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