Basic Electronics.

Course 2022-2023
Degree in Sound and Image Engineering

PROBLEMS THEME 3

1.- Compute the working point of the following transistor and determine its operation
state. Data:   100 , V BEon  1V , VCEsat  0V .
+5V

1k

10k

-5V

2.- Repeat the former problem with the same data but with base resistance equal to 100
k


3.- Obtain the working point of the following PNP transistor operating with automatic
polarization. Data:   100 , V EBon  0.7V , V ECsat  0V .

Vcc = -15V

R1  22k RC  1k 5

-
+
R E  0.5k
R 2  10k

1
 Basic Electronics. Course 2022-2023
Degree in Sound and Image Engineering
4.- The following circuit is used to assess the operation of a transistor in the cut-off,
conduction, and saturation zones,

Vcc = +15V

R1  715k RC  5k

Vs
Ve Q1
R B  10k

Find out, as a function of Ve, the formulas for Vs for each transistor state. Data:
  100 , V BEon  0.7V , VCEsat  0V .

5.- Find out what was asked for in the previous problem but for the following circuit,
Vcc = +15V

R1  100k RC  3k 3
Thevenin

R3  100k A Vs
Ve

R E  2k 2
R2  100k
B

Data:   100 , V BEon  0.7V , VCEsat  0.2V

2
 Basic Electronics. Course 2022-2023
Degree in Sound and Image Engineering
6.-In the following circuit, find out all tensions and intensities. What is the operation
state of each transistor? Data:   50 , VCEsat  0.2V , V BEon  0.7V . The transistors are
identical.
Vcc = +10V

RC1  10k RC 3  1k

R1  100k vB2
Thevenin T2
RC 2  5k
A
T1
v B1
v E1

R 2  6k 8 R E  10k

7.-Find out the operation point of the following scheme with collector-base polarization.
Data:   90 , VCEsat  0.2V , V BEon  0.7V .

Vcc = +10V

RC  4k 7
R B  250k
vo
10F
vi
10F
R E  1k 2

3
 Basic Electronics. Course 2022-2023
Degree in Sound and Image Engineering
8.- Consider the circuit in the following figure, containing a BJT transistor and a LED.
The voltage differential in the LED in conduction state is 2.5V and the maximum
current that it can tolerate is I D max = 30mA. Find out the output voltage v S for each
operation zone of the BJT as a function of v e . In each case check whether the collector
current will damage the LED. Data:   100 , VCEsat  0V , V BEon  0.7V .

Vcc = 15 V

R1  22k RC  1k

vS
ve
RB  10k

9.-Find out the continuous base, collector and emitter voltages for the following
network. Data:   120 , VCEsat  0.2V , V BEon  0.7V .
Vcc = +20V

RC  2k 7
R1  8k 2

vo
10F
vi
10F

R 2  2k 2 R E  1k 8

Vee = -20V

4
 Basic Electronics. Course 2022-2023
Degree in Sound and Image Engineering
10.- Given the following polarization network (right) and the working point specified by
the graph (left), find out VCC , R B y RC . Data: V BEon  0.7V .

VCC
I C (mA)
8

Q I B  40A
RB RC

VCE
20V

11.- In the former circuit, re-compute R B and compute the value of a resistance
between emitter and ground so to obtain VCEQ  7.5V and I CQ  4mA , assuming the
following data:   100 , RC  1k 5 .

12.- Consider the following BJT single-layer amplifier,

Vcc = 20 V

R1 RC
C
C +

ue R E1 RL
R2 us
RE 2
C

Data: Vcc  20V ; R1  22k ; R 2  11k ; RC  1k ; RE1  0.5k ; RE 2  0.5k ;

RL  1k ; VCEsat  0.2V ; VBEon  0.7V ;   100 ; C   ; hie  447 ; hre  hoe  0

Find out the following,

a) Working point of the BJT.
b) Assuming R2  47k , re-compute the working point.

5
 Basic Electronics. Course 2022-2023
Degree in Sound and Image Engineering
us
c) Compute the altern tension gain G  .
ue

13.- Consider the amplifier in the following figure. Do the following,

a) Find out the working point.

b) Draw the continuous load line and its intersections with the I C and
VCE axes.
c) Draw the h-parameters equivalent circuit.
d) Find out the latern tension gain.

Data: Vcc  15V ; R B  150k ; RC  1k ; R E1  220 ; R E 2  470 ; RL  1k ;

VBEon  0.7V ;   100 ; C   ; hie  1k 5 ; hre  hoe  0 ; h fe   .

Vcc

RB RC
C
C +

ue
R E1 RL

us
RE 2
C
-

6
 Basic Electronics. Course 2022-2023
Degree in Sound and Image Engineering
14.- For the amplifier in the following figure, find out RC to achieve the output
v o  2sen( wo t ) .
Data: C  ; CE   R L  1k ; hie  1k ; hre  hoe  0 ; h fe  100 ;
vi  0.05sen( wo t )

Vcc

R1 RC
C
C +

vi RL vo
R2 RE CE

15.- Explain the usefulness of the decoupling capacitor parallell to the emitter resistance
CE. To do this, consider the circuit in the previous problem, comparing the expressions
of the tension gain with and without this capacitor.

16.- For the circuit in the following figure,

a) Find out the load line, plot it, and find out the polarization point.
b) Compute the tension gain.

Vcc

R1 RC
C
C +

vi RL vo
R2 RE C

7
 Basic Electronics. Course 2022-2023
Degree in Sound and Image Engineering
Data: Vcc  15V ; R1  22k ; R 2  6k 8 ; RC  1k ; R E  470 ; RL  1k ;
VBEon  0.7V ;   100 ; C   ; hie  1k 5 ; hre  hoe  0 ; h fe   .

17.- For the following single-layer amplifier, compute,

a) The continuous load line. Determine the working point.

b) The altern tension gain.
Vcc

R1 RC
C
C +
C

ue R E1 RL
R2 RE3
us
RE 2
C

Data: Vcc  15V ; R1  22k ; R2  6.8k ; RC  1k ; R E1  220 ; R E 2  270 ;

R E 3  200 ; RL  1k ; VCEsat  0.2V ; VBEon  0.7V ;   100 ; C   ; hie  1k 5 ;
hre  hoe  0

18.- For the following single-layer amplifier with PNP transistor, find out the altern
tension gain. Data:   100 ; C   ; hie  1k 5 ; hre  hoe  0

Vcc

R1
C
C RC  1k 5

vo
vi
R2 R L  1k
R E  0.5k

8
 Basic Electronics. Course 2022-2023
Degree in Sound and Image Engineering

19.- For the following single-layer BJT amplifier with automatic polarization, find out
the original circuit. Assume decoupling capacitors are used. What set-up is used?

hie

vi + h fe i b hoe RC RL vo
hre u ce

20.- In the situation in problem 19, and taking into account the following small-signal
model, find out
a) The formula of the altern tension gain assuming hre  hoe  0 .
b) The original amplifier circuit.

RS hie

vi + RC RL vo
R1 R2 h fe i b hoe
hre u ce

RE

9
 Basic Electronics. Course 2022-2023
Degree in Sound and Image Engineering

SOLUTIONS

1.- I C  10mA , VCE  0V

2.- I C  4mA , VCE  6V

3.- I C  6.95mA , V EC  1.065V

4.- CUT-OFF: V s  15V , Ve  0.5V

SATURATION: V s  0V , Ve  0.8V
CONDUCTION: V s  40  50Ve , 0.5V  Ve  0.8V

5.- CUT-OFF: V s  15V , Ve  12.9V

SATURATION: V s  6.0583  0.012672Ve , Ve  7.647V
CONDUCTION: V s  9.4469  0.43Ve ,  12.9V  Ve  7.647V

6.- T1 en CUT-OFF: v B1  0.6367V

T2 en SATURATION: I C 2  9.8mA , I B 2  0.93mA

7.- I C  1.063mA , VCE  3.655V .

8.- CUT-OFF: Ve  5.80V , V S  15V

SATURATION: Ve  4.55V , V S  0V
CONDUCTION, -5.80V<Ve<-4.55V, V S  45.5  10Ve
The collector current is maximum in saturation and equal to 12.5mA.

9.- V B  11.59V , V E  12.29V , VC  8.53V

10.- VCC  20V , RC  2k 5 , R B  482.5k

11.- R B  320k , R E  1609 .

12.- a) CONDUCTION, I C  5.51mA , VCE  8.89V

b) SATURATION, VCE  VCEsat  0.2V , I C  9.8mA
c) GV  0.98

13.- a) I C  6.53mA , VCE  3.96V

d) GV  2.11

14.- RC  666.66

16.- a) I C  5.4mA , VCE  7V

10
 Basic Electronics. Course 2022-2023
Degree in Sound and Image Engineering
b) GV  33.3

17.- a) I C  5.44mA , VCE  7V

b) GV  4.31

18.- GV  40

RC R L
 h fe
RC  R L
20.- GV 
1 1
{hie  (1  h fe ) R E }{(  ) R S  1}  R S
R1 R 2

11