- Electronics (Component Base and Operating Principles)

Это окончание невероятной информации про электроника.

...

>


Timing diagrams of the asynchronous RS flip-flop


Synchronous (clocked) RS flip-flop

Symbol


Suppose Q="1" initially; a change to "0" is possible when both R and C are "1"; if Q=1 is needed, then S=1 and C=1, etc.

Timing diagrams

D flip-flop

D flip-flop – (data delay flip-flop) - a synchronous flip-flop whose output state matches the signal that was on its information input (D input) during the previous clock cycle
Symbol and truth table for edge-triggered operation



Timing diagrams


The D flip-flop delays by 1 clock cycle the information present at input D.
Registers can be built from D flip-flops; to fill an 8-bit word, 8 D flip-flops are needed.
Information in D flip-flops is stored until permission arrives to change the information, at which point a new number is written

T flip-flops

A counting flip-flop (T flip-flop) changes its state each time there is an active signal level at its single information input T.

Symbol

The T flip-flop is a frequency divider by 2.

Timing diagrams


Truth table

JK flip-flop (universal)

Symbol Truth table


Timing diagrams




If J and K are connected together, we get a T flip-flop. T flip-flop, at C=1


D flip-flop built from a JK flip-flop

Digital counters

A pulse counter is a device designed to count the number of pulses applied to its input. The counting result is recorded in binary code
A digital circuit that performs the counting function can be assembled from flip-flops and logic elements.
Counters are built on the basis of JK or T flip-flops
The bit width of a counter is determined by the maximum number up to which it counts in binary code
to build a modulo-16 counter, 4 flip-flops are needed

CT2 – a two-bit binary counter


Outputs: 1, 2 – designation of binary digits
C1 – clock input
R - reset to 0

Counters are used:
• for counting numbers, pulses, time intervals,
• for ordering sequences,
• for addressing,
• for building frequency dividers,
• for building memory elements.

Modulo counters
The modulus of a counter shows the number of distinct states the counter passes through during one full counting cycle.

Example 1. For modulo-5 counting, the cycle is the sequence of binary numbers: 000, 001, 010, 011, 100 (i.e. 0,1,2,3,4).

Example 2. At the output of a 4-bit binary counter, 4 binary digits are activated; such a counter counts from 0000 to 1111 (i.e. from 0 to 15).

Asynchronous decimal counter. (Modulo-ten counter)


Counts from 0000 to 1001 (0 to 9), i.e. 4 binary digits are needed - 8 4 2 1; consequently, the counter is implemented on four JK flip-flops and a NAND element. The flip-flops are in toggle mode J=K=1. The clock pulses trigger only the first flip-flop T1, flip-flop T1 triggers T2, and so on.
Such a flip-flop chain is called a serial (ripple-carry) counter.
Each higher-order bit switches twice as slowly as the previous one. The NAND logic element resets all flip-flops to "0" when the tenth pulse arrives (1010), i.e. when logic "1" is applied to D and B on the NAND element, all flip-flops are cleared, and the counter starts counting pulses again; NAND provides the reset.

Synchronous counters

To increase the speed of digital devices, it is necessary for counters to operate synchronously with the clock pulses. This task is implemented in synchronous counters.

Let us consider a 3-bit modulo-8 counter.

JK flip-flops operate in toggle mode and in hold (blocking) mode

Registers

A register is a digital device designed for the temporary storage of a numeric code and its conversion.

Symbol of a four-bit register

L1&L2 – write;
E1∨E2 – output disable;
The basic elements of a register are binary cells, which are D flip-flops.
Registers can be built on synchronous RS flip-flops and JK flip-flops.
To create an 8-bit number, 8 flip-flops are needed

Types of RG registers

Serial RG
• Shift registers are registers in which data is entered bit by bit.
• For example, serially loading the 4-bit combination 0111 into a serial shift register takes 5 clock cycles.

Serial shift register – Timing diagrams


Parallel RG

A parallel register performs parallel data loading, in which all information bits (digits) are entered into the register simultaneously by a single clock pulse. A parallel register can be made into a ring register, in which case information is not lost during movement.

Write at C=1 Qi = Di C = 0

Qi is retained until a new number is written
To write a new number, "1" must be applied to C

Digital-to-analog converters (DACs) and analog-to-digital converters (ADCs)

Main parameters of DACs and ADCs

• Resolution, expressed in bits and characterizing the measurement range of the input quantity.

• Transfer-coefficient error, showing the difference between the actual and prescribed values (in units of the least significant bit).

• Linearity of the characteristic, i.e. the presence of proportionality between the analog reference quantity and the code corresponding to that quantity, in units of the least significant bit.

Possible applications of DACs and ADCs



Digital-to-analog converters


where z0, z1, z2, z3 are coefficients taking the value "0" or "1" depending on whether the corresponding switch is closed or open.
A 4-bit binary code is converted into an output voltage level in the range (0÷15) ΔU, where ΔU is the quantization step. The smaller ΔU is, the higher the resolution of the DAC

Analog-to-digital converters

ADC symbol


An ADC – analog-to-digital converter – is a special type of encoder.
Operating principle of the ADC: the successive counting method is used.

Operating principle of the ADC


The pulse generator PG produces a sequence of pulses, which are converted into a binary code by counter Cnt. This code controls the switches of the DAC. The DAC's output voltage is fed to one of the comparator's inputs. When the voltages Uin and Udac are equal, the comparator outputs a signal that stops the operation of the PG. At the counter's output, a binary code corresponding to the voltage Uin is fixed

Microprocessor

A microprocessor is a universal digital IC (a type of LSI) capable of performing the full range of functions of a computer's central processing unit.

With the advent of microprocessors, there was no longer a need to design a new IC for each new application
Main manufacturers of modern microprocessors: microprocessor manufacturers: IBM, Intel- Pentium, Motorola, AMD, "National Semiconductor", Apple (USA) Macintosh

Block diagram and composition of a microprocessor system


Types of microprocessor buses
Address bus – AB (16);
Data bus – DB (8);
Control bus – CB (10-12), implemented as a set of conductors connecting the main elements of the microprocessor to each other;
CPU – central processing unit


ROM – read-only memory, for storing control programs and a set of initial data for organizing the information-processing process
RAM – random-access memory, for storing the changing part of the data being processed
I/O devices – input/output devices, ports for data exchange between the microprocessor and the keyboard, monitor, printer, etc.

Internal architecture of a microprocessor



ALU – arithmetic logic unit, implementing the set of arithmetic and logic functions of the microprocessor.
Timing and control unit – receives and generates external control signals.
Register set – for temporary storage of instruction codes, data, addresses, and information about the internal state of the microprocessor.
The contents of some internal microprocessor registers can be changed, while another part of them is not accessible to the programmer.

IR (instruction register) – receives the code of the current instruction from the data bus and holds it for the entire time it is executed. The instruction register is not accessible to the programmer.

General-purpose registers – for storing data and intermediate results.

Accumulator – used when the microprocessor has only a single special register for storing data and intermediate results.

Stack pointer register SP – for storing the address of the last unused stack cell (a memory area whose size changes during processing). The stack operates on a "last in, first out" basis.

PC – program counter.

Flag register – status register or condition-code register, programmatically accessible.

All operations in the microprocessor are triggered by clock pulses from an external clock-pulse generator, stabilized by a quartz crystal resonator.

The clock frequency indicates the speed of the microprocessor. The first microprocessors had a speed of 2-4 MHz, modern ones – 3 GHz

Additional tools – coprocessors
• Arithmetic coprocessor;
• Cache memory;
• Microcontroller;
• Graphics microprocessor.

Basic microprocessor instructions
• data transfer instructions (from memory to register and vice versa, transfer from register to register);
• arithmetic and logic instructions (addition, subtraction, etc., and logic operations AND, OR, NOT, etc.);
• branch instructions from one point in the program to another (conditional branches);
• instructions for controlling the operation of the microprocessor; input-output, working with the stack, etc.
Writing, reading, and debugging programs is simplified when using machine-oriented instruction notation – assembly language or BASIC.
8-bit microprocessors are intended for building microcontrollers on their basis – simple microprocessor systems for use in industry, everyday life, as control systems

Digital storage devices

Flip-flop-based memory registers


CS - an always-active inverse chip-select input.
The RG register stores 8 bits of information.
If 8 registers are placed, we get 8 bytes of memory
On the control bus: "0" - read RD, "1" - write WR

Parallel register


If many registers RG1-RG8 are used, a positional decoder DC is needed to select the register for reading or writing information.
For example: to read data from RG1, "1" is applied to input RDRG1, and the address selection code for RG1 goes to the address bus, i.e. "0" appears at output cs1 of the decoder while all others are "1".
By combining a decoder and registers in one IC, a read-only memory chip is obtained.
Such ICs have a single data bus, which is shared using the RD/WR signals, i.e. the data bus is bidirectional.

In processors the data bus always has 8 lines, in computers 16, 32, 64, modern coprocessors 96
In an MSP with 4096 addresses, with a 4096×8 configuration, the control bus usually has 10-50 lines

Example 1: address bus for 64 registers, configuration 64×8


Example 2
Determine the information capacity n of an MSP if the number of address inputs m=10

Main characteristics of memory devices

• information capacity;
• speed;
• data retention time.
1. Information capacity N – the number of memory bits in the storage device;
2. Number of words n – the number of word addresses in the storage device;
3. Word width m – the number of bits per word in the storage device;
4. Number of reprogramming cycles Ncy – the number of write-erase cycles for which the memory device remains operational;
5. Power consumption in steady-state mode Pcc;
6. Power consumption in standby (storage) mode Pccs;
7. Retention time tsg.

Static parameters of memory devices
• Supply voltage Ucc
• Current consumption Icc
• Current consumption in standby mode Iccs
• Supply voltage in standby mode Uccs
• Logic-high voltage Uh
• Logic-low voltage UL

Dynamic parameters of memory devices:
• chip-select time;
• address-access time;
• signal-access time.

Classification of memory devices:

by functional purpose
• RAM
• ROM
• PROM
• EPROM
• EEPROM
• FLASH

by information storage method
Static
Bistable cells (transistors), non-destructive readout of RAM, 10000 cells
Dynamic
The inertial properties of reactive elements (capacitors) are used to store information (periodic recharging/regeneration is required)

by manufacturing technology
• semiconductor memory based on bipolar transistors (TTL, ECL);
• semiconductor memory based on insulated-gate field-effect transistors (CMOS);
• charge-coupled semiconductor memory;
• with MNOS structure (metal - silicon nitride - Si oxide - semiconductor).

by memory array access method
• random access, which allows any order of address sequencing;
• serial access, in which memory elements can only be accessed in increasing or decreasing order of addresses (shift registers)

Random-access memory devices, RAM

A digital information storage device combined with control circuits that ensure write, storage, and read modes of digital information during its processing.
Static and dynamic RAM
Static RAM
• The core is a storage array or memory matrix consisting of individual bistable storage cells.
• Flip-flop cells contain six MOS field-effect transistors.
• Binary information is stored in a cell until it is replaced with different information or until the supply voltage is removed.



Chip selection, row address strobe RAS, columns CAS
• An integrated 1-Kbit RAM chip contains 1024 flip-flops.
• Address-decoding circuits located inside the RAM chip select the specific flip-flop indicated by the address-line signals each time.

Symbol for a static RAM IC (K537RU17)


RAM solves two tasks:
1. Selecting the specific storage cell to which data will be written or from which it will be read
2. Reading or writing information

Dynamic RAM

To increase information capacity, dynamic RAM is used, in which information is stored as the charge on corresponding capacitors.
Capacitor capacitance Cst=0.1 pF, retention time 1ms.
Regeneration (refreshing) of the stored information is required with a period of no more than 1ms.
The memory cell is implemented either on bipolar transistors or on MOS transistors, without a special power source.
Addressing is sequential
Memory cell
• Information is stored on Cgs;
• Power consumption Pcons=50-500mW;
• The whole system operates in switching mode.




Symbol for dynamic RAM ICs

Read-only memory devices, ROM

Types of ROM storage:
• ROM - mask-programmed (memory matrix), programmed at the manufacturing stage.
• PROM - programmable ROM implemented on the basis of storage cells with fusible nichrome links.
• Series K500, K1500, KR556RT.
• The user programs it electrically, using a programmer; PROMs that can be programmed multiple times by the user.
• EPROM - with electrical programming and ultraviolet erasure.
• EEPROM - with electrical programming and electrical erasure.
• SEEPROM (EAPROM) - with electrical programming and selective erasure.
• A common property of ROM chips is multi-bit (word) organization, with reading as the primary mode.

Structure of an electrically programmable ROM


ROM symbol


4 address lines give 256 memory cells

• If a microprocessor system contains several memory modules, a memory map is drawn up, i.e. the distribution of the address space.
• Address space – the bit width of the address bus.
• Width 16 => 65536, the number of addressable registers or memory cells.

ROM K155RE21-24 (ROM)
• Codes of Russian alphabet letters are recorded, RE21
• Codes of Latin alphabet letters are recorded, RE22
• Codes of arithmetic symbols and digits are recorded, RE23
• Codes of additional symbols are recorded, RE24
• Together these ICs form a character generator for 96 characters in a 7×5 format

problems and examples

DIODES, RECTIFIERS

1. Given: the circuit (Fig. 1), U1 = 10 V, U2 = 13 V, U3 = 15 V, U4 = 22 V, R1 = R2 = 1 kOhm. Find Uout



Solution:
When U1 = U2 = U3 = 0, the diodes are closed and the voltage is
Uout = U4•R2 / (R1 + R2) = 11 V.
If we connect U1, Uout will not change – the diodes are closed.
When U2 is connected, Uout becomes 13 V (the diode in the branch with U2 opens).
When U3 is connected, the diode in that branch opens, and Uout becomes 15 V. The other diodes close.

2. Assuming the diodes are ideal, find the current and voltage for the circuits shown in the diagram.


Solution:
For the circuit in the diagram, it is not obvious whether both diodes are conducting. In this case, let us assume both diodes are conducting, perform the calculations, and then check the validity of our assumption.
Assuming both diodes are conducting, it follows for the circuit that
UB = 0; U = 0
The current through diode VD2 can be found from the expression

Writing the current equation for node B, we have

Thus, diode VD1 is indeed conducting, as we initially assumed, and the final results are
I = 1 mA and U = 0 V.

3. Determine the current in the circuit and the voltage across the diodes whose I-V characteristics are given, if Uin = 2.5 V, Rl = 25 Ohm



Solution:
let us plot the combined I-V characteristic of the diodes and the "inverted" I-V characteristic of the load resistor.




Answer: U1 = 0.6 V, U2 = 0.7 V, Ul = 1.2 V, I = 45 mA

Rectifiers
4. In a half-wave rectifier (without a filter), the load voltage is Ulavg = 40 V.
With what Urevmax should the semiconductor diode be chosen?

Solution:


5. In a half-wave rectifier, the voltage on the transformer's secondary winding is U2 = 150 V. The mains frequency is 50 Hz,
Rl = 2 kOhm.
Determine Ulavg, Ifwd, Ilavg, Urevmax, and the ripple factor p.

Solution:
Ulavg =√2•U2/π = 67.36 V
Ilavg = Ulavg/ Rl = 67.36 / 2000 = 0.034 A
Ifwd = Ilavg = 0.034 A
Urev max = √2•U2 = 212 V
p = 1.57
.
6. In a half-wave rectifier with a capacitive filter (Fig. 21), calculate Ulavg and Urevmax if Cf = ∞, U2 = 150 V, mains frequency 50 Hz,
Rl = 2 kOhm, and plot the rectifier's time diagram.


Solution:
Cf = ∞, hence p = 0
Ulavg =√2•U2 / (1 + p) = 212 V
Ilavg = Ulavg/ Rl = 212 / 2000 = 0.106 A
Ifwd = Ilavg = 0.106 A
Urev max = 2√2•U2 = 424 V

Time diagram of the rectifier.


7. In a full-wave bridge rectifier circuit with a filter, the voltage on the transformer's secondary winding is U2 = 150 V. The mains frequency is 50 Hz, Rl = 2 kOhm, given that the filter capacitance is ∞. Determine the average value of the rectified voltage Ulavg, the forward current through the diode Ifwd, Ilavg, Urevmax, and the ripple factor.

Solution:
Cf = ∞, hence p = 0
Ulavg =√2•U2 / (1 + p) = 212 V
Ilavg = Ulavg/ Rl = 212 / 2000 = 0.106 A
Urev max = √2•U2 = 212 V

8. For the half-wave rectifier circuit without a filter (fig.), determine the transformer's turns ratio and the maximum reverse voltage on the diode, if the rectified voltage on the load is 30 V and the voltage on the transformer's primary winding is 220 V (50 Hz).


Solution:
Ulavg =√2•U2/π = 30 V
U2 = 2.22 Ulavg = 66 V
Urev max = √2•U2 = 94 V
N= U1/ U2 = 220/66 = 3.33

9. Determine the filter capacitor's capacitance Cf in a bridge rectifier, if the rectified voltage Ulavg = 12 V, current Ilavg = 10 mA, and the ripple factor must not exceed 0.05.

Solution:


10. In a half-wave rectifier operating into Rl = 250 Ohm, the RMS voltage on the transformer's secondary winding
U2 = 10 V.
Draw the diagram of the half-wave rectifier.
Determine what permissible forward current the semiconductor diode should be chosen for,
Draw the time diagrams of the rectifier's input and output voltages.

Solution:
Ulavg =√2•U2 / π = 4.5 V
Ilavg = Ulavg/ Rl = 4.5 / 250 = 0.018 A
Ifwd = Ilavg = 0.018 A
Ifwd allowed = 1.3• Ifwd = 0.0234 A

Time diagram of the rectifier.


11. In a half-wave rectifier with a capacitive filter (fig.), the voltage on the transformer's secondary winding is U2 = 10 V. The mains frequency is 50 Hz, Rl = 1 kOhm, Cf = 80 μF.
Determine the average value of the rectified voltage Ulavg, Ilavg, Urev max, the ripple factor p, and plot the time diagrams.

Solution:
Ulavg =√2•U2 / (1 + p) = 14.1/(1 + 0.2) = 11.75 V
Urev max = 2√2•U2 = 28.2 V
τdischarge = C•Rl = 0.05 s

We plot the time diagrams:


12. In a half-wave rectifier with a capacitive filter, the load voltage is Ulavg = 40 V, and the ripple factor must not exceed 0.05.
With what Urevmax should the semiconductor diode be chosen?

Solution:

VOLTAGE REGULATOR

1. What is the relative change in output voltage of a parametric voltage regulator, if the zener diode current changed by 2 mA, Ust = 8 V, Rdiff = 16 Ohm?

Solution:


2. A voltage u = Um(1.5 - 2t/T) V is applied to a chain consisting of a series-connected resistor R = 200 Ohm and a KS182 zener diode. Determine the current in the circuit for t = 0.2T, if the differential resistance of the zener diode is Rl = 30 Ohm, Um = 12 V.


Solution:
i = (u – Ust) / (R + Rl)
U(0.2T) = 1.1Um
i(0.2T) = (14 - 8.2)/230 = 0.025 A




3. Draw the characteristic of a zener diode with the parameters:
Ust = 12 V, Ist min = 3 mA, Rdiff = 25 Ohm
Ist max = 50 mA.

Solution:
∆Ist = Ist max - Ist min = 50 – 3 = 47 mA
∆ Ust = ∆Ist •Rdiff = 0.047•25 = 1.175 V
Ust min = Ust - ∆ Ust/2 = 11.42 V
Ust max = Ust + ∆ Ust/2 = 12.59 V

We plot the I-V characteristic of the zener diode as shown in the figure.


4. For the regulator circuit, the zener diode has the parameters:
Ust = 20 V, Ist min = 1 mA, Rdiff = 40 Ohm, Ist max = 71 mA. Determine the current I in the circuit graphically, if Il =20 mA:


Solution:
I = Ist + Il
∆Ist = Ist max - Ist min = 71 - 1 = 70 mA
∆ Ust = ∆Ist •Rdiff = 0.07•40 = 2.8 V
Ust min = Ust - ∆Ust/2 = 18.6 V
Ust max = Ust + ∆Ust/2 = 21.4 V

We plot the I-V characteristic of the zener diode and the I-V characteristic of the resistor. We sum the characteristics. We graphically determine the current of the undivided section of the circuit.
I = 55 mA


5. To stabilize the voltage across a load Rl = 2 kOhm, a parametric voltage regulator is used. The zener diode has the parameters:
Ist min = 1 mA, Ist max = 23 mA, Rdiff = 30 Ohm; the rated output voltage is 11 V, the input voltage is 22 V.
Determine Kst and Rbal.


Solution:


6. Determine the input voltage of the regulator. Zener diode parameters:
Ust = 12 V, Ist min =5 mA, Ist max=35 mA, Rdiff = 20 Ohm
Rbal = 800 Ohm, Rl = ∞.


Solution:
current through the regulator


Since the zener diode and the ballast resistance are connected in series in the circuit,


By Kirchhoff's second law:


7. Determine U2 in the voltage regulator, if U1 = 16 V, R1 = 300 Ohm,
R2 = 1.2 kOhm, Ust min = 12 V, Rst = 15 Ohm.
Note: solve the problem analytically, using the equivalent circuit of the zener diode (an EMF source E = Ust connected in series with a resistor Rst).



Solution:
Let us draw the equivalent circuit of the regulator


We use the two-node method:

U2 = 12.2 V

8. A periodic voltage u changes according to the law u(t) = 24(1 – 2t/T), where T is the period. The zener diode's stabilization voltage is 8 V. R1 = R2 = 1 kOhm.
Plot the graph of the output voltage change.
Assume the diode and zener diode are ideal.


Solution:
During the positive half-cycle, diode VD2 is closed. The output voltage Uout at t = 0 will equal the stabilization voltage. From the moment
t = T/3 to T/2 it changes from 8 V to zero.
During the negative half-cycle, diode VD2 is open. With the zener-diode branch disconnected, the voltage across resistor R2 changes from zero (at t = T/2) to –12 V, at t = T. Connecting the zener-diode branch limits the output voltage to –8 V.

Transistors, amplifiers

1. Determine the base current and the current gain h21e of a BJT connected in a common-emitter configuration, if the increment of the collector current is 17 mA, and of the emitter current is 18 mA. Determine the transistor currents IB, IC, IE and the voltages at its terminals relative to the common bus UB, UC, UE for the circuit.


2. Determine the DC gain of the transistor for
the circuit.


3. Determine the DC gain of the transistor for
the circuit.


4. Determine the current IE in the bipolar-transistor circuit.

5. Determine the current IE and voltage UC for the circuit.


6. Determine the current IE and voltage UC for the circuit.


7. Determine the output power of the voltage amplifier, if Rl = 40 Ohm, the voltage gain is 200, and the input voltage is 0.01 V.


8. Determine the input voltage of the amplifier, if the resistance
Rl = 100 Ohm, Pout = 2.5 W, KU = 50.


9. Determine the input voltage of the amplifier, if the resistance
Rl = 100 Ohm, Pout = 2.5 W, KU = 50.


10. Determine the base current Ib and the voltage drop across the transistor Uce in a voltage amplifier, if Rb = 150 kOhm, Rc = 1.25 kOhm, Ec = 9 V, h21 = 40, Ube = 0


11. Determine the current gain KI and voltage gain KU of a two-stage amplifier, if the load resistance is 10 Ohm, the amplifier's input voltage is 0.1 V, the power delivered to the load is 0.45 W, and the input resistance of the first stage is 100 Ohm.


12. Determine the resistance Rc of the voltage amplifier, if Ec = 10 V, Ucep=4 V, Icp = 5 mA


13. Find the values of Kgain, Rout for the voltage amplifier, if h11 = 900 Ohm, h21 = 50, h22 = 0.00016 S, Rc = 3 kOhm. Use the amplifier's equivalent circuit in the calculation.


14. What is Ec in the voltage amplifier, if Ucep = 7.5 V, Icp = 1 mA, Rout=2.5 kOhm?


15. In a voltage amplifier using a BJT, with a load Rl =250 Ohm, Uin = 0.1 V, Il = 20 mA, Rb = 50 kOhm and transistor parameters h11 = 400, h21 = 40, h22 = 0, determine the voltage gain Ku and the output voltage. Use the amplifier's equivalent circuit in the calculation.


16. The parameters of a transistor connected in a common-emitter (CE) configuration in a single-stage amplifier are as follows: h11 = 250 Ohm, h22 = 0.63•10‾³, h21 = 50, the internal resistance of the signal source being amplified is Rint = 250 Ohm, Ein = 10 mV, the collector-circuit resistance is Rc = 2 kOhm, Rl = 200 Ohm.
Find the current gain, the voltage gain, the power gain, Rout, and Rin of the amplifier. Use the amplifier's equivalent circuit in the calculation.



17. At the quiescent operating point, the collector current IKQ = 2 mA, and the voltage on the collector equals half the supply voltage. The resistor values in the collector circuit R3 = 5 kOhm and emitter circuit R4 = 1 kOhm. Determine the voltage on the base at the quiescent point and the resistances of the base voltage-divider resistors. Take the parameter h21 equal to 40.


18. At the quiescent operating point, the collector current IKQ = 100 μA, and the base voltage Ubq = 2 V. Determine the resistances in the collector and emitter circuits for which the voltage gain equals 5. Parameter h21 = 40.


19. Determine the values of R1, R2, R3, R4 in a common-emitter (CE) amplification stage circuit, if the supply voltage U = 5 V and the stage's gain equals 10. In the calculation, take Ucq=0.4•E V, h21> 30, Icq = 200 μA.


20. Determine the voltage on the transistor's collector for the common-emitter (CE) amplification stage circuit, if the supply voltage E=10 V, h21 = 50, R1 = 50 kOhm, R2 = 5 kOhm, R3 = 3 kOhm, R4 = 0.5 kOhm

Operational amplifiers
1. Determine the output voltage of the summing amplifier (fig.): if
U1 = U2 = U3 = 1V, R1 = 1 kOhm, R2 = 2 kOhm, R3 = 4 kOhm, R4 = 12 kOhm

Fig.

Solution:
Uout = - (Kfb1U1+ Kfb2U2+ Kfb3U3) Kfb = - R4/ Ri
Uout = - (12U1+ 6U2+3U3) = - 1•21= - 21 V

2. What is the output voltage of the inverting amplifier (fig. 3), if
R1 = 500 Ohm, R2 = 5000 Ohm, Uin = 0.2 V

Solution:
Kfb = - R2/ R1 = 10
Uout = Uin• Kfb = 2 V

3. A non-inverting op-amp amplifier operates from a source with voltage Uin =150 mV.
The resistor values are R1=20 kOhm, R2=200 kOhm.
Determine the amplifier's output voltage Uout and gain KU.

Solution:
KU = 1 + R2/ R1 = 11
Uout = Uin• Kfb = 1.650 V

4. Calculation of an inverting amplifier.
Given: Uin = 80 mV, Uout = 4 V, Rl = 10 kOhm, op-amp type 14OUD7
MB=√2 – frequency-distortion coefficient in the high-frequency region. Calculate the values of R1, Rfb, R2 in the inverting amplifier.


Solution:
Determining the resistor values.
Let us draw the amplifier circuit
Let us determine the amplifier's gain
Kfb = (Uout/Uin) = 4•103/80 = 50

We find Rfb from the relation
Rfb>=10•Rl = 10•10 = 100 kOhm

We determine R1:
R1= Rfb/ Kfb = 100/50 = 2 kOhm

Let us calculate the value of R2. This resistor is installed in the amplifier to equalize the input currents.
R2= R1Rfb / ( R1+ Rfb) = (2•100/102) ~ 2 kOhm

5. Calculation of a summing amplifier based on an op-amp.
Given:
Eg1 = - 2 V, Eg2 =3 V, Eg3 = 1 V, Rg1 = 0.5 kOhm, Rg2 = 0, Rg3 = 1 kOhm.

Gains for each input:
Kfb1 = 5, Kfb2 = 2, Kfb3 = 10, Rl = 2 kOhm. Op-amp type 153UD1.

Calculate the resistance values in the summing amplifier and determine Uout


Solution:
We choose Rfb >= 10•Rl, Rfb = 10•2 = 20 kOhm

From the expressions for the gains in the summing amplifier:
Kfb1 = Rfb/(R1 + Rg1) = 5
Kfb2 = Rfb/(R2 + Rg2) = 2
Kfb3 = Rfb/(R3 + Rg3) = 10

We find
R1 = 3.5 kOhm, R2 =10 kOhm, R3 =1 kOhm

We determine Rl, needed to equalize the input currents
Rl = Rfb/(R1 + Rg1)/R2/(R3 + Rg3) = 1.3 kOhm

We find the output voltage of the summing amplifier:
Uout = Eg1• Kfb1 + Eg2•Kfb2 + Eg3• Kfb3 = 10 – 6 – 10 = - 6 V

We determine the output-circuit current:
Iout = Ifb + Il = Uout/Rfb + Uout/Rl = 3.3 mA

This value of Iout does not exceed the output current of the 153UD1-type op-amp, which is 20 mA.

6. Plot the amplifier's Bode magnitude plot, determine the gain, fl and fh, if
R1 = 20 kOhm, R2 = 10 kOhm, R3 = 100 kOhm K = 200000, fu = 3 MHz C1=0.7 μF


Solution: operational amplifier
Kdb = 20 lgK = 20•5.3 = 106 dB
lgfu = 6.5 dec

Input circuit:
τ1 = R1• С1 = 20000•0.7•10-6 = 0.014 s


Break-frequency point of the input circuit's Bode plot:
f1 = 1/2πТ1 = 11.4 Hz lgf1 = lg11.4 = 1.06 dec

Feedback link:

20lg βfb = 20 lg0.09 = - 20.8 dB

Since the amplifier uses "deep" negative feedback, Kfb = 1/β = 11

Plotting the Bode magnitude diagram:

Продолжение:

Часть 1 Electronics (Component Base and Operating Principles)
Часть 2 - Electronics (Component Base and Operating Principles)
Часть 3 - Electronics (Component Base and Operating Principles)