Question 3.SP.14: For the amplifier of Fig. 3-17, CC = 100 μF, RF = 180 kΩ, RL......

For the amplifier of Fig. 3-17, $C_C = 100 μF, R_F = 180 kΩ, R_L = 2 kΩ, R_S = 100 kΩ, V_{CC} = 12 \text{V}$, and $v_S = 4 \sin(20 × 10^3πt) \text{V}$. The transistor is described by the default npn model of Example 3.2. Use SPICE methods to (a) determine the quiescent values $(I_{BQ}, I_{CQ}, V_{BEQ}, V_{CEQ})$ and (b) plot the input and output currents and voltages $(v_S, i_S, v_L, i_L)$.

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(a) The netlist code that follows models the circuit:

 Prb3_14.CIR – CE amplifier vS    1 0 SIN(0V 4V 10kHz) RS   1 2 100kohm CC1 2 3 100uF Q  4 3 0 QNPN RF   3 4 180kohm RC   4 5 2kohm VCC 5 0 12V CC2 4 6 100uF RL   6 0 2kohm .MODEL QNPN NPN() ; Default transistor .DC VCC 12V 12V 1V .PRINT DC IB(Q) IC(Q) V(3) V(4) .TRAN 1us 0.1ms ; Signal values .PROBE .END

Execute   〈Prb3_14.CIR〉   and poll the output file to find $I_{BQ} = IB(Q) = 29.3 μ\text{A}, I_{CQ} = IC(Q) = 2.93 \text{mA}, V_{BEQ} = V(3) = 0.80 \text{V}$, and $V_{CEQ} = V(4) = 6.08 \text{V}$. Since $V_{CEQ} \simeq V_{CC}/2$, the transistor is biased for maximum symmetrical swing.

(b) The Probe feature of PSpice is used to plot $i_S, i_L, v_S$, and $v_L$ as displayed by Fig. 3-18. Notice the 180° phase shift between input and output quantities.

Question: 3.SP.15

Find the value of the emitter resistor RE that, when added to the Si transistor circuit of Fig. 3-17, would bias for operation about VCEQ = 5 V. Let ICEO = 0, β = 80, RF = 220 kΩ, RC = 2 kΩ, and VCC = 12 V. ...

Application of KVL around the transistor terminals...
Question: 3.SP.13

The circuit of Fig. 3-17 uses current- (or shunt-) feedback bias. The Si transistor has ICEO ≈ 0, VCEsat ≈ 0, and hFE = 100. If RC = 2 kΩ and VCC = 12 V, size RF for ideal maximum symmetrical swing (that is, location of the quiescent point such that VCEQ = VCC/2). ...

Application of KVL to the collector-emitter bias c...
Question: 3.SP.12

Collector characteristics for the Ge transistor of Fig. 3-15 are given in Fig. 3-16. If VEE = 2 V, VCC = 12 V, and RC = 2 kΩ, size RE so that VCEQ = -6.4 V. ...

We construct, on Fig. 3-16, a dc load line having ...
Question: 3.SP.10

The Si transistor of Fig. 3-14 is biased for constant base current. If β = 80, VCEQ = 8 V, RC = 3 kΩ, and VCC = 15 V, find (a) ICQ and (b) the required value of RB. (c) Find RB if the transistor is a Ge device. ...

(a) By KVL around the collector-emitter circuit, [...
Question: 3.8

The signal source switch of Fig. 3-9(a) is closed, and the transistor base current becomes iB = IBQ + ib = 40 + 20 sin ωt μA  The collector characteristics of the transistor are those displayed in Fig. 3-9(b). If VCC = 12 V and Rdc = 1 kΩ, graphically determine (a) ICQ and VCEQ, (b) ic and vce, and ...

(a)   The dc load line has ordinate intercept [lat...
Question: 3.7

For the transistor circuit of Fig. 3-8(a), R1 = 1 kΩ, R2 = 20 kΩ, RC = 3 kΩ, RE = 10 Ω, and VCC = 15 V. If the transistor is the generic npn transistor of Example 3.3, use SPICE methods to determine the quiescent values IBQ, VBEQ, ICQ, and VCEQ. ...

The netlist code below models the circuit. EX3_...
Question: 3.SP.28

In the circuit of Fig. 3-8(a), RC = 300 Ω, RE = 200 Ω, R1 = 2 kΩ, R2 = 15 kΩ, VCC = 15 V, and β =  110 for the Si transistor. Assume that ICQ ≈ IEQ and VCEsat ≈ 0. Find the maximum symmetrical swing in collector current (a) if an ac base current is injected, and (b) if VCC is changed to 10 V but ...

(a)  From (3.5) and (3.7), I_{CQ} ≈ I_{EQ} ...
Question: 3.SP.27

In the circuit of Fig. 3-10(a), the transistor is a Si device, RE = 200 Ω, R2 = 10R1 = 10 kΩ, RL = RC = 2 kΩ, β = 100, and VCC = 15 V. Assume that CC and CE are very large, that VCEsat ≈ 0, and that iC = 0 at cutoff. Find (a) ICQ, (b) VCEQ, (c) the slope of the ac load line, (d) the slope of the dc ...

(a)   Equations (3.5) and (3.7), give I_{CQ...
Question: 3.SP.21

The Si Darlington transistor pair of Fig. 3-21 has negligible leakage current, and β1 = β2 = 60. Let R1 = R2 = 1 MΩ, RE = 500 Ω, and VCC = 12 V. Find (a) IEQ2, (b) VCEQ2, and (c) ICQ1. ...

As long as v_S + V_B > 0.7  \text{V}[/la...