Theory and problems of Diodes and transistors

 

The semiconducting diodes and transistors are two of the base of the integrated electronics (integrated circuits, microprocessors, etc) that have allowed a jump very important in the Electronics. There are others not less important, as triac, but in this case we are going away to center in these two. First we will see something of the base of:

Semiconductors

We have two types of semiconductors, with the same raw material, the Silica (sand of the beach).

In the image of the left we have the molecular structure of this material. As it is possible to be appreciated, each Silicon atom is connected with neighboring atoms by a connection (altogether 4 I connect).

 

 

 

 

In order to obtain the P type, we must add atoms of gallium, boron.  (3 valence electrons). Now we have only three electrons for the connections. One is going away to generate a Hollow

In order to construct the first component, we must take to a “piece” of the P type and another one of the N type. This gives rise to

Diode

The diode is a semiconducting component with two terminals (Anode and cathode) so that, according to the polarization, it let pass the current in a single sense. We are going to see what happens when we polarized it.

First of all, we consider the different situations from atoms

In the first one squaring indicates the situation to us of atoms before no union takes place

In the second quadrant, that can happen when it is united a P type and a N type (in particular, what can happen in the Union).

First that we do it is to represent Union P-N without polarizing.

In this situation, just before becoming the union, nuclear barrier does not exist and the matter to both sides of the union is neutral (there are so many electrons as protons)

Once the union becomes, some electrons near the union, will be attracted by the hollows of the border (in side P). This causes that they pass loads from a zone (n) to the other. The consequence is that we positively left to a space in loaded side N and the zone of the border of side P, loaded negatively. In this situation, to electrons it will cost to him but work “to jump the barrier” and therefore will become paralyzed the passage of loads.

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.

We see this in the image of the right

Polarization of union P-N in direct

We are going to connect the anode to the positive pole of the battery and the cathode to the negative. In this case, the electrons that were located in part P of the barrier, will see the positive pole of the battery, being attracted by that field and leaving the place that it had. That causes that the barrier lowers and is taken advantage of so that other electrons happen from N to P. Now we have an electrical current

Polarization of union P-N in Inverse

. Now we are polarizing polarizing zone N with positive tension. The electrons leave the material to arrive at the positive of the battery. The nuclear barrier increases and as soon as they pass electrons from a side to another one. Now we do not have current and the diode says that it is in cuts.

 

We can complete it with this video
 

 

Characteristic line of the diode

In the image of the left we have the diode polarized in direct (quadrant A) and inverse (Quadrant B)

In polarization in direct, we have, once certain past tension (threshold voltage), the diode begins to lead the current without hardly presenting opposition. It is resembled a closed switch.

If we phelp attention to that quadrant, the slope of the curve is very lifted. To small increases of the tension between anode and cathode, great increases in the current take place

In quadrant B, we have the polarization in inverse. In this case, the nuclear barrier of the union is very high and as soon as they pass loads (leakage current).

 

It arrives a little while that the tension of the battery is so high, that an avalanche of loads in the diode takes place, shooting the current (Zone of rupture)

The transistor
Created in the Bell labs of AT&T, it at the time allowed to reduce the electronic systems, since it replaced the emptiness valves.

A transistor is a device that controls and regulates a great current by means of a small signal
 

It has three terminals, that are:

a) →Terminal Base by which the control current enters. If it enters current, the transistor works

b) The Terminal emitter → by where the current enters that we want to control

c) The →Terminal collector who receives the emitter current (with the approval of the base).

 

To the left we have types of transistors NPN and PNP both, with its symbolism.

We see as a transistor of the possible simple way works but

In this case, we must resort to the typical comparison of the faucet, where the small effort of the key, controls great water pressures

 

According to it is the more or less open key, we can have three states. It cuts, conduction and saturation.

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 β of where:

Where IC is the current of Collector and IV the basic current and are fulfilled approximately in the zone of work of the transistor. (not in saturation)

In the following section, we see with a little more formality, how one moves the loads (majority and the minority ones) in the layers of the transistor

In order to diminish the phenomenon of the recombination, the zone of the base one is due to make sufficiently narrows so that the time that arranges in their passage by the base is so small, that few electrons are united to the hollows of the base

 

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

We must tackle a problem about transistor as two circuits (entered and exit) where the relation among them does not give it the current gain, that in this case is 100

First it is to calculate the current that enters the base. As the current is the same that happens through the Rb resistance, we calculated that current.  By the law of Ohm, we have:

 

We replace values and:

 

As the collector current is Beta times the basic current: IC = β*Ib = 100 * 8,6

 

µA = 0,86 mA

 

In order to calculate the BCE, we cleared to tención Vcc, the voltage drop in the collector resistance. We have:BCE = Vcc - Rc*Ic = 12-0,86mA*2K

 

 

 

Ω = 10.86 V

Problems of diodes and transistors

Problem 1: In the circuit of the figure, to calculate the current that leaves the battery if the tension in the union of the diode is 0.6 volts. Solution: 0,84 mA  Problem 2.

We have a diode LED that in the characteristics indicates a maximum intensity to us of work of 20 mA. To calculate the resistance that we must put so that one does not spoil. A tension and voltage drop assumes in the same of 1 volt of the battery of 11 volts.

Problem 3 In the circuit of the figure, to calculate the collector-emitting tension if the gain of the transistor is 20 volts and the tension bases emitter is 1 volt.

Problem 4.  In the circuit of the figure, one assumes that the transistor has a current gain β of 10 and that the voltage drop enters base and emitter is 0.7 volts. To calculate the current of collector and the tension that falls between Collector and emitter Problem 5. To calculate tension BCE, and the currents in each resistance of the following circuit. It is considered that the gain in current is 100 and the tension bases emitter is 0.7 volts. This problem can be done of two forms. The first and most complicated one leaves the solution in connection of down. Second we will see it using the theorem of thevenin  Solution

This exercise also we can see it in the following video

  To show/To hide Solution in video

Theorem of thevenin

You have been difficult the 5? . Certainly yes. In this occasion, to solve a series of equations has assumed it, and if one is not very capable, we will have problems to give with the solution. Luckily, a theorem with the idea exists to make simple a complicated problem.

  We happen to expose it.

We are going to explain it. Basically, the idea is to happen of a muddied circuit, to one simpler, with its generator and its resistance of load. Both equivalents are called if they respond in the same way, that is enters and B we have the same electrical values that enters A' and B'.

For it we suppose two situations

a) The circuit is without load. That is, we have cleared to him what it had connected enters and B. In this case, and with the example of the figure, tension AB is the same that the tension that falls in R3, since there is no current by R2, and therefore, there is voltage drop no. In this case it will be:

If that we have when coming out, that we must when coming out have of 2º circuit. On the other hand, as in 2º circuit is open, the previous tension agrees with the tension of the Thevenin generator

b) If the generators are run out, they would have a tension 0 between tips. Just like a cable

What it is made now is to cortocircuitar ds generating and to calculate the resistance that is seen from the exit

The circuit of the left, now has R1 and R3 in parallel and the result in series with R2

To the right, the resistance that is seen agrees with R thevenin

Therefore:

 We already have our generator and our resistance.

 

Now we make the calculations for problem 5, we replaced it in the 5 and to verify that everything is but simple Problem 6. The circuit of figure 6 uses a Darlington transistor. In this case, the cascade connection causes that the total beta of the system is the product of β1 multiplied by β2. The Vbe tensions each transistor are 0.7 volts,   β1 = 10 and β2 = 20 a) To calculate the current of collector and tension BCE being supposed that both transistors can be replaced by one of gain β1*β2

b) To make same the calculations both whereas clause transistors Ejercicio7: In the circuit of the right, the gain in current is 50 and the tension enters base and emitter 0.7 volts.

To calculate the basic current, the current of collector and the BCE. Exercise 8.  In the circuit of the figure, if the current gain is 50, to calculate: 1º the basic current

2º the Collector-emitting tension

 

   

Exercise 9. To design a circuit so that when illuminating a photodiode, ignites a light bulb. It is necessary to use a transistor.

  Resolution of problems::

•  Problem 7
• a) As the gain is 50, we have IC = β * IV = 50* IV
• b) On the other hand, by the relation between currents that enter and leave (law of José Mota, or law of the hens), we obtain that IC + IV = IE >>>Ie = 50Ib + IV >>

IE = 51*Ib

 For the tensions, and looking for the way that follows the current since it leaves the battery until it arrives at the negative, we have:

 10 V = IV * 10 KΩ + Vbe + IE * 20Ω >>> 10 V = IV * 10 KΩ + 0.7 + 51*Ib * 20Ω.

Reorganizing this expression, we have left that:

From we can here remove the collector current:

IC = 0.844 * 50 = 42.2 mA

IE = 0.844 * 51 = 43.044 mA

 The tension in the circuit of exit:

10 V = 40 * IC + BCE + 20 * IE =>> BCE = 10 - 40*42,2 mA - 20* 43.044 mA = 7.45 V

Problem 5

An base-emitting tension of 0.7 volts and one beta of 100.Vamos is considered to establish the equations of the tensions in the basic circuit:

1º) 10 V = I1x10KΩ + I2 x 2KΩ

2º) 10 V = I1 x 10 KΩ + Vbe + IE x 1 KΩ

Now the formulas for the currents                3º) IC = β x IV          4º) IE = IC + IV

5º) I1 = I2 + IV  

  We replace 3 in the 4 “4º) IE = 100xIb + IV = 101 IV

We develop to the 1 and the 2

1º) 10 V = (I2 + IV) x 10KΩ + I2 x 2KΩ = I2 x 12KΩ + IV x 10 KΩ ““” .

This equation we are going it to replace in the 2  

. 2º) 10 V = (I2 + IV) x 10KΩ + 0.7 + 101 xs IV x 1 KΩ

 

 

 2º) 10 V = I2 x 10KΩ   + IV x 10KΩ + 0.7 + 101 xs IV x 1 KΩ >>

 We already can obtain the rest of values. IC = 9.35 μA * 100 = 0.935 mA

 Of formula 4, we have IE = 101 * IV = 9.35 μA * 101 = 944.35 μA

The tension emitting collector we obtain it from the expression: 10 V = IC * 3K + BCE + IE * 1K

Replacing values: 10 = 0.935 mA * 3 K + BCE + 0.944 mA * 1K