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

Look up the parameters β, Cbc, Cbe or f_T of transistor Q1 and Q2.Static point Q1, Q2.Write equations and draw the load line DCLL and ACLL of Q2. Find maxswing of Vo2.Draw an equivalent smallsignal diagram of the circuit.AV, Zi, Zo, Ai.Draw the output waveform at the output, know that the inputwaveform in the form of sin20000t (mV).Find the lower cutoff frequency of the circuit, drawing the frequency response of the circuit at the low frequency region.Simulate the questions b, f, g. Comment on results and simulations.Know that transistor Q1 and Q2 are D468

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 f T 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 Q 2

180

Cbc= 22 pF

fT= 190 Mhz

b Static point Q 1, Q 2

Analyse DC :

*Transistor Q 1 :

IB1 = V cc−VBE 1

R2+(β +1)R4 = 820+(180+1).120−0,7 = 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 Q 2 :

Rth = R5 R6

R5 +R6 = 82+1082.10 =8,91(kΩΩ)

Vth=R R6

5 +R6.VCC = 82+1010 20= 2,17 (V) IC2 =β.IB=β V th−VBE 2

R th+(β +1) R8=1 80. 2,17−0,7

8,91+(180+1) 1= 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 V o 2

* DCLL:

IC2 = -R 1

7 +R8.VCE2 + R 1

7 +R8 VCC DCLL: IC2 = -12.VCE2 +10 (mA)

* ACLL:

v CE +V CEQ=vCEt

i C .( R7 R L

R7+R L)+v CE =0

 (i Ct−ICQ).( R7 R L

R7+R L)+(v CEt−VCEQ)=0

ACLL: i Ct=

−1

R7 R L

R7+R L

v CEt+I CQ+ 1

R7 R L

R7+R L

V CEQ

i Ct=−2 vCEt+35,83(mA)

i C 2 p Maxswing=Min (1,39;34,44 )=1,39 mA

V CE 2 p Maxswing=Min (17,22;0,7 )=0,7 V V o 2 p Maxswing=VCE2 p Maxswing=0,7 V

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

e A V , Z i , Z o , A i

r e 1=26 mV

I EQ 1 =

26 mV 3,42 mA=7,6(Ω) hie 1=β r e1=180.7,6=1368 (Ω)=1,368 ¿kΩ¿

r e 2=26 mV

I EQ 2 =

26 mV 1,39 mA=18,7(Ω) hie 2=β r e 2=180.18,7=3366 (Ω) = 3,366 (kΩ¿

¿Z i=Zi 1=R2/ ¿ (hie1+β R4)=148,52(kΩΩ)

¿Z o=Z o 2=R7=1 (kΩΩ)

Z o 1=R3=1 kΩΩ

Z i 2=R5 / ¿R6 / ¿ (h ¿¿ie 2)=82 kΩΩ/¿10 kΩΩ/¿3,366 kΩΩ=2,44 kΩΩ¿

¿A v=v o

v i=

v o

v o 1 .

v o 1

v i =A v 2 A v 1

A v1=v o 1

v i =

−hfe 1 ib1.(R3/ ¿R5,6/ ¿Z b 2)

i b1 hie1+i4 R4 =

−(R3/ ¿R5,6/ ¿ (hie 2))

r e1 =−¿ ¿

A v2= v o

v o 1=

−hfe 2 ib2(R7/ ¿R L)

ib2(hie 2) =

−R7/ ¿R L

r e2 =

−500 18,6+1000=−26,7

A v=A v 1 A v 2=−0,704 (−26,7)=18,8

¿A i=A v Z i

R L =

18,8.148,52

f Draw the output waveform at the output, know that the input

waveform in the form of sin20000t (mV).

Ta có:V s=sin 20000 t(mV )

Suy ra : f= 200002 π =3,184 kHz

V i=Vs . Z i1

R1+Z i 1=sin 20000t

148,52 0,5+148,52=0,997 sin 20000 t (mV )

V o=A v .V i=18,8.0,997 sin1000 t=18,7 sin 20000 t (mV )

Dạng sóng V0= 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 response of the circuit at the low frequency region.

- f LC 1= 1

2 π(R1+Z i 1) C1 = 1

2 π(500+148,52.10 3).1 10 −6 =1,07 Hz

- f LC 2= 1

2 π(Z o 1+Z i 2) C2=

1

2 π (1+ 2,44) 103.1 10−6=42,27 Hz

- f LC 3= 1

2 π(Z o 2+R L) C3=

1

2 π (1+1).103.1 10−6=79,58 Hz

- f LC 4= 1

2 π C4 R e 2=

1

2 π 4,7 10−6 20,86=1623 Hz

Với:R e 2= ¿ R8/ ¿ (re 2+R5/ ¿R6

β ) = 1¿ /(18,7 10−3+ 82/ ¿ 10

180 ) =20,86 Ω

- f L=max(f LC 1 , f LC 2 , f LC 3 , f LC 4)=f LC 4=1623 Hz

- Total gain:

A vs= Z i

Z i+R1 A v= 148,52

148,52+0,5.18,8=18,7

V i (mV)

t(us) 0

1

-1

0

V o (mV)

t(us) 18,7

-18,7

13,22

f (kHz)

|A vs| (dB)

18,7

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

Q 2 (1,32mA; 17,353V).

Comment the simulation results

On the 1st floor

Simulation results Q1=(I CQ 1 , V CEQ1)=(3,453 mA ;13,076 V )

Simulation results Q1=(I CQ 1 , V CEQ1)=(3,42 mA ;13,16 V )

Absolute errorγ I

CQ1=|3,453−3,4203,453 |× 100 %=0,96 %

Absolute error γ V CEQ 1=|13,076−13,16013,076 |×100 %=0,64 %

On the 2st floor

Simulation results Q2=(I CQ 2 , V CEQ2)=(1,32 mA ;17,353 V )

Simulation resultsQ2=(I CQ 2 , V CEQ2)=(1,39 mA ;17,22V )

Absolute error γ I CQ2=|1,32−1,391,32 |×100 %=5,3 %

Absolute error γ V CEQ 1=|17,353−17,22017,353 |× 100 %=0,77 %

2 Simulation results f

Output waveform

Vo=16,078 mV

Conment: The output waveform is similar to the calculated result

γ v=|16,078−18,716,078 |× 100 %=16,3 %

3 Simulation results g

Comment: Absolute error γ I CQ2=|1,866−1,6231,866 |× 100 %=13 %

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