Voltage amplifier model (excercise question from Sedra Smith 5th ed.)

[peterpops]

Hi!

I'm doing some exercises in Microelectronic Circuits (Sedra Smith 5th ed) and got stuck on 1.13 (voltage amplifiers)

NOTE! This is not the end-chapter excercises included in the PDF solution manual!

According to SedraSmith a voltage amplifier can be modeled as in the figure.


The question is:
"An amplifier with a voltage gain of +40 dB, an input resistance of 10kΩ, and an output resistance of 1kΩ is used to drive a 1-kΩ load. What is the value of A_vo? Find the value of power gain in dB"
The answer in the book is:
"Ans. 100 V/V; 44 dB"

My solution:
I know
Voltage gain in dB = 40, which means 40 = 20 log(A_v) => A_v = 100

EQ.1: \[{A}_{v}=\frac{{v}_{o}}{{v}_{i}}=100\]

Voltage division gives me \[{v}_{o}={A}_{vo}{v}_{i}\left(\frac{{R}_{L}}{{R}_{L}+{R}_{o}}\right)\]

rewritten \[{A}_{vo}=\left(\frac{{v}_{o}}{{v}_{i}}\right)\left(\frac{{R}_{L}+{R}_{o}}{{R}_{L}}\right)=100\left(\frac{{R}_{L}+{R}_{o}}{{R}_{L}}\right)\]

From here I can go 2 ways:
1. Since I know R_o and R_L I just substitute them in the equation above and the answer is \[{A}_{vo}=100\left(\frac{1000+1000}{1000}\right)=200\]

Or

2. A_vo is the "open-circuit voltage gain" which means I assume that R_L is infinite which gives me \[{A}_{vo}=100\]
Since the answer is 100 V/V I assume solution 2 is the correct one.

Now over to the second part of the question, the power gain in dB.

The current through R_i is (with the help of EQ.1) \[{i}_{i}=\frac{{v}_{i}}{{R}_{i}}=\frac{{v}_{o}}{{100R}_{i}}\]

The current through R_o is \[{i}_{o}=\frac{{v}_{o}}{{R}_{L}}\]

Which gives me the current gain \[{A}_{i}=\frac{{i}_{o}}{{i}_{i}}=\frac{\frac{{v}_{o}}{{R}_{L}}}{\frac{{v}_{o}}{{100R}_{i}}}=\frac{{100R}_{i}}{{R}_{L}}\]
With values inserted A_i = 1000 A/A

Power gain, A_p = A_v*A_i and depending on which voltage gain solution above
A_p = 200 * 1000 = 200000, or A_p = 100 * 1000 = 100000

in dB the power gain is either = 10*log(200000) = 53dB, or = 10*log(100000) = 50dB

So where did I go wrong or is the SedraSmith answer incorrect?

Thanks in advance
/P

 

[barry]

The way I would have solved this would be:

Vo=(Vi*Av0)/2
100*2=Av0
Av0(dB)=20*log(200)=46 dB

You give the input resistance, but that's irrelevant since it's assumed Rs is zero. Perhaps you left out that bit of information??

 

[peterpops]

The way I would have solved this would be:

Vo=(Vi*Av0)/2
100*2=Av0
Av0(dB)=20*log(200)=46 dB

You give the input resistance, but that's irrelevant since it's assumed Rs is zero. Perhaps you left out that bit of information??

So you agree with solution 1?

Is the power gain calculation correct? In that equation I use the input resistance.

 

[albbg]

I think the solution is quite simple:

1st question:
40 dB is the gain of the amplifier alone ("An amplifier with a voltage gain of +40 dB") then Avo=10^(40/20)=100V/V

2nd question:
The power gain is given by Pout/Pin then we need to calculate input and output power.

Pin=Vin²/Ri
Pout=Vout²/RL

Vout=Avo*Vin*RL/(Ro+RL) thus:

Pout=[Avo*Vin*RL/(Ro+RL)]²/RL that is:
Pout=(Avo*Vin)²*RL/(Ro+RL)]²

Then:

Pout/Pin=Avo²*RL*Ri/(Ro+RL)² or, in dB

Pout/Pin(dB)=10*Log[Avo²*RL*Ri/(Ro+RL)²]

Numerically

Pout/Pin(dB)=10*Log[100²*1000*10000/(1000+1000)²]≈44 dB

 

[peterpops]

I think the solution is quite simple:

1st question:
40 dB is the gain of the amplifier alone ("An amplifier with a voltage gain of +40 dB") then Avo=10^(40/20)=100V/V

2nd question:
The power gain is given by Pout/Pin then we need to calculate input and output power.

Pin=Vin²/Ri
Pout=Vout²/RL

Vout=Avo*Vin*RL/(Ro+RL) thus:

Pout=[Avo*Vin*RL/(Ro+RL)]²/RL that is:
Pout=(Avo*Vin)²*RL/(Ro+RL)]²

Then:

Pout/Pin=Avo²*RL*Ri/(Ro+RL)² or, in dB

Pout/Pin(dB)=10*Log[Avo²*RL*Ri/(Ro+RL)²]

Numerically

Pout/Pin(dB)=10*Log[100²*1000*10000/(1000+1000)²]≈44 dB

Thank you for the answer, two questions.

1. Do you use the fact that Avo = "Open-circuit voltage gain" so RL is assumed infinity i.e. Avo = Av = 100 V/V (or 40dB)?

2. Can you explain why my answer to question #2 is incorrect?
Power gain = Pout / Pin

Pout = Vout^2 / RL
Pin = Vin^2 / Ri

same as you, but then I want to use Av = Vout / Vin = 100 (since this info is given for the current system or Rs/Ri/Ro/RL)

so I get Pout / Pin = (Vout^2 / RL) / (Vin^2 / Ri) = (Vout^2 / RL) / (Vout^2 / (10000 * Ri)) = 10000Ri / RL = 100000 W/W or 50dB

 

[albbg]

I read you answer: you forgot the output divider RL/(Ro+RL).

The total voltage gain is Av=Avo*RL/(Ro+RL)

The input current is Iin=Vi/Ri while the output current is Iout=Avo*Vi/(Ro+RL)

Then the current gain will be: Iout/Iin=Avo*Ri/(Ro+RL)

As you said Ap=Av*Ai thus:

Ap=Avo*RL/(Ro+RL)*Avo*Ri/(Ro+RL) that is: Ap=Avo²*RL*Ri/(Ro+RL)²

as per my previous post.