Solve Urgent Amplifier Help: Microelectronics CD Player Problem in Few Hours

[brad sue]

Urgent help with amplifiers please!

Hi,
Please I need help for this problem. It is due in few hours!
The main problem is that I don't 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

 

[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.

 

[berkeman]

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.

 

[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:

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