Design of multistage audio amplifier

[steampunk]

The amplifier has to deliver an average power of 1 W to a 12 Ω speaker from a microphone that produces a 20 mV peak-to-peak sinusoidal signal with rs of 10 kΩ.

How should I approach this considering the optimal solution and design in mind in the cct diagram? With a suitable dc bias current source as well. I am thinking of a single stage CC buffer type cct..

 

[kellys_eye]

Start by deciding which class of amplification you intend to use and the supply to be used.

 

[CircutScoper]

I suggest you start by calculating the peak-to-peak output voltage required to push 1W RMS into your 12 ohm speaker. This, combined with the 20mVpp output of the mike will allow you to calculate the necessary cascaded voltage gain of your gain and output stages, and the minimum supply voltage.

Then you'll have the primary boundary conditions for your design.

 

[Audioguru]

You do not want any DC current in a loudspeaker. Amplifiers usually use a push-pull output and feed audio to the speaker through a coupling capacitor or the output stage uses a power supply with plus ands minus voltages.

Most of us would use a power amplifier IC that is fed from an opamp or transistor mic preamp circuit.
Does the teacher ask for an all-transistors circuit?

If the microphone can hear the speaker then there will be loud acousical feedback howling.

 

[WHONOES]

The calculation for the peak to peak voltage required to achieve 1W into 12Ohms is. (Square root of required power x impedance) x 2.484.
To calculate the required circuit gain, divide the result of previous sum by the output voltage of your microphone.

 

[CircutScoper]

The calculation for the peak to peak voltage required to achieve 1W into 12Ohms is. (Square root of required power x impedance) x 2.484.
To calculate the required circuit gain, divide the result of previous sum by the output voltage of your microphone.

Shouldn't that constant for Vpp from Vrms = 2 x sqr(2) = 2.828?

 

[WHONOES]

Yes you are right it should be 2.8284. Not concentrating properly!

 

[Audioguru]

Since the peak voltage of a sinewave is 1.414 times the RMS then peak-to-peak is 2.848 times, not 2.484 times.
The output push-pull emitter followers have a voltage loss and their biasing also has a voltage loss. The output stage would need at least 12VDC, less if bootstrapping is used.

 

[CircutScoper]

Since the peak voltage of a sinewave is 1.414 times the RMS then peak-to-peak is 2.848 times, not 2.484 times.
The output push-pull emitter followers have a voltage loss and their biasing also has a voltage loss. The output stage would need at least 121VDC, less if bootstrapping is used.

LOL! Don't look now, but methinks WHONOES's 2.484 was considerably closer to correct than your 121. Your 2.848 ain't too terrific, neither.

Meanwhile, a MOSFET H-bridge output topology could probably (just barely) do the trick from only 5VDC.

 

[Audioguru]

My typo said 121V but it should be 12V.

An amplifier that drives both wires of a speaker (H-bridge) is called Bridged or Bridged-Tied-Load. Most car radio amplifiers (14W into 4 ohms when the battery is 13.2V) are like that.

 

[steampunk]

Start by deciding which class of amplification you intend to use and the supply to be used.

The combine gain of the first two stage should not be more than 100. What do you suggest?

 

[steampunk]

I suggest you start by calculating the peak-to-peak output voltage required to push 1W RMS into your 12 ohm speaker. This, combined with the 20mVpp output of the mike will allow you to calculate the necessary cascaded voltage gain of your gain and output stages, and the minimum supply voltage.

Then you'll have the primary boundary conditions for your design.

There is also a catch. I have to make sure the gain after the first two stages should be within 100.

 

[steampunk]

You do not want any DC current in a loudspeaker. Amplifiers usually use a push-pull output and feed audio to the speaker through a coupling capacitor or the output stage uses a power supply with plus ands minus voltages.

Most of us would use a power amplifier IC that is fed from an opamp or transistor mic preamp circuit.
Does the teacher ask for an all-transistors circuit?

If the microphone can hear the speaker then there will be loud acousical feedback howling.

The calculation for the peak to peak voltage required to achieve 1W into 12Ohms is. (Square root of required power x impedance) x 2.484.
To calculate the required circuit gain, divide the result of previous sum by the output voltage of your microphone.

What formula did you use here exactly? And what is the constant called?

 

[Audioguru]

1W into 12 ohms is (square root of 1W x 12 ohms) =3.464V RMS.
The peak voltage is the RMS voltage x the root of 2. Note that the root of 2 is 1.414 and 1.414 x 2= 2.828, not 2.484.
3.464V RMS x (1.414 x 2)= 9.8V peak-to-peak.
The total gain needed= 9.8Vp-p / 20mVp-p= 490 times.
If the first two stages need a gain of 100 then the output stage needs to have a gain of 490/100= 4.9 times.