circuit design with spice module 1. bias point of the

Circuit design with SPICEModule 1. Bias point of the active devices

Lesson 1. The semiconductor diode

To acquire the direct current features, characteristics and device model of the semiconductor diode, the calculation and simulation of the DC bias

The student acquires the material if he/she

• knows the device model of the diode• knows and he can draw theoretical and real voltage-current characteristics of the semiconductor diode• knows and he can apply the mathematical equation of the forward characteristics• knows the different types of diodes and their symbols• can solve simple exercises in practise by counting and with simulation

Acquiring the material takes about 180 minutes.

• semiconductor diode• the ’cut-in’ voltage of a semiconductor diode• forward and reverse DC characteristics• device model• determining bias

For the lesson please turn to the next page Look at the video about assembling the circuit: The video is available from: ZIP file/ sim_1A file

The aim of the lesson

Requirements

Time needed

Keywords

Learning material

Activity:

Revise all information about semiconductor materials, p- and n-type doping, p-n

junction, Ohm’s and Kirchhoff’s circuit laws on the basis of your knowledge taken

before

Contents:

1. Semiconductor diode

The semiconductor p-n junction allows the current to flow only in one direction and

blocks it in the opposite direction. In the n type semiconductor layer the movements of the

electrons and in the p type semiconductor layer the movements of the holes make the current.

Applying positive voltage on the p type layer and at the same time negative voltage on

the n type layer:

electrons enter p type layer

and

holes enter n type layer

the diode is ON

(current flows through the

junction even at low voltage

level)

Applying negative voltage on the p type layer and at the same time positive voltage

on the n type layer:

the number of the carriers

(electrons and holes) is very

limited

the diode is OFF

(the current is very small, only

10-6

-10-10

A)

The device model of the semiconductor diode contains an ideal diode, a DC voltage

source (Uo) and a serial resistor (Rs).

The ideal diode is shortcut in forward direction. The value of the running current can

be arbitrarily small or large while the voltage dropout is zero. The maximum current that can

flow through the junction depends on the surface size of the chip (die). Approximately 2

A/mm2 can be the maximum current on the chip.

At reverse bias the ideal semiconductor diode has zero leakage current.

The ideal diode characteristics

The ’cut-in’ voltage (U0) of a diode is defined as UD voltage when the diode current is

one out of ten of the maximum current applied on the diode

in case of Ge diodes 0.2 - 0.4 V

in case of Si diodes 0.5 - 0.8 V

in case of Schottky diodes 0.3 V

Further on we will suppose that the ’cut-in’ voltage of Si semiconductor

diodes equals to 0.6 V, but we know it is only a simple approximation.

The serial resistance of a semiconductor diode depends on manufacturing technology.

Its value is 1-10 ohm, but if we apply epitaxial layer, we can reach one out of ten of it.

Considering the facts above the voltage-current characteristic of a diode can be

approached with the following linear sections

The voltage-current characteristic of a diode approached with linear sections

A voltage-current characteristic of a real semiconductor diode changes according to

the measurements as it can be seen on the figure below:

A voltage-current characteristic of a real semiconductor diode

Forward characteristic

At forward bias between the voltage (UD) and its current (ID) of the diode the

following empirical formula can be given:

)1( T

D

mU

U

SD eII

IS is the theoretical saturation current of the diode. Its value equals to 100 nA in case

of Ge diodes, and it equals about 10 pA in case of Si diodes.

The value of ’m’ varies between 1 and 2. It shows how far the features of our diodes

are from the ideal ones given by the Shockley theory.

UT is called by thermal voltage. We can determine it with the following formula

q

kTUT where

k is the so called Boltzmann constant. It equals to 1.38 . 10

-23 J/K or VAs/K

T is the temperature given on Kelvin scale

q is the elementary charge, its value is 1.6 . 10

-19 Coulomb.

The formula mentioned above gives 26 mV in room temperature.

Since IS and UT depend on the temperature, in case of constant ID current the UD

voltage will also depend on the temperature. By increasing the temperature by degrees UD

voltage of the diode will decrease by 2-3 mV on constant ID current. Since this statement is

true on a wide range of temperature so the semiconductor diode can be used as a good

thermometer or sensor.

Reverse characteristic

The reverse or leakage current of the diode follows the increase of the reverse voltage

on almost steady value. Theoretically it is the saturation current IS of the diode, but the real

value is much higher.

On higher temperatures this saturation current will be increased. This current will be

doubled if temperature increases 10 degrees step by step.

If the reverse voltage increases further to a defined limit, much current increase can be

experienced even if the voltage changes only a little. This voltage limit is called reverse

breakdown voltage of the diode.

The value of this reverse breakdown voltage can be adjusted in advance with the help

of the technological steps of the manufacturing. It highly depends on the concentration and

the profile of the dopant material.

The relationship between the p-n junction profile and the breakdown voltage is used to

make the so called Zener diodes. The breakdown voltage can be interpreted by tunnel effect

below 10 V, and by avalanche breakdown above 10 V. These two effects often work together

and they can hardly be separated from each other.

The value of the reverse breakdown voltage can reach 10 kV nowadays.

Capacities of a semiconductor diode

At forward bias the diffusion capacity is accumulated in the p-n junction. The volume

of the diffusion capacity is nearly proportional to the current running through the junction and

it can even reach 10-100 nF. It is called diffusion capacity. Remove the diffusion capacity

from the p-n junction take time. So this capacity is disadvantageous if we want the diode to

switch fast.

At reverse bias the width of the p-n junction varies depending on the reverse voltage.

In this case the carriers (electrons and holes) move away from each other. So the p-n junction

is behaving like a capacity. It is called junction capacitance. Its value is about some pF. This

characteristic of the diode can be used to control the frequency of the oscillators by this

reverse voltage. These special diodes are called varicap diodes.

The different types of diodes and their symbols can be seen in the following table:

diode (general)

Schottky diode

Zener diode

Tunnel diode

varicap diode

light-emitting diode (LED)

photodiode (device sensitive to light or

other radiation)

silicon controlled rectifier

transient voltage suppression diode

Activity:

Look for an example to apply for the different diode types above, at least one for each.

EXAMPLES:

1.1.

At room temperature what current flows through the silicon diode, the forward voltage of

which is 0.65 V?

(Data: m = 1.45; IS = 12 pA)

Solution:

Here is the following, well-known formula between the forward voltage (UD) and the current

(ID) of the silicon diode:

)1( T

D

Um

U

SD eII

UT = 26 mV at room temperature, so after substituting data given above we get the following

result for the current of the diode:

mAAeI mV

V

D 369.0)1(1012 2645.1

65.0

12

On the conditions given in this case the current, 0.369 mA is expected on the diode.

1.2.

What voltage can be measured on the silicon diode, on which the flowing current is 1 mA at

20 C?

(Data: m = 1.42; IS [T =20C] = 18 pA)

Solution:

The formula between the forward voltage and the current of the diode will be solved on UD.

In this case we get:

)1ln()1ln(

S

D

S

DTD

I

I

q

Tkm

I

IUmU

After substituting:

VVpA

mA

q

kUD 640.083.1702528.042.1)1

18

1ln(

)2015.273(42.1

The forward voltage on the diode will be 0.64 V, when the current is 1 mA and the

temperature is 20 C.

Examples for circuits:

1.A.

Define the current of the diode in the following circuit:

(D1 is a silicon semiconductor diode with U0= 0.6 V)

On all silicon diodes the voltage drops about 0.6 V at forward bias. We can measure

5 V – 0.6 V = 4.4 V

on the resistor R1. In this case the current is 4.4 V/ 1kΩ = 4.4 mA on both resistance R1 and

diode D1.

1.B.

What voltage can be expected approximately on the point P of the following circuit?

(D1 - D2 - D3 silicon semiconductor diodes with U0 = 0.6 V)

On all three diodes the voltage drops 0.6 V at forward bias.

That is why the voltage on the point P is 15 V – 0.6 V – 0.6 V – 0.6 V = 13.2 V

1.C.

What current flows through the diodes in the circuit below?

(D1 and D2 are silicon semiconductor diodes with U0= 0.6 V)

R_gen, D1, D2 and R_load are placed in serial order in the circuit. That is why the current

(ID) flowing on them is the same. So

DloadDDDgengen IRUUIRU 21

DD IVVIV 8606.06.02203

DIV 10808.1

mAID 666.1

The current expected on the diode equals 1.666 mA.

Exercises for simulations:

1.a.

Assemble the following circuit on SPICE Schematics. By simulating define the value of

R_gen if the current is 1 mA on the diode D1.

The value of R_gen must be a variable like this: Open the video from ZIP file/ sim_1B file

To continue please turn to the next page.

The Start Value is 2000 ohm, the End Value is 25 kΩ.

Save the circuit with the command FILE/Save as ..

After carrying out the simulation (F11) you can get the following result:

On the horizontal axis the resistance, on the vertical axis the current can be seen. By moving

the cursor the detailed bias point can be determined easily. The values can also be seen on the

pop up window.

In this case the value of R_gen must be chosen 4.4 kΩ. The current of the diode results in 1

mA.

To continue please turn to the next page!

The next video is available from ZIP file/ sim_1C file

1.b.

Assemble the following circuit on SPICE Schematics. Set up 1 mA as a forward current of a

diode. Define the forward voltage of the chosen diode depending on the temperature.

Set up the parameter window as follows:

According to these data the forward voltage changing of the diode can be investigated

between –50 and +150 C by 10 C. The simulation results in the following diagram:

As you can see the relation between the temperature and the forward voltage of the diode on a

wide temperature range is linear. So the silicon diode is suitable to measure the temperature.

Use the left and the right buttons of the mouse to display the changes on the small window.


The video is available from ZIP file/ sim_1D!

By rising the temperature by 100 C the forward voltage of the diode will decrease about 200

mV. So the forward voltage of a silicon diode decreases approximately 2 mV/ C.

1. Define what the ‘cut-in’ voltage of a semiconductor diode mean.

2. Choose the correct answer

The forward voltage of a Ge diode is about 0.2-0.4 V.

The leakage current of a Si semiconductor diode is 10 pA theoretically.

The current of the semiconductor diodes increases like a logarithmic function depending on the forward voltage.

According to the tunnel effect and the avalanche breakdown the current of the reverse characteristic of the semiconductor diodes increases suddenlyover a definite bias point.

The forward voltage of a Si semiconductor diode increases approximately 2 mV/ C.

Mutassa a visszajelzést

3. How much is the thermal voltage at room temperature?

0.6 V

equals to the power voltage

0.3 V

0. 026 V

26 µV

4. Match the names to the appropriate symbol.

Tunnel diode

Schottky diode

photodiode (device sensitive to light or otherradiation)

Zener diode

varicap diode

diode (general)

Self-check questions

silicon controlled rectifier

light-emitting diode (LED)

5. Define the value of the thermal voltage at –20 C.

6. Try to graph the forward characteristic of a semiconductor diode with simulation. Plot the change of the resulted characteristic at different temperatures.

Lesson 2. The bipolar transistor

To acquire the direct current features, characteristics and device model of the semiconductor transistors, and set up the bias point of the transistors byusing passive resistors, the calculation and simulation of the DC bias of the different type of bipolar transistors.

The student acquires the material if he/she

• knows the conditions of normal active state of the bipolar transistors;• knows the relations among the currents of the transistor;• can draw the main characteristics of the bipolar transistor by heart;• knows the physical device model of the bipolar transistor and the approximate value of the parameters in the device model;• can evaluate the hij parameters concerning the given bias point;• can design circuits with bipolar transistors in which the transistors are in the normal active bias and can calculate its bias data;• can plot and set up the bias of the bipolar transistors by simulating;


• normal active state of the bipolar transistor• device model of the bipolar transistor• DC characteristics• determining bias

For the lesson please turn to the next page!


Requirements

Time needed

Keywords

Learning material

Activity:

If you studied and heard any information about the bipolar transistor before, repeat it.

Contents:

1.2. The bipolar transistors have got two types: npn or pnp depending on the order of the

layers determining them. Their symbols are as follows:

The bipolar transistor contains two pn junctions. One of them is the base-emitter

diode, the other one is the base-collector diode. The bipolar transistor has got four different

types of region depending on the bias of these two diodes.

region B-E diode B-C diode application

normal active

open

(forward-

biased pn

junction)

closed

(reverse-

biased pn

junction)

linear amplifier

closed closed closed switch-mode

saturation open open switch-mode

inverse active closed open *

* used rarely, it is favourable in case of extremely small reverse current we need

The following biases are necessary for the normal active state:

Two diodes switched serial in opposite direction don’t still create a usable bipolar

transistor. The thickness of the basis layer must be smaller than the diffusion length of the

electrons or holes.

In the bipolar transistors the closed base-collector junction is controlled with the open

base-emitter junction.

The current – called ICB0 – flowing through the closed base-collector junction consists

of mainly electrons in case of npn transistors. The base layer of an npn bipolar transistor is

more highly doped than the collector layer. So the minority carriers are the electrons in the

closed base-collector junction.

The current of closed base-collector junction can be increased by injecting electrons

into the base layer. It can be reached by forward biasing the base-emitter junction. In case of

npn transistors the emitter n+ layer is doped the most highly. That is why the current flowing

through the open base-emitter junction contains mainly electrons.

The electrons flowing through the open base-emitter junction cannot meet the holes

and recombination cannot happen, because the width of the base layer is very small. These

electrons pass through the base into the collector. The volume of this diffusion current

depends on the concentration of the electrons. This electron concentration is large near the

emitter because of the open junction and it is small near the collector layer because of the

closed junction.

Neglecting the role of the holes in the current of the base-emitter diode, furthermore

supposing the recombination rate is small in the base layer, the collector current of the bipolar

transistor can be described as follows:

ECBC IAII 0

where A is the DC current amplification factor. The value of „A” is smaller than 1, but it is

very close to 1. Its typical value ranges from 0.95 to 0.99. On the right side of the equation the

value of the ICB0 is much smaller than A.IE, that is why it is often neglected. In this case

EC IAI

In case of the direction of the DC currents given in the figure, the relations among the

currents of the bipolar transistor can be described as follows:

1

1

)1(

B

I

B

II

IAIB

BIBI

IBI

III

ECB

EEBC

BE

BCE

Activity:

Draw the device (npn transistor) above on a different sheet of paper and solve the

currents of the bipolar transistor from each other on your own. Repeat it with a pnp

transistor, too.

B used in the formulas above is the so called DC current amplification factor

concerning the base current.

In case of small, AC signals the relations among the currents of the transistor can be

given with similar forms of formulas. The differences are as follows: small „i”-s are used for

current of the transistor and is used instead of B for current amplification factor:

1

)1(

ECB

BE

BCE

iii

ii

iii

If the emitter of the bipolar transistor is connected to the ground, the input to the base

and the output is connected to the collector, the characteristics featuring the bipolar transistor

can be measured. The relation between the input voltage (UBE) and the input current (IB) can

be seen in the quarter III. Its name is input characteristic. This input characteristic is defined

with the constant value of the UCE voltage.

The output characteristic can be found in the quarter I. This characteristic shows the

related UCE – IC pairs with the constant value of the IB current.

The H parameters are suitable the best to describe the bipolar transistor.

The H parameters of an ordinary four-pole can be determined as follows:

2221212

2121111

uhihi

uhihu

Each H parameter can be counted and measured with the formulas given below:

0

1

111 2 u

i

uh short-circuit input impedance

0

2

112 1 i

u

uh open-circuit reverse voltage gain

0

1

221 2 u

i

ih short-circuit forward current gain

0

2

222 1 i

u

ih open-circuit output admittance

The H parameter device model (equivalent circuit) of an ordinary four-pole (two-port)

is given below:

The following model shows the hybrid or Giacoletto model of a bipolar transistor.

This model can often be used to substitute the bipolar transistors in the simulations.

The meaning of the parameters used in the model above is as follows:

rBB' – the resistance between the outer and inner base points of the model. Its value is between

5 and 50 .

re – the differential resistance of the base-emitter diode. Its value can be approached from the

emitter current:

E

Te

I

Ur

Its value (re) is equal to 26 if the emitter current is 1 mA. (The thermal voltage UT can

be regarded 26 mV at room temperature.)

B, or – forward current-amplification factor on the basis of the base current. Its value varies

depending on the application of the transistor.

In case of low frequency, low power transistors (the maximum collector current can be

100 mA), the forward current-amplification factor is between 50 and 500.

In case of low frequency, high power transistors (the maximum collector current can be

some amperes), the forward current-amplification factor is between 20 and 50.

In case of high frequency (the maximum transit frequency can be at least 1 GHz), the

forward current-amplification factor is usually between 50 and 100.

In case of superbeta transistors (in which base layer is extremely thin), the forward

current-amplification factor can reach 1000-5000.

– reverse voltage gain. It means in what extent the change of the collector-emitter voltage

affects the voltage of the base-emitter diode. Its value is usually 10-4

-10-5

. In case of

transistors built in integrated circuits the reverse voltage gain can be as low as 10-6

.

gm – transconductance. It shows in what extent the change of the base-emitter voltage can

vary the collector current of the transistor. Depending on the current of the transistor, gm

is between 10-500 mS. It can be counted as follows:

11

21

h

hgm

There are obvious relations among the parameters used in the physical model and the

H parameters. These relations are essential for calculating features of the amplifiers made

from bipolar transistors. These essential formulas are as follows:

.''0

1

111 )1(

)1()1(

2 constu

B

BEeBB

eeBBu CEI

Urr

rrr

i

uh

The parameter h11 (short-circuit input impedance) is the gradient of the tangent of the

input characteristic at the given bias point which can be found in the quarter III. Its value is

approximately some k.

.0

2

112

111

1

)1()1(

)1(1 constI

CE

BE

ee

ei Bu

u

rr

r

u

uh

The parameter h12 (open-circuit reverse voltage gain, or ) is the gradient of the

tangent of the curve at the given bias point which is in the quarter IV. Its value can be

neglected in most cases.

)1()1()1(

)1(

1

10

1

221 2

e

eme

ur

ri

gri

i

ih

.21 constU

B

C

CEI

Ih

The parameter h21 (short-circuit forward current gain or ) is the gradient of the

tangent of the curve at the given bias point which is in the quarter II.

ee

m

e

me

e

e

e

irr

gru

gr

r

ru

ru

ih

2

2

0

2

222

)1()1(

)1(

1

.22

2)1( constI

CE

C

eeBU

I

rrh

The parameter h22 (open-circuit output admittance) can be calculated with the sum of

two admittances. The parameter h22 is the gradient of the tangent of the output characteristic at

the given bias point which can be found in the quarter I. Its value is approximately 10-100 S.

EXAMPLES:

1.3.

The base current of a given bipolar transistor is 350 A. Calculate the value of the emitter and

collector current if the current amplification factor is B = = 90.

Solution:

On the basis of the relations among the currents of the bipolar transistors the formulas below

can be described:

BC

BE

IBI

IBI

)1(

After substituting the real values the following results can be received:

mAAAI

mAAAI

C

E

5.31105.311035090

85.311085.3110350)190(

36

36

1.4.

Define the parameters H of the given bipolar transistor.

(Data: IB = 12 A; B = = 240; rBB’ = 35 ; = 2,210-5

)

Solution:

First the emitter current must be calculated:

mAAIBI BE 892.21012)1240()1( 6

Supposing the device is at room temperature, the thermal voltage will be 26 mV. In this case:

99.8892.2

26

mA

mV

I

Ur

E

TE

After this the parameters H can be calculated as follows:

krrh EBB 2.26.220199.824135)1('11

5

12 102.21

h

24021 h

Srr

hEE

894.41

10894.499.8

102.222)1( 6

5

22

kh

3.20410894.4

116

22

1.5.

Define the bias currents, voltages of the pins and the parameters H of the bipolar transistor in

the circuit below. (UBE = 0.6 V)

(Data: = B = 120; rBB’ = 25 ; = 1.510-5

)

Solution:

The Kirchhoff law says:

tEEBEBB UIRUIR

After exchanging the emitter current for the base current, the base current IB can be

calculated.

tBEBEBB UIBRUIR )1(

After substituting the real values the following results can be received:

VIVIk BB 12)1120(3306.056

VIB 4.11)3993056000(

AIB 84.118

The emitter and the collector currents can be calculated as follows:

mAIBI BE 38.14)1( mAIBI BC 26.14

The voltages on the pins of the transistor can be defined with the help of the Ohm’s law:

VmAIRU EEE 745.438.14330

VVVAkVIRUU BBtB 345.5655.61284.1185612

or

VVVUUU BEEB 345.56.0745.4

VVVmAVIRUU CCtC 15.885.31226.1427012

The base-collector diode must be closed in the normal active region:

VVVUUU CBBC 805.215.8345.5

Finally define the parameters H of the bipolar transistor:

8.24381.112125)1()1( ''11

E

TBBEBB

I

Urrrh

( 81.138.14

26

mA

mV

I

Ur

E

TE )

5

12 105.1 h

12021 h

Sr

hE

57.1681.1

105.122 5

22

In the end the reciprocal value of the parameter h22:

kh

3.20410894.4

116

22

1.6.

Define the detailed parameters of the bias point in the following circuits. All the currents of

the different components and all the voltages of the nodes must be given.

(Data: B = = 350)

Solution:

Supposing BII 0 , which means that the current of R2 is much bigger than the base current,

the following relation gets true:

VVkk

kU

RR

RU tB 515

510

5

21

2

The voltage of the open base-emitter diode will be approximately 0.6 V. So the voltage of the

emitter can be calculated like this:

VVVUUU BEBE 4.46.05

The emitter current can be defined with the Ohm’s law:

mAk

V

R

UI

E

EE 2

2.2

4.4

The base current can be calculated if the emitter current and the current amplification factor is

known:

AAmA

B

II E

B 7.510698.5351

2

1

6

Supposing BII 0 , the current of R1 and R2 will be 1 mA:

mAk

V

RR

UI t 1

15

15

21

0

As it can be seen our starting presumption is true and the calculation is real. The data of the

collector are also needed:

mAmAmAIB

BI EC 2994.12

351

350

1

(vagy

mAAIBI BC 27.5350 )

VmAkVIRUUUU CCtRCtC 8.726.315

Finally check if the collector-emitter voltage is positive. This is the condition for the normal

active state of our npn bipolar transistor:

04.34.48.7 VVVUUU ECCE

1.7.

Draw the currents and their directions in the following circuit. Define the base current with

parameters.

Solution:

On the basis of the Kirchhoff’s law:

tEEBEBBCC UIRUIIRIIIR )()( 010

EEBE IRUIR 02

Using, that

BE IBI )1( or BC IBI

)]()([)1(

)(

22121

212

ECE

CBEtB

RRRRRRBRR

RRRURUI

Supplementary material

In case of B = = 120, calculate the base UB and the collector UC voltages.

)]56.06.8(5.1)6.815(56.0[1216.815

)5.16.815(6.06.812

kkkkkkkk

kkkVkVIB

AAmAmAIB 269948.25676.3390

14.88

]74.13216.13[121129

06.152.103

After evaluating IB, the other parameters of the bias point can be solved more easily:

mAIIBI BBE 145.3121)1(

mAIBI BC 119.3

VmAIRU EEE 761.1145.3560

VVVUUU BEEB 361.26.0761.1

Ak

V

R

UI B 5.274

6.8

361.2

2

0

AIIB 5.3000

mAIII BC 419.30

In the end define the collector voltage, too:

VVVmAkVIIIRUU BCCtC 87.613.512419.35.112)( 0

Examples for circuits:

1.D.

Resistances are usually used to achieve the normal active region in case of bipolar

transistors. Some kinds of solutions can be seen below:

Activity:

Draw the circuits above on a different sheet of paper. Try to draw these circuits by

heart without any help.

Draw the solutions to achieve the normal active region with pnp bipolar transistors,

too.

Think over what you can do if you have negative supply voltage.

Examples for simulations:

1.c.

Assemble the following circuit on SPICE Schematics.

The next video is available from ZIP file/1_2c file

Vary the collector voltage between 0 and 5 V and raise the base current up to 100 µA by 10 µA step by step at the same time as you can see in the setupwindows:


The next video is available from ZIP file/1_2c_2 file

After carrying out the simulation running with the values above, the following output

characteristic of the bipolar transistor can be seen:

1.d.

In the following circuit below set up the collector voltage by varying the value of R1. Define

the value of R1 when the collector voltage equals 12 V.

Solution:

Assemble the circuit

The next video is available from ZIP file/1_2_R file

Save the circuit. Don’t forget about the voltage marker on the collector.

Set up the varying values as follows:

Carry out the simulation. By means of cursor give the right value of R1 when the collector

voltage is just 12 V.


The next video is available from ZIP fájl/1_2_2_probe

As you can see at the window above, the value of R1 is equal to 124.54 kΩ when the collector

voltage is 12 V.

6. Simulate and show the output characteristic of a p-channel field-effect transistor with different gate-source voltages by the help of PSPICE. Choose onefrom j-FET or MOS-FET p-channel transistors as you like.

Module 2. Asymmetrical amplifiersLesson 1. Asymmetrical amplifiers with bipolar transistors

To get to know the setting up, the ac four-pole equivalent circuits of the asymmetrical amplifiers applying bipolar transistors.To define the most important parameters of the circuits by calculation and SPICE simulation.

The student acquires the material properly if he/she

• knows and can draw the circuits of asymmetrical amplifiers (common emitter, common base and common collector amplifiers) by heart,• is able to assemble, draw these amplifier circuits with npn or pnp transistors, with positive or negative supplier voltage,• is able to draw the ac four-pole equivalent circuits of the asymmetrical amplifiers applying bipolar transistors,• can calculate the most important parameters (voltage amplification rate, current amplification rate, input and output resistance, poweramplification rate),• knows the relations to calculate the parameters above using the ac four-pole equivalent circuits of the asymmetrical amplifiers,• can apply these relations on examples,• can change the circuits by calculation and simulation to set the expected parameters.


• asymmetrical amplifiers• common emitter amplifier• common base amplifier• common collector amplifier• small signal ac equivalent circuits• calculating and setting the working parameters• simulating and setting the working parameters

For the lesson please turn to the next page!

The next video can help you to set up the simulation parameters:You can reach from ZIP file


Requirements

Time needed

Keywords

Learning material

Activity:

Revise all about the bipolar transistors, their device model, the parameters of the

device model, their calculation and setting the bias point of the bipolar transistors.

Refresh your studies about the four-pole equivalent circuits, especially the hybrid

parameters.

Contents:

2.1. Asymmetrical amplifiers with bipolar transistors

In this chapter we are dealing with linear, small signal, active and asymmetrical amplifier

circuits. Their equivalent circuits are as follows:

outfoutiRu

The ordinary equivalent circuit of asymmetrical amplifiers

The most important parameters and their definitions of the asymmetrical amplifiers

can be seen below:

input resistance in

in

ini

uR

output resistance shortcutout

loadlessout

outi

uR

transfer impedance fR

in

out

Ai

uZ

transfer admittance fR

in

out

Au

iY

voltage amplification rate fR

in

out

Uu

uA

current amplification rate fR

in

out

Ii

iA

power amplification rate IUPAAA

Using the relations above the following relation can be given between the voltage

amplification rate and the current amplification rate of the asymmetrical amplifiers:

f

in

U

f

in

in

out

in

in

f

out

in

in

in

out

out

out

in

out

IR

RA

R

R

u

u

R

u

R

u

iu

u

iu

u

i

iA

The bipolar transistor can be controlled only through its base-emitter diode. Therefore

the input signal can be connected with the base or the emitter terminal. The output signal can

be received from the collector or/and the emitter terminal of the transistor. The third terminal

is always connected with the ground in the small signal equivalent circuit. This third terminal

can be any of the three ones of the transistors. Accordingly there are three basic asymmetrical

amplifiers. In case of

common emitter amplifier:

the emitter is connected with the ground from the point of view of ac small signal equivalent

circuits,

the input signal goes to the base of the bipolar transistor,

the output signal comes from the collector.

common collector amplifier:

the collector is connected with the ground from the point of view of ac small signal equivalent

circuits,

the input signal goes to the base of the bipolar transistor,

the output signal comes from the emitter.

common base amplifier:

the base is connected with the ground from the point of view of ac small signal equivalent

circuits,

the input signal goes to the emitter of the bipolar transistor,

the output signal comes from the collector.

2.1.A Common emitter amplifier (with emitter capacitor parallel with the emitter

resistor)

Applying positive supplier voltage and using an npn bipolar transistor as an active

device for the amplifier, the following circuit can be assembled. While calculating the

parameters of this amplifier all the capacitor of the circuit should be ideal. (They should be

considered shortcut in the middle of the frequency range.)

Common emitter amplifier (with emitter capacitor Ce parallel with the emitter resistor Re)

The following figure shows the ac small signal equivalent circuit of the common

emitter amplifier:

The ac small signal equivalent circuit of the common emitter amplifier

(the emitter capacitor Ce short-circuits the emitter resistor Re, therefore the emitter terminal of

the transistor connects directly with the ground)

Supposing the reverse voltage gain is small enough, it will be neglected in the

calculations of the most important working parameters of the circuit. (h12 = = 0)

Let us start our calculations with defining of the voltage amplification rate, Au:

)1

(

)1

(

2211

2122

21

11

fC

in

fC

in

in

out

uRR

hh

h

u

RRh

hh

u

u

uA

Let us give the value of h21/h11 in the following way:

m

eeBB

grrrh

h

)1('11

21

Let us define the value of the load-resistance Rt:

tfCRRR

In most cases the value of 1/h22 is significantly larger than the value of the load-

resistance Rt, therefore:

tmtmuRgR

hgA )

1(

22

Since the value of the transconductance gm of the bipolar transistor is about 10-100 mS

and the value of the load-resistance Rt is about 1-10 k, the voltage amplification rate Au of

the common emitter amplifier is approximately a few hundreds.

The negative sign shows that the common emitter amplifier turns the phase, which

means, that there is a 180-degree phase shift between the phase of the input and output

signals.

The next very important working parameter is the input resistance R in, that can be

defined in the following way:

1121'211121

1 hRrhrRRhRRRBeBBBBBBin

As it can be seen from the relation above, besides the voltage devider R1-R2 the value

of the input resistance is determined by the parameter h11, which is in close connection with

the emitter current of the bipolar transistor at the bias point. Since re is the differential

resistance of the base-emitter diode and its value can be approached from the emitter current,

E

T

eI

Ur and

erh )1(

11 , the input resistance will be a few k in this circuit.

The output resistance of the circuit can be defined with the following formula:

22

1

hRR

Cout

In case of the common emitter amplifier, the value of the output resistance Rout is of

the order of RC.

Now let us calculate the current amplification rate of this circuit:

1

11

11

22

22

121

)1

(

1

ihR

h

RRh

Rh

ih

i

iA

B

fC

C

in

out

i

11

21

11

11

21hR

R

RR

Rh

h

hR

RR

RhA

B

B

fC

CB

fC

C

i

During the calculation we must notice that there are two current dividers in the circuit.

Besides this the value of 1/h22 is supposed to be much larger than the value of RC.

The input voltage uin can be calculated in two ways: with the help of the input current

iin and with the help of the current i1 which is the input base current of the transistor:

1111121)( ihihRRu

inin

1

11

11i

hR

hi

B

in

It can be seen from the formula above, that the current amplification rate of this circuit

is smaller than the forward current-amplification factor of the bipolar transistor, but the

relation between them is linear.

The current amplification rate can also be expressed in the following way:

)(11

hRRR

RgA

B

fC

C

mi

Of course, the same result can be got if the following relation is substituted:

f

B

fC

fC

m

f

B

tm

f

in

uiR

hR

RR

RRg

R

hRRg

R

RAA

1111)(

)(11

hRRR

RgA

B

fC

C

mi

2.1.B Common emitter amplifier (without emitter capacitor)

Common emitter amplifier (without emitter capacitor)

To get the equivalent circuit of this common emitter amplifier, first draw the device

model of the bipolar transistor and then complete it with the other parts of the circuit.

Remember that the capacitors are ideal, they are considered to be short-circuited in the studied

frequency region. Furthermore, the two terminals of the voltage supply are also regarded to be

short-circuited in the equivalent circuit:

The ac small signal equivalent circuit of the common emitter amplifier (without emitter

capacitor)

Give the working parameters of the circuit. The calculation of the voltage

amplification rate is as follows:

Em

tm

E

fC

in

out

uRg

Rg

iRhih

iRRh

u

uA

1)1(

)(

121111

121

During the calculation the value of 1/h22 is supposed to be much larger than the value

of Rt. The value of gmRE in the denominator is often much higher than 1 which is neglected in

the practice. Therefore the result is simpler and usable but unpunctual.

E

t

uR

RA

This common emitter amplifier also turns the phase. The sign of the voltage

amplification rate is negative.

The input resistance of the circuit can be calculated like this:

EeBBBEBBin

RhrhrRRhhRRR 1112121'211121

The value of the output resistance is approximately equal to the value of RC:

CoutRR

The current amplification rate of the circuit can be described with the following

formula:

fC

C

EBB

BB

iRR

Rh

RhhRR

RRA

21

211121

21

1

Knowing the values of the voltage amplification rate Au and the input resistance Rin

and using the formula below, the same result can be got for the current amplification rate:

f

be

uiR

RAA

After substituting the formulas for the voltage amplification rate Au and the input

resistance Rin, the following formula is resulted:

f

EB

Em

tm

iR

RhhR

Rg

RgA

1)

1( 2111

fEB

EB

fC

fC

ERRhhR

RhhR

RR

RR

Rhh

h 1

1

1

2111

2111

2111

21

fC

C

EBB

BB

RR

Rh

RhhRR

RR

21

211121

21

1

2.1.C Common collector amplifier

In case of a common collector amplifier the input comes to the base and the output

goes from the emitter. The collector of the bipolar transistor connects with the ground in the

ac small signal equivalent circuit.

Common collector amplifier

The ac small signal equivalent circuit of the common collector amplifier

The working parameters of the common collector amplifier can also be described with

formulas:

22

22

22

121111

22

121

11

1

1)1(

1)1(

hRRg

hRRg

hRRihih

hRRih

u

uA

fEm

fEm

fE

fE

in

out

u

The formula above is too complicated for everyday use. It can be simplified if the

value of 1/h22 is much larger than the value of the emitter resistance RE and the value of the

load Rf. Furthermore:

fEtRRR

So the formula above for the voltage amplification rate Au of the common collector

amplifier can become simpler like this:

tm

tm

uRg

RgA

1

Since the value of gmRt is larger than a few hundreds, considering it the value of 1 is

neglected. Therefore, the value of the voltage amplification rate of the common collector

amplifier is practically a unit or a bit lower, about 0.98 - 0.99.

The common collector amplifier doesn’t turn the phase. The sign of the voltage

amplification rate is positive.

The value of the input resistance of the circuit is as follows:

tBfEBBin

RhhRh

RRhhRRR

1

11

2111

22

211121

To get a large input resistance for the circuit it is advisable to choose the values of the

base voltage divider resistors RB1 and RB2 high enough. They are often higher than 100 kΩ. In

this case the value of the current of the base divider is likely to be as small as the value of the

base current of the transistor. Remember that in this case the base divider RB1 and RB2 gets

loaded and the calculation of the bias becomes complicated.

The output resistance of the circuit is as follows:

121

2111

h

RRRhRR

gBB

Eout

Using a small generator resistance (Rg 0), the value of gBBRRR

21 in the

numerator can be neglected. Furthermore, the numerator can be changed as it can be seen in

the formula below:

eeBBrhrhrh )1()1(

2121'11

In this approximation the value of the output resistance Rout of the circuit can be given

very simply:

eeE

e

EoutrrR

h

rhRR

1

)1(

21

21

As it can be seen from the results above, the common collector amplifier has a

large input resistance

small output resistance and

unit voltage amplification rate.

Because of these three features the common collector amplifier is an excellent voltage

follower circuit.

The formula for the current amplification rate can be calculated in the way below:

1

22

211121

22

2111

22

22

121

11

11

1

1

)1(

i

Rh

RhhRR

Rh

Rhh

Rh

R

hR

ih

i

iA

fE

fE

fE

E

in

out

i

In most practical cases the value of 1/h22 is much larger than the value of RE.

Therefore its effect can be neglected in the circuit.

Let us introduce R0

fER

hRhhR

22

21110

11

In this case the formula for the current amplification rate can be described more

simply:

fE

E

B

B

iRR

Rh

RR

RA

)1(

21

0

To calculate the current amplification rate, the formulas below must be used:

10021iRiRRRu

inin that is

1

0

0i

RR

Ri

B

in

2.1.D Common base amplifier

In case of a common base amplifier the input comes to the emitter of the bipolar

transistor and the output goes from the collector. Therefore the base connects with the ground

in the ac small signal equivalent circuit.

Common base amplifier

The ac small signal equivalent circuit of the common base amplifier

Define the working parameters of this circuit. (During the calculation the effect of the

value of 1/h22 is supposed to be neglected because its value is much larger than the value of

the collector resistance.)

tm

fC

in

out

uRg

ih

RRih

u

uA

111

121

where

fCtRRR

For this reason the voltage amplification rate of the common base amlpifier and that

one of the common emitter amplifier are equal. The difference between them is that the

common base amplifier doesn’t turn the phase.

The input resistance of the circuit is as follows:

eeE

eBB

EEinrrR

h

rhrR

h

hRR

1

)1(

121

21'

21

11

In case of a real circuit make sure whether the last approximation is correct.

The value of the output resistance of the circuit is equal to the value of the collector

resistance with a good approach.

CoutRR

The current amplification rate of the common base amplifier is:

fC

CfC

C

in

out

iRR

R

ih

RR

Rih

i

iA

121

121

)1(

The calculation can also be done in a different way:

fC

C

fmfC

fC

m

f

e

tm

f

in

uiRR

R

RgRR

RRg

R

rRg

R

RAA

1

As it can be seen, the formulas calculated in different ways result in the same values.

The current amplification rate of the common base amplifier is defined practically by the

current divider assembled from RC and Rf. However, the value of this current “amplification”

rate is smaller than 1.

Activity:

Think over and repeat the working parameters and their values of the amplifiers listed

above. Place these circuits in order according to their voltage amplification rate, input

and output resistance and current amplification rate.

Look for typical applications for all amplifiers where their beneficial features can be

utilized the best.

EXAMPLES:

2.1.

Define the bias and the working parameters of the circuit below.

Supposing that UBE = 0.6 V.

(Data: B = = 50; = 10-4

; rBB’ = 16 )

Solution:

Currents running at each resistance and terminal of the bipolar transistor can be seen on the

circuit below:

The voltage level of the base terminal can be given in three different ways (from the power

supply, the ground and the emitter):

0011.2220 IkVIRUU

tB

)(2.78)(002 BBB

IIkIIRU

VIBkVUIRUUBBEEEtB

6.0)1(2.220

The solution of the system of equations above is given below:

AIAIVUBB

41.2947.1761.160

The current running through the resistance R2:

AAAIIB

88.20541.2947.176)(0

Knowing the base current, the values of the emitter and the collector current are as follows:

mAAIBIBE

5.141.2951)1(

mAIBIBC

47.1

The voltages of the emitter and the collector terminal are given below:

orVVVUUUBEBE

7.166.01.16

VVVmAkVIRUUEEtE

7.163.3205.12.220

VmAkIRUCCC

1047,18,6

To calculate the working parameters of the amplifier, the h-parameters of the bipolar

transistor are needed:

9003.175116)1()1(''11

E

T

BBEBBI

Urrrh

3.175.1

26

mA

mV

I

Ur

E

T

E

4

12101

h

5021

h

kh

Sr

h

E

67.861

54.11

3.17

1022

22

4

22

Now the calculation of the working parameters (AU, Rin, Rout, AI, AP,) can be done.

)1

(

22h

RRgAfCmu

mSrh

hg

e

m56

11

21

kkkkh

RRfC

67.567.86568.61

22

31867.556 kmSAu

( dBAAu

dB

u50318log20log20

)( )

The input resistance is not really large:

8559002.781.221121

kkhRRRin

The output resistance depends mainly on the collector resistance RC:

kkkh

RRCout

3.667.868.61

22

The current amplification rate Ai can be given by describing the current dividers of the circuit:

1121

11

22

22

21

1

1

hRR

h

Rh

R

hR

hA

fC

C

i

Knowing the relation between the transistor parameters: h21/h11 = gm:

85.4855.03.62

3.656)(

1

1

1121

22

22

kk

kmShRR

Rh

R

hR

gA

fC

C

mi

The power amplification rate can be defined by knowing the voltage and the current

amplification rates:

5.154586.4)318( iup

AAA

)89.315.1545lg10lg10()(

dBAAAiu

dB

p

2.2.



(Data: B = = 49; = 0; rBB’ = 0 )

Solution:

According to the Kirchhoff’s current law the current entering the base of the transistor is

equal to the current leaving the base junction:

0)1(

)(

21

00

E

BEBBBt

BBRB

UU

R

U

R

UUIIII

After substituting the real values of the parts of the circuit:

02.2)149(

6.0

56220

2.18

k

VU

k

U

k

UVBBB

According to the equation above, the voltage level of the base:

VUB

8.2

The current I0 running through the resistor R2 can be defined easily:

Ak

V

R

UI

B 5056

8.2

2

0

Going on the calculation for the voltage level of the emitter:

VVVUUUBEBE

2.26.08.2

mAk

V

R

UI

E

E

E1

2.2

2.2

Using the relation between the emitter and the base currents of the transistor, the base current

can be calculated like this:

AmA

B

II E

B20

50

1

1

The current running through the resistor R1 results in the sum of the currents I0 and IB. Its

value is equal to 70 A.

Further data:

mAAIBIBC

98.02049

VmAkIRUCCRC

43.898.06.8

VVVUUURCtC

77.943.82.18

The bipolar transistor is in the normal active region, because:

057.72.277.9 VVVUUUECCE

The h-parameters of the bipolar transistor are as follows:

261

26

mA

mV

I

Ur

E

T

e

krrhEBB

3.12650)1('11

012

h

4921

h

22

22

10

2

hS

rh

E

The value of the transconductance gm of the transistor must be known for the calculation of

the working parameters of the amplifier:

mSmV

mA

U

I

h

hg

T

E

m7.37

26

11

11

21

The voltage amplification rate Au of the circuit:

)3.10(26.394.83

1.274

2.27.371

)476.8(7.37

1dB

kmS

kkmS

Rg

RgA

Em

tm

u

During the calculation the effect of the 1/h22 is neglected. The input resistance of the circuit:

kkkkRhRRREin

2.21493.15622011121

kkkkRin

86.313.11156220

The output resistance depends mainly on the collector resistance RC:

kRRg

hRR

CEmCout6.8)1(

1

22

The value of the current amplification rate of this circuit:

fC

C

E

iRR

Rh

RhhRR

RRA

21

211121

21

)1(

kk

k

kkkk

kkA

i476.8

6.849

2.2503.156220

56220

2.2476.8

6.849

9.155

64.44

kk

k

k

kA

i

Finally, the power amplification rate can be given in the following way as usual:

)6.8(2.72.2)26.3( dBAAAiup

2.3.



(Data: B = = 100; = 10-4

; rBB’ = 35 )

The equations for calculating the bias of the transistor are as follows:

0013300 IkIRU

B

)(125)(002 BBtB

IIMVIIRUU

VIBkUIRUBBEEEB

6,0)1(7,40

The solution of the system of equations above is given below:

AIAIVUBB

75.797.1228.40

The current running through the resistance R1 is equal to the current I0, that is 12.97 A, while

the current running through the resistance R2 is the sum of I0 and IB, that is 20.72 A.

The voltage level on the emitter terminal is 0.6 V higher than the voltage level on the base

terminal of the transistor (in case of a pnp transistor):

VVVUUUBEBE

68.36.028.4

The emitter and the collector current at the bias point of the transistor can be seen below:

AAIBIBE

8.78275.7101)1( or

Ak

V

R

UI

E

E

E8.782

7.4

68.30

and

AAIBIBC

77575.710

The voltage level on the collector terminal is equal with the voltage level of the power supply.

That is -25 V.

The h-parameters are as follows:

krrheBB

39.32.3310135)1('11

2.33

8.782

26

A

mV

I

Ur

E

T

e

4

12101

h

10021

h

kh

Sr

h

E

1661

24.62.33

1022

22

4

22

Furthermore, the transconductance gm of the transistor is also necessary:

mSU

I

h

hg

T

E

m30

11

21

This circuit is a common collector amplifier, therefore its voltage amplification rate Au is

expected to be very close to a unit:

992,028,4301

28,430

1

kmS

kmS

Rg

RgA

tm

tm

u

)28.4166687.41

(

22

kkkkh

RRRfEt

The input resistance of the amplifier is a bit higher than before:

kkMkRhRRRtin

28.410139.3133011121

kkMkRin

15867.4351330

The value of the output resistance is decreasing significantly:

1

1

21

2111

22

h

RRRh

hRR

g

Eout

The inner resistance of the input generator is not given in the exercise, therefore let us use the

following approach Rg 0. In this case:

3.3356.331667.4101

39.31667.4 kk

kkkR

out

As it can be seen well:

2.330 eRout

rRg

The current amplification rate of the amplifier can be calculated the most simply with the

general formula:

)25.7(3.268

158992.0 dB

k

k

R

RAA

f

be

ui

Finally, the power amplification rate can be given in the following way:

)58.3(28.23.2992.0 dBAAAiup

2.4.



(Data: B = = 50; = 0; rBB’ = 50 )

The voltage level of the base terminal can be calculated with the following equations:

)(200)(0001 BBB

IIkIIRU

00210020 IkVIRUU

tB

VIBkVUIRUUBBEEEtB

6.0)1(120

The solution of the system of equations containing the three unknown values (UB, IB and I0)

gives the results below:

AIAIVUBB

29.3255.5177.160

In consequence of the solution the current running through the resistance R1 is equal to I0+IB,

that is 83.84 A, while the current running through the resistance R2 is equal to I0 = 32.29 A.

The voltage level of the emitter is given below:

VVVUUUBEBE

37.176.077.16

The following results show the value of the emitter current at the bias point of the transistor:

ormAk

VV

R

UU

R

UI

E

tE

E

RE

E63.2

1

)20(37.17

mAAIBIBE

63.255.5151)1(

The voltage level and the current of the collector can be calculated in the following way:

mAAIBIBC

58.255.5150

VmAkIRUCCC

06.1058.29.30

Based on the level of the terminal voltages of the transistor, the active device is in the normal

active region:

VVVVUUUECCE

031.7)37.17(06.10

After defining the bias parameters, the working parameters of the amplifier are the next steps

in the calculation. First the h-parameters must be given at the bias point:

krrhEBB

55488.95150)1('11

88.9

63.2

26

mA

mV

I

Ur

E

T

E

012

h

5021

h

22

22

10

2

hS

rh

E

Only the value of the transconductance gm of the transistor should be calculated:

mSh

hg

m25.90

554

50

11

21

The first working parameter to be calculated is the voltage amplification rate of the amplifier:

)6.50(7.338)1009.3(25.90)( dBkkmSRRgRgAfCmtmu

Then come the values of the input and the output resistances:

78.988.91krRReEin

kRRCout

9.3

The current amplification rate is calculated like this:

)6.29(033.0100

78.97.338 dB

R

RAA

f

be

ui

Finally, the power amplification rate can be given in the following way:

)5.10(2.11033.07.338 dBAAAiup

Exercises for simulations:

2.a.

Assemble the following circuit on SPICE Schematics.

To control the common emitter amplifier, sinus wave generator should be used with 10 mV

amplitude and 1 kHz frequency at the input of the circuit:

The values of the bias of the transistor are as follows:

As it can be seen the transistor works in the normal active region because the base-emitter

diode of the transistor is open and the base-collector diode is closed at the same time.

The voltage amplification rate Au of the amplifier is evaluated in the frequency range with the

help of a Bode-diagram as usual. The AC analysis must be set up as it can be given in the

windows:

Clicking to the windows „AC Sweep…” the details of its setup can also be adjusted:

After executing the simulation the next graph can be seen:

The amplitude of the output signal is 3.36 V in the wide frequency range. Since the amplitude

of the input signal level is 10 mV, the voltage amplification rate of the amplifier is the ratio of

these two values, that is |AU| = 336 (50.52 dB).

After drawing the input and output signals in the same window, you can make sure that the

circuit turns the phase. The phase shift is 180 degrees between them.

The setup parameters for the graph above are as follows:

To plot the input and the output signals at the same time, the command Plot/Add Plot to

Window must be used. The parameters of the axes can be adjusted to the values wanted with

the help of the command Plot/Axis Settings.

Remember that the difference between the scales on the vertical axes is two orders of

magnitude.

After this let us examine what will happen if the value of the input signal is increased from 10

mV to 200 mV as follows:

In case of larger output signal levels the output signal is limited between +10 V and -13 V.

The shape of the signal is distorted significantly.

Clicking to the button „FFT”, the increasing number and level of overtones (at 2 kHz, 3 kHz

… 15 kHz … etc.) are seemed to appear beside the 1 kHz signal. This phenomenon shows the

increased value of the distortion rate.

To continue please turn to the next page !

Using a longer examination time, a more punctual shape of the spectrum can be achieved. In

this case, of course, the computing period of time needed to carry out the simulation

increases. The suggested parameters of the setup can be seen in the following window:

2.b.

Using the same previous circuit, remove the capacitor CE from the circuit. By doing this

change, the bias parameters of the bipolar transistor and the values of the voltage and current

levels of the whole circuit remain unchanged.

Let us examine what effect this change has for the voltage amplification rate of the modified

amplifier.

The voltage amplification rate Au of the amplifier decreases significantly.

If the frequency of the input signal is tuned to 1 kHz and its level is 10 mV, the level of the

output signal is equal to 62.1 mV. These values correspond with |AU| = 6.21 (15.86 dB).

The amplification rate with the value of more than 300 received in the previous exercise falls

as low as 10.

circuit design with spice module 1. bias point of the

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