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Homework Help: Urgent help with amplifiers

  1. Sep 5, 2007 #1
    Urgent help with amplifiers plz!!

    Please I need help for this problem. It is due in few hours!
    The main problem is that I dont succeed to visualize the problem to make a drawing.
    please, can someone help me?

    (the context of the course is microelectronics on amplifier)

    A compact disc (CD) player laser pick-up provides a signal output of 10[mV]peak to peak (pp) and has an output resistance of 10[kW]. The pick-up is to be connected to a speaker whose equivalent resistance is 8[W].

    a) Calculate the voltage that would be delivered to the speaker if the speaker were connected directly to the pick-up.

    b) Assume that the speaker needs 20[V]pp to deliver clear acoustical output. Design an equivalent circuit for an amplifier that would deliver this output when connected between the pick-up and the speaker.

    I have a possible solution for the question b) given by the teacher as follow:
    One possible solution would be a transconductance amplifier with Gmsc = 900; Ri = 10[kOhm]; Ro = 10[Ohm]

    But I cannot figure out how they get those values. I suspect they chose some values for Ro and/or Ri

    Thank you
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 14, 2007 #2
    a) Resistive divider

    For a), think of an ideal voltage source with 10mV p-p hooked via a series resistor of [tex]10\mathrm{k}\Omega[/tex] to the load resistor of [tex]8\Omega[/tex]. This is a simple resistive divider. The speaker voltage is the voltage across the [tex]8\Omega[/tex] resistor.
  4. Sep 14, 2007 #3


    User Avatar

    Staff: Mentor

    The amplifier's input resistance is chosen to match the laser circuit's output resistance -- this gives maximum power transfer between those two stages. The output impedance of the amplifier should closely match the speaker input impedance for the same reason.
  5. Sep 15, 2007 #4

    Well, a transconductance amplifier is an amplifier which controls its output current by the input voltage.

    Many real-life transconductance amplifiers have a pretty high input impedance, so for audio frequency use this is a good enough assumption.

    So, let's assume that there's no current flowing into the input of the amplifier,

    [tex]R_i=\infty\ \Omega[/tex]
    [tex]I_i=0\ \mathm{A}[/tex]

    we can ignore the source impedance - there's no current flowing through it, so there's no voltage drop.

    Good enough, we have 10mV RMS as the input voltage.

    Real life amplifiers often have some non-zero output resistance, so we can pick a small number there. To make life easy, say it's 8 Ohms, so that the speaker will see half of the output voltage (it's a 1:1 voltage divider).

    [tex]R_{out}=8\ \Omega[/tex]

    To get 20V rms across the speaker, we need 40V rms on the output, or

    [tex]I_{out\ rms} = \frac{V_{out\ rms}}{R_{total}} = \frac{40\ \mathrm{V_{rms}}}{16\ \Omega}=2.5\ \mathrm{A}_{rms}[/tex]

    By definition, transconductance

    [tex]g_m=\frac{I_{out\ rms}}{V_{in\ rms}}=\frac{2.5\ \mathrm{A}_{rms}}{10\ \mathrm{mV}_{rms}}=250\ \mathrm{S}[/tex]

    So, the circuit to get this has following parameters:

    [tex]R_{in}=\infty\ \Omega,\ \ R_{out}=8\ \Omega,\ \ g_m=250\ \mathrm{S}[/tex]
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