Study of the operation of a single-stage amplifier based on a bipolar transistor in linear mode. A single-stage amplifier based on a bipolar transistor in a common-emitter circuit. Fixed base current, bridge rectifier Preparation for the lesson

Laboratory work

"Single stage low frequency amplifier"

Polyanchev S., Korotkov R.

Goal of the work: Study of the circuit of a resistive-capacitive amplifier stage on a bipolar transistor and experimental determination of the main characteristics of the amplifiers.

Theoretical part.

Main characteristics of amplifiers

The amplifier increases the energy of the control signal due to the energy of the auxiliary source. Although any amplifier increases the signal power, in practice there are three groups of amplifiers: voltage, current and power. In accordance with this division, amplification factors are distinguished by direction, current and power.

Voltage gain or, also called voltage transfer coefficient K u is the ratio of the output voltage

b out =U out of the amplifier to its input voltage b input =U input
:

K u =
=
=K u e jφ ,

where K u =
- is called the amplitude-frequency response of the amplifier, and φ(ω)=φ 2 -φ 1 – the phase-frequency response.

The current and power gain factors are entered in the same way:

K I =
; K P =
.

One of the important characteristics of the amplifier stage is its amplitude characteristic: the dependence of the amplitude of the output signal on the amplitude of the input signal. Figure 1 shows the frequency (a), phase (b) and amplitude (c) characteristics of the ideal (dashed curve) and real (solid line) amplifiers.

In Fig. 1a Δω=ω B -ω H is called the bandwidth of the amplifier stage.

The amplifier stage is an element of some radio circuit - a signal source is connected to the amplifier input, and a load is connected to the output. To match the amplifier with the named elements, as well as to analyze the operation of the cascade, it is necessary to know the input and output impedance of the amplifier.

The input impedance of an amplifier is the resistance between its input terminals when exposed to the amplified signal, i.e.

Zin =, similarly the output resistance of the cascade Zout =
.

2. Setting the transistor operating mode for direct current with a fixed base current in circuits with a common emitter

The circuit of a single-stage amplifier suitable for practical use is shown in Fig. 2.

Capacitors 1 and 2 serve to separate the amplifier and the signal source by direct current. Resistances Rb and Rk are required to create the necessary constant voltages between the electrodes of the transistor. In this case, the absolute values ​​of the voltages at one or another terminal, as a rule, do not play a significant role; only the relative values ​​are important. Only after creating the required constant voltages between the individual electrodes of the transistor, or, as they say, the DC circuit mode, is normal operation of the amplifier stage possible.

To select the operating mode of a transistor, you need to know the family of its output characteristics, i.e. dependence of the collector current Ik on the emitter-collector voltage U eq, for various fixed values ​​of the base current Ib. You will also need the value of the base current transfer coefficient for the selected transistor:

β=
.

Figure 3 shows the family of output current-voltage characteristics (CV characteristics) of the transistor and the load straight line.

The intersection points of the load line, the equation of which is given by the choice of E and R to, determine the constant currents and voltages in the circuit for a fixed base current. In Fig. 3, the selected values ​​are marked with asterisks. Point B is called the operating point. In the vicinity of this point, changes in voltages and currents will occur when an alternating signal is applied to the input of the amplifier.

The choice of operating point is dictated by obtaining minimal nonlinear distortion and maximum dynamic range of input signal amplification.

The magnitude of the power source voltage E is determined by the specified value of the variable component U out. Since U out =U 0out cosωt, it should be E>2 U 0out. From above, the voltage value of the power source is limited by the maximum permissible value U ke max > E ≥ (2U 0out +1).

The choice of Rk must be made so that the working section of the load line does not fall into the region of unacceptably high dissipation powers and the region of electrical breakdown. After the choice of F and R k has been made, the base current I b * should be fixed using resistance R b * so that the operating point is point B, which, as can be seen from Fig. 3, satisfies the requirement of minimal nonlinear distortion and maximum dynamic range of the amplifier. Point B corresponds to U ke *= and I to *=
. Let's find the value of R b required for the selected operating mode of the transistor. The emitter junction is connected in the forward direction, the collector junction in the reverse direction.

Typically, in circuits with OE on low-power bipolar transistors, U be is tenths of a volt with values ​​of E units of volt, so we can calculate with great accuracy

I b = .

Since I k = β I b, then

R b = =
=
=2βR k.

Finally, R b =2βR k, it is clear that R b depends on the transistor parameter β.

A circuit with a fixed base current requires a minimum of parts and is characterized by low current consumption from the power source, since R b is large.

However, due to the spread of transistor parameters (β), when changing a transistor, R b must also be recalculated. Another significant disadvantage is the low temperature stability of the circuit.

The introduction of a small resistance R e, R e into the Emitter circuit makes it possible to quite simply adjust the values: R in (increases with the introduction of R e), K U and K p (decreases with the introduction of R e). The position of the operating point practically does not change (if R e

3. Feedback in amplifiers

The connection that ensures the return of part of the signal energy from the output of the amplifier to its input is called feedback (FE). The block diagram of the amplifier with OS is as follows:

Here K is an amplifier with gain K; g – feedback circuit; U x – feedback voltage; U g – input signal; Uin is the voltage that controls the transistor.

That part of the circuit that produces U in from U g and U x is called a summing node.

The feedback coefficient is the ratio:

w=
.

If the phases of the input signal and the feedback voltage coincide, then such feedback is called positive, if the phases of the named voltages are opposite - negative. Amplifiers implement negative feedback. Let's consider the effect of feedback on the gain of an amplifier. A-priory,

K=; K OS =
; f=.

Using these definitions, we can write the following chain of equalities

=k(U g +U l)=k(U g + lU out).

Where does it come from?

U out (1-kzh)=kU g

The value (1- kzh) is called the feedback depth.

From the last equality it follows that with negative feedback k = -kzh and K OS =
, i.e., negative feedback reduces the gain.

Practical part

Exercise 1: Investigate the amplitude characteristic of the amplifier U out = f(U in).

R k = 1 kOhm; R e = 100 Ohm; R 1 = 10 kOhm; R 2 = 750 Ohm; C p = 1 µF; f = 1 kHz.

Tables for the graph:

Graph 1. Dependence of U out on U in for an amplifier

Task 2: Investigate the dependence of the gain K ac on frequency for an amplifier.

1) Without capacitor.

Tables for the graph:

2) Ccapacitor(S E = 10 µF)

Tables for the graph:

Graph 2. Dependence of Kus on f with a capacitor (top) and without a capacitor (bottom)

Task 3: Investigate the dependence of the gain on the resistance value R E for the middle frequency (f = 1 kHz).

Chart table:

Graph 3. Dependence of Кус on R E

Graph 4. Dependence of Кус on RК

Conclusion: In this work, we became familiar with the principle of constructing an RC amplifier stage, its main characteristics and the purpose of the elements. The equivalent circuits of the transistor and amplifier were examined in detail. The main characteristics of the transistor low-frequency amplifier are obtained. The influence of emitter and collector resistances on the gain was studied. There are no discrepancies with theories.

Literature

1. V.N. Ushakov. “Fundamentals of radio electronics and radio engineering devices.” M., "Higher School", 1976.

2. E.I. Manaev. “Fundamentals of radio electronics.” M., “Radio and Communications”, 1985.

Introduction

A transistor is a semiconductor electronic device that controls current in an electrical circuit by changing the input voltage or current. But in essence, this is an ordinary switch that turns the current on and off, on which, by the way, the computer code is based, where 1 means that there is current, and 0 means that there is no current. We owe the invention of this device to the American laboratory Bell Labs, where William Shockley, John Bardeen and Walter Brattain created it back in 1947. But as always happens with great inventions, initially it was not noticed by the public, and only 9 years later scientists received the Nobel Prize in physics. The very name “transistor” was coined by their colleague John Pierce, who combined it from 2 words - “transfer” - transfer and “resistance” - resistance.

The first to notice the invention were radio amateurs who used them to amplify the signal. Feeling that the invention could bring profit, the laboratory decided to license the use of transistor technology. Success was not long in coming, and already in 1956 the first portable radio receiver appeared, which was previously impossible due to the use of bulky lamps, and compact transistors easily coped with this task, which now made it possible to always carry music with you. The invention of such a portable device showed the importance and relevance of the new technology, which began to attract new inquisitive minds of inventors to this area. And 2 years later, Jack Kilby and Robert Noyce took a giant step in the development of transistors, using their new technology they combined them into one chip. This revolutionary step introduced Noyce to Gordon Moore, with whom he created Intel in 1968.

It was the microcircuit based on transistors that marked the beginning of a new stage in electronics, and it was it that made the emergence of modern computers possible. In 1965, one of the publications formulated “Moore’s Law,” which stated that the number of transistors in a microcircuit should double every year. This law is constantly predicted to die, but for more than forty years it continues to work. For example, the first Intel 4004 processor, released in 1971, had 2,300 transistors, and by 1989, the Intel 486 already had 1,200,000 of them. Thus, bypassing many obstacles along the way and constantly improving, the latest Intel Core 2 Extreme processor surpassed the mark of 820,000,000 transistors.

Thus, for more than sixty years, one small invention continues to move technology forward, constantly raising it to a new level. And it is probably impossible to imagine what the world would look like without this small device.

Coursework assignment

Determine the nodal potentials in the circuit. Construct the transfer characteristic of the circuit in the base-collector section of the transistor) and plot the operating point on it. Indicate on the characteristics the area of ​​operation of the transistor.

Estimate by calculation the main small-signal parameters of the circuit under consideration.

Determine the operating area of ​​the amplifier without nonlinear distortion based on the input and output current-voltage characteristics of the transistor.

Construct a circuit diagram with node potentials, transfer, transition, families of input and output current-voltage, amplitude-frequency characteristics using an application program for computer modeling and research of electronic circuits (Electronics Workbench, Multisim, Micro-Cap).

Compare the results with those obtained by calculation.

Fig.1

Table 1 Initial data

			Transistor type

Parameters of the KT3102G transistor.

Silicon transistor, n-p-n structure.

Table 2 Parameters of the KT3102G transistor

Designation	Meaning	Parameter
		Static current transfer coefficient of a bipolar transistor in a common emitter circuit (the range of permissible values ​​is given)
		Amplifier cutoff frequency
		Collector junction capacitance (Ck) at collector voltage (Ukb)
Uke.us/(Ik/Ib),		Collector-emitter saturation voltage (Uke.s) of a bipolar transistor at a given collector current (Ik) and a given base current (Ib)
Ube.us/(Ik/Ib),		Collector-emitter saturation voltage (Ube.sat) of a bipolar transistor at a given collector current (Ik) and a given base current (Ib)
		Reverse collector current
		Emitter reverse current
		Maximum permissible emitter-base DC voltage
		Maximum permissible DC collector-emitter voltage
		Maximum permissible collector current
		Maximum permissible power dissipation at the collector

Table 3 Series of nominal values ​​of parameters of typical radioelements (GOST 2825-67)

Row index	Numerical odds multiplied by 10

Since the course work will use the Workbench 5.12 application, which does not contain the KT3102G transistor, we will instead use its foreign analogue BC109C, which is similar in parameters. Therefore, the calculated values ​​may differ from the values ​​obtained using the Workbench application.

The static gain of the base current is chosen to be equal to 500.

Since the transistor is silicon, the contact difference of both junctions is equal - the value of the base-emitter voltage. Since it is a constant voltage source, the circuit can be simplified by removing all capacitors and unnecessary resistors. We also remove the alternating voltage source from the circuit and get the circuit shown in Fig. 2

Fig.2

Let's assume that the transistor is in the normal active region. Considering that the operating point is in class A, let's calculate the collector voltage.

For small-signal circuits, the voltage across Re is 5-30% of the voltage Ek, so we’ll choose 10%.

Let's determine the resistance and, for this, calculate the emitter current, using for this the amplification factor of the emitter current, expressed through the amplification factor of the base current:

By condition b=500, then

Let's calculate the base current in the same way:

We get:

If we neglect the base current, then in the section A-B a current flows equal to the ratio:

From expressions (2) and (3) it follows that

Let's find the base resistance. To do this, we need the coefficient of the unstable operating point of the cascade, expressed as:

From here we calculate the resistance value RB, which is also equal to the parallel connection of resistors R1 and R2.

Solving the system of equations (4) and (5) we find R2 and R1

We get:

We take the nominal values ​​of the resistors in accordance with the E24 series, then we get:

Task 2

Let's consider the node potentials in the circuit. Construct the transfer characteristic of the circuit in the base-collector section of the transistor) and plot the operating point on it. Indicate on the characteristics the area of ​​operation of the transistor.

Let's consider the nodal potentials in the circuit shown in Fig. 3.

Fig.3

Let's find the potential difference at the emitter:

single stage amplifier bipolar transistor

Let's find the potential difference on the base:

Let's find the potential difference across the collector:

We obtained the nodal potentials:

To construct the transfer characteristic, we will use the Workbench 5.12 application. In order to build a relationship, you need to install two voltmeters in the circuit: the first one is for reading the base potential, placed between the base and the ground, the second one is for reading the collector potential, placed between the collector and the ground. Also, in order to regulate the base potential, an EMF source connected to the base is introduced into the circuit (Fig. 4).

Fig.4

Fig.5

The transfer characteristic (Fig. 5) shows the operating point (PT) corresponding to the values:

Task 3

Estimate by calculation the main small-signal parameters of the circuit under consideration. And also at what amplitude of the input signal nonlinear distortions will occur in the circuit.

Fig.6

Initial data:

For a transistor, the resistance of the p-n junction is:

We accept

Let's calculate the input resistance in a circuit with a common emitter:

Let's calculate the current gain:

Let's find the resistance when the load is connected in parallel with the collector resistance:

Let's calculate the voltage gain:

Let's calculate the power gain:

Let's calculate the input impedance of the circuit:

Let's calculate the output resistance of the circuit:

Let's calculate:

Task 4

It is necessary to find out at what amplitude of the input signal nonlinear distortions will occur in the circuit. The output signal amplitude cannot be greater than.

Let's find the effective value of the amplitude of the input signal:

Let's plot the output current-voltage characteristics of the transistor - (we take it from the reference book in electronic form) (Fig. 7).

Fig.7

At the output I-V characteristics of the transistor we will plot the operating point, as well as the load line for direct (A-B) and alternating current.

We will construct a DC load line based on two extreme cases.

First case (A): transistor fully open

Second case (B): the transistor is completely closed

In order to construct the operating point on the current-voltage characteristic, a straight line should be drawn at the level before the intersection with the static load straight line. This intersection will be the working point.

The direct line in alternating current has a slope and passes through the operating point. Since the scale of the OY axis is in miles Amperes, the resulting value b must be multiplied by 1000.

Task 5

Based on information about the lower limit frequency of the amplifier's passband, taking into account data on the load resistance and the signal source, determine the capacitance of the isolation and blocking capacitors.

Considering that

Let's find the capacitances of the separation (Cp1 and Cp2) and blocking (Sbl) capacitors.

When calculating the time constant φ for each capacitor, we will take into account only this capacitor, considering that other capacitors replace the corresponding points in the circuit.

We obtain the following equivalent circuits for calculating time constants.

Fig.8

First, let's calculate the time constant for the lower frequency:

Let us assume that all time constants are equal to each other:

Let's calculate the values ​​of and, as well as:

We get:

We take the nominal values ​​of the resistors in accordance with the E24 series, then we get:

Task 6

Construct the amplifier's frequency response and phase response, from which the limiting frequencies of the amplifier's passband can be determined.

Let's calculate the upper limit frequency of the amplifier passband. For this we need the at parameter.

The upper limit frequency of any amplifier stage is determined by formula (8).

The G coefficient for a cascade with a common emitter is determined by formula (10).

Let us determine the average lifetime of minority charge carriers in the base:

Let us determine the equivalent capacitance of the collector junction:

Junction capacitance at zero bias;

Contact potential difference, which is equal to 0.7 V;

Transition voltage.

Let's find the bandwidth:

Let's build the frequency response and phase response for a single-stage amplifier. To do this, we will use the Workbench 5.12 application. You need to add a pulse generator (Function Generator) to the circuit, and you also need to connect the Bode Plotter to the circuit in such a way that its input is connected to one of the terminals on the input of the circuit, and the output to one of the output terminals of the circuit (Fig. 9).

Fig.11

Conclusion

In the course of the course work, calculations were made of the main parameters of the single-stage amplifier BC109C. We determined the resistance of the resistors included in the circuit, the separating capacitances Cp1 and Cp2 and the blocking capacitor Sbl. As well as small-signal circuit parameters Kuo, Kio, Kp, Rin, Rout.

Bibliography

1) Gusev V.G., Gusev M.Yu. Electronics. -M.: “Higher School”. 1991 -622 pp.: ill.

2) Rekus G. G., Chesnokov V. N. Laboratory work on electrical engineering and fundamentals of electronics: Textbook. manual for non-electrical engineers. specialist. universities - M.: Higher. school, 1989. - 240 pp.: ill.

3) Lachin V.I., Savelov N.S. Electronics: Textbook. allowance. - Rostov n/a publishing house "Phoenix", 2000. - 448 p. Application Software: Electronic Workbench Pro Edition

In amplifiers based on bipolar transistors, three transistor connection schemes are used: with a common one, with a common emitter, with a common collector.

In a transistor circuit with a common emitter, the amplifier provides voltage, current, and power amplification. Such an amplifier has average values ​​of input and output resistance compared to switching circuits with a common base and a common collector.

Transistor parameters largely depend on temperature. A change in ambient temperature leads to a change in the operating mode of the transistor in a simple amplifier circuit when the transistor with a common emitter is turned on.

To stabilize the operating mode of the transistor when the temperature changes, emitter stabilization circuits are used to stabilize the operating mode of the transistor.

Figures 5.14 and 5.15 show circuits of single-stage amplifiers based on n-p-n and p-n-p bipolar transistors with emitter temperature stabilization of the transistor operating mode.

Let's trace the circuits through which direct currents flow in the amplifier according to the diagram in Figure 5.14. The direct current of the voltage divider flows through the circuit: plus the power supply, resistors R1, R2, minus the power supply. The direct current of the base of transistor VT1 flows through the circuit: plus of the power supply, resistor R1, base-emitter junction of transistor VT1, resistor Re, minus of the power supply. The direct collector current of transistor VT1 flows through the circuit: plus of the power supply, resistor RK, collector-emitter terminals of the transistor, resistor Re, minus of the power supply. The bipolar transistor as part of the amplifier operates in a mode where the base-emitter junction is biased in the forward direction, and the base-collector junction is biased in the reverse direction. Therefore, the constant voltage across resistor R2 will be equal to the sum of the voltage at the base-emitter junction of transistor VT1 and the voltage across resistor Re:UR2=Ube+URe. It follows that the constant voltage at the base-emitter junction will be equal to Ube = UR2 - URe.

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In a transistor circuit with a common emitter, the amplifier provides voltage, current, and power amplification. Such an amplifier has average values ​​of input and output resistance compared to switching circuits with a common base and a common collector.

In rest mode, i.e. in the absence of an input signal (U input = 0), the direct current I BO under the influence of E K passes through the circuit + E K – E- B- R B - -E K. The magnitude of this current by selecting the values ​​of R B is set such that the transistor is half open, i.e. the voltage across it would be approximately half E K. In turn, with a large base current, the transistor opens completely, i.e. its resistance between the emitter and collector is very small, the voltage U EC is almost zero, and at I B = 0 the transistor is completely closed, i.e. The resistance is high and it practically does not allow current I K to pass through.

Capacitor C p1 serves to connect a source of variable input EMF E in, with internal resistance R in, to the base circuit. The coupling capacitor C p2 serves to isolate the alternating component of the collector voltage at the load Rn.

18. Determination of the initial conditions that ensure the specified operating mode of the amplifier with OE

Let's consider an RC amplifier in which the transistor is connected to a circuit with a common emitter and emitter stabilization of the initial operating mode is used.

Currents in the circuit are found using the formulas:

Suppose that i B = i B2, then:

Let us assume that the supply voltage Ek is given and it is required to ensure the initial operating mode at a given initial current I K N.

Considering that i E » i K:

The current i division of the voltage divider on resistors R 1 and R 2 is selected, flowing when the transistor base is disconnected from the divider.

An important parameter is the voltage gain of the amplifier, which is found using the formula:

19. Operational amplifiers (op-amps): areas of application, conventional graphical representation, block diagram. Purpose of block diagram elements

A typical circuit of an amplifier stage based on a transistor with an OE is shown in Fig. 11.5. The input amplified alternating voltage UВХ is supplied to the input of the transistor through the isolation capacitor CP1. Capacitor CP1 prevents the transfer of the constant voltage component of the input signal to the amplifier input, which can cause a violation of the direct current operating mode of the VT transistor. The amplified alternating voltage generated at the collector of the transistor VT is supplied to an external load with resistance RN through an isolation capacitor CP2. This capacitor serves to separate the output collector circuit from the external load by the constant component of the collector current IK0. The values ​​of IK0 and other constant components of current and voltage in the transistor circuits depend on the initial operating mode (initial position of the operating point), set in the absence of a signal.

Fig. 11.5. Bipolar transistor amplifier with OE

The operating point of the transistor is the point of intersection of the dynamic characteristics (load straight line) with one of the static current-voltage characteristics. This position is determined on the characteristics by a set of direct components of currents and voltages in the output IK0, UKE0 and input IB0, UBE0 circuits.

The operation of the amplification stage is illustrated in Fig. 11.6.

Fig. 11.6. Graphic illustration of the operation of an amplifier stage on a transistor with OE

The process of signal amplification can be reflected by the following relationship of electrical quantities:

UВХm→IБm→IКm→IКmRК→(UКЭm= EPIT - IКmRК) = UOUTm.

The figure shows that the input signal voltage with amplitude UВХm =UBЭm changes the value of the base current in phase. These changes in the base current cause proportional changes in the collector current and collector voltage in the collector circuit, and the amplitude of the collector voltage turns out to be significantly greater than the amplitude of the base voltage. The signal voltages at the input and output of the cascade are shifted in phase by 180º, i.e. are in antiphase. When the transistor is operating in active (amplifying) mode, the operating point should be approximately in the middle of the segment AB of the load straight line. The maximum changes in the base input current must be such that the operating point does not go beyond the limits of the segment AB. Figure 11.7 shows the timing diagrams of the operation of the transistor stage with the correct choice of the rest point and the magnitude of the input signal. It is very important to ensure that not only the magnitude of the input signal is correct, but also the quiescent current. With a small initial quiescent current and a minimum signal, the transistor will not open and will be in cutoff mode; with a large bias and a high signal level, it can go into saturation. Rice. 11.8. shows the voltage at the collector of the transistor: a - with insufficient bias current; b - with excess bias current; c - when the input signal is excessive.

Fig. 11.7. Timing diagrams of the operation of a transistor amplifier in a circuit with an OE

The initial position of the operating point is provided by a voltage divider consisting of resistors R1 and R2, the resistance values ​​of which are determined from the ratios: R1 = (EK - UBE0 - URE) / (ID + IB0); R2 = (UBE0 + URE) / ID, where ID = (2…5) IB0 is the current in the divider circuit.

When ensuring the operating mode of the transistor, it is necessary to carry out temperature stabilization of the operating point position (reduce the influence of temperature on the initial position of the operating point). For this purpose, a resistor RE is introduced into the emitter circuit, which creates an OOS voltage for direct current URE.

Fig. 11.8. Timing diagrams of collector voltage under incorrect conditions

The OOS in this circuit operates as follows: when, for example, the temperature of the transistor changes, the collector current increases. This causes a corresponding increase in the emitter current and voltage drop across it. Consequently, the voltage UBE = UB - UE, which is the control voltage for the transistor, decreases, the transistor is switched off, the collector current decreases and returns to the specified mode. The introduction of OOS reduces the gain of the circuit. In order for the feedback to act only on direct current and to eliminate the negative feedback on alternating current, the resistor RE is shunted with a capacitor SE, the resistance of which at the frequency of the amplified signal should be insignificant. When analyzing the circuit, we can assume that there is no environmental protection for alternating current. In this case, the current stage gain