Joined Dec 29, 2004
83
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.

(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

#### Distort10n

Joined Dec 25, 2006
429
Well, incorrect units aside you could use a simple op-amp in a non-inverting configuration. I do not understand why your professor told you that a transconductance (voltage to current) amplifier is needed. The output of the mic pic-up is voltage, and voltage will be delivered to the load.
Since the mic-pic up has a high output impedance, then this will cause insignificant % error if the input is into the ideally infinite input impedance of an amplifier's non-inverting input. Since the input is 10mVpp, then set the gain to be 2000.

#### bloguetronica

Joined Apr 27, 2007
1,453
That was a bad example of an exercise. Digital input from the CD is not converted "directly" to an analog signal for the speaker to use. Please, send your teacher to school again. Also, 20Vpp produces a very loud sound on any casual speaker, assuming the speaker is not destroyed

Your professor is an example of what is theory without practice...

#### spar59

Joined Aug 4, 2007
57
I agree that a transconductance amplifier is an unusual suggestion however there may be some merits with it:-

The force produced on the speaker cone is proportional to the ampere-turns in the voice coil (turns being constant for a given coil) and the magnetic flux (constant for a given magnet).

Since the impedance of the voice coil varies with frequency the current in it (and hence cone deflection) for a given applied voltage would vary with frequency, hence driving from a current rather than a voltage source may have an advantage.

However most hi-fi amplifiers are pure voltage amplifiers not transconductance.

Steve.

#### Eduard Munteanu

Joined Sep 1, 2007
86
That was a bad example of an exercise. Digital input from the CD is not converted "directly" to an analog signal for the speaker to use. Please, send your teacher to school again. Also, 20Vpp produces a very loud sound on any casual speaker, assuming the speaker is not destroyed

Your professor is an example of what is theory without practice...
Regular audio CDs are analog media. That's why CDROMs have sockets for special cables going directly to the soundcard, and that's why the actual ripping takes much longer than on data CDs (which had been digitally written to).

Loud? $$P_{avg}={ V^2_{rms} \over R } = { V^2_p \over {2 R} } = { V^2_{pp} \over {8 R} }$$, which yields 12.5W for a 4ohm speaker. Well, maybe it's bad for small speakers, like desktop ones.

#### bloguetronica

Joined Apr 27, 2007
1,453
Regular audio CDs are analog media. That's why CDROMs have sockets for special cables going directly to the soundcard, and that's why the actual ripping takes much longer than on data CDs (which had been digitally written to).

Loud? $$P_{avg}={ V^2_{rms} \over R } = { V^2_p \over {2 R} } = { V^2_{pp} \over {8 R} }$$, which yields 12.5W for a 4ohm speaker. Well, maybe it's bad for small speakers, like desktop ones.
Are you sure about that? And what about the track data? And CD text in newer CD's? Any CD will present microscopic holes, representing ones ande zeros. They are binary, and therefore, digital. The info is to be converted to analog before it's used.
And actually the ripping takes time because you need to convert the *.cda files in your CD to *.mp3 files. Although, you can copy these directly from the CD to the computer and listen them.

#### bloguetronica

Joined Apr 27, 2007
1,453
Regular audio CDs are analog media. That's why CDROMs have sockets for special cables going directly to the soundcard, and that's why the actual ripping takes much longer than on data CDs (which had been digitally written to).
Are you sure about that? And what about the track data? And CD text in newer CD's? Any CD will present microscopic holes, representing ones and zeros. They are binary, and therefore, digital. The info is to be converted to analog before it's used.
And actually the ripping takes time because you need to convert the *.cda files in your CD to *.mp3 files. Although, you can copy these directly from the CD to the computer and listen them.

Loud? $$P_{avg}={ V^2_{rms} \over R } = { V^2_p \over {2 R} } = { V^2_{pp} \over {8 R} }$$, which yields 12.5W for a 4ohm speaker. Well, maybe it's bad for small speakers, like desktop ones.
Well, it is loud for me. Let's say that 0.5W already produces a clear acoustical signal. No need for deafening.
Also, you should redo your calculations, since 20Vpp gives 14.142V RMS, which yields 24.99W for a 8Ohm speaker or 49.999W for a 4Ohm speaker. Don't forget that Vpp = Vrms x sqrt (2) for a sinusoidal wave.

#### techroomt

Joined May 19, 2004
198
analog laser pick-ups - thats a new one on me. can you explain please?

#### Eduard Munteanu

Joined Sep 1, 2007
86
Are you sure about that? And what about the track data? And CD text in newer CD's? Any CD will present microscopic holes, representing ones and zeros. They are binary, and therefore, digital. The info is to be converted to analog before it's used.
And actually the ripping takes time because you need to convert the *.cda files in your CD to *.mp3 files. Although, you can copy these directly from the CD to the computer and listen them.
No, just extracting the PCM data (as in ripping to WAV files) from audio tracks takes more time on low-end and even mid-level drives. No MP3 encoding or anything else.

The CD-Text part is indeed digital.

And those .CDAs are not real files. Windows' VFS makes them appear there .

And that's why Windows has a properties panel for CDROMs that contains an option named "Digital audio extraction". These let you play audio CDs without that CDROM->soundcard cable.

Well, it is loud for me. Let's say that 0.5W already produces a clear acoustical signal. No need for deafening.
Also, you should redo your calculations, since 20Vpp gives 14.142V RMS, which yields 24.99W for a 8Ohm speaker or 49.999W for a 4Ohm speaker. Don't forget that Vpp = Vrms x sqrt (2) for a sinusoidal wave.
No, Vp = Vrms x sqrt(2) for a sine wave.

P.S.: Analog is also a reason why CDROM benchmarks test the Digital Audio Extraction (DAE) quality. Data CDs are read perfectly, as they were written, but audio CDs are not.

#### bloguetronica

Joined Apr 27, 2007
1,453
I have to confirm that.

No, Vp = Vrms x sqrt(2) for a sine wave.
Sorry for the typo, just forgot that Vpp is peak-to-peak voltage and it is the double of Vp.

#### Distort10n

Joined Dec 25, 2006
429
I agree that a transconductance amplifier is an unusual suggestion however there may be some merits with it:-

The force produced on the speaker cone is proportional to the ampere-turns in the voice coil (turns being constant for a given coil) and the magnetic flux (constant for a given magnet).
Are there any good websites that discuss this?