Bjt Ac Analysis

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