Transcript
BJT AC Analysis 1 of 38 The re Transistor model
Remind Q-poiint re = 26mv/IE
BJT AC Analysis 2 of 38 Three amplifier configurations, Common Emitter Common Collector (Emitter Follower) Common Base
BJT AC Analysis 3 of 38 Process Replace transistor with small-signal model. Replace capacitors with short-circuits (at midband frequency caps have relatively low impedance) Replace DC voltage sources with short-circuits. Replace DC current sources with open-circuits).
BJT AC Analysis 4 of 38
R1
1840k
RC
4k
Q1
Q2N2222
C1
10u12
V212Vdc
0
0
C2
10u12
RL
1000000
Vo
Vb
V3
FREQ = 10000
VAMPL = 1mV
VOFF = 0
The simulation results include the following, IB = 6.172µA IC = 0.9932mA IE = 0.999mA VC = 8.027V VB = 0.6433mV
BJT AC Analysis 5 of 38 Output voltage, vo; i.e., collector voltage. The peak voltages are +142.692mV and -147.4mV, an average of about 145mV.
Time
4.0ms 4.2ms 4.4ms 4.6ms 4.8ms 5.0msV(RL:1)
-200mV
0V
200mV
(4.2245m,-147.400m)
(4.2745m,142.692m)
BJT AC Analysis 6 of 38 Avb = vcp/vbp where ‘”p” means the peak value. vcp = -icp(Rc//ro) = -βibp(Rc//ro) = -βibpRc when ro is >> Rc vbp = ibp (1 + β)re) Avb = vcp/vbp ~ -Rc/re and = 26mV/0.999mA = 26 ohms |Avb| = 4000/26 = 153.8 which is close to the simulation value, 145 which is the result of 145mV/1mV.
BJT AC Analysis 7 of 38 Aib = icp/ibp = β where ‘”p” means the peak value. The AC collector current peak-to-peak is about 1.0302mA – 0.957434mA = 0.072766mA, so the peak is about 0.36383mA.
Time
4.0ms 4.2ms 4.4ms 4.6ms 4.8ms 5.0msIC(Q1)
0.950mA
0.975mA
1.000mA
1.025mA
1.050mA
(4.2765m,957.434u)
(4.2244m,1.0302m)
The AC base current is nearly identical to current passing through the capacitor, C1, and the peak is about 220nA.
Time
4.0ms 4.2ms 4.4ms 4.6ms 4.8ms 5.0msI(C1)
-400nA
0A
400nA
(4.2708m,-222.757n)
(4.2195m,217.707n)
Using these values the transistor beta is 165.
BJT AC Analysis 8 of 38 Zib = vbp/ibp = ibp(1 + β)re/ ibp = (1 + β)re Using the beta information from the simulation, Zib = 4300 ohms Zi = RB// ZibIn this circuit since R is so large, Zi ~ Zib Output Impedance, Zo Turn off the input signal, Vs = 0. The input current is zero so the collector current is zero. ib = 0 and ic = 0 Connect a test signal generator to the circuit output
Vxro Rc
ix
Zo = vx/ix ix = vx/(Rc//ro) Zo = Rc//ro
BJT AC Analysis 9 of 38 Common Emitter Amplifier Circuit – 4 resistor bias and gain stabilizing emitter resistor.
RE1
200
R1
100k
RC
4k
Q1
Q2N2222
C1
10u12
Vcc12Vdc
0
0 C2
10u12
RL
1000000
Vb
Vo
V3
FREQ = 10000
VAMPL = 1mV
VOFF = 0
RS
50
R2
50k
RE
4k
C310u
12
The simulation results include the following, IB = 4.053µA IR1 = µA IC = 0.7565mA IE = 0.7614mA VC = 8.974V VB = 3.835mV
BJT AC Analysis 10 of 38 Output voltage, vo; i.e., collector voltage. The peak voltages are +16.721mV and -16.753mV, an average of about 16.4mV.
Time
4.0ms 4.2ms 4.4ms 4.6ms 4.8ms 5.0msV(C2:2)
-20mV
0V
20mV
(4.3247m,-16.753m)
(4.2735m,16.721m)
BJT AC Analysis 11 of 38 Avb = vcp/vbp where ‘”p” means the peak value. vcp = -icp(Rc//ro) = -βibp(Rc//ro) = -βibpRc when ro is >> Rc vbp = ibp (1 + β)(RE1 + re) Avb = vcp/vbp ~ -Rc/(RE1 + re) and = 26mV/0.7614mA = 31.15 ohms |Avb| = 4000/(200 + 31.15) = 17.31 which nearly identical to the simulation value, which is the result of 17.3. Rit = vb/ib = (1 + βac)(re + Re1) = 121(26.597 + 200) = 27.418k Avb = vc/vb ve = ib(1 + βac)Re1 vb = ib(1 + βac)(re + Re1) vc = -βac(ib)Rc Avb = vc/vb = -βac(ib)Rc . ib(1 + βac)(re + Re1) = -βac(Rc) .
BJT AC Analysis 12 of 38 (1 + βac)(re + Re1) = -120*3.9k . 121*226.597 = -17.069 Notice that the gain is stabilized by including the Re1 resistor. Without Re1 the gain is very dependent on the value of the transistor gain, β, which varies quite a bit and results in variations in base current which in turn changes the value of re.
BJT AC Analysis 13 of 38 Rot
ib
0
(1+B)re
2021
Rc
3.9k
v x
Re1
200
B
vx - ix (Rc) = 0 Rot = vx/ix = Rc = 3.9k
BJT AC Analysis 14 of 38 ROL
ib
Rc
3.9k
(1+B)re
2021
0
Re1
200
v x
RL
1k
B
RL’ = Rc//RL = Rc*RL . = 795.92 Rc + RL ROL = RL’
BJT AC Analysis 15 of 38 With RL attached the voltage gain is reduced, Avb = vc/vb = -βac*RL’ . = 3.482 which is a low level of gain (1 + βac)(re + Re1)
BJT AC Analysis 16 of 38 Avs .
Rs
50
Rth
7.5k
Bib
0
(1+B)re
2021
vb
Re1
200
vc
Vs
RL
1k
Rc
3.9k
RL
1k
Vs'
Rc
3.9k
Rs'
49
Bvb
ib
0
vc
(1+B)re
2021Re1
200
Vs’ = Vs*Rth . Rs + Rth
BJT AC Analysis 17 of 38 Rs’ = Rs*Rth . Rs + Rth Vs’ = ib[Rs’ + (1 + βac)(re + Re1)] Vs*Rth . = ib[Rs’ + (1 + βac)(re + Re1)] Rs + Rth Vc = ib*RL’ = Vs * Rth . * RL . [Rs + Rth] [Rs’ + (1 + βac)(re + Re1)]
BJT AC Analysis 18 of 38 Another way to tackle the gain calculation simplifies the math,
Rs
50
Rth
7.5k
vb
Rit
27.418k
Vs
Vb/Vs = Rth//Rit . = 5889 . = 0.99158 Rth//Rit + Rs 5889 + 50 From earlier analysis, Avb = Vc/Vb = -βac*RL’ . (1 + βac)(re + Re1) Combining the above two equations, (Vb/Vs) * (Vc/Vb) = Vc/Vs so that, Vc/Vs = Rth//Rit . * -βac*RL’ . = 0.99158 * 3.482 = 3.4528 Rth//Rit + Rs (1 + βac)(re + Re1) High-frequency analysis will use this approach of breaking down the problem into pieces followed by multiplication of the individual gains.
BJT AC Analysis 19 of 38 To summarize, 1. Compute Rit 2. Compute the gain Vs/Vb using the value of Rit. 3. Compute the gain Vc/Vb. 4. Multiply the gains to get the overall gain Vc/Vs.
Avb A1
BJT AC Analysis 20 of 38 Graphical Analysis DC & AC load lines on characteristics Inverting nature of amplifier -
Increase input voltage, increases base and collector current, decreases collector voltage because of more drop across load resistance.
BJT AC Analysis 21 of 38 Emitter Follower Amplifier (Common-Collector)– Basic Operation The emitter follow circuit acts as a buffer between stages.
R1
30k
R2
14.3k
RE
10k
Q1
Q2N2222
Rs
50
C1
112
V215Vdc
0
0
Vb
Vs
FREQ = 1000
VAMPL = 1mV
VOFF = 0 Ve
C2
112
RL
10k
V
V
BJT AC Analysis 22 of 38 The input and output voltage waveforms are nearly identical; almost independent of the value of the load resistor.
Time
5.0ms 5.5ms 6.0ms 6.5ms 7.0ms 7.5ms 8.0ms 8.5ms 9.0ms 9.5ms 10.0msV(Rs:1) V(C2:2)
-1.0mV
0V
1.0mV
BJT AC Analysis 23 of 38 Emitter Follower Amplifier (Common-Collector)– AC Analysis
C3
1n
R1
30k Vcc20Vdc
C1
1n
RL
500
Re
40k
0
Re1
200
Q2
Q2N3904
R2
10k
Rs
50 Q2N3904
Re2
4.1k
C2
1u
Vs
FREQ = 10000VAMPL = 0.001VOFF = 0
Rc
3.9k
βac = 120 assume ro is very large re2 = 66.93
BJT AC Analysis 24 of 38 mitter-Follower Stage E
RL
500
Rc
3.9k
ic1
Re
40k
C2
1u
Q2
Q2N3904
BJT AC Analysis 25 of 38 Without the load resistor,
+
-
v c1 vbRa
40k
Rc
3.9k
ib(1+B)re2
121*66.39
0
B
vb - ib*(1 + βac)re2 - ib(1 + βac)Ra = 0 Rit = vb/ib = (1 + βac)(re2 + Ra) = 121(66.39 + 40000) = 4.848M
BJT AC Analysis 26 of 38 With the load resistor,
(1+B)re2
121*66.39
Rc
3.9k
B
+
vbRa//RL
493.9
-
0
v c1
ib
Rit = vb/ib = (1 + βac)(re2 + Ra//RL) = 121(66.39 + 493.9) = 67.8k ve/vb = (1 + βac)(Ra//RL)/(1 + βac)(re2 + Ra//RL)
= Ra//RL/(re2 + Ra//RL)
BJT AC Analysis 27 of 38 Output Resistance
B
vb
ib
v xRa//RL
493.9
(1+B)re2
121*66.39
Rc
3.9k
-
+
0 Ro = vx / ix Sum currents entering the emitter node. ix + βac*ib2 + ib2 - vx / (Ra//RL) = 0 ix + (1 + βac)ib2 - vx / (Ra//RL) = 0 ib2 = - vx / [Rc + (1 + βac)*re2)] ix + (1 + βac)(- vx) / [Rc + (1 + βac)*re2)] - vx / (Ra//RL) = 0
BJT AC Analysis 28 of 38 ix - vx [ 1 . + 1 . ] = 0 Rc/(1 + βac) + re2 Ra//RL Two resistors in parallel, (Rc/(1 + βac) + re2 ) and Ra//RL equals Ro.
BJT AC Analysis 29 of 38
BJT AC Analysis 30 of 38 Common-Base Configuration
Zo
VoIoVoIo
Zo ∞
re 26 mVIEIE
Zi ViIiViIi
Zi rere
BJT AC Analysis 31 of 38 Common Base Amplifier
Av
VoViVoVi Ai
IoIiIoIi
RL
Vo Ic RL.Ic RL
Ai α Ie.
Ieα IeIe
Vo α Ie. RL.α Ie RL Vi Ie re.Ie re Ai αα
Av αRLre
.αRLre
Example IE = 4 mA α = 0.98 RL = 560 ohms Vi = 2 mV Compute re, Ii, Vo, Av, Ai
BJT AC Analysis 32 of 38 Common-Base Configuration
BJT AC Analysis 33 of 38 Zi = RE//re Zo = Rc Av = Vo/Vi
Vo = IC RCVi = IE re
Av = αIE RC = αRC
IE re re
Ai = α
BJT AC Analysis 34 of 38 Darlington Curren
BJT AC Analysis 35 of 38
BJT AC Analysis 36 of 38 Current Source Circuit
BJT AC Analysis 37 of 38 Current Mirror Circuit
BJT AC Analysis 38 of 38
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