Calculate the voltage gain, current gain, input impedance, and output impedance for the amplifier shown in Figure 19.131 in the textbook.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 19, Problem 92P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the amplifier shown in Figure 19.131 in the textbook.

Expert Solution & Answer

Answer to Problem 92P

The voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.131 in the textbook for the amplifier circuit.

hie=4 kΩhre=10−4hfe=100hoe=30 μS

From the given amplifier circuit, the internal resistance (Rs) is 1.2 kΩ, emitter resistance (Re) is 240 Ω, and the load resistance (RL) is 4 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a amplifier in terms of hybrid parameters as follows:

Av=VcVb (1)

Here,

Vc is the collector voltage, which is the output voltage and

Vb is the base voltage.

Write the expression for current gain of the amplifier as follows:

Ai=IcIb (2)

Here,

Ib is the base current and

Ic is the collector current.

Write the expression for input impedance of the amplifier as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 1

From Figure 1, write the expression for emitter current as follows:

Ie=Ib+Ic (4)

Write the expression for base voltage from the circuit in Figure 1 as follows:

Vb=hieIb+hreVc+(Ib+Ic)Re (5)

Write the expression for collector current as follows:

Ic=hfeIb+VcRe+1hoe (6)

Write the expression for collector voltage as follows:

Vc=−IcRL (7)

From Equation (7), substitute (−IcRL) for Vc in Equation (6) as follows:

Ic=hfeIb+(−IcRL)Re+1hoe(1+hoeRLRehoe+1)Ic=hfeIb

Rearrange the expression as follows:

IcIb=hfe1+hoeRLRehoe+1=hfe(1+Rehoe)1+Rehoe+hoeRL

IcIb=hfe(1+Rehoe)1+(Re+RL)hoe (8)

From Equation (2), substitute Ai for IcIb as follows:

Ai=hfe(1+Rehoe)1+(Re+RL)hoe

Substitute 100 for hfe, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ai=(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)=(100)[1+(240 Ω)(30×10−6 S)]1+(240 Ω+4000 Ω)(30×10−6 S)=(100)(1.0072)1.1272=89.3542

From Equation (6), substitute (hfeIb+VcRe+1hoe) for Ic in Equation (8) as follows:

hfeIb+VcRe+1hoeIb=hfe(1+Rehoe)1+(Re+RL)hoe

Rearrange the expression as follows:

hfeIb+VcRe+1hoe=hfe(1+Rehoe)1+(Re+RL)hoeIb[hfe(1+Rehoe)1+(Re+RL)hoe−hfe]Ib=VcRe+1hoe[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe1+(Re+RL)hoe]Ib=hoeVc1+Rehoe

Ib=[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] (9)

Rearrange the expression in Equation (5) as follows:

Vb=hieIb+hreVc+(IbRe+IcRe)

Vb=(hie+Re)Ib+hreVc+IcRe (10)

From Equations (7) and (9), substitute (−VcRL) for Ic and [1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] for Ib as follows:

Vb=(hie+Re){[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]}+hreVc+(−VcRL)Re={(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}Vc

Rearrange the expression as follows:

VcVb=1{(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VcVb=1((4 kΩ+240 Ω)[1+(240 Ω+4 kΩ)(30 μS)](30 μS)[1+(240 Ω)(30 μS)]{(100)[1+(240 Ω)(30 μS)]−100−(240 Ω+4 kΩ)(30 μS)(100)}+10−4−240 Ω4 kΩ)=1((4000 Ω+240 Ω)[1+(240 Ω+4000 Ω)(30×10−6 S)](30×10−6 S)[1+(240 Ω)(30×10−6 S)]{(100)[1+(240 Ω)(30×10−6 S)]−100−(240 Ω+4000 Ω)(30×10−6 S)(100)}+10−4−240 Ω4000 Ω)=1{(4240 Ω)(1.1272)(30×10−6 S)(1.0072)[(100)(1.0072)−100−112.72]+10−4−0.06}=1{0.1434(1.0072)(−112)+10−4−0.06}

Simplify the expression as follows:

VcVb=−16.3476

From Equation (1), substitute Av for VcVb to obtain the voltage gain of the amplifier.

Av=−16.3476

From Equation (9), substitute (1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb for Vc and from Equation (8), substitute hfe(1+Rehoe)1+(Re+RL)hoeIb for Ic in Equation (10) as follows:

Vb={(hie+Re)Ib+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb+hfe(1+Rehoe)1+(Re+RL)hoeIbRe}={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}Ib

Rearrange the expression as follows:

VbIb={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VbIb={(4 kΩ+240 Ω)+(10−4)[1+(240 Ω)(30 μS)][(100)(1+(240 Ω)(30 μS))−100−(240 Ω+4 kΩ)(30 μS)(100)][1+(240 Ω+4 kΩ)(30 μS)](30 μS)+(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)(240 Ω)}=4240 Ω+(10−4)(1.0072)(100.72−100−12.72)33.816 μS+(100.72)(240 Ω)1.1272=4240 Ω−35.7417 Ω+21445 Ω=25649 ΩVbIb≅25.649 kΩ

From Equation (3), substitute Zin for VbIb to obtain the input impedance of the amplifier.

Zin=25.649 kΩ

Consider output voltage (Vc) as 1 V and redraw the circuit in Figure 1 as Figure 2 to find the output impedance of the amplifier.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 2

Apply KVL to the input loop for the circuit in Figure 2 as follows:

Ib(Rs+hie)+hreVc+Re(Ib+Ic)=0

Substitute 1 for Vc as follows:

Ib(Rs+hie)+hre+Re(Ib+Ic)=0

Ib(Rs+hie+Re)+hre+ReIc=0 (11)

Apply KCL at the output node for the circuit in Figure 2 as follows:

Ic=VcRe+1hoe+hfeIb=hoeVc1+hoeRe+hfeIb

Substitute 1 for Vc as follows:

Ic=hoe1+hoeRe+hfeIb

Rearrange the expression as follows:

Ic=hoe+(1+hoeRe)hfeIb1+hoeReIb=(1+hoeRe)Ic−hoe(1+hoeRe)hfe

Ib=Ichfe−hoe(1+hoeRe)hfe (12)

From Equation (12), substitute [Ichfe−hoe(1+hoeRe)hfe] for Ib in Equation (11) as follows:

[Ichfe−hoe(1+hoeRe)hfe](Rs+hie+Re)+hre+ReIc=0[(Rs+hie+Re)hfe+Re]Ic=(Rs+hie+Re)hoe(1+hoeRe)hfe−hre[(Rs+hie+Re)+hfeRehfe]Ic=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)hfeIc=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)(Rs+hie+Re)+hfeRe

Substitute 100 for hfe, 4 kΩ for hie, 1.2 kΩ for Rs, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ic=(1.2 kΩ+4 kΩ+240 Ω)(30 μS)−[1+(30 μS)(240 Ω)](100)(10−4)[1+(30 μS)(240 Ω)](1.2 kΩ+4 kΩ+240 Ω)+(100)(240 Ω)=0.1632−(1.0072)(100)(10−4)(1.0072)(5440 Ω)+24000 Ω=0.153129479 Ω=5.1935×10−6 S

Write the expression for output impedance of the amplifier as follows:

Zout=VcIc

Substitute 1 for Vc and 5.1935×10−6 S for Ic to obtain the output impedance of the amplifier.

Zout=15.1935×10−6 S=192500 Ω=192.5 kΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

a. Calculate the synchronous speed of a 3-phase, 12-pole induction motor that is excited by a 60 Hz source. b. What is the nominal speed if the slip at full-load is 6 percent?

A 3-phase, 75 hp, 440 V induction motor has a full-load efficiency of 91 percent and a power factor of 83 percent. Calculate the nominal current per phase.

Please answer all the questions a) What is the minimum required transformer rating for each transformer in kVA? b) Find the voltage required at Bus 1. c) The loadcentre is to be powered by PV panels that have the same I-V curve (ISC = 120 A, VOC = 60 V). Identify the configuration that uses the minimum number of panels to provide enough power to Bus 1. You can assume that the inverter converts the power at 95% efficiency and requires V DC input to generate the same VRMS AC output.

Answer 2

Question

Chapter 19, Problem 92P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the amplifier shown in Figure 19.131 in the textbook.

Expert Solution & Answer

Answer to Problem 92P

The voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.131 in the textbook for the amplifier circuit.

hie=4 kΩhre=10−4hfe=100hoe=30 μS

From the given amplifier circuit, the internal resistance (Rs) is 1.2 kΩ, emitter resistance (Re) is 240 Ω, and the load resistance (RL) is 4 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a amplifier in terms of hybrid parameters as follows:

Av=VcVb (1)

Here,

Vc is the collector voltage, which is the output voltage and

Vb is the base voltage.

Write the expression for current gain of the amplifier as follows:

Ai=IcIb (2)

Here,

Ib is the base current and

Ic is the collector current.

Write the expression for input impedance of the amplifier as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 1

From Figure 1, write the expression for emitter current as follows:

Ie=Ib+Ic (4)

Write the expression for base voltage from the circuit in Figure 1 as follows:

Vb=hieIb+hreVc+(Ib+Ic)Re (5)

Write the expression for collector current as follows:

Ic=hfeIb+VcRe+1hoe (6)

Write the expression for collector voltage as follows:

Vc=−IcRL (7)

From Equation (7), substitute (−IcRL) for Vc in Equation (6) as follows:

Ic=hfeIb+(−IcRL)Re+1hoe(1+hoeRLRehoe+1)Ic=hfeIb

Rearrange the expression as follows:

IcIb=hfe1+hoeRLRehoe+1=hfe(1+Rehoe)1+Rehoe+hoeRL

IcIb=hfe(1+Rehoe)1+(Re+RL)hoe (8)

From Equation (2), substitute Ai for IcIb as follows:

Ai=hfe(1+Rehoe)1+(Re+RL)hoe

Substitute 100 for hfe, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ai=(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)=(100)[1+(240 Ω)(30×10−6 S)]1+(240 Ω+4000 Ω)(30×10−6 S)=(100)(1.0072)1.1272=89.3542

From Equation (6), substitute (hfeIb+VcRe+1hoe) for Ic in Equation (8) as follows:

hfeIb+VcRe+1hoeIb=hfe(1+Rehoe)1+(Re+RL)hoe

Rearrange the expression as follows:

hfeIb+VcRe+1hoe=hfe(1+Rehoe)1+(Re+RL)hoeIb[hfe(1+Rehoe)1+(Re+RL)hoe−hfe]Ib=VcRe+1hoe[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe1+(Re+RL)hoe]Ib=hoeVc1+Rehoe

Ib=[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] (9)

Rearrange the expression in Equation (5) as follows:

Vb=hieIb+hreVc+(IbRe+IcRe)

Vb=(hie+Re)Ib+hreVc+IcRe (10)

From Equations (7) and (9), substitute (−VcRL) for Ic and [1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] for Ib as follows:

Vb=(hie+Re){[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]}+hreVc+(−VcRL)Re={(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}Vc

Rearrange the expression as follows:

VcVb=1{(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VcVb=1((4 kΩ+240 Ω)[1+(240 Ω+4 kΩ)(30 μS)](30 μS)[1+(240 Ω)(30 μS)]{(100)[1+(240 Ω)(30 μS)]−100−(240 Ω+4 kΩ)(30 μS)(100)}+10−4−240 Ω4 kΩ)=1((4000 Ω+240 Ω)[1+(240 Ω+4000 Ω)(30×10−6 S)](30×10−6 S)[1+(240 Ω)(30×10−6 S)]{(100)[1+(240 Ω)(30×10−6 S)]−100−(240 Ω+4000 Ω)(30×10−6 S)(100)}+10−4−240 Ω4000 Ω)=1{(4240 Ω)(1.1272)(30×10−6 S)(1.0072)[(100)(1.0072)−100−112.72]+10−4−0.06}=1{0.1434(1.0072)(−112)+10−4−0.06}

Simplify the expression as follows:

VcVb=−16.3476

From Equation (1), substitute Av for VcVb to obtain the voltage gain of the amplifier.

Av=−16.3476

From Equation (9), substitute (1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb for Vc and from Equation (8), substitute hfe(1+Rehoe)1+(Re+RL)hoeIb for Ic in Equation (10) as follows:

Vb={(hie+Re)Ib+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb+hfe(1+Rehoe)1+(Re+RL)hoeIbRe}={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}Ib

Rearrange the expression as follows:

VbIb={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VbIb={(4 kΩ+240 Ω)+(10−4)[1+(240 Ω)(30 μS)][(100)(1+(240 Ω)(30 μS))−100−(240 Ω+4 kΩ)(30 μS)(100)][1+(240 Ω+4 kΩ)(30 μS)](30 μS)+(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)(240 Ω)}=4240 Ω+(10−4)(1.0072)(100.72−100−12.72)33.816 μS+(100.72)(240 Ω)1.1272=4240 Ω−35.7417 Ω+21445 Ω=25649 ΩVbIb≅25.649 kΩ

From Equation (3), substitute Zin for VbIb to obtain the input impedance of the amplifier.

Zin=25.649 kΩ

Consider output voltage (Vc) as 1 V and redraw the circuit in Figure 1 as Figure 2 to find the output impedance of the amplifier.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 2

Apply KVL to the input loop for the circuit in Figure 2 as follows:

Ib(Rs+hie)+hreVc+Re(Ib+Ic)=0

Substitute 1 for Vc as follows:

Ib(Rs+hie)+hre+Re(Ib+Ic)=0

Ib(Rs+hie+Re)+hre+ReIc=0 (11)

Apply KCL at the output node for the circuit in Figure 2 as follows:

Ic=VcRe+1hoe+hfeIb=hoeVc1+hoeRe+hfeIb

Substitute 1 for Vc as follows:

Ic=hoe1+hoeRe+hfeIb

Rearrange the expression as follows:

Ic=hoe+(1+hoeRe)hfeIb1+hoeReIb=(1+hoeRe)Ic−hoe(1+hoeRe)hfe

Ib=Ichfe−hoe(1+hoeRe)hfe (12)

From Equation (12), substitute [Ichfe−hoe(1+hoeRe)hfe] for Ib in Equation (11) as follows:

[Ichfe−hoe(1+hoeRe)hfe](Rs+hie+Re)+hre+ReIc=0[(Rs+hie+Re)hfe+Re]Ic=(Rs+hie+Re)hoe(1+hoeRe)hfe−hre[(Rs+hie+Re)+hfeRehfe]Ic=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)hfeIc=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)(Rs+hie+Re)+hfeRe

Substitute 100 for hfe, 4 kΩ for hie, 1.2 kΩ for Rs, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ic=(1.2 kΩ+4 kΩ+240 Ω)(30 μS)−[1+(30 μS)(240 Ω)](100)(10−4)[1+(30 μS)(240 Ω)](1.2 kΩ+4 kΩ+240 Ω)+(100)(240 Ω)=0.1632−(1.0072)(100)(10−4)(1.0072)(5440 Ω)+24000 Ω=0.153129479 Ω=5.1935×10−6 S

Write the expression for output impedance of the amplifier as follows:

Zout=VcIc

Substitute 1 for Vc and 5.1935×10−6 S for Ic to obtain the output impedance of the amplifier.

Zout=15.1935×10−6 S=192500 Ω=192.5 kΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 19, Problem 92P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the amplifier shown in Figure 19.131 in the textbook.

Expert Solution & Answer

Answer to Problem 92P

The voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.131 in the textbook for the amplifier circuit.

hie=4 kΩhre=10−4hfe=100hoe=30 μS

From the given amplifier circuit, the internal resistance (Rs) is 1.2 kΩ, emitter resistance (Re) is 240 Ω, and the load resistance (RL) is 4 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a amplifier in terms of hybrid parameters as follows:

Av=VcVb (1)

Here,

Vc is the collector voltage, which is the output voltage and

Vb is the base voltage.

Write the expression for current gain of the amplifier as follows:

Ai=IcIb (2)

Here,

Ib is the base current and

Ic is the collector current.

Write the expression for input impedance of the amplifier as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 1

From Figure 1, write the expression for emitter current as follows:

Ie=Ib+Ic (4)

Write the expression for base voltage from the circuit in Figure 1 as follows:

Vb=hieIb+hreVc+(Ib+Ic)Re (5)

Write the expression for collector current as follows:

Ic=hfeIb+VcRe+1hoe (6)

Write the expression for collector voltage as follows:

Vc=−IcRL (7)

From Equation (7), substitute (−IcRL) for Vc in Equation (6) as follows:

Ic=hfeIb+(−IcRL)Re+1hoe(1+hoeRLRehoe+1)Ic=hfeIb

Rearrange the expression as follows:

IcIb=hfe1+hoeRLRehoe+1=hfe(1+Rehoe)1+Rehoe+hoeRL

IcIb=hfe(1+Rehoe)1+(Re+RL)hoe (8)

From Equation (2), substitute Ai for IcIb as follows:

Ai=hfe(1+Rehoe)1+(Re+RL)hoe

Substitute 100 for hfe, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ai=(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)=(100)[1+(240 Ω)(30×10−6 S)]1+(240 Ω+4000 Ω)(30×10−6 S)=(100)(1.0072)1.1272=89.3542

From Equation (6), substitute (hfeIb+VcRe+1hoe) for Ic in Equation (8) as follows:

hfeIb+VcRe+1hoeIb=hfe(1+Rehoe)1+(Re+RL)hoe

Rearrange the expression as follows:

hfeIb+VcRe+1hoe=hfe(1+Rehoe)1+(Re+RL)hoeIb[hfe(1+Rehoe)1+(Re+RL)hoe−hfe]Ib=VcRe+1hoe[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe1+(Re+RL)hoe]Ib=hoeVc1+Rehoe

Ib=[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] (9)

Rearrange the expression in Equation (5) as follows:

Vb=hieIb+hreVc+(IbRe+IcRe)

Vb=(hie+Re)Ib+hreVc+IcRe (10)

From Equations (7) and (9), substitute (−VcRL) for Ic and [1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] for Ib as follows:

Vb=(hie+Re){[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]}+hreVc+(−VcRL)Re={(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}Vc

Rearrange the expression as follows:

VcVb=1{(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VcVb=1((4 kΩ+240 Ω)[1+(240 Ω+4 kΩ)(30 μS)](30 μS)[1+(240 Ω)(30 μS)]{(100)[1+(240 Ω)(30 μS)]−100−(240 Ω+4 kΩ)(30 μS)(100)}+10−4−240 Ω4 kΩ)=1((4000 Ω+240 Ω)[1+(240 Ω+4000 Ω)(30×10−6 S)](30×10−6 S)[1+(240 Ω)(30×10−6 S)]{(100)[1+(240 Ω)(30×10−6 S)]−100−(240 Ω+4000 Ω)(30×10−6 S)(100)}+10−4−240 Ω4000 Ω)=1{(4240 Ω)(1.1272)(30×10−6 S)(1.0072)[(100)(1.0072)−100−112.72]+10−4−0.06}=1{0.1434(1.0072)(−112)+10−4−0.06}

Simplify the expression as follows:

VcVb=−16.3476

From Equation (1), substitute Av for VcVb to obtain the voltage gain of the amplifier.

Av=−16.3476

From Equation (9), substitute (1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb for Vc and from Equation (8), substitute hfe(1+Rehoe)1+(Re+RL)hoeIb for Ic in Equation (10) as follows:

Vb={(hie+Re)Ib+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb+hfe(1+Rehoe)1+(Re+RL)hoeIbRe}={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}Ib

Rearrange the expression as follows:

VbIb={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VbIb={(4 kΩ+240 Ω)+(10−4)[1+(240 Ω)(30 μS)][(100)(1+(240 Ω)(30 μS))−100−(240 Ω+4 kΩ)(30 μS)(100)][1+(240 Ω+4 kΩ)(30 μS)](30 μS)+(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)(240 Ω)}=4240 Ω+(10−4)(1.0072)(100.72−100−12.72)33.816 μS+(100.72)(240 Ω)1.1272=4240 Ω−35.7417 Ω+21445 Ω=25649 ΩVbIb≅25.649 kΩ

From Equation (3), substitute Zin for VbIb to obtain the input impedance of the amplifier.

Zin=25.649 kΩ

Consider output voltage (Vc) as 1 V and redraw the circuit in Figure 1 as Figure 2 to find the output impedance of the amplifier.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 2

Apply KVL to the input loop for the circuit in Figure 2 as follows:

Ib(Rs+hie)+hreVc+Re(Ib+Ic)=0

Substitute 1 for Vc as follows:

Ib(Rs+hie)+hre+Re(Ib+Ic)=0

Ib(Rs+hie+Re)+hre+ReIc=0 (11)

Apply KCL at the output node for the circuit in Figure 2 as follows:

Ic=VcRe+1hoe+hfeIb=hoeVc1+hoeRe+hfeIb

Substitute 1 for Vc as follows:

Ic=hoe1+hoeRe+hfeIb

Rearrange the expression as follows:

Ic=hoe+(1+hoeRe)hfeIb1+hoeReIb=(1+hoeRe)Ic−hoe(1+hoeRe)hfe

Ib=Ichfe−hoe(1+hoeRe)hfe (12)

From Equation (12), substitute [Ichfe−hoe(1+hoeRe)hfe] for Ib in Equation (11) as follows:

[Ichfe−hoe(1+hoeRe)hfe](Rs+hie+Re)+hre+ReIc=0[(Rs+hie+Re)hfe+Re]Ic=(Rs+hie+Re)hoe(1+hoeRe)hfe−hre[(Rs+hie+Re)+hfeRehfe]Ic=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)hfeIc=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)(Rs+hie+Re)+hfeRe

Substitute 100 for hfe, 4 kΩ for hie, 1.2 kΩ for Rs, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ic=(1.2 kΩ+4 kΩ+240 Ω)(30 μS)−[1+(30 μS)(240 Ω)](100)(10−4)[1+(30 μS)(240 Ω)](1.2 kΩ+4 kΩ+240 Ω)+(100)(240 Ω)=0.1632−(1.0072)(100)(10−4)(1.0072)(5440 Ω)+24000 Ω=0.153129479 Ω=5.1935×10−6 S

Write the expression for output impedance of the amplifier as follows:

Zout=VcIc

Substitute 1 for Vc and 5.1935×10−6 S for Ic to obtain the output impedance of the amplifier.

Zout=15.1935×10−6 S=192500 Ω=192.5 kΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 19, Problem 92P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the amplifier shown in Figure 19.131 in the textbook.

Expert Solution & Answer

Answer to Problem 92P

The voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.131 in the textbook for the amplifier circuit.

hie=4 kΩhre=10−4hfe=100hoe=30 μS

From the given amplifier circuit, the internal resistance (Rs) is 1.2 kΩ, emitter resistance (Re) is 240 Ω, and the load resistance (RL) is 4 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a amplifier in terms of hybrid parameters as follows:

Av=VcVb (1)

Here,

Vc is the collector voltage, which is the output voltage and

Vb is the base voltage.

Write the expression for current gain of the amplifier as follows:

Ai=IcIb (2)

Here,

Ib is the base current and

Ic is the collector current.

Write the expression for input impedance of the amplifier as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 1

From Figure 1, write the expression for emitter current as follows:

Ie=Ib+Ic (4)

Write the expression for base voltage from the circuit in Figure 1 as follows:

Vb=hieIb+hreVc+(Ib+Ic)Re (5)

Write the expression for collector current as follows:

Ic=hfeIb+VcRe+1hoe (6)

Write the expression for collector voltage as follows:

Vc=−IcRL (7)

From Equation (7), substitute (−IcRL) for Vc in Equation (6) as follows:

Ic=hfeIb+(−IcRL)Re+1hoe(1+hoeRLRehoe+1)Ic=hfeIb

Rearrange the expression as follows:

IcIb=hfe1+hoeRLRehoe+1=hfe(1+Rehoe)1+Rehoe+hoeRL

IcIb=hfe(1+Rehoe)1+(Re+RL)hoe (8)

From Equation (2), substitute Ai for IcIb as follows:

Ai=hfe(1+Rehoe)1+(Re+RL)hoe

Substitute 100 for hfe, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ai=(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)=(100)[1+(240 Ω)(30×10−6 S)]1+(240 Ω+4000 Ω)(30×10−6 S)=(100)(1.0072)1.1272=89.3542

From Equation (6), substitute (hfeIb+VcRe+1hoe) for Ic in Equation (8) as follows:

hfeIb+VcRe+1hoeIb=hfe(1+Rehoe)1+(Re+RL)hoe

Rearrange the expression as follows:

hfeIb+VcRe+1hoe=hfe(1+Rehoe)1+(Re+RL)hoeIb[hfe(1+Rehoe)1+(Re+RL)hoe−hfe]Ib=VcRe+1hoe[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe1+(Re+RL)hoe]Ib=hoeVc1+Rehoe

Ib=[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] (9)

Rearrange the expression in Equation (5) as follows:

Vb=hieIb+hreVc+(IbRe+IcRe)

Vb=(hie+Re)Ib+hreVc+IcRe (10)

From Equations (7) and (9), substitute (−VcRL) for Ic and [1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] for Ib as follows:

Vb=(hie+Re){[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]}+hreVc+(−VcRL)Re={(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}Vc

Rearrange the expression as follows:

VcVb=1{(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VcVb=1((4 kΩ+240 Ω)[1+(240 Ω+4 kΩ)(30 μS)](30 μS)[1+(240 Ω)(30 μS)]{(100)[1+(240 Ω)(30 μS)]−100−(240 Ω+4 kΩ)(30 μS)(100)}+10−4−240 Ω4 kΩ)=1((4000 Ω+240 Ω)[1+(240 Ω+4000 Ω)(30×10−6 S)](30×10−6 S)[1+(240 Ω)(30×10−6 S)]{(100)[1+(240 Ω)(30×10−6 S)]−100−(240 Ω+4000 Ω)(30×10−6 S)(100)}+10−4−240 Ω4000 Ω)=1{(4240 Ω)(1.1272)(30×10−6 S)(1.0072)[(100)(1.0072)−100−112.72]+10−4−0.06}=1{0.1434(1.0072)(−112)+10−4−0.06}

Simplify the expression as follows:

VcVb=−16.3476

From Equation (1), substitute Av for VcVb to obtain the voltage gain of the amplifier.

Av=−16.3476

From Equation (9), substitute (1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb for Vc and from Equation (8), substitute hfe(1+Rehoe)1+(Re+RL)hoeIb for Ic in Equation (10) as follows:

Vb={(hie+Re)Ib+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb+hfe(1+Rehoe)1+(Re+RL)hoeIbRe}={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}Ib

Rearrange the expression as follows:

VbIb={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VbIb={(4 kΩ+240 Ω)+(10−4)[1+(240 Ω)(30 μS)][(100)(1+(240 Ω)(30 μS))−100−(240 Ω+4 kΩ)(30 μS)(100)][1+(240 Ω+4 kΩ)(30 μS)](30 μS)+(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)(240 Ω)}=4240 Ω+(10−4)(1.0072)(100.72−100−12.72)33.816 μS+(100.72)(240 Ω)1.1272=4240 Ω−35.7417 Ω+21445 Ω=25649 ΩVbIb≅25.649 kΩ

From Equation (3), substitute Zin for VbIb to obtain the input impedance of the amplifier.

Zin=25.649 kΩ

Consider output voltage (Vc) as 1 V and redraw the circuit in Figure 1 as Figure 2 to find the output impedance of the amplifier.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 2

Apply KVL to the input loop for the circuit in Figure 2 as follows:

Ib(Rs+hie)+hreVc+Re(Ib+Ic)=0

Substitute 1 for Vc as follows:

Ib(Rs+hie)+hre+Re(Ib+Ic)=0

Ib(Rs+hie+Re)+hre+ReIc=0 (11)

Apply KCL at the output node for the circuit in Figure 2 as follows:

Ic=VcRe+1hoe+hfeIb=hoeVc1+hoeRe+hfeIb

Substitute 1 for Vc as follows:

Ic=hoe1+hoeRe+hfeIb

Rearrange the expression as follows:

Ic=hoe+(1+hoeRe)hfeIb1+hoeReIb=(1+hoeRe)Ic−hoe(1+hoeRe)hfe

Ib=Ichfe−hoe(1+hoeRe)hfe (12)

From Equation (12), substitute [Ichfe−hoe(1+hoeRe)hfe] for Ib in Equation (11) as follows:

[Ichfe−hoe(1+hoeRe)hfe](Rs+hie+Re)+hre+ReIc=0[(Rs+hie+Re)hfe+Re]Ic=(Rs+hie+Re)hoe(1+hoeRe)hfe−hre[(Rs+hie+Re)+hfeRehfe]Ic=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)hfeIc=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)(Rs+hie+Re)+hfeRe

Substitute 100 for hfe, 4 kΩ for hie, 1.2 kΩ for Rs, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ic=(1.2 kΩ+4 kΩ+240 Ω)(30 μS)−[1+(30 μS)(240 Ω)](100)(10−4)[1+(30 μS)(240 Ω)](1.2 kΩ+4 kΩ+240 Ω)+(100)(240 Ω)=0.1632−(1.0072)(100)(10−4)(1.0072)(5440 Ω)+24000 Ω=0.153129479 Ω=5.1935×10−6 S

Write the expression for output impedance of the amplifier as follows:

Zout=VcIc

Substitute 1 for Vc and 5.1935×10−6 S for Ic to obtain the output impedance of the amplifier.

Zout=15.1935×10−6 S=192500 Ω=192.5 kΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 19, Problem 92P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the amplifier shown in Figure 19.131 in the textbook.

Expert Solution & Answer

Answer to Problem 92P

The voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.131 in the textbook for the amplifier circuit.

hie=4 kΩhre=10−4hfe=100hoe=30 μS

From the given amplifier circuit, the internal resistance (Rs) is 1.2 kΩ, emitter resistance (Re) is 240 Ω, and the load resistance (RL) is 4 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a amplifier in terms of hybrid parameters as follows:

Av=VcVb (1)

Here,

Vc is the collector voltage, which is the output voltage and

Vb is the base voltage.

Write the expression for current gain of the amplifier as follows:

Ai=IcIb (2)

Here,

Ib is the base current and

Ic is the collector current.

Write the expression for input impedance of the amplifier as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 1

From Figure 1, write the expression for emitter current as follows:

Ie=Ib+Ic (4)

Write the expression for base voltage from the circuit in Figure 1 as follows:

Vb=hieIb+hreVc+(Ib+Ic)Re (5)

Write the expression for collector current as follows:

Ic=hfeIb+VcRe+1hoe (6)

Write the expression for collector voltage as follows:

Vc=−IcRL (7)

From Equation (7), substitute (−IcRL) for Vc in Equation (6) as follows:

Ic=hfeIb+(−IcRL)Re+1hoe(1+hoeRLRehoe+1)Ic=hfeIb

Rearrange the expression as follows:

IcIb=hfe1+hoeRLRehoe+1=hfe(1+Rehoe)1+Rehoe+hoeRL

IcIb=hfe(1+Rehoe)1+(Re+RL)hoe (8)

From Equation (2), substitute Ai for IcIb as follows:

Ai=hfe(1+Rehoe)1+(Re+RL)hoe

Substitute 100 for hfe, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ai=(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)=(100)[1+(240 Ω)(30×10−6 S)]1+(240 Ω+4000 Ω)(30×10−6 S)=(100)(1.0072)1.1272=89.3542

From Equation (6), substitute (hfeIb+VcRe+1hoe) for Ic in Equation (8) as follows:

hfeIb+VcRe+1hoeIb=hfe(1+Rehoe)1+(Re+RL)hoe

Rearrange the expression as follows:

hfeIb+VcRe+1hoe=hfe(1+Rehoe)1+(Re+RL)hoeIb[hfe(1+Rehoe)1+(Re+RL)hoe−hfe]Ib=VcRe+1hoe[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe1+(Re+RL)hoe]Ib=hoeVc1+Rehoe

Ib=[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] (9)

Rearrange the expression in Equation (5) as follows:

Vb=hieIb+hreVc+(IbRe+IcRe)

Vb=(hie+Re)Ib+hreVc+IcRe (10)

From Equations (7) and (9), substitute (−VcRL) for Ic and [1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] for Ib as follows:

Vb=(hie+Re){[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]}+hreVc+(−VcRL)Re={(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}Vc

Rearrange the expression as follows:

VcVb=1{(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VcVb=1((4 kΩ+240 Ω)[1+(240 Ω+4 kΩ)(30 μS)](30 μS)[1+(240 Ω)(30 μS)]{(100)[1+(240 Ω)(30 μS)]−100−(240 Ω+4 kΩ)(30 μS)(100)}+10−4−240 Ω4 kΩ)=1((4000 Ω+240 Ω)[1+(240 Ω+4000 Ω)(30×10−6 S)](30×10−6 S)[1+(240 Ω)(30×10−6 S)]{(100)[1+(240 Ω)(30×10−6 S)]−100−(240 Ω+4000 Ω)(30×10−6 S)(100)}+10−4−240 Ω4000 Ω)=1{(4240 Ω)(1.1272)(30×10−6 S)(1.0072)[(100)(1.0072)−100−112.72]+10−4−0.06}=1{0.1434(1.0072)(−112)+10−4−0.06}

Simplify the expression as follows:

VcVb=−16.3476

From Equation (1), substitute Av for VcVb to obtain the voltage gain of the amplifier.

Av=−16.3476

From Equation (9), substitute (1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb for Vc and from Equation (8), substitute hfe(1+Rehoe)1+(Re+RL)hoeIb for Ic in Equation (10) as follows:

Vb={(hie+Re)Ib+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb+hfe(1+Rehoe)1+(Re+RL)hoeIbRe}={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}Ib

Rearrange the expression as follows:

VbIb={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VbIb={(4 kΩ+240 Ω)+(10−4)[1+(240 Ω)(30 μS)][(100)(1+(240 Ω)(30 μS))−100−(240 Ω+4 kΩ)(30 μS)(100)][1+(240 Ω+4 kΩ)(30 μS)](30 μS)+(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)(240 Ω)}=4240 Ω+(10−4)(1.0072)(100.72−100−12.72)33.816 μS+(100.72)(240 Ω)1.1272=4240 Ω−35.7417 Ω+21445 Ω=25649 ΩVbIb≅25.649 kΩ

From Equation (3), substitute Zin for VbIb to obtain the input impedance of the amplifier.

Zin=25.649 kΩ

Consider output voltage (Vc) as 1 V and redraw the circuit in Figure 1 as Figure 2 to find the output impedance of the amplifier.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 2

Apply KVL to the input loop for the circuit in Figure 2 as follows:

Ib(Rs+hie)+hreVc+Re(Ib+Ic)=0

Substitute 1 for Vc as follows:

Ib(Rs+hie)+hre+Re(Ib+Ic)=0

Ib(Rs+hie+Re)+hre+ReIc=0 (11)

Apply KCL at the output node for the circuit in Figure 2 as follows:

Ic=VcRe+1hoe+hfeIb=hoeVc1+hoeRe+hfeIb

Substitute 1 for Vc as follows:

Ic=hoe1+hoeRe+hfeIb

Rearrange the expression as follows:

Ic=hoe+(1+hoeRe)hfeIb1+hoeReIb=(1+hoeRe)Ic−hoe(1+hoeRe)hfe

Ib=Ichfe−hoe(1+hoeRe)hfe (12)

From Equation (12), substitute [Ichfe−hoe(1+hoeRe)hfe] for Ib in Equation (11) as follows:

[Ichfe−hoe(1+hoeRe)hfe](Rs+hie+Re)+hre+ReIc=0[(Rs+hie+Re)hfe+Re]Ic=(Rs+hie+Re)hoe(1+hoeRe)hfe−hre[(Rs+hie+Re)+hfeRehfe]Ic=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)hfeIc=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)(Rs+hie+Re)+hfeRe

Substitute 100 for hfe, 4 kΩ for hie, 1.2 kΩ for Rs, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ic=(1.2 kΩ+4 kΩ+240 Ω)(30 μS)−[1+(30 μS)(240 Ω)](100)(10−4)[1+(30 μS)(240 Ω)](1.2 kΩ+4 kΩ+240 Ω)+(100)(240 Ω)=0.1632−(1.0072)(100)(10−4)(1.0072)(5440 Ω)+24000 Ω=0.153129479 Ω=5.1935×10−6 S

Write the expression for output impedance of the amplifier as follows:

Zout=VcIc

Substitute 1 for Vc and 5.1935×10−6 S for Ic to obtain the output impedance of the amplifier.

Zout=15.1935×10−6 S=192500 Ω=192.5 kΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Answer 6

Chapter 19, Problem 92P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the amplifier shown in Figure 19.131 in the textbook.

Expert Solution & Answer

Answer to Problem 92P

The voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.131 in the textbook for the amplifier circuit.

hie=4 kΩhre=10−4hfe=100hoe=30 μS

From the given amplifier circuit, the internal resistance (Rs) is 1.2 kΩ, emitter resistance (Re) is 240 Ω, and the load resistance (RL) is 4 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a amplifier in terms of hybrid parameters as follows:

Av=VcVb (1)

Here,

Vc is the collector voltage, which is the output voltage and

Vb is the base voltage.

Write the expression for current gain of the amplifier as follows:

Ai=IcIb (2)

Here,

Ib is the base current and

Ic is the collector current.

Write the expression for input impedance of the amplifier as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 1

From Figure 1, write the expression for emitter current as follows:

Ie=Ib+Ic (4)

Write the expression for base voltage from the circuit in Figure 1 as follows:

Vb=hieIb+hreVc+(Ib+Ic)Re (5)

Write the expression for collector current as follows:

Ic=hfeIb+VcRe+1hoe (6)

Write the expression for collector voltage as follows:

Vc=−IcRL (7)

From Equation (7), substitute (−IcRL) for Vc in Equation (6) as follows:

Ic=hfeIb+(−IcRL)Re+1hoe(1+hoeRLRehoe+1)Ic=hfeIb

Rearrange the expression as follows:

IcIb=hfe1+hoeRLRehoe+1=hfe(1+Rehoe)1+Rehoe+hoeRL

IcIb=hfe(1+Rehoe)1+(Re+RL)hoe (8)

From Equation (2), substitute Ai for IcIb as follows:

Ai=hfe(1+Rehoe)1+(Re+RL)hoe

Substitute 100 for hfe, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ai=(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)=(100)[1+(240 Ω)(30×10−6 S)]1+(240 Ω+4000 Ω)(30×10−6 S)=(100)(1.0072)1.1272=89.3542

From Equation (6), substitute (hfeIb+VcRe+1hoe) for Ic in Equation (8) as follows:

hfeIb+VcRe+1hoeIb=hfe(1+Rehoe)1+(Re+RL)hoe

Rearrange the expression as follows:

hfeIb+VcRe+1hoe=hfe(1+Rehoe)1+(Re+RL)hoeIb[hfe(1+Rehoe)1+(Re+RL)hoe−hfe]Ib=VcRe+1hoe[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe1+(Re+RL)hoe]Ib=hoeVc1+Rehoe

Ib=[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] (9)

Rearrange the expression in Equation (5) as follows:

Vb=hieIb+hreVc+(IbRe+IcRe)

Vb=(hie+Re)Ib+hreVc+IcRe (10)

From Equations (7) and (9), substitute (−VcRL) for Ic and [1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] for Ib as follows:

Vb=(hie+Re){[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]}+hreVc+(−VcRL)Re={(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}Vc

Rearrange the expression as follows:

VcVb=1{(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VcVb=1((4 kΩ+240 Ω)[1+(240 Ω+4 kΩ)(30 μS)](30 μS)[1+(240 Ω)(30 μS)]{(100)[1+(240 Ω)(30 μS)]−100−(240 Ω+4 kΩ)(30 μS)(100)}+10−4−240 Ω4 kΩ)=1((4000 Ω+240 Ω)[1+(240 Ω+4000 Ω)(30×10−6 S)](30×10−6 S)[1+(240 Ω)(30×10−6 S)]{(100)[1+(240 Ω)(30×10−6 S)]−100−(240 Ω+4000 Ω)(30×10−6 S)(100)}+10−4−240 Ω4000 Ω)=1{(4240 Ω)(1.1272)(30×10−6 S)(1.0072)[(100)(1.0072)−100−112.72]+10−4−0.06}=1{0.1434(1.0072)(−112)+10−4−0.06}

Simplify the expression as follows:

VcVb=−16.3476

From Equation (1), substitute Av for VcVb to obtain the voltage gain of the amplifier.

Av=−16.3476

From Equation (9), substitute (1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb for Vc and from Equation (8), substitute hfe(1+Rehoe)1+(Re+RL)hoeIb for Ic in Equation (10) as follows:

Vb={(hie+Re)Ib+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb+hfe(1+Rehoe)1+(Re+RL)hoeIbRe}={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}Ib

Rearrange the expression as follows:

VbIb={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VbIb={(4 kΩ+240 Ω)+(10−4)[1+(240 Ω)(30 μS)][(100)(1+(240 Ω)(30 μS))−100−(240 Ω+4 kΩ)(30 μS)(100)][1+(240 Ω+4 kΩ)(30 μS)](30 μS)+(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)(240 Ω)}=4240 Ω+(10−4)(1.0072)(100.72−100−12.72)33.816 μS+(100.72)(240 Ω)1.1272=4240 Ω−35.7417 Ω+21445 Ω=25649 ΩVbIb≅25.649 kΩ

From Equation (3), substitute Zin for VbIb to obtain the input impedance of the amplifier.

Zin=25.649 kΩ

Consider output voltage (Vc) as 1 V and redraw the circuit in Figure 1 as Figure 2 to find the output impedance of the amplifier.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 2

Apply KVL to the input loop for the circuit in Figure 2 as follows:

Ib(Rs+hie)+hreVc+Re(Ib+Ic)=0

Substitute 1 for Vc as follows:

Ib(Rs+hie)+hre+Re(Ib+Ic)=0

Ib(Rs+hie+Re)+hre+ReIc=0 (11)

Apply KCL at the output node for the circuit in Figure 2 as follows:

Ic=VcRe+1hoe+hfeIb=hoeVc1+hoeRe+hfeIb

Substitute 1 for Vc as follows:

Ic=hoe1+hoeRe+hfeIb

Rearrange the expression as follows:

Ic=hoe+(1+hoeRe)hfeIb1+hoeReIb=(1+hoeRe)Ic−hoe(1+hoeRe)hfe

Ib=Ichfe−hoe(1+hoeRe)hfe (12)

From Equation (12), substitute [Ichfe−hoe(1+hoeRe)hfe] for Ib in Equation (11) as follows:

[Ichfe−hoe(1+hoeRe)hfe](Rs+hie+Re)+hre+ReIc=0[(Rs+hie+Re)hfe+Re]Ic=(Rs+hie+Re)hoe(1+hoeRe)hfe−hre[(Rs+hie+Re)+hfeRehfe]Ic=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)hfeIc=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)(Rs+hie+Re)+hfeRe

Substitute 100 for hfe, 4 kΩ for hie, 1.2 kΩ for Rs, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ic=(1.2 kΩ+4 kΩ+240 Ω)(30 μS)−[1+(30 μS)(240 Ω)](100)(10−4)[1+(30 μS)(240 Ω)](1.2 kΩ+4 kΩ+240 Ω)+(100)(240 Ω)=0.1632−(1.0072)(100)(10−4)(1.0072)(5440 Ω)+24000 Ω=0.153129479 Ω=5.1935×10−6 S

Write the expression for output impedance of the amplifier as follows:

Zout=VcIc

Substitute 1 for Vc and 5.1935×10−6 S for Ic to obtain the output impedance of the amplifier.

Zout=15.1935×10−6 S=192500 Ω=192.5 kΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Answer 7

Chapter 19, Problem 92P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the amplifier shown in Figure 19.131 in the textbook.

Answer 8

Expert Solution & Answer

Answer to Problem 92P

The voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given Data:

Refer to Figure 19.131 in the textbook for the amplifier circuit.

hie=4 kΩhre=10−4hfe=100hoe=30 μS

From the given amplifier circuit, the internal resistance (Rs) is 1.2 kΩ, emitter resistance (Re) is 240 Ω, and the load resistance (RL) is 4 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a amplifier in terms of hybrid parameters as follows:

Av=VcVb (1)

Here,

Vc is the collector voltage, which is the output voltage and

Vb is the base voltage.

Write the expression for current gain of the amplifier as follows:

Ai=IcIb (2)

Here,

Ib is the base current and

Ic is the collector current.

Write the expression for input impedance of the amplifier as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 1

From Figure 1, write the expression for emitter current as follows:

Ie=Ib+Ic (4)

Write the expression for base voltage from the circuit in Figure 1 as follows:

Vb=hieIb+hreVc+(Ib+Ic)Re (5)

Write the expression for collector current as follows:

Ic=hfeIb+VcRe+1hoe (6)

Write the expression for collector voltage as follows:

Vc=−IcRL (7)

From Equation (7), substitute (−IcRL) for Vc in Equation (6) as follows:

Ic=hfeIb+(−IcRL)Re+1hoe(1+hoeRLRehoe+1)Ic=hfeIb

Rearrange the expression as follows:

IcIb=hfe1+hoeRLRehoe+1=hfe(1+Rehoe)1+Rehoe+hoeRL

IcIb=hfe(1+Rehoe)1+(Re+RL)hoe (8)

From Equation (2), substitute Ai for IcIb as follows:

Ai=hfe(1+Rehoe)1+(Re+RL)hoe

Substitute 100 for hfe, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ai=(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)=(100)[1+(240 Ω)(30×10−6 S)]1+(240 Ω+4000 Ω)(30×10−6 S)=(100)(1.0072)1.1272=89.3542

From Equation (6), substitute (hfeIb+VcRe+1hoe) for Ic in Equation (8) as follows:

hfeIb+VcRe+1hoeIb=hfe(1+Rehoe)1+(Re+RL)hoe

Rearrange the expression as follows:

hfeIb+VcRe+1hoe=hfe(1+Rehoe)1+(Re+RL)hoeIb[hfe(1+Rehoe)1+(Re+RL)hoe−hfe]Ib=VcRe+1hoe[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe1+(Re+RL)hoe]Ib=hoeVc1+Rehoe

Ib=[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] (9)

Rearrange the expression in Equation (5) as follows:

Vb=hieIb+hreVc+(IbRe+IcRe)

Vb=(hie+Re)Ib+hreVc+IcRe (10)

From Equations (7) and (9), substitute (−VcRL) for Ic and [1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe] for Ib as follows:

Vb=(hie+Re){[1+(Re+RL)hoe]hoeVc(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]}+hreVc+(−VcRL)Re={(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}Vc

Rearrange the expression as follows:

VcVb=1{(hie+Re)[1+(Re+RL)hoe]hoe(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe]+hre−ReRL}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VcVb=1((4 kΩ+240 Ω)[1+(240 Ω+4 kΩ)(30 μS)](30 μS)[1+(240 Ω)(30 μS)]{(100)[1+(240 Ω)(30 μS)]−100−(240 Ω+4 kΩ)(30 μS)(100)}+10−4−240 Ω4 kΩ)=1((4000 Ω+240 Ω)[1+(240 Ω+4000 Ω)(30×10−6 S)](30×10−6 S)[1+(240 Ω)(30×10−6 S)]{(100)[1+(240 Ω)(30×10−6 S)]−100−(240 Ω+4000 Ω)(30×10−6 S)(100)}+10−4−240 Ω4000 Ω)=1{(4240 Ω)(1.1272)(30×10−6 S)(1.0072)[(100)(1.0072)−100−112.72]+10−4−0.06}=1{0.1434(1.0072)(−112)+10−4−0.06}

Simplify the expression as follows:

VcVb=−16.3476

From Equation (1), substitute Av for VcVb to obtain the voltage gain of the amplifier.

Av=−16.3476

From Equation (9), substitute (1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb for Vc and from Equation (8), substitute hfe(1+Rehoe)1+(Re+RL)hoeIb for Ic in Equation (10) as follows:

Vb={(hie+Re)Ib+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoeIb+hfe(1+Rehoe)1+(Re+RL)hoeIbRe}={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}Ib

Rearrange the expression as follows:

VbIb={(hie+Re)+hre(1+Rehoe)[hfe(1+Rehoe)−hfe−(Re+RL)hoehfe][1+(Re+RL)hoe]hoe+hfe(1+Rehoe)1+(Re+RL)hoeRe}

Substitute 100 for hfe, 4 kΩ for hie, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

VbIb={(4 kΩ+240 Ω)+(10−4)[1+(240 Ω)(30 μS)][(100)(1+(240 Ω)(30 μS))−100−(240 Ω+4 kΩ)(30 μS)(100)][1+(240 Ω+4 kΩ)(30 μS)](30 μS)+(100)[1+(240 Ω)(30 μS)]1+(240 Ω+4 kΩ)(30 μS)(240 Ω)}=4240 Ω+(10−4)(1.0072)(100.72−100−12.72)33.816 μS+(100.72)(240 Ω)1.1272=4240 Ω−35.7417 Ω+21445 Ω=25649 ΩVbIb≅25.649 kΩ

From Equation (3), substitute Zin for VbIb to obtain the input impedance of the amplifier.

Zin=25.649 kΩ

Consider output voltage (Vc) as 1 V and redraw the circuit in Figure 1 as Figure 2 to find the output impedance of the amplifier.

Fundamentals of Electric Circuits, Chapter 19, Problem 92P , additional homework tip 2

Apply KVL to the input loop for the circuit in Figure 2 as follows:

Ib(Rs+hie)+hreVc+Re(Ib+Ic)=0

Substitute 1 for Vc as follows:

Ib(Rs+hie)+hre+Re(Ib+Ic)=0

Ib(Rs+hie+Re)+hre+ReIc=0 (11)

Apply KCL at the output node for the circuit in Figure 2 as follows:

Ic=VcRe+1hoe+hfeIb=hoeVc1+hoeRe+hfeIb

Substitute 1 for Vc as follows:

Ic=hoe1+hoeRe+hfeIb

Rearrange the expression as follows:

Ic=hoe+(1+hoeRe)hfeIb1+hoeReIb=(1+hoeRe)Ic−hoe(1+hoeRe)hfe

Ib=Ichfe−hoe(1+hoeRe)hfe (12)

From Equation (12), substitute [Ichfe−hoe(1+hoeRe)hfe] for Ib in Equation (11) as follows:

[Ichfe−hoe(1+hoeRe)hfe](Rs+hie+Re)+hre+ReIc=0[(Rs+hie+Re)hfe+Re]Ic=(Rs+hie+Re)hoe(1+hoeRe)hfe−hre[(Rs+hie+Re)+hfeRehfe]Ic=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)hfeIc=(Rs+hie+Re)hoe−(1+hoeRe)hfehre(1+hoeRe)(Rs+hie+Re)+hfeRe

Substitute 100 for hfe, 4 kΩ for hie, 1.2 kΩ for Rs, 10−4 for hre, 30 μS for hoe, 240 Ω for Re, and 4 kΩ for RL as follows:

Ic=(1.2 kΩ+4 kΩ+240 Ω)(30 μS)−[1+(30 μS)(240 Ω)](100)(10−4)[1+(30 μS)(240 Ω)](1.2 kΩ+4 kΩ+240 Ω)+(100)(240 Ω)=0.1632−(1.0072)(100)(10−4)(1.0072)(5440 Ω)+24000 Ω=0.153129479 Ω=5.1935×10−6 S

Write the expression for output impedance of the amplifier as follows:

Zout=VcIc

Substitute 1 for Vc and 5.1935×10−6 S for Ic to obtain the output impedance of the amplifier.

Zout=15.1935×10−6 S=192500 Ω=192.5 kΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the amplifier are −16.3476_, 89.3542_, 25.649 kΩ_, and 192.5 kΩ_, respectively.

Answer 12

Ch. 19.2 - Find the z parameters of the two-port network in...Ch. 19.2 - Calculate I1 and I2 in the two-port of Fig. 19.11....Ch. 19.3 - Obtain the y parameters for the T network shown in...Ch. 19.3 - Obtain the y parameters for the circuit in Fig....Ch. 19.4 - Determine the h parameters for the circuit in Fig....Ch. 19.4 - Find the impedance at the input port of the...Ch. 19.4 - For the ladder network in Fig. 19.30, determine...Ch. 19.5 - Find the transmission parameters for the circuit...Ch. 19.5 - Prob. 9PP Ch. 19.6 - Determine [y] and [T] of a two-port network whose...

Calculate the voltage gain, current gain, input impedance, and output impedance for the amplifier shown in Figure 19.131 in the textbook.

Calculate the voltage gain, current gain, input impedance, and output impedance for the amplifier shown in Figure 19.131 in the textbook.

Answer to Problem 92P

Explanation of Solution

Want to see more full solutions like this?

Chapter 19 Solutions