plates (+q on one surface and q on the othersurface). The equation for the amount of charge stored is given as

Q = C * v

The above equation can be easily used to define Farad: which is the unit of capacitance.

1 farad = 1 coulomb/ 1 volt

There is no net charge on the capacitor as the positive charge accumulated onone plate is balanced by negative charge on the other plate.

Current and Voltage Relationship for Capacitor :

Since Q = C * v

And current = rate of change of charge w.r.t time

i.e. i = dQ/ dt

Putting the value of Q in the above equation, we get

i = d/ dt (C* v)


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Let us suppose for the simplicity that C is constant w.r.t time and voltage acrossthe capacitor

i = C * dv/ dt

dv = 1/ C * i * dt

Capacitors are linear devices as they obey both the properties of homogeneityand superposition (short cut to proof: the above two equations for i and v arelinear; as the derivative and integral are linear operators)

Power and energy for Capacitor :

Since power = voltage * current

P = v * i

P = v * C * dv/ dt (as i = c * dv/ dt)

P = d/dt (1/2 * C * v 2)

Since power is rate of change of energy w.r.t time

So Energy = E = 1/2 * C * v 2

Capacitor and a Current Source :

Consider the current source i(t) of square wave of time period T withmagnitude I and let the initial voltage of capacitor v(t 0).

I i (t) C

+

v (t)-


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Analyzing an RC circuit:

The reason to analyze this kind of a circuit is that, if we have linear elements in the

circuit with the independent current and voltage sources, we make the rest of circuit asTHEVNIN equivalent and analyze our element of interest out of it (which is capacitor in

this case). It makes Analysis easy.

Let us consider a series RC circuit with DC voltage source v I (t)=V I ,and voltage on

capacitor when time is zero is V 0 i.e. v C (t=0)=V 0 Applying node method at point 1:

(vC V I )/ R + C * dv C /dt = 0 (since i C = C * dv C / dt)

R * C * dv C / dt+ v C = V I A

Given

i (t)

I

t 0 t T

v(t)

t

v(t)

t 0 t 1 t 1+T

q=I*

v=q

v=I* v


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R*C has the units of time.

An Example of RC Circuit:

Let us assume vC (0) = V 0 (given)

v I (t) = V I

R * C * dv C / dt + v C =V I from equation A

(WE will use Method of Homogenous and Particular Solutions to solve this differential

equation. It comprises of the following steps

1. Find the particular solution2. Find the homogenous solution

3. The total solution is the sum of particular and homogenous solution. Then use theinitial conditions to solve for the remaining constant)

Now solve the equation B, by the method of homogenous and particular solution

1. Particular Solution:

V Cp = any solution that satisfies original equation A

We use trial and error method

R * C * dv Cp / dt + v Cp = V I

Let v Cp = V I (guess)

So R * C * dV I / dt + V I = V I

Since dV I / dt=0 (as V I is constant w.r.t time)

(t)


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V I = V I

So vCp = V I (so our guess is correct )

2. vCH : Solution to the Homogenous Equation by setting Drive to Zero:

R * C * dv CH / dt + v CH = 0

Let v CH =A * est (guess)

R * C * d/dt(A * e st ) + A * e st = 0

R * C * s *A * e st + A * e st =0

R * C * s + 1 = 0 characteristic equation

S = - 1/ (R * C)

vCH = A * e-1/(R * C) * t

RC is called time constant

vCH = A * e-t/

3. Total Solution:

vC = v Cp + v CH

Putting the value of vCp and vCH

vC = V I + A * e-1/(R * C)*t A

Now find the value of A using the initial condition

vC = v 0 at t=0

As t=0 so e -1/ (R * C)*t = 1

So equation A becomes

v0 = V I + A

A = v 0 - V 1

vC = V I + (v 0 - V I ) * e-1/(R * C) * t (vC is actually vC (t) )


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Since i C = C * dv C / dt = C * d/ dt(V I + (v 0 - V I ) * e-1/(R * C) * t )

iC = -1/ R * (v 0 - V I ) * e-1/(R * C) * t (iC is actually iC (t) )

Plot Solution:

vC = v C (0) * e-t/(R * C) + V I * (1-e

t/(R * C) )

dvC / dt = -(v C - V I )/ (R * C) * e-t/(R * C)

= (V I - v0)/(R * C) slope at t=0(represented by green line)

Series and Parallel CONNECTIONS:

1. Series Connection:

Consider two capacitors connected side by side with a voltage source in series.Next using KVL method

t=0 t=0

After long period of time voltage on thecapacitor becomesequal to that of source


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v(t)=v 1(t)+v 2(t)

Since same charge q (or current i) passes through the circuit

So q/ C eq = q/ C 1 + q/C 2

1/C eq = 1/ C 1 + 1/ C 2

2. Parallel Connection:

Let us consider two capacitors and a voltage source connected in parallel to eachother

Same voltage is across both the capacitor

The charge (or current) that comes out of the source is the sum of charges onboth the capacitors

q = q 1 + q 2

C eq * v = C 1 * v1+ C 2 * v2

C eq = C 1 + C 2 (since v=v 1=v 2)

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Capacitor and First-Order Systems