#### peterpops

##### Junior Member level 3

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