Tải bản đầy đủ (.docx) (13 trang)

Bài tập Điện tử cơ bản- SPKT STT 43 (10đ)

Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (315.5 KB, 13 trang )

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY AND EDUCATION
FACULTY OF ELECTRICAL AND ELECTRONICS ENGINEERING
--------

BIG EXERCISE
SUBJECT: BASIC ELECTRONICS
TOPIC 43

Teacher: BUI THI TUYET DAN
Student:
Student ID:

Ho Chi Minh City, May 12th, 2017


TOPIC 43: Give a circuit as shown, find:
a. Look up the parameters β, Cbc, Cbe or of transistor Q1 and Q2.
b. Static point Q1, Q2.
c. Write equations and draw the load line DCLL and ACLL of Q2. Find
maxswing of Vo2.
d. Draw an equivalent small-signal diagram of the circuit.
e. AV, Zi, Zo, Ai.
f. Draw the output waveform at the output, know that the input
waveform in the form of sin20000t (mV).
g. Find the lower cutoff frequency of the circuit, drawing the frequency
response of the circuit at the low frequency region.
h. Simulate the questions b, f, g. Comment on results and simulations.
Know that transistor Q1 and Q2 are D468


a. Look up the parameters of transistor Q1 và Q2




 min   max 120  240

 180
2
2

Cbc= 22 pF
fT= 190 Mhz
b. Static point Q1, Q2
Analyse DC :
*Transistor Q1:

IB1 = = = 0,019 (mA)
→IC1 = β.IB1 = 180.0,019 = 3,42 (mA)


VCE1 = VCC – IC1.(R3 + R4)
=20 – 3,42.(1 +1)= 13,16 (V)
 Q1 = (3,42 mA ; 13,16 V)
*Transistor Q2:

Rth = =
Vth=.VCC = .20= 2,17 (V)
IC2 =β.IB=β.== 1,39 (mA)
VCE2= VCC – IC2.(R7+R8)=20 -1,39.(1+1)= 17,22 (V)
 Q2 = (1,39mA ; 17,22V)



c. Write equations and draw the load line DCLL and ACLL of Q2. Find
maxswing of
* DCLL:
IC2 = -.VCE2 + . VCC
DCLL: IC2 = -.VCE2 +10 (mA)
* ACLL:


ACLL:



d. Draw an equivalent small-signal diagram of the circuit.

e. AV, Zi, Zo, Ai
k
= 3,366 (k


f. Draw the output waveform at the output, know that the input
waveform in the form of sin20000t (mV).
Ta có:
Suy ra : f= =3,184 kHz

Dạng sóng = 18,7sin20000t (mV)
=> Output signal in phase with input signal and amplitude: vmaxp=18,7 (mV)


f. Find the lower
cutoff frequency of the circuit, drawing the frequency

Vi(mV)
response of the circuit at the low frequency region.
-

=

1
0

t(us)

Với:-1= 1 =20,86 Ω

Vo(mV)

- Total gain:
18,7

0

(dB)

t(us)

-18,7
18,7
13,22

1,623


f (kHz)


g. Simulate the questions b, f, g. Comment on results and simulations
1. Simulation results b

From simulation data, we have a static workpoint Q1(3,453mA; 13,076V) and
Q2(1,32mA; 17,353V).
Comment the simulation results
On the 1st floor
Simulation results
Simulation results
Absolute error
Absolute error
On the 2st floor


Simulation results
Simulation results
Absolute error
Absolute error


2. Simulation results f
Output waveform
Vo=16,078 mV

Conment: The output waveform is similar to the calculated result



3. Simulation results g
Comment: Absolute error

Summary: There is a difference between the results of the simulation and the
results of calculations due to the operating conditions of the transistor as
temperature, ... is the cause of the error between the results of calculation and
simulation. However, the insignificant error should be acceptable



×