Condensers

A condenser is a passive component whose purpose is to accumulate a certain electrical charge. It is constituted by two separated plates of a dielectric (an insulating material, that can be ceramic, paper, plastic, etc).

- The unit of measurement is the farad, but as a condenser of a farad is very great and nothing usual in electronics, we used sub-multiple, as:
- Milifaradio, that is thousand times but small.
- The microfarad, a million of times but small one.
- Nanofaradio, a smaller trillion, that is 10
^{-9}Farads.

Picofaradio, or 10^{-12} Farads.

We will see a good video on the principles of the condensing work of Eugene Khutoryansky and adapted by our student Pablo Moreno Gómez

- We have several methods to read the capacity, next to the tolerance and the maximum tension of work, according to the type of condenser.
- Condensers with band of colors. In this case, the same criterion is applied that stops the resistance (table 2.7). For the tensions, there is a criterion also based on the colors (nonexact) so that, 5º color indicates in hundreds the maximum tension of work, so that if 5º color is black, we have 100 volts or yellow 400 volts.
- Marked alphanumeric. There is a letter that indicates the tolerance to us (M of the 20%, K of the 10% and j of 5%). Tolerance + takes to the rule value + maximum Tension. It is necessary to consider that the value that occurs is always in picoF, except it takes to comma decimal and it does not take unit, that occurs in microfarads, or, if it takes the unit, that in that case, the letter of the unit also does of comma.

In the case of the left figure, we have a condenser with letter j (5%) of tension 2A (by table talks about to a tension of 100 volts and value 10 but 4 zero, that is 100,000 pF, or 100 nF or 0,1 µF.

Activities number 1. To realise the exercises of page 38

Activity 2. It realises the exercises of page 40 As we saw in the subject of the resistance, when they were put in series added one after another one because major was offered difficulty to that they pass electrons.

In the case of the condensers, it happens something similar but to the inverse one, that is:

If **we associated the condensers in parallel**, which we do is to increase the “effective surface to keep the electrical charge” and therefore, when we have association in parallel, the total capacity is the sum of each capacity.

If embargo, to associate the **condensers in series**, we obtain the formula of the resistance of the parallel, but changing R by C, therefore:

Activity 3. In series to calculate the capacity of 5 condensers of 2, 3, 4, 5 and 6 milifaradios

Activity 4. If we have 101 condensers in parallel of 60 picofaradios, to calculate the capacity in nanofaradios of the equivalent condenser

- Activity 5. To solve the following circuits. In the second case, to simplify the circuit, with the substitution of the condensers by its equivalent Solutions.
- 1º Circuit. C (equi) = 7.93 μF

2º c (equi) = 5 μF

The capacity of the condensers depends on several factors, as it is the surface of the plates, among others factors. Nevertheless, the load that reaches a condenser (that is moderate in Coulombs and it imagines by the letter Q) depends on the tension that there is in the plates of the condenser and the value of the capacity, that is:

Example. To calculate the load of a condenser of 20 mF if the tension in the plates is of 10 volts. To calculate whatever is increased the load if the tension increases to 100 volts.

In the first case, clearing load Q, we have Q = V x C = 10 20 xs mF = 0.2 Coulombs.

In the second case, when increasing 10 times the tension, increases to 10 times the load and, therefore, the load is 2 Coulombs.

Once it is understood that the load is the “amount of electrons that can store”, we will see how it is the process of load and unloading of the condenser. In order to load a condenser, one is used a battery and a resistance that the electron passage limits. Of that form we have a circuit as the shown one the left. In this case we have chosen to put a lifted resistance of 1MΩ.

When we filled a water demi-john with a faucet, the speed at which it fills is always the same. For example, 10 liters every minute.

In the case of the condenser, the thing changes since the load that is storing generates an electric field that is against that more loads enter. For this reason, the graph that shows how the load evolves with time responds to an exponential curve of the type: All this load experiment we have realised it with the program cocodile V3.5.

If you want it you can unload here.

The representation of the tension in the plates of the condenser with respect to the time comes given by the mathematical formula:

- where:
- Vc it is the tension of the condenser for every temporary moment t
- Vf is the maximum tension that can reach. In this case, the tension of the battery

RC is the product of the resistance of load by the value of C

For the unloading, the formula that represents the value of the tension in the plates of the condenser with respect to the time is

- Where:
- Vc it is the tension of the condenser for every temporary moment t
- I saw is the tension that has the condenser at moment zero. If it were loaded to the maximum, it will be the tension of the battery.

RC is the product of the resistance of load by the value of C

Activity 6. To calculate the tension that a condenser of 8 reaches μF to the 20 seconds if it is fed by a battery of 10 volts and one resistance of load of 1 MΩ.Activity 7.

R mounts a circuit RC with the values of = 990 KΩ and one capacity of 10 Microfaradios. Obtén the embarkation schedule of the condenser.

The pulser that we put in parallel with the condenser is to cause the unloading of the same and to have the load curve from value 0 If we see the graph, we have the sensation of which not load to the maximum never.

This is truth and, for this reason, tenth that the condenser has been loaded when it has passed an equal time to 5 times the value of R through C, this it is:

For the previous exercise, we have them 5 Tao is equal a:

If we diminished the resistance, we have the time of load falls like if the capacity of the condenser falls

Activity 8.

In the previous circuit, to add to the pulsers or switches necessary to obtain a circuit that loaded and unloads the condenser. To put the clamp of test in the condenser to obtain the unloading and followed load curve

Activity 9.

A system pyrotechnics needs to activate with a tension of 20 volts so that the fireworks jump. We have a battery of that tension and a condenser of 400 microfarads. To calculate the load resistance that I must put so that the system activates to the 20 seconds

Activity 10.

We only have in the factory condensing of 20 mF and a 5 condenser of mF. For a system programmer, we required a condenser of 15 mF. What solution we can give?

PRACTICE 2. In the circuit of the figure, we left from the totally empty condenser and closed the switch.

- To calculate: 1º theoretical Part.
- To calculate the time that takes the condenser in loading itself
- To calculate the time that takes in unloading
- What tension reaches to the 10 seconds, moment at which, the commutator is pressed so that the condensing pass to unloading.

2º

Simulator part. Using our program of simulation (proteus or cocrodile), to make points 1, 2, 3 and 4, using for it the unloading and embarkation schedules. Once they obtain each of them, to obtain in the same, the times that are requested.

We can put a switch in parallel with the condenser so that the unloading becomes fast.

3º the values of RC are such that the load and unloading have been long time. What we must do so that 100 times are reduced. To modify the circuit and to make screenshot of the load and unload.

PRACTICE 3. Advancing events. We have seen that the condenser is based on two plates, separated, where the passage of electrical charges among them is not allowed. But, aside from loading and unloading a condenser, why it is worth in a circuit if it does not let pass the loads.

The answer is:

The loads do not happen from a plate to another one, but it causes that by the cable they can go and come, if we put an alternating power supply to him, where there are positive cycles and negative and those cycles cause that the electrons move in the cable.

The condenser will lie down to load itself and to unload cosntantemente.

This factor is going to depend on the frequency of the generator and the capacity on the same. Without entering, the practice consists seeing in detail how it varies the luminosity of a light bulb when we changed to the frequency of the generator and the capacity of the same.

1º Montar the circuit that is in the image, where R are a resistance of 5 Ohms, L1 and L2 is two light bulbs and C is the condenser, whose value is necessary to vary. The generator has a frequently variable tension of 7 volts. L1 is had as luminosity witness.

2º Seleccionar a value of the condenser, ** home by 1 nF**, so that the L2 light bulb illuminates just as the L1. Once selected the value, to fit the frequency above and down to verify as the luminosity of L2 behaves 3º

To do the same with the condenser. Once found an suitable frequency, it varies C above and down to verify as it varies the luminosity

4º Hacer 4 screenshots for each one of the situations