# Amplifier

Discussion in 'Homework Help' started by tjjam2003, Apr 18, 2006.

1. ### tjjam2003 Thread Starter Member

Apr 5, 2006
22
0
An amplifier has an input voltage of 300mV. If the voltage gain is 15dB, what is the output voltage?

Is the correct formula (AV) decibels=20 Log V out / V in ??

I know I have to solve for V out. I keep getting the wrong answer.

The answer is V out = 1686 but that is not the answer I get.

Maybe I am not rewriting the formula the way it should be?

(I am a bit weak in algebra.)

2. ### paultwang Well-Known Member

Mar 8, 2006
80
0
1. What answer did you get?

2. What is 15 dB in linear scale?

3. What is 300 mV times [the answer of #2] ?

Apr 5, 2006
22
0

4. ### tjjam2003 Thread Starter Member

Apr 5, 2006
22
0
15 db = 20 LOG (V out/V in)

15/20 = LOG (V out/ V in)

.75 = Log (V out/ 300mV)

Take the anti Log of each side:

10^.75 = Vout/300mV

5.62 = Vout/300

1687 = V out

Got it....THANKS... U guys are the GREATEST !!!

5. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
1,229
Recheck your 1687 = V out.

6. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
That was cool how 300 mV suddenly became 300 Volts. I'd like to get me somma that!

ROFL

7. ### rodn.m Member

May 3, 2006
12
0
i am doing a similar course(industrial electronics) and i noticed the formula looked familiar. but as the last person said, it should be 0.3 volts not 300 volts.

anyhow, i am unfamiliar with the term antilog.

i have a casio fx=100c calculator

how did you get from .75 to 5.62 ?

curios and confuesed

8. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
The term "antilog" refers to the inverse of the common logarithm function. In modern terms, the exponential function with 10 as the base, and the antilog are the same thing. The term was more familiar when school children were taught how to do multiplication by adding logarithms and taking the antilog of the result. As an undergraduate I remember the poorest of the engineering students, who could not afford \$30 for a Eugene Dietzen Log-Log Duplex Decitrig Slide Rule. At exam time they would pull out a little card with a table of four place logarithms, and use that to solve their problems.

If you look carefully at the post you will see the following line
Code ( (Unknown Language)):
1.
2. 10 ^ .75 = Vout/300mV
3.
Ten, the base of common logarithms, raised to the .75 or 3/4 ths power is 5.62 You can obtain the same result by cubing 10 and taking the square root twice. The correct development from this point would be
Code ( (Unknown Language)):
1.
2. 10 ^ (3/4) = Vout/300mV
3.
4. 5.62 = Vout/300mV
5.
6. (5.62)*(300mV) = Vout
7.
8. 1.687 = Vout
9.
I hope this helps.