Series connections - Groupings in parallel     
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CONDENSERS    “2nd Part”

CHARGE AND DISCHARGES FROM A CONDENSER

Let us examine the circuit of figure 3 in which the condenser is represented by its graphic symbol.

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As soon as the condenser (C) is connected to the pile, it occurs the phenomenon already previously analyzed, namely that a certain number of electric charges pass from a reinforcement to the other. 

This displacement constitutes an electrical current which, on figure 3, is directed according to the conventional direction. This current is called current of load of the condenser.

The charging current persists until the quantity of electricity arrived on the reinforcements of the condenser generates, between those, a potential difference equal to the tension of the pile. The condenser is then known as charged.

Once the charged condenser, it circulates no current in the circuit, since the tension created at the boundaries of (C) equal but is opposed to the tension of the pile.

The discharge of the condenser can easily be observed. It is enough to withdraw the condenser and to connect it for example, at the boundaries of a resistance, like illustrated.

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The tension present at the terminals of the condenser makes circulate a current in the resistance R which, according to conventional direction's, is directed positive reinforcement towards the negative reinforcement. 

This current due to the electric charges accumulated on the reinforcements of the condenser lasts only one short moment. It ceases when the loads present in excess on a reinforcement joined the reinforcement on which they are missing. This operation carried out, the condenser known as is discharged and the current created by this discharge is called current of discharge of the condenser.

If the once withdrawn condenser of its load circuit (figure 3) is not connected to a resistance, it preserves on its reinforcements the accumulated loads.

The condenser would remain indefinitely charged and the dielectric one being between its reinforcements was insulating perfect. In practice, that never arrives and the dielectric one lets gradually pass the electric charges of a reinforcement to the other, which discharges the condenser slowly.

The fact most important to retain of what we have just seen is :

that a condenser, after having taken care, prevents any later circulation of the current provided by a pile.

The CONDENSER AND ELECTRIC POWER

We have just seen how we can charge a condenser by means of a pile and to then discharge it in a resistance. It is easy to understand that at the time of these two operations, of the electric power is brought into play. 

For that, it is enough to observe the heating effect generated in resistance by the current of discharge of the condenser. This heating effect is done inevitably at the price of an electric consumption of power.

This power consumption was obviously provided by the condenser which, itself, had received it pile. 

If the condenser is able to yield this energy to resistance, it is that it dissipated it but not stored before.

The condenser with the property to store the electric power, it is thus a preserving element of energy to the difference of the resistance which is a wasteful element. 

Let us see now how it is possible to quantify the electric power stored by a condenser and how this one was stored.

ENERGY STORED BY A CONDENSER

To charge a condenser, the pile must move a quantity of electricity (Q) from one reinforcement to another of this component. This quantity of electricity is determined by the product of the tension (V) of the battery by the capacity (C) of the condenser. To produce this phenomenon, a certain energy (W) is provided by the pile and this energy is equal to the product of the quantity of electricity (Q) by the tension (V).

However, this energy (W) is not stored completely by the condenser, actually, the condenser stores only half of the energy (W) provided by the pile. Second half is dissipated in heat in internal resistance of the pile and possibly in other resistances of the circuit.

To be convinced some, let us examine figure 5 where resistance (R) represents the total resistance of the circuit, i.e. the internal resistance of the pile plus that of the electric connections and the reinforcements of the condenser.

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At the time when the condenser is connected to the circuit, tension Vc on its terminals is null, therefore all the tension V of the pile is found at the boundaries of resistance R (Vr = V). Then progressively with the load of the condenser, tension Vc increases, while Vr decreases. When the load is finished, all the tension of the pile is found at the boundaries of C, while Vr is null.

Tensions Vc and Vr have a similar but opposite pace thus since one is increasing and the other decreasing one. Since the electrical current (I) crosses at the same time R and C, the quantity of electricity (Q) provided by the pile to the circuit is divided well into two equal parts between R and C. energy Wc stored by the condenser is thus equal to the Wr energy dissipated in resistance.

Total energy W provided by the pile is equal to Wr + Wc but like Wr = Wc, we also have Wr = Wc / 2.

Energy provides by a pile to charge a condenser is given by the formula W = Q X V ; this enables us to quantify the energy really stored by the condenser :

Wc = Q x V / 2

Or knowing that the quantity of electricity accumulated by a condenser is given by the product Q = C x V, we can replace Q in the preceding formula by its value and we obtain :

Wc = C x V² / 2

All energy Wc stored by the condenser is then completely restored by this one during its discharge.

BRACKET : It is interesting to see according to which laws vary the terminal voltage of the condenser and the current which circulates in the circuit during the load of the condenser. These paces are respectively deferred figure 6-b and 6-c while the figure 6-a gives the electric circuit taken as example :

X8.gif 

At the initial moment t0, where the electric connection is established, it circulates of the pile to the condenser a current I = V / R equal to that which would circulate permanently if we not had a condenser but a simple wire not having any resistance (fig. 6-c). Just after t0, the condenser begins its load and tension Vc on its terminals grows (fig. 6-b). Consequently, the current (I) starts to decrease until being cancelled when (C) is charged: the terminal voltage of (C) is maximum.

The variations of the current and the tension take a form known as exponential, the equations of such curves are as follows :

I = V / R . (e - t / RC)

Vc = V (1 - e - t / RC)

equations in which (e = 2,72 approximate, represent the base of the logarithms natural or Napierian and the value given by product RC (Resistance in ohm and capacity in farad) constitutes the time-constant of the circuit measured in seconds. From these laws rises that in any interval of time equal to RC (of 0 to RC, RC with 2 RC, etc…), the value of the current (of load or discharge) always decreases in the same ratio by 2,72.

Example : Let us take the interval of 0 at RC, therefore (t = RC)

        I = V / R . e t / RC

however     t = RC              I = V / R . e - RC / RC = V / R . e- 1

 e-1 = 1 / e                   I = V / R . 1 / e = V / R / e = V / R / 2,72

The current (I) decreased well by 2,72 times since to t = 0, its value were V / R and that at time t = RC, its value is V / R / 2,72.

It is followed from there that theoretically the current is never cancelled and that the time of load or discharge of the condenser is infinitely large. However, in practice, we note that after a time equal to 5 times the constant RC, the current are worth 0,7 % of its initial value and we can consider that the load (or discharges it) of the condenser is finished.

THE ELECTRIC FIELD

We now will analyze how the condenser stores energy.

Let us suppose that we charge a condenser with air and imagine that one of the positive loads present while exceeding on the positive reinforcement is detached from this one and is in the dielectric one (figure 7-a).

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This load is pushed back by the positive reinforcement whereas on the contrary, it is attracted by the negative reinforcement. On this load thus a force acts having the direction given by the arrow on the figure 7-a. This same force would act on any other positive load being detached from the positive reinforcement.

If we trace the courses followed by a certain number of loads, we obtain the various trajectories represented in arrow features (figure 7-b).

These lines are called fields tension because the force which determines the displacement of the positive loads acts along this one. The whole of the tension fields delimits the zone of the space in which an electric charge is subjected to a force. The zone thus determined represents an electric field of force or more simply an electric field.

Any positive load which is in the field is subjected to the effect of a force and tends to move. This force achieves a work which is given by the product of the intensity of the force by the length of the displacement of the load. Any work is obtained at the price of a consumption of energy. In the case of the condenser, the power consumption to produce work is the electric power stored by the condenser.

We thus understand that the energy stored by the condenser decreases with each time a load is detached from the positive reinforcement and passes on that negative. With each transfer, small of energy is transformed into work achieved by the force which generates the displacement of the load. If all the loads are detached from their reinforcement, the totality of the energy stored by the condenser is transformed into work and this one discharges completely.

Actually, no load can be detached from the positive reinforcement since the dielectric one is insulating almost perfect. On the other hand, these loads can move at the same time as the reinforcement.

Let us analyze the consequences of a bringing together of the two reinforcements, consequences which we know but to which no precise answer was brought.

We suppose that the condenser of the figure 8-a has a capacity of 3 µF and that it is charged by a pile of 4 volts. The quantity of electricity (Q) present on its reinforcements is given by the formula Q = C x V is 3 µF x 4 V = 12 µC.

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So now the condenser is disconnected pile, it preserves obviously the quantity of electricity of 12 µC and a tension between its reinforcements of 4 V.

Let us imagine that the positive reinforcement approaches the negative reinforcement and that the distance between them is reduced by half (figure 8-b).

If the positive reinforcement comes in contact with the negative reinforcement, in other words if it moves of a distance d, all the energy stored by the condenser is transformed into work. We thus understand that if the positive reinforcement moves d / 2, the stored energy is reduced by half. However, as the two reinforcements are insulated, the quantity of electricity cannot decrease; consequently, it is the tension between reinforcements which is reduced by half and takes for value 2 volts (figure 8-b). The value of the condenser increases and becomes:

C = Q / V = 12 µC / 2 = 6 µF

We thus have just brought a concrete explanation to the fact that the capacitance of a capacitor increases when the distance between its reinforcements decreases and that, in particular, it doubles when the distance is reduced by half.

At the beginning of this lesson, we had brought closer the reinforcements while leaving the condenser connected with the pile (figure 1) and we had seen that the pile provided an additional charging current. We can say now that this current is used to preserve the terminal voltage of the condenser equal to the power provided by the pile.

Let us introduce now between the reinforcements of the condenser in charge of the figure 8-a, but not connected to the pile, a dielectric solid (figure 8-c). If this dielectric A a relative permittivity er of 2, the capacity of the condenser is doubled. The introduction of this dielectric place the condenser under the same conditions as into the figure 8-b, after a bringing together of its reinforcements.

In this case also, half of the stored energy is transformed into work, but since there is not displacement of reinforcements, this work is inevitably produced differently. To explain that, it is necessary to remember the principle of polarization of dielectric exposed appears 1-a and 1-b (polarization of dielectric). The offsetting of the electronic orbits is determined by the intensity of the field which tends to attract the electrons towards the positive reinforcement of the condenser. The work achieved by this force on each electron is very weak, however the number of the electrons of dielectric being considerable, the sum of the various forces leads to the consumption of half of the energy stored by the condenser.

DIELECTRIC RIGIDITY

We know now that the electric field is characterized by tension fields of which we know already the direction and the direction of their action (figure 7-b), but to be complete on the electric field, it is also necessary to know its intensity.

The intensity of the electric field acting in the dielectric one of a condenser is obtained by dividing the tension existing between its reinforcements by the distance which separates them.

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Note : Not to confuse this symbol with that of the electromotive force of a pile (f.e.m.) from where the presence of the arrow on the symbol indicating that it is about a vector.

The unit of the intensity of the electric field is the volt per meter (symbol V / m).

It is easy to understand that the intensity of the electric field increases when the tension between the reinforcements increases or when the distance which separates them decreases. This fact involves a practical consequence of first importance. However, the intensity of the electric field cannot increase indefinitely and, arrived at a certain threshold, the value of the intensity becomes such as the electric charges can cross the dielectric one of a reinforcement to the other.

This passage of current appears in the form of a violent electric discharge, a kind of flash which perforates the dielectric one and establishes an irreversible contact between the reinforcements of the condenser. The condenser is then known as in short-circuit and becomes unusable. We can regard this discharge of the condenser or breakdown as an instantaneous transformation into heat of all the energy stored by this one.

The value of the intensity of the field for which there is breakdown is the dielectric rigidity of material constituting insulator. This value is different for each type of material.

Each dielectric material is thus characterized not only by its relative permittivity but also by its dielectric rigidity.

The table of figure 9 gives the dielectric rigidity of materials already enumerated in figure 1.1 for their relative permittivity.

Material

Dielectric rigidity in kV / cm

Dry air

21

Special paper for condenser (KRAFT)

200 to 400

Mica

600 to 1800

Magnesia titanate

50 to 100

Glow, Rutile-zirconias, Titanate of calcium

40 to 80

Barium titanates and Zirconates

40 to 60

Polystyrene (Styroflex)

400

Polytetrafluoretylene (PTFE, Teflon)

400 to 800

Polymonochlorotrifluoretylene (PCFTE)

1000 to 2000

Ethylene polyterephthalate (Polyester, Mylar)

1000 to 2000

Electrolytic with aluminum

approximately 10 000

Electrolytic with tantalum

approximately 10 000

    9. - Dielectric Rigidity of various materials

For dielectric materials usually used, the value of dielectric rigidity is extremely high and for this reason, we not use as unit V / cm but kV / cm as in figure 9.

For example, for a condenser whose two reinforcements are distant of 1cm and having polystyrene like dielectric, breakdown occurs for a tension of about 400 kV.

If the distance between the reinforcements is only 1 mm, same breakdown occurs to 40 kV. There are condensers where the thickness of dielectric is only of some thousandth of millimetre and we thus include/understand that their breakdown occurs even for low tensions, tensions which we meet in the electric or electronic circuits where the condensers are used. For this reason, each condenser carries an indication of tension called tension of service : value which one should not exceed under penalty of damaging the component following a breakdown.

You recall, on this subject : that a condenser is characterized not only by its capacity but also by its tension of service.

Even the air can lose its dielectric properties following a breakdown, thus you note that the air has a dielectric rigidity which is 21 kV / cm for the dry air.

The flashes which we observe at the time of the storms are the manifestation of the breakdown of the air. Indeed, of the electric charges accumulate in the clouds which behave then as the plates of a capacitor.

It is thus established an electric field between two clouds which are with different potentials or between a cloud and the ground. When the intensity of the electric field exceeds the rigidity of the air, which in addition strongly decreases when the air is wet, there is an electric discharge between the clouds or the ground and the cloud.

The second case is very dangerous and to prevent that the electric discharge does not produce a damage on the dwellings endangering the life of its occupants, the buildings are protected by a lightning conductor.

The lightning conductor being the highest place of the building, it is exposed thus to the electric discharge. The lightning conductor, connected to the ground, transmits the electric discharge to him. To facilitate the contact with the ground, it is necessary to note the presence of an iron plate buried in the ground (obligatory).

HIGH OF PAGE SERIES CONNECTIONS - GROUPINGS IN PARALLEL

We do not have, for the moment, considered circuits having one condenser, but these components as resistances can form various groupings.

GROUPINGS IN PARALLEL

Figure 10 represents a parallel grouping of two condensers called for the circumstance C1 and C2.

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C1 and C2 have each one a reinforcement connected to the positive pole of the pile and the other braces connected to the negative pole of the same pile, so that at the boundaries of each condenser, there is the same tension.

This last characteristic is common to any parallel grouping as that was already known as during the analysis of the groupings of resistances.

Since between the reinforcements of C1 and C2 we apply the same tension, each condenser takes care with a quantity of all the more large electricity as its capacity is high.

Let us see how we can determine the equivalent capacity (Ceq) presented by such a circuit, and this knowing the value of C1 and C2.

To this end, let us imagine to bring closer the two condensers until putting in contact their reinforcement connected to the same pole, like illustrated appears 11-a. That is possible insofar as the reinforcements are connected to the same electric potential.

H13.gif

The two condensers thus joined together constitute a single condenser called Ceq in the figure 11-b. This condenser has the same dielectric one, the same distance between reinforcements as C1 and C2. The only difference lies in the increase in the surface of the reinforcements.

Taking into account the formula giving the capacitance of a capacitor :

C = e0 x er x (S / d)

We know that if surface S increases, the capacity C of the condenser also increases and this in the same proportions.

In the case of the figure 11-a, surface S of Ceq is equal to the sum of surfaces of C1 and C2. We thus deduce that the capacity of the condenser are equivalent Ceq of the figure 11-b is equal to the sum of the capacities of C1 and C2.

The capacity equivalent to two or several condensers connected in parallel is equal to the sum of the capacities of each condenser :

Ceq = C1 + C2 + C3 +….

SERIES CONNECTIONS

Let us consider the C1 condensers and C2 of figure 12. To facilitate our explanations, the reinforcements of these condensers are called A, B, C and D.

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At the time of the load of C1 and C2, reinforcement A takes care positively and D braces it negatively.

The reinforcements B and C not connected to the pile constitute with the driver which connects them, a simple metal body. This body takes care by induction with the signs represented figure 12. On the reinforcement B appears a quantity of equal electricity but of sign opposed to that presents on A, while on C appears a quantity of equal electricity but of sign opposed to that presents on D.

If we call (+ Q) the quantity of electricity present on A, we have (- Q) on B, and Si we call (- Q) the quantity of electricity present on D, we have + Q on C.

The condensers C1 and C2 thus store the same quantity of electricity Q.

As in any series connection, the tension V provided by the pile is divided into two tensions V1 and V2 respectively at the boundaries of the condensers C1 and C2. In figure 13 is deferred the same circuit with the various tensions present.

H15.gif

The V1 tension at the boundaries of C1 is equal to :

V1 = Q / C1

The V2 tension at the boundaries of C2 is equal to :

V2 = Q / C2

As it is known as previously V = V1 + V2, therefore :

H16.gif 

We can replace in figure the 13 condensers C1 and C2 by an equivalent condenser which we call Ceq (figure 14).

H17.gif

(Ceq) equivalent with C1 and C2, stores the same quantity of electricity Q as C1 and C2.

We can write the relation (2) :

V = Q x 1 / Ceq --------------> (2)

ndeed  : Ceq = Q / V              V = Q / Ceq = Q x 1 / Ceq

The relations (1) and (2) are equal since they give both the value of the tension V.

(1) = (2) ---------> Q x (1 / C1 + 1 / C2) = Q x 1 / Ceq

While simplifying (1) and (2) by Q, we obtain the value of Ceq :

H18.gif

We thus determined the value of Ceq according to C1 and C2. Extended to the general case several series condensers, this formula becomes:

H19

Thus, to calculate the equivalent capacity (Ceq) of two or several series condensers, the three operations following are to be carried out :

When two condensers only are in series, we adopt the following formula which derives from the general formula:

Ceq = (C1 x C2) / (C1 + C2)

For a quantified practical example, let us apply what we have just seen. 

That is to say to calculate the capacity equivalent to the circuit represented figure 15.

To calculate Ceq, let us carry out the three necessary operations :

1 / C1 = 1 / 5 = 0,2

1 / C2 = 1 / 10 = 0,1

1 / C3 = 1 / 2 = 0,5

1 / C1 + 1 / C2 + 1 / C3 = 0,2 + 0,1 + 0,5 = 0,8 = -------> 1 / Ceq

Ceq = 1 / 0,8 = 1,25 nF

The condensers C1, C2 and C3 in series are thus equivalent to a single capacity of 1,25 nF.

To carry out a comparison between the two types of associations, it should be noted that in the case of condensing in parallel, the value of the equivalent condenser is always higher than the value of each condenser while in the case of an association in series, the value of the equivalent condenser is in all the cases, lower than the value of each condenser and even better, it is lower than smallest capacities.

The formulas presented are also used for more complex calculations of circuits born from the combination of the two types of associations.

Let us see for example, how to calculate the total capacity of the circuit represented figure 16.

H21.gif

In this circuit, we see that C1, C2 and C3 constitute a parallel grouping connected in series with C4.

Let us calculate initially, the condenser equivalent to C1, C2 and C3 in parallel, condenser which we call C123.

C123 = C1 + C2 + C3 = 1 µF + 5 µF + 2 µF = 8 µF

Let us replace in figure 16, C1 C2 and C3 by their condenser is equivalent C123, we obtain the figure 17-a).

H22.gif

With the figure 17-a, we are in the presence of two condensers (C123 and C4) connected in series. The calculation of the condenser equivalent (Ceq) to this assembly gives the condenser equivalent to the circuit of figure 16:

Ceq = C123 x C4 / C123 + C4 = 8 x 12 / 8 + 12 = 96 / 20 = 4,8 µF

In the presence of complex circuits like that of figure 16, it is always necessary to gradually simplify the circuit until not obtaining more that only one condenser whose capacity represents the total capacity of the outgoing circuit.

Thus the analysis of the groupings of condensers finishes.

As we said, it is important to know that a once charged condenser prevents any circulation of current provided by a pile.

In the next lessons, we will see that there are other generators providing of the currents different from that provided by a pile. With respect to these currents, the condensers react differently. This property is used when we wish to separate, in the same circuit, two types of different currents.

This property will be analyzed in detail at the time of the next following lessons.

We finish this lesson with a summary table of the electric quantities relating to the condenser like their unit and their formula so necessary (figure 18).

Electric quantities

Measuring unit

Denomination

Symbol

Denomination

Symbol

FORMULAS

Absolute permittivity

e

Farad per meter

F / m

 

Capacity

C

Farad

F

C = er x e0 x (S / d)

Quantity of stored electricity

Q

Coulomb

C

Q = C x V

Stored energy

W

Joule

J

W = C x V² / 2

18. - Electric quantities relating to the condenser.

In the next lesson, we will examine the third fundamental component of the electronic circuits : inductance, as all the phenomena which it generates when it is introduced into a circuit, for example electromagnetism.

 

     

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