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] ?
     
  3. tjjam2003

    Thread Starter Member

    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
    3,373
    1,159
    Recheck your 1687 = V out.
     
  6. Papabravo

    Expert

    Feb 24, 2006
    10,148
    1,791
    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.

    please explain?

    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
    10,148
    1,791
    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.
     
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