Basic electronics 4 THAT

Basic electronics

Before this subject, the review of the subject of electricity of 3 would be advisable that

In this first subject we are going to know some the basic components of

electronics and we are going to begin by simplest. We speak of the resistance. They are made specifically to make difficult the passage of the current by the circuits and thus to assure that it does not pass an excess of electrons through the components. It has a system of bands, that in the image of the example is orange, black, red and golden. The three first, that is to say, indicate the value of the resistance in Ohms and the quarter the tolerance what can be turned aside of the value that indicates the 3 first colors. For this example, we have, when looking in the table of colors, 3 0 00 Ω with 5% of deviation. The maximum value is 3000 Î© but 5%, that is 3150 Ω and for the minimum is 3000 €“ 150 = 2850Ω Vamos to practice a little with the following examples a) Red, red black silver   b) green, blue, red, gold   c) gray red brown silver
Exercise. We are going to realise with the companions of Computer science a program in php so that, when selecting the colors show to the value of the resistance next to the maximum and minimum to us.  To puncture in Program to calculate value of the resistance

 2 Variable resistors

Within this block we have all the resistance that by action mechanical (we move an axis) or by physical variations (the light, the temperature changes. ) the value of the resistance varies.

2,1 Potentiometer.

Resistance of the potentiometer




In the figure of the left we have this component, employee for example in the equipment of music to give but volume to the loudspeakers. In the part inferior is the scheme where the green strip represents the coal film that for of resistance. If we turned the axis so that the midpoint (that makes contact with the film) is but to the left supposes that the resistance from point 1 to the m is smaller. If we turned to the right, there is but film and therefore but resistance, reason why the resistance that there is between 1 and m it increases. Example.   That value we have between 1 and does patita€ of the center if the value of the resistiva film is 2000 Î© and the dark green zone represent 30% of all the resistance? . The dark green zone is 2000 * 0.3 = 600 Î© and the other part (between m and 2) that comes as green the clear one is 2000 * 0.7 = 1400 Ω.

2,2 Variable resistors of the temperature (NTC and PTC).


As it indicates the name, the value of the resistance varies when raising or lowering the temperature. NTC (Negative coefficient temperature). In this case to the being a negative coefficient, means that the low resistance if the temperature raises and vice versa. To see diagram PTC (Positive coefficient temperature). In this case it raises if the temperature raises and low if the low temperature (blue line). It is necessary to say that the graph is idealized (the real one differs in some shade).

 2.3 ldr

The resistance falls when it affects but light exceeds she. An application is in the lights of the streets, that activate the lampposts when the night arrives. In order to approach this component we are going to make a small practice in factory where we are going to control a light bulb (light of the lamppost) when the night becomes. In this circuit we used a transistor (still not seen) and a LDR. When it is at night, the light does not arrive at the LDR and its resistance is high (1 M). The electrons that come from the battery and which they happen through the 10 resistance of K, when arriving at the entrance of the LDR take the way from the base from the transistor that offers less resistance. This causes that the transistor works and the light ignites. In this case the day arrives, the light increases and the resistance of the LDR lowers. The electrons that come from the resistance of 10 K happen now through the LDR and therefore the transistor, without basic electrons, does not work and the light bulb goes out.

 3. Condensers

The purpose that they have is the one to store an electrical charge and to use it later, that is to say, is something as well as a small battery.

She is compound of two separated metallic plates of a dielectric material

The capacity of the condenser is moderate in Farads and comes as the relation between the load that stores and the tension that appears in its terminals, that is

\ dpi {120} \ fn_cs \ large C = \ tailcoat {Q} {V}

Circuit of load of the condenser One of the applications of the condensers is as timer, so that the load is increasing gradually and controlled by a resistance. We show the circuit. The load curve of the condenser is the following one: In the graph it is considered that for 5 t, the condenser practically are loaded.

We have a condensing one takes the tension from the battery (9 volts in this example) when 5 times are pin the product of R X C, that is t = 5 XS R X C

 We are going to make a small practice basing us on the following video. It is necessary to consider that is no resistance in the system of load and unloading and therefore we must put a suitable resistance.

 4. Semiconductors

The semiconductors are materials that not being good drivers of the electricity, when adding other elements become very good drivers. The material semiconducting but used base is the silicon, that we found it in the silica of the beach sand and therefore, a material that a priori we thought that it can be very cheap and abundant. As we can see in the image, it is connected with neighboring atoms by a connection (altogether 4 I connect) to obtain the commercial semiconductors is to add to an extra element (gallium, boron. ) that contributes three I connect and is free one (a hollow is free) or others as arsenic or phosphorus (5 valence electrons) contributing an electron extra and created the material N type, therefore If + Gallium = Material P (excess of positive charge) If + Arsenic = Material N (excess of electrons)

We show cases both, where to first we have added to the Boron and to the second phosphorus to him


We will see now that it happens when we united the two parts. The electrons of the N type but next to the border move and occupy the hollows of the P type, so that spent a time a barrier is created that it prevents that but electrons pass the barrier.

It is necessary to consider that the matter to the left one was neutral. If they arrive electrons become material with negative polarity that repels the arrival of new electrons

If now we polarized the P Type negatively and the N type positively, (denominated polarization in inverse), the barrier grows because the connections of the P type are occupied by electrons of the battery.

Nevertheless, if we polarized in direct (P to positive and N to negative) the electrons that pass the barrier follow€ until the end of the material until arriving at the terminal of the Anode. (To see figure of down).

We can complete it with this video

 We will make some practices with the diodes and for it we can use diodes leds, where we are going to verify that a polarization in inverse prevents that the current pass and therefore the light does not ignite.

As review of the subject of the relay, we are going to use one to detect the presence of an intruder in our house. For it we must consider that once we stepped on the detector, the alarm must sound until we deactivate it. Bond that the intruder raises the foot here and has not passed anything, not no, yes that has happened

We will see in this practice the refeeding concept. How one takes place. To mount the circuit and to reason where it is the refeeding. We use two pulsers, first that are normally open and will be the one that detects the presence of the intruder. The second, in our control center, the one that deactivates the alarm. That happens if the intruder breaks cables of the sensor, it continues sounding the alarm?

Once we have seen the transistor, we are going to replace the relay (expensive element) by two transistors and some resistance (but economic). The circuit to mount is this.

5 the transistor

The Transistor is a device of three terminals that is born in the Bell labs from

AT&T.  It was an authentic revolution since it came to replace the old valves of emptiness which allowed to a considerable saving of space and energy. (if we had to make a computer with valves, we would need an institute to lodge all the components and a fortune to pay the invoice of the electricity).

He is made up of the base, the emitter and the collector and of simplified way can be shelp that

a) The base is the terminal by which the control current enters. If it enters current, the transistor works

b) The emitter is the terminal that sends the current that we want to control

c) The collector is the terminal that picks up (with the approval of the base) the current of emitter

In the previous image we have types NPN and PNP both, this means that we have a layer N, in center a P and to finish another N for the case of the NPN. We will see as it works of simplified way. It is necessary to say that they are polarized of different form, although in this subject always we are going to use the NPN where the collector must go positively polarized and the emitter negatively.

If by the base we entered a current, the barrier of the union base-collector falls allowing the current flow. To but current basic but current of collector. These currents are related to the parameter Î² of where:

\ dpi {120} \ fn_cs \ large \ beta = \ tailcoat {IC} {IV}

 Being IC the current of Collector and IV the basic current. (This is fulfilled for small basic currents where the transistor does not enter saturation)

We are going to see as a simple circuit with a transistor is analyzed.

When the transistor well is polarized we have is a current that goes from the battery, enters by the base and leaves by emitter. We can make that branch of the following form.


We eliminate the transistor and we put

diode simulating the base-emitting union. The tension in that diode when it leads is 0.7 volts, therefore we have current it that leaves the battery is\ dpi {120} \ fn_cs \ large IV = \ tailcoat {(9-0,7) V} {10K \ Omega} = 0,83 mA

Once we have the current of the base, the current of the Collector is:

\ dpi {120} \ fn_cs \ large IC = \ beta x IV = 100 83 Xs 0,83 mA = mA

Problem: 1 Calcular the current of diode LED if the voltage drop between Anode and cathode is 2voltios.

2 we have bought a diode LED and in the characteristics it indicates to us that so that gives the optimal light without spoiling is necessary that it has 3 volts between anode and cathode, moment at which mA circulates a 30 intensity of. To calculate the resistance that we must put in series if the battery is of 10 volts.

3 Indica in the following circuits the light bulbs that they ignite reasoning the answer

4 the circuit of the following figure represents a bridge rectifier, used to rectify the tension. If we consider that the signal of the generator is sine, it explains that signal we are going to have in the left leg of the light bulb.

Smoothing circuit

5 Calcular the current of collector of a transistor if the parameter Î² is 100 and the basic current of 2 milliamperes.


Circuit 1
Problem 6


6 Calcular the current of the light bulb if Î² is 100 and the tension between base-emitting is 0,7 volts

Solution: 830 mA




circuit-two-light bulbs
Problem 7

7 In the circuit of the figure, to calculate the tension that falls in each light bulb and the Collector-emitting tension if we know that Î² of the transistor is of 100 and the voltage drop enter base and emitter is 0.7 volts. The light bulb has a resistance of 100 Î©.

To see Resolution at the end of the page


Problem 8

8 In the circuit of the left we have two 10 resistance of k and one light bulb in the collector of 100 Î©. Î’ is the 120 and tensions BCE = 2 volts and the Vbe = of 1 volt. To calculate the tension of the battery of the basic circuit

Problem 9



Problem 9 Analizar completely the circuit of figure 9, that is to say, BCE, the tension that falls in the light bulb, the tension that falls in each resistance, if the current gain is 100, the resistance of the light bulb is of 100 Ohms and the voltage drop enters base and emitter is 1 volt. In the analysis to suppose that the collector current is equal to the emitter current.

circuit two transistors
Problem 10

Problem 10: In the circuit of the figure we have two transistors with gain of current of 100 (β = 100) and one basic resistance of 100,000 Î© in 1 transistor and exit of emitter entering the base of the following one directly. If the resistance of the light bulb is 100 Î© and the Vbe tension of each transistor is 0.7 volts, to calculate the current that happens through the light bulb, as well as, that is to say, to calculate the total gain of current of the circuit the relation between the Current of the light bulb and the current that to him transistor enters the base of 1.


Valve: Used electronic component stops for commutation and amplification of signals by means of electric fields that they influence in the movement of electrons in an empty€ space to very low pressure.


 Periodic signal that it is alternating the positive and negative cycle with a certain frequency. The one of the domestic network is 50 Hertz, this means that the current changes of sense 50 times per second.


Exercise 7: They request the tension to me of each light bulb. I know that the tension is the Current of collector x the resistance of the light bulb, that is:

  • Vb = IC * Rb.  But, I do not know the Current of the light bulb. We must obtain it by the relation that is between the IC and the IV of the transistor by the relation of the gain, that is:
  • β = IC/IV, therefore, IC = Î² * IV. We follow without being able to calculate nothing because we do not have the IV, but, when having all the data in the circuit of the base we are going easily to obtain this IV and of there all the data that ask to us. We happen to calculate the IV.
  • IV = Vb/Rb = (4 €“ 0.7)/10,000 = 0.33 mA
  • Now the collector current: IC = Î² * IV = 100 * 0.33 = 33 mA
  • Now the Tension in each light bulb: Vb = IC * Rb = 33 mA * 100 = 3300 mV = 3.3 V
  • As 2 light bulb has the same resistance, the voltage drop is the same, that is 3.3 Volts
  • Now it lacks the fall that is between collector and emitter. As from top to bottom we have 9 Volts and fall 3.3 but 3.3 volts in the light bulbs, we have left in between collector and emitting she falls 9 + 3.3 €“ 3.3 volts = 2.4 Volts