Urgent help with amplifiers

1. Sep 5, 2007

brad sue

Urgent help with amplifiers plz!!

Hi,
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. Sep 14, 2007

kuba

a) Resistive divider

For a), think of an ideal voltage source with 10mV p-p hooked via a series resistor of $$10\mathrm{k}\Omega$$ to the load resistor of $$8\Omega$$. This is a simple resistive divider. The speaker voltage is the voltage across the $$8\Omega$$ resistor.

3. Sep 14, 2007

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.

4. Sep 15, 2007

kuba

Transconductance

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,

$$R_i=\infty\ \Omega$$
$$I_i=0\ \mathm{A}$$

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).

$$R_{out}=8\ \Omega$$

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

$$I_{out\ rms} = \frac{V_{out\ rms}}{R_{total}} = \frac{40\ \mathrm{V_{rms}}}{16\ \Omega}=2.5\ \mathrm{A}_{rms}$$

By definition, transconductance

$$g_m=\frac{I_{out\ rms}}{V_{in\ rms}}=\frac{2.5\ \mathrm{A}_{rms}}{10\ \mathrm{mV}_{rms}}=250\ \mathrm{S}$$

So, the circuit to get this has following parameters:

$$R_{in}=\infty\ \Omega,\ \ R_{out}=8\ \Omega,\ \ g_m=250\ \mathrm{S}$$

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