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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, at high current base, 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, conditional graphic image, structural scheme. Purpose of block diagram elements
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Principle of operation transistor amplifier is based on the fact that with the help small changes voltage or current in the input circuit of the transistor, you can get significantly larger changes in voltage or current in its output circuit.
A change in the voltage of the emitter junction causes a change in the transistor currents. This property of a transistor is used to amplify electrical signals.
To convert changes in the collector current that occur under the influence of input signals into a changing voltage, a load is connected to the collector circuit of the transistor. The load is most often a resistor or an oscillating circuit. In addition, when amplifying alternating electrical signals between the base and emitter of the transistor, it is necessary to turn on a constant voltage source, usually called a bias source, with the help of which the operating mode of the transistor is set. This mode is characterized by the flow through its electrodes in the absence of an input electrical signal of some direct currents of the emitter, collector and base. With the use of an additional source, the size of the entire device increases, its weight increases, the design becomes more complicated, and two sources cost more than one. At the same time, you can get by with one source used to power the collector circuit of the transistor. One such amplifier circuit is shown in the figure.
In this circuit, the load of the amplifier is resistor R K, and using resistor R b, the required base current of the transistor is set. If the operating mode of the transistor is set (it is often said that the operating point on the characteristics of the transistor is set), the base current and voltage U BE become known, and the resistance of the resistor R b, which provides this current, can be determined by the formula:
R b =(G K -U BE)/I B.
Since U BE is usually no more than 0.2...0.3 V for germanium transistors and 0.6...0.8 V for silicon ones, and the voltage G K is measured in units or even tens of volts, then U BE<
R b ≈G K /I B.
From the expressions it follows that, regardless of the type of transistor VT, its base current will be constant: I B = G K / R b. Therefore, this scheme was called common emitter (CE) circuits and fixed base current.
The operating mode of the transistor in the amplifier stage at constant currents and voltages of its electrodes is called the initial, or resting mode.
Including a load in the collector circuit of the transistor results in a voltage drop across the load resistance equal to the product I K R K .
As a result, the voltage acting between the collector and emitter Uke of the transistor turns out to be less than the voltage G K of the power source by the amount of the voltage drop across the load resistance, i.e.:
U KE =G K -I K R K .
If this dependence is displayed graphically on a family of static output characteristics of the transistor, then it will look like a straight line. To construct it, it is enough to determine only two points belonging to it (since only one straight line can be drawn through two points). Each point must be specified by two coordinates: I K and U CE.
Having given a specific value for one of the coordinates, the second coordinate is determined by solving the equation U KE = G K -I K R K . A straight line constructed in accordance with an equation on a family of static output characteristics of a transistor is called a load straight line.
The load line shown in Figure (a) is constructed for the case when G K = 10 V and R K = 200 Ohm.
1st point: =0;U KE =G K -0R K =G K =10 V;
2nd point: I K =30 mA; U KE =10—30-10^3-200=10—6=4 V.
If in the initial mode (rest mode) the base current is 2 mA, this mode will be determined by point A lying on the load line at the point where it intersects with the static output characteristic obtained at I BO = 2 mA. In this case, I KO = 20 mA; U BEO =5.8 V. If we move point A to the family of input characteristics (Fig., b), we can find U BEO. It is equal to 0.25 V.
When an alternating voltage with an amplitude of 50 mV (0.05 V) is applied to the input of the amplifier, on the voltage axis of the input characteristics relative to the voltage U BEO = 0.25 V, segments corresponding to a voltage of 0.05 V are laid on both sides, and perpendiculars are restored from their ends to the U axis of the BE until it intersects with the static characteristic, on which point A is located, indicating the rest mode of the amplifier. At the points of intersection of perpendiculars with the characteristic, the letters B and C are placed. Thus, when an alternating voltage is supplied to the input, the operating mode will no longer be determined by point A, but by its movements between points B and C. In this case, the base current varies from 1 to 3 mA. In other words, an alternating voltage at the input of the amplifier leads to the appearance of an alternating component in its input current - the base current. In this example, the amplitude of the alternating component of the base current, as can be seen from the figure, is 1 mA.
Points B and C can be transferred to the family of output characteristics. They will be located at the intersection of the load characteristic with the static ones obtained at base currents equal to 1 and 3 mA. From this figure, it is clear that in the load mode an alternating component of the collector voltage appeared. Otherwise, the collector voltage no longer remains constant, but changes synchronously
with changes in input voltage. Moreover, the change in collector voltage ΔU BE =7.5-4.3=3.2V turns out to be 32 times greater than the change in input voltage ΔU BE =0.3-0.2=0.1V; i.e., the input voltage was amplified by 32 times.
Since the voltage of the power source G K is constant, the change in the collector voltage is equal to the change in voltage across the collector load resistor, i.e. ΔU KE = ΔI K R K. From this expression it is clear that the greater the resistance of the resistor R K, the more the voltage on it changes and the greater the gain will be. However, it is possible to increase the resistance of resistor R K only to a certain limit, exceeding which can even lead to a decrease in gain and the appearance of large distortions of the amplified signal.
In the amplifier, the circuit of which is shown in the top figure, the operating mode of the transistor is determined by the base current, which is set by resistor R b. The operating mode of the transistor can also be set by applying voltage from the divider R1R2 to its emitter junction.
The divider current I D flowing through resistors R1 and R2 causes a voltage drop across the resistance of resistor R2, which is applied to the emitter junction of the transistor and biases it in the forward direction. This voltage is determined mainly by the ratio of the resistances of resistors R1, R2 and the current I D flowing through them and is almost independent of the type of transistor. Therefore, such a circuit is sometimes called a fixed-bias circuit.
Ministry of Education of the Republic of Belarus
GOMEL STATE TECHNICAL UNIVERSITY
them. P.O.SUKHOY
name of the faculty _______AIS __________________
"APPROVED"
head department _____________
"______" _____________2002
EXERCISE
in course design
Student Ilyin E.V. PE-21
1. Project theme Single-stage amplifier based on a bipolar transistor in a switching circuit
with a common emitter. Fixed base current, bridge rectifier ________________
2. Deadline for the student to submit the completed project May-2002 __________________________
3. Initial data for the project._________________________________________________ _____
_________________U n m=8.7 IN .__________________________________________________ ______
_________________R n = 19 0 Ohm .________________________________________________ ______
R To = 190 0 Ohm .____ _____________________________________________ ______
R G =240 Ohm__________________________________________________________ ______
fn=45 Hz_______________________________________________________________
1. Determine the coordinates, Ek. Draw load lines. Select transistor__
2. Identify the elements that ensure rest mode ._ ________________
3. Graphic-analytical calculation of amplifier parameters _________________________________
5. Determine the parameters of the amplifier Rin, K u , TO i through h-parameters.______________________
9. Construct timing diagrams of signals (frequency 1 kHz)___________________________
a) Er(t), Uin(t), Ub(t), Ue(t); b) I b (t), I G (t); c) Iк(t), In(t), Ipit(t);_____________________
c) Ub(t), Ue(t), Uk(t), Un(t), Ek; e) U2(t), Uв(t), Ust(t)=Ek(t).________________________
11. Draw a circuit diagram of the electrical device____________________
5. List of graphic material. ___________________________________________________
Load lines, static I-V characteristics of the transistor, timing diagrams of signals,____ ________
electrical circuit diagram of the device._________________________________________
___________________________________________________________________________________
6. Project consultants (indicating project sections).______________________________
______________________________________________________________________________________________________________________________________________________________________
7. Schedule of work on the project for the entire design period _______________
__________________________________________________________________________________
Supervisor ______________
The task was accepted for execution.
___________________________________________ (date and student signature)
Initial data
1 Class A voltage amplifier based on a bipolar transistor in a common-emitter circuit.
N =8 - option number, r =3 - fixed base current circuit.
Unm=8.7B- voltage amplitude across the load;
Rн=1900 Ohm- load resistance;
Rк=1900 Ohm- resistance of the collector resistor;
Rg=240Om- resistance of the generator (source of harmonic signal);
3 Bridge rectifier and filter.
Calculation of an amplifier in a switching circuit with OE
1 Determine the coordinates of the rest point 0, supply voltage Ek. Construct static and dynamic load lines. Determine the requirements for the transistor based on limiting parameters and current-voltage characteristics. Select transistor.
We calculate currents
Load current amplitude
Resistor current amplitude Rk
Collector current amplitude
Check to exclude opening error:
We determine the equivalent resistance in the collector circuit for the variable component I to R kn =R to êêR n = (R to R n)/(R to +R n) and the amplitude of the collector current Iкm=Uкm/Rкн.
AC Resistance 
Collector current amplitude 
The quiescent current is selected from the condition Iok>Iкm or Iok=Iкm+D I, where D I =1¸3 mA is the minimum collector current.
Collector quiescent current Iok = Iкm +D I = 12 +2= 14 mA.
To exclude the saturation mode, the quiescent voltage is determined from the condition Uoke>Uкm or Uoke=Uкm+D U, where D U =2¸3 V is the minimum voltage.
Collector-emitter quiescent voltage Uoke=Uкm+DU= 3+1= 4 V.
Determine the supply voltage:
Ek = Uoke + Iok Rk =4 + 0.014· 620 = 12.68 » 13 V.
The static load line (SLL) passes through the points with coordinates, And .
Voltage U A is the point of dynamic load, the straight line that passes through [ Uoke; Iok]
U A =Uoke+IokRkn= 4 + 0.014 · 250 = 7.5V.
The dynamic load line (DLL) passes through the points with coordinates, And .
After constructing the load lines, the limiting parameters of the transistor are determined:
Ik max > UA/Rkn or Ik max > Iok + Ikm, Ukemax > Ek, Rkmax > Iok × Uoke.
Based on the calculated data, we select the transistor from the reference book.
The transistor is selected according to the following principle:
I to max >I ok + I km = 14 + 12 =26 mA
U ke max >Ek=13 V
P to max > U oke × I ok = 14 × 4 = 56 mW
The calculated data is satisfied by the KT312A transistor (for the main parameters, see Appendix A).
Let's build static and dynamic load lines on a separate sheet, having previously transferred the input and output characteristics of the selected transistor.
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. Temperature change environment leads to a change in the operating mode of the transistor in a simple amplifier circuit when a 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 bipolar npn transistors And p-n-p types 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. D.C 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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