Most Important Purpose electronic devices- amplification of electrical signals. Devices designed to perform this task are called electronic amplifiers.

An amplifier is an electronic device that controls the energy supplied from a power source to a load. Moreover, the power required for control is, as a rule, much less than the power supplied to the load, and the shapes of the input (amplified) and output (at the load) signals are the same.

Schematic representation of amplifier operation

Amplification devices are widely used in automation and telemechanics, in monitoring, control and regulatory systems, counting and solving computers, instrumentation, household radio equipment, etc.

The most important technical indicators are: gain (voltage, current and power), input and output resistance, output power, range of amplified frequencies, frequency, phase and nonlinear distortions.

Most amplified signal sources develop very low voltage. It makes no sense to supply it directly to the power amplification stage, because with a weak control voltage, it is impossible to obtain any significant changes in the output current, and, consequently, the output power. Therefore, the composition block diagram The amplifier, in addition to the output stage that delivers the required power of the useful signal to the load, includes preliminary amplification stages.

These cascades are usually classified according to the nature of the load resistance in the output circuit of the transistor. The most widely used are resistive amplifier stages, the load resistance of which is a resistor.

In pre-amplifier stages based on bipolar transistors, a circuit with common emitter(OE), which has a high voltage and power gain, a relatively high input resistance and allows the use of one common power source for the base and collector circuits.

Resistive cascade on a bipolar transistor

The simplest scheme A resistive amplifier stage with a common emitter and powered from a single source is shown in Fig. The input signal enters the base and changes its potential relative to the grounded emitter. This leads to a change in the base current, and, consequently, to a change in the collector current and the voltage across the load resistance RK. The decoupling capacitor Cp1 serves to prevent the flow of the DC component of the base current through the input signal source. Using capacitor Cp2, an alternating voltage component Uke is supplied to the output of the cascade, varying according to the law of the input signal, but significantly exceeding it in magnitude. An important role is played by the resistor RB in the base circuit, which ensures the selection of the initial operating point on the characteristics of the transistor and determines the operating mode of the DC cascade.

The simplest circuit of a resistive amplifier stage with a common emitter

To clarify the role of resistor RB, let us turn to the figure illustrating the process of signal amplification by a circuit with a common emitter. In principle, the amplification process can be reflected by the following relationship of electrical quantities.

Graphical explanation of the process of amplifying a signal with a common emitter circuit

Um IN I B m IK m IK m RK (Um KE = EK - IK m RK) = U m OUT

Indeed, by first examining Fig. a and then Fig. b, one can be convinced that the voltage of the input signal with amplitude (Um VX = UBE m) changes the value of the base current in phase. These changes in the base current cause collector circuit proportional changes in the collector current and voltage on the collector, and the amplitude of the collector voltage (taking into account the scale along the x-axis) turns out to be significantly greater than the amplitude of the voltage at the base. It should be noted that the signal voltages at the input and output of the cascade are shifted in phase by 180°, i.e., they are in antiphase.

This means that the cascade under consideration, without violating the law of signal change (in our particular case, the signal changes according to a sinusoidal law), at the same time rotates its phase by 180°.

To obtain the least distortion of the amplified signal, the operating point (rest point) P should be located in the middle of the segment AB of the load straight line, constructed in the family of output characteristics of the transistor (class A amplification mode). From Fig. b it is clear that the position of the operating point P corresponds to the bias current in the IBP base circuit. To obtain the selected mode, it is necessary to provide the amplifier with the required amount of bias current in the base circuit. This is what resistor RB is used for in the circuit in the first figure.

The circuit shown in Fig. is called a circuit with a fixed base current. Fixed base current biasing features a minimal number of parts and low current consumption from the power supply. In addition, the relatively large resistance of the resistor RB (tens of kOhms) practically does not affect the value of the input resistance of the cascade. However, this biasing method is suitable only when the cascade operates with small fluctuations in the temperature of the transistor. In addition, the large scatter and instability of the parameter in even for transistors of the same type make the cascade operating mode very unstable when changing the transistor, as well as over time.

More efficient is a circuit with a fixed bias voltage on the base. In this circuit, resistors R"B and R"B, connected in parallel to the power source EK, form a voltage divider.

This increases the stability of the circuit's operating mode, since changes in current in the emitter and collector circuits of the transistor have little effect on the bias voltage.

The resistance R"B of the divider is connected in parallel to the input resistance of the transistor. In addition, neglecting the small internal resistance of the power source, we can assume that R"B and R"B are connected in parallel to each other. Therefore, the divider formed by resistors R"B and R"B should have a sufficiently high resistance (on the order of several kOhms), otherwise the input resistance of the cascade will be unacceptably small.

When constructing circuits transistor amplifiers it is necessary to take measures to stabilize the position of the operating point on the characteristics. The main destabilizing factor that disrupts the stable operation of a transistor circuit is the influence of temperature. Exist various ways thermal stabilization of the operating mode of transistor cascades.

The most widely used scheme for thermal stabilization of the regime is shown in Fig.

Circuit of a resistive stage with a fixed bias voltage

In this circuit, in opposition to the fixed forward bias voltage removed from the resistor R"B, the voltage that appears on the resistor RE when the emitter current passes through it is switched on. Let for some reason, for example, with an increase in temperature, the constant component of the collector current increases. Since IE = IK + IB, then an increase in the current IK will lead to an increase in the emitter current IE and a voltage drop across the resistor RE. As a result, the voltage between the emitter and the base UBE will decrease, which will lead to a decrease in the base current IB, and therefore the current IK.

On the contrary, if for any reason the collector current decreases, then the voltage across the resistor RE will also decrease, and forward voltage UBE will increase. This will increase the base current and collector current.

In most cases, the resistor RE is shunted by the capacitor SE sufficiently large capacity(on the order of tens of microfarads). This is done to remove the alternating component of the emitter current from the resistor RE.

Data for calculation.

Circuit diagram of an amplifier stage based on a bipolar transistor with OE

Main characteristics

Transistor

Uke/(Ik/Ib)V/(mA/mA)

Uke/RV/kOhm

Ikm/IknmA/mA

Legend electrical parameters bipolar transistor KT312V:

Designation:	Parameter
	static current transfer coefficient
	limiting frequency of current transfer coefficient
	collector junction capacitance (Ck) and voltage at the collector (Uk) at which it is measured
	emitter junction capacitance (Ce) and emitter/base voltage (Ue) at which it is measured
Rb*Ck psec	circuit time constant feedback at high frequency
Uke/(Ik/Ib) V/(mA/mA)	collector-emitter saturation voltage (Uke) of a bipolar transistor at a given collector current (Ik) and a given base current (Ib)
	reverse collector current
	maximum permissible constant pressure collector-base
Uke/R V/kOhm	maximum permissible constant collector-emitter voltage (Uke) for a given resistance value connected between the base and emitter (R)
	maximum permissible emitter-base DC voltage
Ikm/Ikn mA/mA	maximum permissible constant (Ikm) collector current maximum permissible collector current in saturation mode (Ikn) or in pulse
	maximum permissible continuous power dissipation at the collector

The common emitter amplifier used to be the basic circuit of all amplification devices.

In the last article we talked about the simplest transistor bias circuit. This scheme (figure below) depends on , and it in turn depends on temperature, which is not good. As a result, distortion of the amplified signal may appear at the output of the circuit.

To prevent this from happening, a couple more are added to this circuit and the result is a circuit with 4 resistors:

Let's call the resistor between the base and emitter R bae, and the resistor connected to the emitter will be called R uh. Now, of course, main question: “Why are they needed in the diagram?”

Let's start, perhaps, with R uh.

As you remember, it was not in the previous scheme. So let's assume that along the circuit + Upit—->R to —–> collector—> emitter—>R e —-> ground runs electricity, with a power of several milliamps (if you do not take into account the tiny base current, since I e = I k + I b) Roughly speaking, we get the following chain:

Consequently, we will have some voltage drop across each resistor. Its value will depend on the current in the circuit, as well as on the value of the resistor itself.

Let's simplify the diagram a little:

Rke is the resistance of the collector-emitter junction. As you know, it mainly depends on the base current.

As a result, we get a simple voltage divider, where

We see that on the emitter there is already WILL NOT voltage to zero volts, as was the case in the previous circuit. The voltage at the emitter will already be equal to the voltage drop across the resistor R e.

What is the voltage drop across R e? Let's remember Ohm's law and calculate:

As we can see from the formula, the voltage at the emitter will be equal to the product of the current in the circuit and the resistance value of the resistor R e. This seems to have been sorted out. We'll look at why all this fuss is a little lower.

What function do resistors perform? R b And R bae?

These two resistors are again a simple voltage divider. They set a certain voltage to the base, which will change if it changes +Upit, which happens extremely rarely. In other cases, the voltage at the base will be dead.

Let's go back to R e.

It turns out that he does the most main role in this scheme.

Suppose that, due to the heating of the transistor, the current in this circuit begins to increase.

Now let's look at what happens after this step by step.

a) if the current in this circuit increases, then the voltage drop across the resistor also increases R e.

b) voltage drop across the resistor R e- this is the voltage at the emitter U e. Therefore, due to the increase in current in the circuit U e it got a little bigger.

c) at the base we have a fixed voltage U b formed by a resistor divider R b And R bae

d) the voltage between the base and emitter is calculated by the formula U be = U b – U e. Hence, U bae will become smaller because U e increased due to the increased current, which increased due to the heating of the transistor.

e) Once U bae decreased, which means the current strength I b, passing through the base-emitter also decreased.

f) Derived from the formula below I to

I k =β x I b

Consequently, when the base current decreases, the collector current also decreases;-) The operating mode of the circuit returns to its original state. As a result, we got a circuit with negative feedback, which was played by a resistor R uh. Looking ahead, I will say that ABOUT negative ABOUT brotherly WITH ligation (OOS) stabilizes the circuit, and positive, on the contrary, leads to complete chaos, but is also sometimes used in electronics.

Calculation of the amplifier stage

1) First of all, we find from the datasheet the maximum permissible power dissipation that the transistor can dissipate into the environment. For my transistor this value is 150 milliWatts. We will not squeeze all the juice out of our transistor, so we will reduce our power dissipation by multiplying by a factor of 0.8:

P race = 150x0.8 = 120 milliwatts.

2) Determine the voltage on U ke. It should be equal to half the voltage Upit.

Uke = Upit / 2 = 12/2 = 6 Volts.

3) Determine the collector current:

I k = P race / U k e = 120 × 10 -3 / 6 = 20 milliamps.

4) Since half the voltage has dropped at the collector-emitter U ke, then another half should fall on the resistors. In our case, 6 Volts drop across the resistors R to And R e. That is, we get:

R k + R e = (Upit / 2) / I k = 6 / 20x10 -3 = 300 Ohm.

R k + R e = 300, A R k =10R e, because K U = R k / R e and we took K U =10 ,

then we make a small equation:

10R e + R e = 300

11R e = 300

R e = 300 / 11 = 27 Ohm

R k = 27x10=270 Ohm

5) Determine the base current I base from the formula:

We measured the beta coefficient in the previous example. We got it around 140.

Means,

I b = I k / β = 20x10 -3 /140 = 0.14 milliamps

6) Voltage divider current I cases, formed by resistors R b And R bae, is generally chosen to be 10 times greater than the base current I b:

I div = 10I b = 10x0.14 = 1.4 milliamps.

7) Find the voltage at the emitter using the formula:

U e = I to R e = 20x10 -3 x 27 = 0.54 Volts

8) Determine the voltage at the base:

U b = U b e + U uh

Let's take the average of the base-emitter voltage drop U be = 0.66 Volt. As you remember, this is the voltage drop across the P-N junction.

Hence, U b =0.66 + 0.54 = 1.2 Volts. This is exactly the voltage that will now be present at our base.

9) Well, now, knowing the voltage at the base (it is equal to 1.2 Volts), we can calculate the value of the resistors themselves.

For ease of calculations, I am attaching a piece of the cascade diagram:

So from here we need to find the resistor values. From the formula of Ohm's law we calculate the value of each resistor.

For convenience, let us have a voltage drop of R b called U 1, and the voltage drop is R bae will U 2.

Using Ohm's law, we find the resistance value of each resistor.

R b = U 1 / I div = 10.8 / 1.4x10 -3 = 7.7 KiloOhm. We take 8.2 KiloOhm from the nearest row

R be = U 2 / I div = 1.2 / 1.4x10 -3 = 860 Ohm. We take 820 Ohm from the series.

As a result, we will have the following denominations on the diagram:

Checking the operation of the circuit in hardware

You won’t be satisfied with theory and calculations alone, so we assemble the circuit in real life and test it in practice. I got this diagram:

So, I take mine and attach probes to the input and output of the circuit. The red waveform is the input signal, the yellow waveform is the output signal amplified signal.

First of all, I apply a sinusoidal signal using my Chinese frequency generator:

As you can see, the signal has amplified almost 10 times, as expected, since our gain factor was equal to 10. As I already said, the amplified signal in the OE circuit is in antiphase, that is, shifted by 180 degrees.

Let's give another triangular signal:

It seems to be buzzing. If you look closely, there are slight distortions. The nonlinearity of the input characteristic of the transistor makes itself felt.

If you recall the oscillogram of a circuit with two resistors

then you can see a significant difference in gain triangle signal

Conclusion

The circuit with OE was used as the most popular during the peak of popularity of bipolar transistors. And there is an explanation for this:

Firstly, this circuit amplifies both current and voltage, and therefore power, since P=UI.

Secondly, its input impedance is much greater than its output impedance, making this circuit an excellent low-power load and an excellent signal source for loads following it.

Well, now some cons:

1) the circuit consumes a small current while in standby mode. This means that it makes no sense to power it with batteries for a long time.

2) it is already morally outdated in our age of microelectronics. In order to assemble an amplifier, it is easier to buy a ready-made microcircuit and make it based on it

A. Bepsky
RM. HF-VHF. 1/2002

When designing transistor power amplifiers, radio amateurs often do not perform a complete calculation of the circuit due to the complexity and large volume of calculations. Computer methods modeling radio engineering devices undoubtedly facilitates the design process, but the acquisition and development of such programs also causes certain problems, therefore graphical methods calculations for some radio amateurs may be the most acceptable and accessible, for example, the method described in.

One of the main goals when designing power amplifiers is to obtain maximum output power. However, when choosing the amplifier supply voltage, the condition must be met - Uke max of the output transistor should not exceed by more than 10% the value given for it in the reference book. When designing, it is also necessary to take into account the reference values ​​of Ik max and Pk max of the transistor and, in addition, know the value of the coefficient b.

The meaning of the notation used is illustrated in Fig. 1. Using the reference parameters of the transistor, a coordinate system Uk, Ik is constructed on graph paper, and the straight lines Ik max, Uke max and the maximum power curve Pk max are drawn on it (Fig. 2). The operating point of the transistor is located inside the area limited by the straight lines Ik max and Uke max and the hyperbola Pk max.

Fig.1

The output power of the cascade will be greater, the closer to the hyperbola Pk max the load straight line passes.

Maximum power is achieved when the hyperbola touches a straight line. Maximum output voltage is ensured if the load line leaves the point Uke max. For the simultaneous fulfillment of both mentioned conditions, the straight line emanating from the point Uke max must touch the hyperbola Pk max.

Sometimes it becomes necessary to obtain a large current through the output transistor. In this case, it is necessary to draw a load straight line from the point Ik max tangent to the hyperbola Pk max. The transistor will operate in class A mode.

Let us choose the operating point of the MP transistor so that the output voltage is maximum and symmetrical. From the working point we draw straight lines parallel to the axes Uk and Ik. At the point of intersection with the Uk axis we obtain the value of the cascade supply voltage, and at the point of intersection with the Ik axis - the value of the transistor quiescent current (Iko). After this, knowing the coefficient in the transistor, you can determine the base current Ibo for the selected operating point. In addition, you can calculate other cascade parameters that are important for the developer. It should be borne in mind that the resistance of the resistor Re must be chosen as low as possible (in the extreme case, equal to zero).

To illustrate the described method for calculating the limiting parameters of power amplifiers, consider the algorithm for developing an output stage on a 2N3632 transistor (approximate analogue - KT907).

For this transistor: Uke max = 40V; Pk max=23 W; Ik max=3 A; b=50...110 (for calculations we take b=100); ft=400 MHz.

Graphically we obtain the following data: Up=16 V; Iko=1.36 A; Uout=30 V: Iкm=2.8A.

Determine the base current:

Current through divider:

Resistance of divider resistors.

Ministry of Education Russian Federation

Izhevsk State Technical University

Department of Design radio-electronic equipment»

Course work

“Calculation of ULF on BPT”

Completed by: student of group 671

A.N. Kirdyashkin

Checked by: S. A. Derendyaev

Izhevsk 2003

1. Technical task.
2. Goal of the work.
3. Schematic diagram of the cascade.
4. Determining the type of transistors.
5. Amplifier equivalent circuit.
6. Calculation of amplifier frequency response and phase response.
7. Conclusion .
8. Literature.

Work assignment:

• Gain not less than 30dB;
• Bandwidth from 10 Hz to 10 KHz;
• Permissible unevenness frequency response: Mn=Mv=1.41;
• Input signal amplitude 10 mV;
• Input impedance not less than 10 Kom;
• Load resistance no more than 10 Kom;
• Load capacitance 50 pF;
• Source voltage - + 9V.

Purpose of work: Learn to calculate ULF on BPT.

Requirements for the amplifier.

In order to design an amplifier. Need to know: amplifier output power P out. , output voltage U out., or load resistance R n . Permissible harmonic distortion K g, operating frequency range ( f n and f c), frequency distortion at the lowest and highest operating frequencies Mn. dB and MV dB; input data: input voltage U in, internal resistance of the signal source R and.

TO In addition to the specified basic data, the purpose of the amplifier and its operating conditions must be known (for example, temperature measurement range environment etc.), type of power source (rectifier, battery, galvanic cell and etc.).

Amplifier calculation sequence.

Designing an amplifier begins with drawing up a block diagram and selecting its elements based on the requirements for the amplifier. A typical block diagram of an amplifier with input and output devices, preliminary and power amplifiers is shown in the figure.

When choosing a block diagram, they decide whether the designed amplifier has input output devices, a powerful amplifier, or a pre-amplifier. Having drawn up a block diagram of the amplifier, choose circuit diagrams input and output devices (rheostatic-capacitive, transformer), cascade powerful amplifier(single-act, push-pull, transformer, transformerless), pre-amp stages (feedforward, rheostat, transformer, inverse, etc.). After this, transistors are selected for all amplifier stages and the number of stages is found based on the given output power or output voltage and voltage of the signal source, approximately determining the gain required from the stages. After that, a schematic diagram of the amplifier is drawn up and the specified frequency distortions are distributed among the circuits and stages that introduce these distortions. Distribution Mn and Mv are carried out separately at the lowest and highest operating frequencies, then proceed to the selection of operating modes of the transistors and the electrical calculation of the circuit parts. The amplifier is calculated starting from the final stage, then the pre-final stage is calculated, etc.

Selecting the final stage circuit, transistor for it, operating mode and switching method.

In transistor amplifiers audio frequency the final stage is usually a cascade powerful gain must deliver the specified signal power to the load with the least power consumption from power supplies and acceptable level nonlinear and frequency distortions. When designing the final cascade, first of all, it is decided whether the cascade will be single-acting or push-pull. At the same time, it is taken into account that the push-pull cascade delivers twice as much more power than a one-act. Has lower harmonic distortion, an output transformer without permanent magnetization and allows three to five times the power supply ripple, but requires two transistors, an output transformer with twice the number of turns of the primary winding and midpoint, as well as the inverse circuit of the previous cascade. In addition, the push-pull circuit allows the use of an economical mode. This greatly reduces the required power of the amplifier's power supply. When switched on with a common emitter and a common collector, the transistors in the arms of a push-pull circuit must be selected with the same values, and also, if possible, with the same cutoff frequency.

A single-ended stage has one transistor and can only be used in A, which increases the power of the power source. It does not require an inverse circuit in the previous stage, allows less power supply ripple, and has a higher harmonic distortion. The dimensions of the output transformer for such a stage are larger due to the presence of permanent magnetization.

Electrical circuit diagram of ULF

Rice. 1

Calculation problems.

To calculate a transistor amplification stage, you must have the following data: output power P. out, load resistance R .n., permissible harmonic distortion coefficient K. g, lower and higher operating frequencies f n and f in, permissible frequency distortion coefficients of the cascade Mn and Mv, lowest and highest temperature environment T ambient Max. And Tokr. min. In addition, the type of power source (AC, dry cell batteries, rechargeable batteries) must be known. The calculation of the amplifier stage includes: selection of the power source voltage, if it is not specified, selection of the quiescent point (quiescent current of the output circuit), current and bias voltage of the input circuit, load resistance of the output circuit alternating current, check on weekends dynamic characteristics(load line) supplied by the power cascade P-, determining the amplitude of the current and voltage of the input signal (input power) and the input resistance of the cascade, calculating the harmonic coefficient of the cascade Kg, calculating the resistances that set the bias, and the stabilization circuit, if necessary. The calculation of the amplifier stage also includes the calculation of the electrical data of the output transformer, its structural calculation and the calculation of the radiator that cools the transistor of the powerful amplification stage.

The design of radiators that cool cascade transistors can be different. The radiator is made of metal with high heat dissipation, usually aluminum.

Determining the type of transistors.

For the amplifier stage, the transistor is selected according to three parameters: upper limit frequency f , the magnitude of the collector quiescent current I K0 , and the highest permissible collector voltage U CE add. .

Limit frequency of base current transmission f must be more than 5 times the specified upper frequency of the amplifier f V :

f  5 f V = 50000 Hz.

The collector quiescent current is selected from condition I To add. > I К0 > 1.5 I Н, , where k =20 lg (U Н / U 1 ), U Н =100 mB , I Н = U Н / R Н =100mV/10kOhm=10 µA. I To add. > I K0 > 1.5*10 mkA =15 mkA.

Amplifier supply voltage E To should be selected based on the value of the highest permissible collector voltage, i.e. less than 0.8 U CE add. .

U KE additional =30V, let's set E K =9V<0,8* U КЭ доп =0,8*30=24В.

The imported transistor meets the requirements Q 2 N 3904.

Its parameters:

f  = 250 MHz

I To add. = 100 mA >> 1.5 I K = 1 mA

U CE add. = 25 V. Let's set E K = 9V< 0.8U КЭ доп. = 24В.

Selecting the DC operating mode of the transistor and calculating the ratings of the amplifier elements.

Calculation of the output stage with a common emitter:

Based on the family of output and input characteristics of the transistor, we select the operating point,

To do this, we will construct a load line: select the value of the collector current I K, I K0 =10m A, U KE =1/2* E K =4.5 B.

DesignLab R 1  E P / I kmax =418 Ohm.

Having set the parameters of the circuit, we will construct a graph in Figure 2.

Rice. 2

K0 =10m A, U KE =4.5 B.  I B = 25 μ A .

Let's plot the input characteristic of the transistor. 3.

rice. 3

Calculation.

Load Resistance: R H = 10 kOhm.

Finding the amplitude of the output signal: K =20 lg (U N / U 1), we express U N, U N =250 mB.

Quiescent collector current: I K0 =10m A .

According to the input characteristic Fig. 3 we find: quiescent base current, quiescent voltage between base and emitter: U BE0 =0.667 B, I B0 =0.05m A.

r ВХ =  U /  I =(0.680-0.654)/(0.078-0.03)=0.8 kOhm.

Resistance in the collector circuit R K calculated: R K = (E P - U KE)/ I K0 = (9-4.5) V/10 m A = 450 Ohm.

Calculate the resistance in the emitter circuit R E . To do this, first of all, let’s set the voltage drop across it: U Re =0.2 E P =1.8V

Hence R E2 = U R E / I E0  U R E / I K 0 = 180 Ohm.

R 4 : R 4 =(E P - U E0 - U BE0)/(I B0 + I D), where U E0 =0.2 E P =1.8V. I D =(2-5) I B0 =0.15m A.

R 4 =(9-1.8-0.667)/(0.05+0.15)=32.6 kOhm.

Divider resistance: R 5 = (U E0 + U BE0 )/ I D = (1.8 + 0.667)/0.15 = 16 kOhm.

R OUT =450 Ohm.

R IN = [(R 4 R 5 )/(R 4 + R 5 )]* r IN /[(R 4 R 5 )/(R 4 + R 5 )]+ r IN =(10.8*0.8)/(10.8+0.8)=0.7 kOhm.

Calculation of the input stage with a common collector:

Based on the family of output and input characteristics of the transistor, we will select an operating point; for this we will construct a load straight line: we select the value of the collector current I K, I K0 =5m A, U KE =1/2* E K =4.5 B.

Let's construct the output characteristic of the transistor, for this purpose in DesignLab Let’s implement a circuit for connecting a transistor with a common emitter, where R 1  E P / I kmax =850Ohm. Having set the parameters of the circuit, we will draw a graph, Figure 2.

rice. 4

The working point has the following coordinates I K0 =5m A, U KE =4.5 B,  I B =25mk A.

Let's plot the input characteristic of the transistor Fig. 4.

Calculation

Load Resistance: R H =0.7 kOhm.

Quiescent collector current: I K0 =5m A .

According to the input characteristic Fig. 4 we find: quiescent base current, quiescent voltage between base and emitter: U BE0 =0.650 V, I B0 =0.025mA.

Collector Quiescent emitter voltage: U KE0 =(0.4-0.45) E P =0.4*9=3.6V.

The input resistance of the transistor is characterized by the resistance of the base emitter circuit: r VХ =  U /  I = 1 kOhm.

Emitter resistance R 3: R 3 =(E P - U KE0)/ I E =(9-3.6)/5m=1kOhm.

Calculating the resistance of the divider R 2 : R 2 =(E P - U KE0 - U BE0 )/ I B0

R 2 =(9-3.6-0.650)/0.025=190 kOhm.

Stage output impedance: R OUT = R E  r K(E), where r E =  T /(I K0 + I B0), r E =26/(10+0.025)=2.6 Ohm, R OUT =1000*2, 6/(1000+2.6)=2.6 Ohm.

Cascade input impedance: R ВХ =(1+ )(R 3 * R Н)/(R 3 + R Н).

rice . 5

where  =  h 21 E min * h 21 E max =  400 * 1000 = 632. R VХ =(1+632)(1*0.7)/(1+0.7)=260 kOhm.

Input impedance of the amplifier stage: R ВХ = R ВХ  R 2 = (260*190)/(260+190) =110 kOhm.

Calculation of containers C 1, C 2, C 3, C 4.

To calculate coupling capacitors C 1, C 2, C 3 it is necessary to set the frequency distortion coefficient at the lower operating frequency M HP introduced by this capacitor, distributing the given permissible distortions M N = 1.41 dB between separation. WITH R. and blocking C 4 capacitors.

Blocking capacitor. S: S E = (10 20)/2  f N R 7, where f N = 10 Hz.

S E = 10/6.28*10*180=884 µF.

Coupling capacitors C 1, С 2, С 3: С 1 =1/(2  f Н *(R 1 + R ВХ)*  М Н 2 -1), where М Н =1.41, С 1 =1/(6.28*10*142010*0.994)=112nF.

C 2 =1/(2  f N *(R OUT + R IN )*  M N 2 -1), C 1 =1/(6.28*10*8405*0.994)=1.2 µF.

C 3 =1/(2  f N *(R OUT + R IN )*  M N 2 -1), C 1 =1/(6.28*10*10002.6*0.994)=1.6 µF.

Amplifier equivalent circuit.

			Rк1	rin1	rin2	Rout1	Rout2	Cр1	Cр2	h21Ib1	h21Ib2
0.2 MOhm	32.6 kOhm	16kOhm	0.4kOhm	1kOhm	0.8kOhm	2.6Ohm	450Ohm	112nF	1.2uF	10mA	20mA

Calculation of amplifier frequency response and phase response.

To construct the frequency response and phase response characteristics, in DesignLab Let’s implement the circuit of the low-frequency amplifier stage, which is shown in Fig. 6

Rice. 7

Having set the values ​​of the elements, go to the dialog box and select the menu Setup Mode Analysis (set parameters). On the menu Setup Mode Analysis We build graphs that are presented in Figure 7.

We observe in Figure 7 that the bandwidth is slightly narrower (does not match technical specifications), in order to expand the bandwidth we will change the capacitance of the coupling capacitors, i.e. increase. We will also change the amplitude, for this we will change the resistance R 7. Fig. 8

Frequency response phase response

rice. 8

We display graphs with new parameters.

Conclusion.

In the course project, I learned how to: calculate the frequency response and phase response of an amplifier using the obtained function, create an equivalent circuit, calculate the ratings of passive elements, and compare the results.

Bibliography.

1. Yu. A. Bulanov, S. N. Usov “Amplifiers and radio receivers” Moscow “Higher School” 1980.
2. I. P. Zherebtsov “Fundamentals of Electronics” ENERGOATOMIZDAT 1985
3. G.V. Voishvillo “Amplification devices” Moscow “Radio and Communications” 1983.
4. I. P. Stepanenko “Fundamentals of the theory of transistors and transistor circuits” “Energy” Moscow 1967.
5. A. V. Tsykina “Design of transistor amplifiers” “Communication” Moscow 1965.

Siberian State Automobile and Highway Academy

Department of APP and E

COURSE PROJECT

“CALCULATION OF A TRANSISTOR AMPLIFIER

ACCORDING TO A COMMON EMITTER SCHEME”

in the discipline: “Electrical Engineering”

Option-17

Completed: art. gr. 31AP

Tsigulev S.V.

Checked by: Denisov V.P.

1. Basic concepts

2. Purpose of elements and principle of operation of the amplifier stage according to the circuit with OE

3. Work assignment

4. The procedure for calculating a transistor amplifier according to a circuit with OE

Bibliography

1. Basic concepts

Amplifiers are one of the most common electronic devices, used in automation systems and radio circuits. Amplifiers are divided into pre-amplifiers (voltage amplifiers) and power amplifiers. Transistor preamplifiers, like tube amplifiers, consist of one or more amplification stages. Moreover, all amplifier stages have common properties; the difference between them can only be quantitative: different currents, voltages, different meanings resistors, capacitors, etc.

For preamplifier stages, resistive circuits (with rheostatic-capacitive coupling) are the most common. Depending on the method of supplying the input signal and obtaining the output signal, the amplifier circuits received the following names:

1) with common base OB (Fig. 1, a);

2) with a common collector OK (emitter follower) (Fig. 1, b);

3) with a common emitter - OE (Fig. 1, c).

The most common is the OE scheme. Scheme with OB in preamplifiers is rare. The emitter follower has the highest input resistance and the lowest output resistance of all three circuits, so it is used when working with high-resistance converters as the first amplifier stage, as well as for matching with a low-resistance load resistor. In table 1 comparison is given various schemes turning on transistors.

Table 1

2. Purpose of elements and principle of operation of the amplifier stage according to the circuit with OE

There are many options for implementing an amplifier stage circuit using an OE transistor. This is mainly due to the peculiarities of setting the rest mode of the cascade. We will consider the features of amplification stages using the example of circuit Figure 2, which received greatest application when implementing a cascade on discrete components.

The main elements of the circuit are the power supply

, the controlled element is a transistor and a resistor. These elements form the main circuit of the amplifier stage, in which, due to the flow of collector current controlled through the base circuit, an amplified AC voltage at the output of the circuit. The remaining elements of the cascade play a supporting role. Capacitors are separating. The capacitor prevents the input circuit of the cascade from being shunted by the DC input source circuit, which allows, firstly, to eliminate the leakage direct current through the input signal source through the → → circuit and, secondly, to ensure independence from internal resistance of this voltage source at the base in rest mode. The function of a capacitor is reduced to passing an alternating voltage component into the load circuit and retaining a direct component.

Resistors

and are used to set the rest mode of the cascade. Because the bipolar transistor controlled by current, the quiescent current of the controlled element (in in this case current) is created by setting the corresponding value of the quiescent base current. A resistor is designed to create a current flow circuit. Together with the resistor, it provides the initial voltage at the base relative to the “+” terminal of the power source.

Resistor

is a negative feedback element designed to stabilize the rest mode of the cascade when the temperature changes. The temperature dependence of the rest mode parameters is determined by the dependence of the rest collector current on temperature. The main reasons for this dependence are changes in temperature of the initial collector current, voltage and coefficient. Temperature instability specified parameters leads to a direct dependence of current on temperature. In the absence of measures to stabilize the current, its temperature changes cause a change in the rest mode of the cascade, which can lead, as will be shown below, to the mode of operation of the cascade in the nonlinear region of the transistor characteristics and distortion of the shape of the output signal curve. The likelihood of distortion increases as the amplitude of the output signal increases.

Manifestation of negative feedback and its stabilizing effect on current