Calculate the voltage gain, current gain, input impedance, and output impedance for the transistor network shown in Figure 19.132 in the textbook.

Question

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Answer 1

Question

Chapter 19, Problem 93P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the transistor network shown in Figure 19.132 in the textbook.

Expert Solution & Answer

Answer to Problem 93P

The voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.132 in the textbook for the transistor network circuit.

hie=2 kΩhre=2.5×10−4hfe=150hoe=10 μS

From the given transistor network circuit, the internal resistance (Rs) is 1 kΩ, emitter resistance (Re) is 0.2 kΩ, which is 200 Ω, and the load resistance (RL) is 3.8 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a transistor network 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 transistor network 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 transistor network as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ai=(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)=(150)[1+(200 Ω)(10×10−6 S)]1+(200 Ω+3800 Ω)(10×10−6 S)=(150)(1.002)1.04=144.5192

Ai≅144.5

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VcVb=1((2 kΩ+200 Ω)[1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)[1+(200 Ω)(10 μS)]{(150)[1+(200 Ω)(10 μS)]−150−(200 Ω+3.8 kΩ)(10 μS)(150)}+(2.5×10−4)−200 Ω3.8 kΩ)=1((2000 Ω+200 Ω)[1+(200 Ω+3800 Ω)(10×10−6 S)](10×10−6 S)[1+(200 Ω)(10×10−6 S)]{(150)[1+(200 Ω)(10×10−6 S)]−150−(200 Ω+3800 Ω)(10×10−6 S)(150)}+(2.5×10−4)−200 Ω3800 Ω)=1{(2200 Ω)(1.04)(10×10−6 S)(1.002)[(150)(1.002)−150−6]+(2.5×10−4)−0.0526}=1{−0.004+(2.5×10−4)−0.0526}

Simplify the expression as follows:

VcVb=−17.7462≅−17.74

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

Av=−17.74

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VbIb={(2 kΩ+200 Ω)+(2.5×10−4)[1+(200 Ω)(10 μS)][(150)(1+(200 Ω)(10 μS))−150−(200 Ω+3.8 kΩ)(10 μS)(150)][1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)+(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)(200 Ω)}=2200 Ω+(2.5×10−4)(1.002)(150.3−150−6)10.4 μS+(150.3)(200 Ω)1.04=2200 Ω−137.2933 Ω+28904 Ω=30967 ΩVbIb=30.967 kΩ≅31 kΩ

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

Zin=31 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 transistor network.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 2 kΩ for hie, 1 kΩ for Rs, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ic=(1 kΩ+2 kΩ+200 Ω)(10 μS)−[1+(10 μS)(200 Ω)](150)(2.5×10−4)[1+(10 μS)(200 Ω)](1 kΩ+2 kΩ+200 Ω)+(150)(200 Ω)=0.032−(1.002)(150)(2.5×10−4)(1.002)(3200 Ω)+30000 Ω=−0.005633206 Ω=−0.16864×10−6 S

Write the expression for output impedance of the transistor network as follows:

Zout=VcIc

Substitute 1 for Vc and (−0.16864×10−6 S) for Ic to obtain the output impedance of the transistor network.

Zout=1−0.16864×10−6 S=−5.9296×106 Ω=−5.9296 MΩ≅−6 MΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

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Answer 2

Question

Chapter 19, Problem 93P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the transistor network shown in Figure 19.132 in the textbook.

Expert Solution & Answer

Answer to Problem 93P

The voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.132 in the textbook for the transistor network circuit.

hie=2 kΩhre=2.5×10−4hfe=150hoe=10 μS

From the given transistor network circuit, the internal resistance (Rs) is 1 kΩ, emitter resistance (Re) is 0.2 kΩ, which is 200 Ω, and the load resistance (RL) is 3.8 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a transistor network 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 transistor network 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 transistor network as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ai=(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)=(150)[1+(200 Ω)(10×10−6 S)]1+(200 Ω+3800 Ω)(10×10−6 S)=(150)(1.002)1.04=144.5192

Ai≅144.5

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VcVb=1((2 kΩ+200 Ω)[1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)[1+(200 Ω)(10 μS)]{(150)[1+(200 Ω)(10 μS)]−150−(200 Ω+3.8 kΩ)(10 μS)(150)}+(2.5×10−4)−200 Ω3.8 kΩ)=1((2000 Ω+200 Ω)[1+(200 Ω+3800 Ω)(10×10−6 S)](10×10−6 S)[1+(200 Ω)(10×10−6 S)]{(150)[1+(200 Ω)(10×10−6 S)]−150−(200 Ω+3800 Ω)(10×10−6 S)(150)}+(2.5×10−4)−200 Ω3800 Ω)=1{(2200 Ω)(1.04)(10×10−6 S)(1.002)[(150)(1.002)−150−6]+(2.5×10−4)−0.0526}=1{−0.004+(2.5×10−4)−0.0526}

Simplify the expression as follows:

VcVb=−17.7462≅−17.74

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

Av=−17.74

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VbIb={(2 kΩ+200 Ω)+(2.5×10−4)[1+(200 Ω)(10 μS)][(150)(1+(200 Ω)(10 μS))−150−(200 Ω+3.8 kΩ)(10 μS)(150)][1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)+(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)(200 Ω)}=2200 Ω+(2.5×10−4)(1.002)(150.3−150−6)10.4 μS+(150.3)(200 Ω)1.04=2200 Ω−137.2933 Ω+28904 Ω=30967 ΩVbIb=30.967 kΩ≅31 kΩ

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

Zin=31 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 transistor network.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 2 kΩ for hie, 1 kΩ for Rs, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ic=(1 kΩ+2 kΩ+200 Ω)(10 μS)−[1+(10 μS)(200 Ω)](150)(2.5×10−4)[1+(10 μS)(200 Ω)](1 kΩ+2 kΩ+200 Ω)+(150)(200 Ω)=0.032−(1.002)(150)(2.5×10−4)(1.002)(3200 Ω)+30000 Ω=−0.005633206 Ω=−0.16864×10−6 S

Write the expression for output impedance of the transistor network as follows:

Zout=VcIc

Substitute 1 for Vc and (−0.16864×10−6 S) for Ic to obtain the output impedance of the transistor network.

Zout=1−0.16864×10−6 S=−5.9296×106 Ω=−5.9296 MΩ≅−6 MΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

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Answer 3

Question

Chapter 19, Problem 93P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the transistor network shown in Figure 19.132 in the textbook.

Expert Solution & Answer

Answer to Problem 93P

The voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.132 in the textbook for the transistor network circuit.

hie=2 kΩhre=2.5×10−4hfe=150hoe=10 μS

From the given transistor network circuit, the internal resistance (Rs) is 1 kΩ, emitter resistance (Re) is 0.2 kΩ, which is 200 Ω, and the load resistance (RL) is 3.8 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a transistor network 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 transistor network 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 transistor network as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ai=(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)=(150)[1+(200 Ω)(10×10−6 S)]1+(200 Ω+3800 Ω)(10×10−6 S)=(150)(1.002)1.04=144.5192

Ai≅144.5

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VcVb=1((2 kΩ+200 Ω)[1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)[1+(200 Ω)(10 μS)]{(150)[1+(200 Ω)(10 μS)]−150−(200 Ω+3.8 kΩ)(10 μS)(150)}+(2.5×10−4)−200 Ω3.8 kΩ)=1((2000 Ω+200 Ω)[1+(200 Ω+3800 Ω)(10×10−6 S)](10×10−6 S)[1+(200 Ω)(10×10−6 S)]{(150)[1+(200 Ω)(10×10−6 S)]−150−(200 Ω+3800 Ω)(10×10−6 S)(150)}+(2.5×10−4)−200 Ω3800 Ω)=1{(2200 Ω)(1.04)(10×10−6 S)(1.002)[(150)(1.002)−150−6]+(2.5×10−4)−0.0526}=1{−0.004+(2.5×10−4)−0.0526}

Simplify the expression as follows:

VcVb=−17.7462≅−17.74

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

Av=−17.74

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VbIb={(2 kΩ+200 Ω)+(2.5×10−4)[1+(200 Ω)(10 μS)][(150)(1+(200 Ω)(10 μS))−150−(200 Ω+3.8 kΩ)(10 μS)(150)][1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)+(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)(200 Ω)}=2200 Ω+(2.5×10−4)(1.002)(150.3−150−6)10.4 μS+(150.3)(200 Ω)1.04=2200 Ω−137.2933 Ω+28904 Ω=30967 ΩVbIb=30.967 kΩ≅31 kΩ

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

Zin=31 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 transistor network.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 2 kΩ for hie, 1 kΩ for Rs, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ic=(1 kΩ+2 kΩ+200 Ω)(10 μS)−[1+(10 μS)(200 Ω)](150)(2.5×10−4)[1+(10 μS)(200 Ω)](1 kΩ+2 kΩ+200 Ω)+(150)(200 Ω)=0.032−(1.002)(150)(2.5×10−4)(1.002)(3200 Ω)+30000 Ω=−0.005633206 Ω=−0.16864×10−6 S

Write the expression for output impedance of the transistor network as follows:

Zout=VcIc

Substitute 1 for Vc and (−0.16864×10−6 S) for Ic to obtain the output impedance of the transistor network.

Zout=1−0.16864×10−6 S=−5.9296×106 Ω=−5.9296 MΩ≅−6 MΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

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Answer 4

Question

Chapter 19, Problem 93P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the transistor network shown in Figure 19.132 in the textbook.

Expert Solution & Answer

Answer to Problem 93P

The voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.132 in the textbook for the transistor network circuit.

hie=2 kΩhre=2.5×10−4hfe=150hoe=10 μS

From the given transistor network circuit, the internal resistance (Rs) is 1 kΩ, emitter resistance (Re) is 0.2 kΩ, which is 200 Ω, and the load resistance (RL) is 3.8 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a transistor network 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 transistor network 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 transistor network as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ai=(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)=(150)[1+(200 Ω)(10×10−6 S)]1+(200 Ω+3800 Ω)(10×10−6 S)=(150)(1.002)1.04=144.5192

Ai≅144.5

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VcVb=1((2 kΩ+200 Ω)[1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)[1+(200 Ω)(10 μS)]{(150)[1+(200 Ω)(10 μS)]−150−(200 Ω+3.8 kΩ)(10 μS)(150)}+(2.5×10−4)−200 Ω3.8 kΩ)=1((2000 Ω+200 Ω)[1+(200 Ω+3800 Ω)(10×10−6 S)](10×10−6 S)[1+(200 Ω)(10×10−6 S)]{(150)[1+(200 Ω)(10×10−6 S)]−150−(200 Ω+3800 Ω)(10×10−6 S)(150)}+(2.5×10−4)−200 Ω3800 Ω)=1{(2200 Ω)(1.04)(10×10−6 S)(1.002)[(150)(1.002)−150−6]+(2.5×10−4)−0.0526}=1{−0.004+(2.5×10−4)−0.0526}

Simplify the expression as follows:

VcVb=−17.7462≅−17.74

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

Av=−17.74

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VbIb={(2 kΩ+200 Ω)+(2.5×10−4)[1+(200 Ω)(10 μS)][(150)(1+(200 Ω)(10 μS))−150−(200 Ω+3.8 kΩ)(10 μS)(150)][1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)+(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)(200 Ω)}=2200 Ω+(2.5×10−4)(1.002)(150.3−150−6)10.4 μS+(150.3)(200 Ω)1.04=2200 Ω−137.2933 Ω+28904 Ω=30967 ΩVbIb=30.967 kΩ≅31 kΩ

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

Zin=31 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 transistor network.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 2 kΩ for hie, 1 kΩ for Rs, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ic=(1 kΩ+2 kΩ+200 Ω)(10 μS)−[1+(10 μS)(200 Ω)](150)(2.5×10−4)[1+(10 μS)(200 Ω)](1 kΩ+2 kΩ+200 Ω)+(150)(200 Ω)=0.032−(1.002)(150)(2.5×10−4)(1.002)(3200 Ω)+30000 Ω=−0.005633206 Ω=−0.16864×10−6 S

Write the expression for output impedance of the transistor network as follows:

Zout=VcIc

Substitute 1 for Vc and (−0.16864×10−6 S) for Ic to obtain the output impedance of the transistor network.

Zout=1−0.16864×10−6 S=−5.9296×106 Ω=−5.9296 MΩ≅−6 MΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

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Answer 5

Chapter 19, Problem 93P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the transistor network shown in Figure 19.132 in the textbook.

Expert Solution & Answer

Answer to Problem 93P

The voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.132 in the textbook for the transistor network circuit.

hie=2 kΩhre=2.5×10−4hfe=150hoe=10 μS

From the given transistor network circuit, the internal resistance (Rs) is 1 kΩ, emitter resistance (Re) is 0.2 kΩ, which is 200 Ω, and the load resistance (RL) is 3.8 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a transistor network 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 transistor network 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 transistor network as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ai=(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)=(150)[1+(200 Ω)(10×10−6 S)]1+(200 Ω+3800 Ω)(10×10−6 S)=(150)(1.002)1.04=144.5192

Ai≅144.5

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VcVb=1((2 kΩ+200 Ω)[1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)[1+(200 Ω)(10 μS)]{(150)[1+(200 Ω)(10 μS)]−150−(200 Ω+3.8 kΩ)(10 μS)(150)}+(2.5×10−4)−200 Ω3.8 kΩ)=1((2000 Ω+200 Ω)[1+(200 Ω+3800 Ω)(10×10−6 S)](10×10−6 S)[1+(200 Ω)(10×10−6 S)]{(150)[1+(200 Ω)(10×10−6 S)]−150−(200 Ω+3800 Ω)(10×10−6 S)(150)}+(2.5×10−4)−200 Ω3800 Ω)=1{(2200 Ω)(1.04)(10×10−6 S)(1.002)[(150)(1.002)−150−6]+(2.5×10−4)−0.0526}=1{−0.004+(2.5×10−4)−0.0526}

Simplify the expression as follows:

VcVb=−17.7462≅−17.74

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

Av=−17.74

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VbIb={(2 kΩ+200 Ω)+(2.5×10−4)[1+(200 Ω)(10 μS)][(150)(1+(200 Ω)(10 μS))−150−(200 Ω+3.8 kΩ)(10 μS)(150)][1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)+(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)(200 Ω)}=2200 Ω+(2.5×10−4)(1.002)(150.3−150−6)10.4 μS+(150.3)(200 Ω)1.04=2200 Ω−137.2933 Ω+28904 Ω=30967 ΩVbIb=30.967 kΩ≅31 kΩ

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

Zin=31 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 transistor network.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 2 kΩ for hie, 1 kΩ for Rs, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ic=(1 kΩ+2 kΩ+200 Ω)(10 μS)−[1+(10 μS)(200 Ω)](150)(2.5×10−4)[1+(10 μS)(200 Ω)](1 kΩ+2 kΩ+200 Ω)+(150)(200 Ω)=0.032−(1.002)(150)(2.5×10−4)(1.002)(3200 Ω)+30000 Ω=−0.005633206 Ω=−0.16864×10−6 S

Write the expression for output impedance of the transistor network as follows:

Zout=VcIc

Substitute 1 for Vc and (−0.16864×10−6 S) for Ic to obtain the output impedance of the transistor network.

Zout=1−0.16864×10−6 S=−5.9296×106 Ω=−5.9296 MΩ≅−6 MΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

Answer 6

Chapter 19, Problem 93P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the transistor network shown in Figure 19.132 in the textbook.

Expert Solution & Answer

Answer to Problem 93P

The voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.132 in the textbook for the transistor network circuit.

hie=2 kΩhre=2.5×10−4hfe=150hoe=10 μS

From the given transistor network circuit, the internal resistance (Rs) is 1 kΩ, emitter resistance (Re) is 0.2 kΩ, which is 200 Ω, and the load resistance (RL) is 3.8 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a transistor network 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 transistor network 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 transistor network as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ai=(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)=(150)[1+(200 Ω)(10×10−6 S)]1+(200 Ω+3800 Ω)(10×10−6 S)=(150)(1.002)1.04=144.5192

Ai≅144.5

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VcVb=1((2 kΩ+200 Ω)[1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)[1+(200 Ω)(10 μS)]{(150)[1+(200 Ω)(10 μS)]−150−(200 Ω+3.8 kΩ)(10 μS)(150)}+(2.5×10−4)−200 Ω3.8 kΩ)=1((2000 Ω+200 Ω)[1+(200 Ω+3800 Ω)(10×10−6 S)](10×10−6 S)[1+(200 Ω)(10×10−6 S)]{(150)[1+(200 Ω)(10×10−6 S)]−150−(200 Ω+3800 Ω)(10×10−6 S)(150)}+(2.5×10−4)−200 Ω3800 Ω)=1{(2200 Ω)(1.04)(10×10−6 S)(1.002)[(150)(1.002)−150−6]+(2.5×10−4)−0.0526}=1{−0.004+(2.5×10−4)−0.0526}

Simplify the expression as follows:

VcVb=−17.7462≅−17.74

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

Av=−17.74

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VbIb={(2 kΩ+200 Ω)+(2.5×10−4)[1+(200 Ω)(10 μS)][(150)(1+(200 Ω)(10 μS))−150−(200 Ω+3.8 kΩ)(10 μS)(150)][1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)+(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)(200 Ω)}=2200 Ω+(2.5×10−4)(1.002)(150.3−150−6)10.4 μS+(150.3)(200 Ω)1.04=2200 Ω−137.2933 Ω+28904 Ω=30967 ΩVbIb=30.967 kΩ≅31 kΩ

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

Zin=31 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 transistor network.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 2 kΩ for hie, 1 kΩ for Rs, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ic=(1 kΩ+2 kΩ+200 Ω)(10 μS)−[1+(10 μS)(200 Ω)](150)(2.5×10−4)[1+(10 μS)(200 Ω)](1 kΩ+2 kΩ+200 Ω)+(150)(200 Ω)=0.032−(1.002)(150)(2.5×10−4)(1.002)(3200 Ω)+30000 Ω=−0.005633206 Ω=−0.16864×10−6 S

Write the expression for output impedance of the transistor network as follows:

Zout=VcIc

Substitute 1 for Vc and (−0.16864×10−6 S) for Ic to obtain the output impedance of the transistor network.

Zout=1−0.16864×10−6 S=−5.9296×106 Ω=−5.9296 MΩ≅−6 MΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

Answer 7

Chapter 19, Problem 93P

To determine

Calculate the voltage gain, current gain, input impedance, and output impedance for the transistor network shown in Figure 19.132 in the textbook.

Answer 8

Expert Solution & Answer

Answer to Problem 93P

The voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given Data:

Refer to Figure 19.132 in the textbook for the transistor network circuit.

hie=2 kΩhre=2.5×10−4hfe=150hoe=10 μS

From the given transistor network circuit, the internal resistance (Rs) is 1 kΩ, emitter resistance (Re) is 0.2 kΩ, which is 200 Ω, and the load resistance (RL) is 3.8 kΩ.

Formula used:

Refer to Equation 19.73 in the textbook and write the expression for voltage gain of a transistor network 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 transistor network 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 transistor network as follows:

Zin=VbIb (3)

Calculation:

Redraw the given circuit as shown in Figure 1.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ai=(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)=(150)[1+(200 Ω)(10×10−6 S)]1+(200 Ω+3800 Ω)(10×10−6 S)=(150)(1.002)1.04=144.5192

Ai≅144.5

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VcVb=1((2 kΩ+200 Ω)[1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)[1+(200 Ω)(10 μS)]{(150)[1+(200 Ω)(10 μS)]−150−(200 Ω+3.8 kΩ)(10 μS)(150)}+(2.5×10−4)−200 Ω3.8 kΩ)=1((2000 Ω+200 Ω)[1+(200 Ω+3800 Ω)(10×10−6 S)](10×10−6 S)[1+(200 Ω)(10×10−6 S)]{(150)[1+(200 Ω)(10×10−6 S)]−150−(200 Ω+3800 Ω)(10×10−6 S)(150)}+(2.5×10−4)−200 Ω3800 Ω)=1{(2200 Ω)(1.04)(10×10−6 S)(1.002)[(150)(1.002)−150−6]+(2.5×10−4)−0.0526}=1{−0.004+(2.5×10−4)−0.0526}

Simplify the expression as follows:

VcVb=−17.7462≅−17.74

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

Av=−17.74

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 150 for hfe, 2 kΩ for hie, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

VbIb={(2 kΩ+200 Ω)+(2.5×10−4)[1+(200 Ω)(10 μS)][(150)(1+(200 Ω)(10 μS))−150−(200 Ω+3.8 kΩ)(10 μS)(150)][1+(200 Ω+3.8 kΩ)(10 μS)](10 μS)+(150)[1+(200 Ω)(10 μS)]1+(200 Ω+3.8 kΩ)(10 μS)(200 Ω)}=2200 Ω+(2.5×10−4)(1.002)(150.3−150−6)10.4 μS+(150.3)(200 Ω)1.04=2200 Ω−137.2933 Ω+28904 Ω=30967 ΩVbIb=30.967 kΩ≅31 kΩ

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

Zin=31 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 transistor network.

EE 98: Fundamentals of Electrical Circuits - With Connect Access, Chapter 19, Problem 93P , 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 150 for hfe, 2 kΩ for hie, 1 kΩ for Rs, (2.5×10−4) for hre, 10 μS for hoe, 200 Ω for Re, and 3.8 kΩ for RL as follows:

Ic=(1 kΩ+2 kΩ+200 Ω)(10 μS)−[1+(10 μS)(200 Ω)](150)(2.5×10−4)[1+(10 μS)(200 Ω)](1 kΩ+2 kΩ+200 Ω)+(150)(200 Ω)=0.032−(1.002)(150)(2.5×10−4)(1.002)(3200 Ω)+30000 Ω=−0.005633206 Ω=−0.16864×10−6 S

Write the expression for output impedance of the transistor network as follows:

Zout=VcIc

Substitute 1 for Vc and (−0.16864×10−6 S) for Ic to obtain the output impedance of the transistor network.

Zout=1−0.16864×10−6 S=−5.9296×106 Ω=−5.9296 MΩ≅−6 MΩ

Conclusion:

Thus, the voltage gain, current gain, input impedance, and output impedance for the transistor network are −17.74_, 144.5_, 31 kΩ_, and −6 MΩ_, 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 transistor network shown in Figure 19.132 in the textbook.

Calculate the voltage gain, current gain, input impedance, and output impedance for the transistor network shown in Figure 19.132 in the textbook.

Answer to Problem 93P

Explanation of Solution

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