Bjt Ac Analysis

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  • Topic: Electronic amplifier, Common emitter, Common collector
  • Pages : 18 (1702 words )
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  • Published : December 22, 2012
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BJT AC Analysis The re Transistor model

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Remind Q-poiint re = 26mv/IE

BJT AC Analysis Three amplifier configurations, Common Emitter Common Collector (Emitter Follower) Common Base

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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

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12Vdc

V2

RC

R1 1840k

4k
C2
2
Vo

0
1

10u

C1
2
Vb

Q1

1
VOFF = 0 V3 VAMPL = 1mV FREQ = 10000

10u

Q2N2222
RL 1000000

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
Output voltage, vo; i.e., collector voltage. The peak voltages are +142.692mV and -147.4mV, an average of about 145mV. 200mV (4.2745m,142.692m)

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0V

(4.2245m,-147.400m)

-200mV 4.0ms V(RL:1)

4.2ms

4.4ms Time

4.6ms

4.8ms

5.0ms

BJT AC Analysis Avb = vcp/vbp where ‘”p” means the peak value. vcp = -icp(Rc//ro) = -βibp(Rc//ro) = -βibpRc vbp = ibp (1 + β)re) Avb = vcp/vbp ~ -Rc/re and = 26mV/0.999mA = 26 ohms when ro is >> Rc

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|Avb| = 4000/26 = 153.8 which is close to the simulation value, 145 which is the result of 145mV/1mV.

BJT AC Analysis Aib = icp/ibp = β where ‘”p” means the peak value.

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The AC collector current peak-to-peak is about 1.0302mA – 0.957434mA = 0.072766mA, so the peak is about 0.36383mA. 1.050mA (4.2244m,1.0302m) 1.025mA

1.000mA

0.975mA (4.2765m,957.434u)

0.950mA 4.0ms IC(Q1)

4.2ms

4.4ms Time

4.6ms

4.8ms

5.0ms

The AC base current is nearly identical to current passing through the capacitor, C1, and the peak is about 220nA. 400nA (4.2195m,217.707n)

0A (4.2708m,-222.757n)

-400nA 4.0ms I(C1)

4.2ms

4.4ms Time

4.6ms

4.8ms

5.0ms

Using these values the transistor beta is 165.

BJT AC Analysis Zib = vbp/ibp = ibp(1 + β)re/ ibp = (1 + β)re Using the beta information from the simulation, Zib = 4300 ohms Zi = RB// Zib In 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 ix

Vx
ro
Rc

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Zo = vx/ix ix = vx/(Rc//ro) Zo = Rc//ro

BJT AC Analysis
Common Emitter Amplifier Circuit – 4 resistor bias and gain stabilizing emitter resistor.

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12Vdc

Vcc RC R1 100k 4k C2 2 1 10u Vo

0

RS 50 1

C1 2 10u Vb

Q1

Q2N2222 RL 1000000

VOFF = 0 V3 VAMPL = 1mV FREQ = 10000 R2 50k RE1 200

RE 1 4k

2 C3 10u

0

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
Output voltage, vo; i.e., collector voltage. The peak voltages are +16.721mV and -16.753mV, an average of about 16.4mV. 20mV

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(4.2735m,16.721m)

0V

(4.3247m,-16.753m)

-20mV 4.0ms V(C2:2)

4.2ms

4.4ms Time

4.6ms

4.8ms

5.0ms

BJT AC Analysis Avb = vcp/vbp where ‘”p” means the peak value. vcp = -icp(Rc//ro) = -βibp(Rc//ro) = -βibpRc vbp = ibp (1 + β)(RE1 + re) Avb = vcp/vbp ~ -Rc/(RE1 + re) and = 26mV/0.7614mA = 31.15 ohms when ro is >> Rc

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|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 (1 + βac)(re + Re1)
= = -120*3.9k . 121*226.597 -17.069

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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...
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