Module 4

Capacitance4.1 Charge

Q. What is charge?

A. A parameter we use to describe an excess or lack of electrons in matter.

One electron has a charge of 1.602 x 10-19 Coulombs.

It takes 6.242 x 1018 electrons to give a charge of 1 Coulomb.

Volts =Joules

CoulombsAmperes =

Coulombssecond

(1)

A charged body produces an ‘Electric Flux’ around it. (Like an aura around the body which attracts oppositely charged bodies.)

4.2 Electric Fields

To help us visualise the electric flux we draw electric field lines from points of positive charge to points of negative charge.

Figure 4.1

• For a given charge a body has a set number of field lines.

• A field line wants desperately to get from a positive charged body to a negatively charged body.

• Electric field lines like to keep away from each other.

The resultant is field lines will congregate on the side of a body that is closest to an opposite charge.

In figure 4.1 there are many more field lines between the two charged bodies than anywhere else.

The density of the field lines tells us the strength of the flux at that point. The density at point A is much higher than that at point B.

D =y

A(2)

where D = Flux density Y = Electric flux A = Area

The more we charge a body, (ie. add or subtract electrons from it) the greater will be the electric flux. The amount of flux depends directly on the amount of charge.

y ≡ Q (C) (3)

It is easier for us to deal with flux if we can see its effect. The effect we can observe is force. This gives rise to the electric field strength.

E =F

Q2(N/C) (4)

where E = Electric field strengthF = Force exerted on a small positive test charge (Q2)

Q2 = Positive test Charge

Q2 Q1

F

r

Figure 4.2

We can expand this to see the absolute force applied between bodies of any charge at any distance.

F =kQ1Q 2

r2 (N) (5)

where k = 9 x 109 N.m2/C2

By substituting (5) into (4) we see that the electric field strength at any point which is distance r from a charge Q is

E =

FQ

(4)

E =kQ1

r 2 (N/C) (6)

Why do we need to know about electric fields when looking at electronic components.

4.3 Capacitance

We have so far looked at small charged points in space. What if we use large flat conductors instead.

Q. How do we make the plates have a charge?

A. This is what a voltage source does. Let’s use a battery.

R

V

Figure 4.3

• When we connect the battery the negative terminal will try to push out electrons with a pressure of V volts.

• For every electron that goes out the negative terminal one must return to the positive terminal. What happens?

= electron

R

V

• One of the plates will have an excess of electrons (-ve charge) and the other will have a lack of electrons (+ve charge).

• Eventually equilibrium is reached, no more current will flow from the battery and the charge on the plates remains constant.

Q. What will be the voltage on the plates at this time?

Our plates form what is called a capacitor.

A capacitor has the unit of Farads. A capacitor that has 1 Coulomb of charge produced by a 1 volt potential has a capacitance of 1 Farad.

This is expressed mathematically as

C =QV (F) (7)

where C = capacitance in FaradsQ = charge in CoulombsV = potential in Volts

Note: 1 Farad is a very large amount of capacitance. We more normally use sub-multiples.

Unit Value PronouncedµF 10-6 Farads micro FaradnF 10-9 Farads nano FaradpF 10-12 Farads pico Farad (puff)

Table 4.1.

What is happening between the plates when we connect the battery?

Figure 4.4

• Charge between the two plates produces electric flux.

• If the plates are a uniform distance apart the density of the lines will be constant across the plates.

• At the edges of the plates we will see fringing fields. This area is very much less than the area of the plate so is usually ignored.

Now that we know the potential on our two charged plates we can redefine the electric field strength.

E =Vd

(V/m) (8)

where V = Voltage across the platesd = distance between the plates

Example 4.1A capacitor has 5 Volts across the plates. The plates are separated by 1 mm. What is the electric field strength created within this capacitor?

Solution

E =

Vd

=5

0.001= 5, 000 V/ m

So far we have assumed that there is air between the plates of the capacitor. Is this OK?

Remember air is an insulator. Insulators have the property that the electrons within the material cannot leave the atom they are in. Thus no current can flow through the material. Let us place another insulator between the plates.

Eair

Ediel

Eresult

V(volts) V(volts)

(a) (b)

Figure 4.5

When an insulator is placed between our capacitor plates and a voltage is applied the atoms will be distorted into what we call dipoles. That is each atom will have opposing poles, positive and negative.

We say the material is polarised.

The charge on the bulk of the insulating material is neutral. (Fig 4.5a)

The edges of the insulating material do carry charge. (Fig 4.5b)

The electric field in the insulating material opposes the electric field between the plates leaving us with a reduced resultant electric field.

We call the insulating material a dielectric (di - oppose).

We still have a problem. The resultant electric field has decreased due to the dielectric. Equation (8) says this is not possible. What has happened?

The dielectric allows more charge to accumulate on the plates thus keeping the electric field strength constant. More charge means more capacitance.

C ↑=Q ↑V (9)

Different dielectric materials will allow different amounts of charge to be deposited. A better dielectric material gives

• More charge

• More flux y ≡ Q

• Larger flux density D =y

A

The ratio of flux density to electric field strength in a dielectric material is called the permittivity (e) of the material.

e =

DE

(F/m) (10)

You may think of permittivity as how easily the dielectric will ‘permit’ electric field lines within it.

A vacuum has a permittivity of 8.85 x 10-12 F/m and it is given the symbol eo.

Other materials usually have their permittivity quoted as relative permittivity. ie. it is the materials permittivity relative to the permittivity of a vacumm.

e r =e

e o(no units) (11)

The true permittivity for a material is therefore

e = ereo (12)

Dielectric. Rel Perm (Av)Vacuum 1.0Air 1.0006Teflon 2.0Paper 2.5Rubber 3.0Transformer oil 4.0Mica 5.0Porcelain 6.0Glass 7.5Distilled water 80.0Ceramic 7500.0

Table 4.2

We now know the effect of dielectric material on the final capacitance.

What else do we need to know to design a capacitor.

e =

DE

=Y / AV/d

=Q/ AV/d

=QdVA

(13)But

C =QV

So

C = eAd (F) (14)

We can calculate the capacitance of a capacitor from the area of the plates (A), the separation of the plates (d) and the permittivity of the dielectric (e).

Example 4.2A capacitor has a dielectric of mica, a plate area of 100 mm2 and a plate separation of 0.001 mm. What is its capacitance?

Solution

C = eAd

= 5.0 x 8.85 x 10 -12 x0.0001 m 2

0.000001 m= 4.42 x10-9 F = 4.42nF

How do we make better capacitors? What is better?

• Large values• Small size• High dielectric strength• Low leakage current

Large values and small size go hand in hand. We need

• Large plate area (rolled foils)• High permittivity dielectrics (ceramics etc)• Small plate separations (thin dielectric layers)

4.4 Dielectric strength

If a sufficiently large potential is applied across a dielectric the bonds within the material may be broken and a current will flow. This is observed with lightning.

Different materials have different dielectric strengths.

Dielectric Dielectric strength Volts/mm

Rel. Perm.

Air 3,000 1.006Ceramic 3,000 7500Porcelain 8,000 6.0Transformer oil 16,000 4.0Rubber 28,000 3.0Paper 50,000 2.5Teflon 60,000 2.0Glass 120,000 7.5Mica 200,000 5.0

Table 4.3

To make useful capacitors we need a dielectric material that has a high permittivity and high dielectric strength. It also must be able to be formed into the appropriate shapes.

When trying to reduce the size of capacitors we often try to use thin dielectric layers to increase the capacitance. This of course increases the chance of the dielectric breaking down when a large voltage is applied.

Example 4.3

What voltage could the capacitor from the previous example be expected to handle?

Solution

d=0.001 mmMica has a dielectric strength of 200,000 volts/mm

Maximum voltage = 200,000 x 0.001 = 200 volts.

4.5 Leakage current

C

R

Figure 4.6

The dielectrics used in capacitors are very good insulators. They are not however perfect.

Impurities in the material will cause a finite leakage current to flow through the capacitor. This is modelled by a parallel resistance across the capacitor.

This resistance may be 10’s to 100’s of MW for a good capacitor.

4.6 Symbols

Normal Polarized Polarized Variable Trimmer

Figure 4.7

Some capacitors are polarised. This means they must have DC voltages applied in one direction only.

SELF LEARNING: Read about types of capacitors (Boylestad 10.6)

4.7 Series and Parallel Capacitors

C1 C2 C3 CNVb

Figure 4.8

When capacitors are placed in parallel they all have the same voltage across them.

Vb = V1 = V2 = V3... = Vn (15)

Each capacitor will store charge according to its capacitance and the voltage.

C =QV

\Q = CV

The total charge will be the sum of the charge on all the capacitors.

Q T = Q1 + Q2 + Q3 ... + Q N (16)

CTVb = C1V1 + C2V2 + C3V3... + C nVn (17)

V1 = V2 = VN = Vb

Therefore the total capacitance will be the sum of all the capacitors.

C T = C1 + C2 + C 3... + CN (18)

C1 C2 C3 CNVb

Figure 4.9

When capacitors are placed in series the charge on each capacitor will be the same. Remember the same current must flow through all capacitors when they are charged. Therefore

Q T = Q1 = Q2 = Q3 ... = Q N (19)

The voltage on each will be different if they have different values of capacitance.

C =QV

\V =QC

But the total voltage must equal the sum around the loop. (Kirchoff)

Vb = V1 + V2 + V3... + Vn (20)

\QTC T

=Q1C1

+Q2C2

+Q3C3

... +QNC N (21)

All the Q’s are the same so we may divide both sides by Q

C T =1

1C1

+1

C2+

1C3

...+ 1C N

(22)

Note: This is a similar equation for resistors in parallel.

4.8 Step Response

R

CVb ncic

Figure 4.10

Q. What happens to the voltage on a capacitor when we apply a DC voltage?

A. A capacitor is like a bucket into which we can pour electrons. The more we pour in, the higher the level rises. The level of charge determines the voltage across the capacitor.

R

CVb ncic

• When the switch connects to the voltage source, current will flow through the resistor into the capacitor.

• Initially the voltage across the capacitor is zero so the current will be Vb/R.

• As charge is stored in the capacitor the voltage across it will rise.

• When the voltage across it rises the current through the resistor will decrease

• i c =Vb - Vc

R(23)

• This continues until Vc = Vb at which point the current through R equals zero. The capacitor is fully charged.

This function is an exponential function and the charge current through the R and C can be written as

iC(charg e) =Vb

Re -t / RC Charge current (A) (24)

1.0

0.8

0.6

0.4

0.2

0.0

i c

543210

time constants

Charge

Figure 4.11

The value of R and C determines how quickly the capacitor will charge. We define

t = RC =Vi

QV

=Q

Q / t= t Time constant (s) (25)

1.0

0.8

0.6

0.4

0.2

0.0

n c

543210

time constants

Charge

Figure 4.12

As discussed the voltage on the capacitor increases with time also in an exponential fashion as

VC(ch arg e) = Vb (1- e- t / t ) (26)

Easy to think of capacitor charging in terms of the water analogy

Example 4.4An RC circuit has an R of 100 W and a C of 0.1 mF and a supply voltage of 5 volts. How far does the capacitor voltage rise in 5 ms?

Solutiont = 100 x 0.1 x 10-6 = 10 ms

VC(ch arg e) = Vb (1- e- t / t ) = 5(1- e-5x10-6 /10x10-6) = 1.96volts

Note: It is handy to remember that in one time constant a capacitor will charge to 0.632 of the supply voltage.

R

C ncic

Vb

Figure 4.13

When the switch is changed to the other position the capacitor will be discharged. The rate of discharge will be controlled by the size of the capacitor and resistor.

†

iC (disch arge) = -Vb

Re-t /t (27)

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

i c

543210

time constants

Discharge

Figure 4.14

VC(disch arg e) = Vbe- t / t (28)

1.0

0.8

0.6

0.4

0.2

0.0

n c

543210

time constants

Discharge

Figure 4.14

Can look at discharge as water as well

What if we reduce the pipe size

4.9 Capacitors and AC Signals

SELF LEARNING: Revise complex numbers and phasors, Hambley 5.1 to 5.4, Boylestad 14.6 to 14.12We will be dealing with sinusoidal signals of which we need to know the frequency and amplitude. Mostly deal with frequency in radians (w)

w = 2pf radians/s (29)

Firstly what happens when we apply a sinusoidal voltage to a resistor?

RVmsinwt

i

Figure 4.15

Ohms law is obeyed and a current is produced proportional to the voltage.

-2

-1

0

1

2

350300250200150100500degrees

Resistor

Voltage

Current

Figure 4.16

The current produced is ‘in phase’ with the voltage from the supply.

The capacitor is a bucket into which we pour current. It follows that we may calculate the voltage on the capacitor by summing the amount of current we have put in. This is an integral

u c =1C

i cdtÚ (30)

Take the derivative of both sides and multiply by C and we have an expression for the current.

i c = Cducdt (31)

i c =dQdt

=dCnc

dt= C dnc

dt

Let us apply a sinusoidal voltage to an RC circuit. The current through the resistor and capacitor must be the same, they are in series!

R

CVmsinwt

i

Figure 4.17

The voltage across the resistor will be in phase with the current through it as before. Thus a sinusoidal current

The voltage across the capacitor will be

u c =1C

icdtÚ =1C

sinwt dtÚ = -1

wCcoswt (32)

The voltage across the capacitor is a - cos function whilst the current through it is a sin function.

-2

-1

0

1

2

350300250200150100500degrees

Capacitor

current

voltage

Figure 4.18

There is a 90° phase shift between the current and voltage.

The current leads the voltage by 90°or

The voltage lags the current by 90°

I can’t understand why this happens!!!!!!!A capacitor is filled as current pours into it. The voltage across it keeps increasing as more current pours in. With a sinusoidal voltage supply the current is coming in until the sine crosses the zero axis. At this point the capacitor voltage will be a maximum. Similarly the voltage on the capacitor will now discharge as the current goes negative. It will completely discharge then charge in the reverse direction. The reverse current will finish when the sine crosses zero and this will coincide with the maximum negative charge on the capacitor. Thus a 90° phase shift.

Q. Is there a law like Ohm’s law that applies to capacitors for AC signals?A. Yes but a capacitor does not have resistance.

u c =1C

icdtÚ =1C

sin wt dtÚ = -1

wCcoswt (32)

Look back to equation (32) see that a current of sin wt produced a voltage of 1

wCcos wt .

The 1wC term is the effective AC resistance of the capacitor. We call it the

capacitive reactance. X c =1

wC (33)

This describes the magnitude of opposition to current but not the phase shift. To add this information we make use of the j operator.

j = -1 (34)

- jXc =1

jwC (35)

We can now work out the ‘AC resistance’ of a circuit containing resistors and capacitors. This is called the impedance of the circuit, symbol Z. It is the sum of all the resistances and reactances in a circuit.

Example 4.5

200W

10mF1 volt100 Hz Z

Calculate the input impedance for the series circuit above.

Solution Z = Z R + Z C

ZR = R = 200W ZC = - jXC

X c =1

wC=

12 p100Hz 10x10 -6 F

=1

6.28 x10-3

= 159.1W

Z = R +(-jXc ) = 200 - j 159.1W

= 255.5 W – - 38.5°

Example 4.6

200W1 volt100 Hz Z 10mF

Calculate the input impedance for the parallel circuit above.

Solution 1Z

=1

Z R+

1Z C

R = 200W

X c =1

wC = 159.1W

Z = R //(- jXc ) =1

1R

+1

- jXc

=1

1200

+1

- j159.1

=1

0.005 + j 6.3x10 -3 =1

8.03 x10-3 –51.5°

= 77.5 - j97.4W = 124.5– - 51.5°

All of this information can be represented with phasor diagrams. Lets revisit Example 4.5.

R = 200W

Xc = 159.1W

f = -38.5˚

|Z|=255.5

Figure 4.19

Q. What about voltage and current?

A. Remember the two components are in series so both have the same current. Use this as a reference.

i =VZ

=1

255.5= 3.91mA

Vsupply = 1 V

VR = 783 mV

VC = 622.7mV

f = -38.5˚

i = 3.9 mA

Figure 4.20

What about parallel components? Look at Example 4.6.

The voltage across both components is the same. Use this as a reference.

Vsupply = 1 V iR = 5.0 mA

iC = 6.28 mAf = 51.5˚

i = 8.03 mA

Figure 4.21Can't remember the phase angles

i t =Vs

Z t=

1V–0°124.5W– - 51.5°

= 8.03mA–51.5°

4.10 Filters using capacitors

R

CInput Output

Low Pass

Figure 4.22

Because of the change in reactance of a capacitor with frequency we may use them to make filters.

The circuits in figure 4.22 and figure 4.25 are simple voltage dividers except we are using a resistor and a capacitor instead of two resistors.

The output will be a fraction of the input (as for any voltage divider) and is calculated in the normal manner.

Low Pass

R

CInput Output

Low Pass

Vout = Vin- jXc

R - jXc (36)

VoutVin

=1

R- jXc

+ 1 (37)

VoutVin

=1

jwRC + 1 (38)

This is the transfer function or gain of the filter.

Q. What does this equation mean?

VoutVin

=1

jwRC + 1A. When the frequency of the input signal (w) gets high the gain gets small.When the frequency of the input signal gets small the gain gets large.

This is a low pass filter. It only passes low frequencies.

1

0.707

wcPass band Stop band w

gain

Figure 4.23How can the circuit do this?

Remember the capacitor has a reactance X c =1

wC , as w (2pf) becomes larger the reactance becomes smaller. Thus the output from the voltage divider becomes less.

Q. At what point do we say the filter has cutoff the signal?

A. Substitute w =1

RC into equation (38).

VoutVin

=1

j1 + 1 (39)

1

j1

Mag = 1.414

f = 45˚

Figure 4.24

VoutVin

=1 –0°

1.414 –45°

VoutVin

= 0.707 – - 45°

At this point the output signal is 0.707 of the input and it is -45° out of phase.

We call this the half power point, cutoff frequency or breakpoint of the filter.

Note: The power of a signal is P =V 2

Z.

If the output voltage is reduced to 0.707 then the output power must be reduced to 0.5 (0.707 2).

Thus the cutoff frequency occurs when

wc =1

RC (40)

Example 4.7Plot the gain and phase of the output signal for the low pass filter circuit when R = 100W and C = 1.6 mF. fc =

12pRC

Solution

At 1 HzVoutVin

=1

j 2p1 Hz100W 1.6x10 -6 F + 1

VoutVin

=1

j1x10-3 + 1ª 1 –0°

Gain = 1 Phase = 0°

At 1000 HzVoutVin

=1

j2p 1000Hz 100W1.6x10-6 F + 1

VoutVin

=1

j1+ 1 = 0.707 – - 45°

Gain = 0.707 Phase = -45°

At 1 MHzVoutVin

=1

j 2p1x106 Hz 100W1.6x10 -6 F + 1

VoutVin

=1

j1000 + 1 ª 0.001 – - 90°

Gain = 0.001 Phase = -90°

10.707

1kHz1Hz 1 MHz

1kHz1Hz 1 MHz

Phase

Gain

0̊

-45˚

-90˚

Freq

Freq

These plots are called Bode plots.

High Pass

R

CInput Output

High Pass

Figure 4.25

Calculate the gain for this filter.

Vout = VinR

R - jXc(41)

VoutVin

=1

1 -jXcR

=1

1-j

wRC(42)

Q. What does this equation mean?

A. When the frequency of the input signal (w) gets high the gain gets large.

When the frequency of the input signal gets small the gain gets small.

This is a high pass filter. It only passes high frequencies.

1

0.707

wc Pass bandStop band w

gain

Figure 4.26

The function of this filter is similar to the lowpass except the arms of the divider are reversed.

The breakpoint (cutoff frequency) is calculated exactly the same as for a low pass filter.

Substitute w =1

RC into equation (42).

VoutVin

=1

1 - j1ª 0.707 –45°

This is the half power point as demonstrated for the low pass.

Repeating Example 4.7 but with the R and C reversed in the high pass configuration we find.

At 1 HzGain = 0.001 Phase = 90°

At 1000 HzGain = 0.707 Phase = 45°

At 1 MHzGain = 1.0 Phase = 0°

10.707

1kHz1Hz 1 MHz

1kHz1Hz 1 MHz

Phase

Gain

0̊

45˚

90˚

Freq

Freq

Figure 4.27