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Amplifying circuits with transistors and op amps

amplifying circuit

We have already seen that a transistor presents a current gain and, therefore, when introducing a variable current by the base, we when coming out have a signal with a greater intensity (we are amplifying the current)

Figure 1: Amplifying circuit with transistor NPN

Amplifier with transistor NPN.

We will see how the signal of exit of a circuit as the represented one in image 1 is obtained, where we have other elements that we are going to try together.

The function that has the condensers is the one to offer an almost null impedance to the alternating signals that we want to amplify but an infinite impedance to the DC.

We will see the first Theorem of Thevenin. This theorem allows to simplify feeding systems us, in simple others but for a later analysis.

 Normally in the amplifying circuits, the feeding of the basic circuit is taken of the same battery, but this same one causes that the system can be complicated facing an electrical analysis. The theorem that we are going to see simplifies the basic circuit in simpler other.

The branch of two resistance to polarize the base, it we will be able to replace by a generator and a resistance.

Theorem of Thevenin. calculation thevenin

A system made up of several generators and resistance seen from two points and B, can be replaced by a single generator (Vt) and an equivalent resistance Rt.

To show/To hide Development of the Thevenin Theorem

 Development of the Thevenin theorem

In order to obtain this theorem, Thevenin supposed two extreme situations.

If the objective is in the same way to create a simple circuit but that bearing that the original one, in extreme situations of work, both it must tolerate the same

calculation circuit thevenin a) Circuit with an impedance of Infinite exit \ boldsymbol {Vsalida = 10 * \ 20 20 tailcoats {} {20 +} = 5 Volts}

We are going to leave both to circuits with infinite load. In this one case, the current that circulates around the Rth is zero, since we do not have load (secondly circuit), being the tension when coming out the same that the tension of the Vth generator. If we calculated the tension that there is when coming out (we do it in 1º circuit) we will have the tension thevenin. simplified circuitThe tension of exit in 1º circuit is the voltage drop in the resistance of 20 Ω, since through the resistance of 10 Ω does not pass current. The S-value:

Therefore, 5 volts are going to be the Vth

b) In 2º case, we must calculate the short circuit current, where both circuits must have the cortocircuitada exit now.

 This is a variant of the postulated second, but she is more sensible Icc = \ tailcoat {V (TV)}{Rt} = \ 5 tailcoats {} {Rt}

In circuit 2, current Icc is:

This Icc exactly must be just as the Icc in 1º circuit that we happened to calculate

V (r=10) = \ 20//10 10 tailcoats {} {20 +} *20//10 The Icc are the current that happens through the resistance of 10 Ω. First we calculated the tension that has that resistance:

where 20//10 are the value of the two resistance in parallel, that gives a value us of 6,666. V (r=10) = \ 6.666 10 tailcoats {} {20 +} *6,666 = 2.5 volts

Therefore:

The current that happens through the 10 Ohms is: 2,5/10 = 0.25 To, that it is the short circuit current that it must also happen through circuit 2.

0,25 = \ 5 tailcoats {} {Rth} \ Rightarrow Rth = \ 20 0.25 5 tailcoats {} {} = \ boldsymbol {\ Omega}  

 Replacing values:

 

If we applied the theorems, we will see that the values with the shown ones by this system agree “both Of putting under circuits the same situation (open Circuit and circuit in Short).

Vab = \ 10 tailcoats {9V * K} {50 k + 10 k} = \ boldsymbol {1.5 Volts}

Example of application:

Vab = \ 10 tailcoats {9V * K} {50 k + 10 k} = \ boldsymbol {1.5 Volts}

As it is shelp in the figure, to calculate the Thevenin tension, the tension is taken that offers the circuit of study and the load takes off to him. In this case, when clearing to our circuit the load (current that enters the base) has a voltage divider whose S-value:

Rth = \ 8.33 50 50 tailcoats {10 *} {10 +} = \ boldsymbol {K \ Omega}

 IV = \ tailcoat {(1.5 - 0.7) 8333 volts} {\ Omega} = \ boldsymbol {96 \ mu To}The Thevenin Resistance is obtained to the supposition that the generators are in short, therefore, have two resistance in value parallel:

 Now we will see where it is working within the characteristic line of exit. We will look for point Q. Process

1º two points are chosen to draw up the load straight line. One of the points is when there is no Tension BCE. In this case, the current that circulates around the collector is 9 9V/1K = mA

 

The other value happens when IC = 0. In that case, tension BCE is the current of the battery.

 

Point Q is in the intersection of the straight line of load with the line of the basic current. That point indicates to us what current passes through the transistor and the tension between Collector and emitter

 

 characteristic line transistorWe will see a video of review of the transistor where it is explained how is point Q  

 

 

 

Exercise 1: To locate to point Q as intersection between the straight line of load and the curve of IV.

Exercise 2: In the circuit of the previous practical exercise, the transistor is 2N3055.

1º Buscar in Internet the characteristic lines. On the same, to make the straight line of load and to locate point Q

op amp2º Simular in some program (proteus or cocodrile) to locate the values of point Q

Op amps

 The Op amp, in future AO, is an integrated circuit with some characteristic electrical that make ideal him to make tasks such as:

* Operator to make mathematical tasks as sum or reduces subtraction

* Construction of active filters

* Converters IV (Current - tension)

* Accomplishment of converters analogical-digitalises, etc
The Characteristics of the ideal Op amp are:
1º Gain of tension in way Infinite differential
2º Gain of tension in common way zero
3º Resistance of infinite entrance
4º Resistance of null exit
5º Current of null entrance (zero).
6º Ancho of infinite band
7º Infinitely rank of voltage available in the exit.

In addition to these characteristics, we must consider the factor of:

Virtual earth

In the situation where the AO works in normal conditions (without being saturated), she is had difference enters the tension the investing entrance (v) and the earth terminal is almost 0 volts. This situation takes to consider that the investing entrance acts as a virtual earth, which facilitates much the calculations of certain circuits.

It is necessary to consider that if the AO is in saturation, the exposed thing before is not valid, since it appears a tension in V (-) and earth.

With these theorems, we are going to analyze our first circuit that uses the AO.

AO working in investing way

AO as investing amplifier

The AO working as amplifying investor has entrance + connected to earth and the connected v to the resistance R1 and R2. As the impedance of the A.O is infinite (in the ideal case) and very high in the real case, we have current it that comes from I saw, agrees with the current that comes from Vo, or what is the same, I1 = - I2 \ boldsymbol {I1 = \ tailcoat {Vi} {R1}}

On the one hand, we have: \ boldsymbol {I2 = \ tailcoat {Vo} {R2}}

 In the circuit of exit: {\ color {DarkRed} \ boldsymbol {\ tailcoat {Vi} {R1} = - \ tailcoat {Vo} {R2} \ Rightarrow Vo = - \ tailcoat {R2} {R1} *Vi}}

Of this situation,

Noninvesting A.O

Noninvesting AO

 \ mathbf {V2 = \ tailcoat {Vo} {R1 + R2} * R1}

In this situation and working the AO in normal operation, we have it tension in entrances V (+) and V (-) are practically equal (V1 = V2). This way, we can obtain the value of V (+) as the tension in a value voltage divider: I saw = \ tailcoat {Vo*R1} {R1 + R2} \ Rightarrow Vo = I saw * \ tailcoat {R1 + R2} {R1} = \ boldsymbol {Vi* (1 + \ tailcoat {R2} {R1})}

Therefore, clearing the V2 tension and changing it by I saw, we have:

Of the previous formula it is deduced that, in a noninvesting AO, the tension gain is, always, major that 1. 

Using a AO with an feeding of + - 10 volts, we want a circuit that does not invest the signal and has a gain of tension of 21.

a) To design the circuit, being calculated the values of the components that are needed

b) If the entrance signal is 40 mV (sen ωt), to calculate and to draw the signal of entrance and exit in the same graph.

c) To mount in Cocodrile the circuit for the previous signal of entrance and to represent the graph with the exit tension

d) To do the same but in this occasion, the entrance signal is of 2V (sen ωt).  You observe something rare?

AO adder

AO in way adder

The AO can also add two signals (or but). For it the superposition theorem is used, by which:

 The answer of the linear systems with several entrances is equal to the sum of the answer of the system for each one of the entrances.

First of all, for a V1 entrance with V2= 0, the exit is V0 = -10/10 * V1

For 2º case and coming in the same way, V0 = -10/10 * V2 => V0 = - V2.

 The sum of both gives us: V0 = - (V1 + V2).

We can change the values of the resistance if we want in addition that the signals leave amplified, etc.

AO in way comparator

Op amp Comparator \ boldsymbol {Vr2 = \ tailcoat {V} {R1 + R2} *R2}

In this case, the A.O compares the values of the entrance, giving a high or low exit, according to the value of the entrances. For example, we suppose that entrance V (+) is connected to a certain sensor. When that the tension of the sensor arrives at a value, the system activates. In the circuit of the figure, we have a AO in way comparator, so that when the entrance signal I saw superior to another one arrives at a value from Vref reference, the exit she will be high and activate some other device.  The tension that we have in the R2 resistance is:

When the tension I saw is superior to this value, the system activates.

Activities. Problems of amplifiers amp-not-investor1º In the circuit of the figure, R1 = 20 K and R2 =40 K.

a) To calculate the tension gain

amplifying circuitb) To draw the exit signal if the entrance signal is 2 sen (wt) Volts Problem 2

In the circuit of the figure, to make the following tasks Vbe whereas clause = 0.7 V and beta=100:

1º Hacer the Thevenin circuit

2º Calcular the collector current

 

3º Calcular the signal of exit for an entrance of 20 sen (wt) mV

Activity 3

A greenhouse of Almeria works products of high economic but too sensible value to the low temperatures. To design a complete system so that arrival a critical temperature, activates a system heating engineer.

adder with AO

Activity 4

To mount in proteus an adder where the 10 resistance are of K, coming from the following form
1º In proteus, search AO TL064. To add
2º Ir to generator - > DC. Añadir two for the positive and negative feeding of the AO
3º Now we must add two alternating sources that will be our entrances. We take in Generator, selection SINE
4º Añadir the resistance (equals) as one is in the previous circuit

5º Configurar the alternating feeding to 2 volts of tip and frequencies of 1 KHz and 2º 2 signal of Khz. The feeding of the A.O is 9 and -9 volts

6º Visualizar the signal in the oscilloscope, that one is due to show as one is in the figure. As the A.O is in investing way, we have invested the exit signal so that is better the result.

7º Probar with God signals of the same 180 out of phase frequency and º. What is obtained?

8º Probar with other three pairs of signals and to show the result.

Design PCB