Power output is where amp design always starts.

Power is most comprehensibly expressed as continuous sin wave power, this is what is meant when a reputable manufacturer advertises an amp as, say, 100 Watts. This is also called RMS power, and it is the same measure as that employed by the electricity company when they charge you for power. Although there are other measures such as peak instantaneous power and PMPO we will ignore them for the moment.

RMS power permits the common Ohms law equation (V = I * R) to be extended to cover power in AC and DC circuits in an exactly equivalent fashion.

In a DC circuit Watts=Amps*Volts (W=I*V). Since Volts=Current*Resistance we can substitute I*R for V in the power equation. Thus W=I*V becomes W=I*I*R or I^2*R (I squared R). I'm going to start omitting some of the multiplication signs (*) now. Rearranging V=IR gives us I=V/R. We can also substitute for I in W=IV to get W=V^2/R (V squared upon R). We now have 2 equations which tell us the required voltage or current in a particular load (speaker) to give a required power output.

In order to fully appreciate the situation is is necessary to distinguish (in AC) between peak, peak-to-peak and RMS measurements. RMS measurements are required to bring the actual power (rate of work) into line with DC measurements.

AC voltages and currents are sin waves. Because they vary around 0 (zero), their average value is 0. An AC voltage with a maximum positive excursion of 1V (and a negative excursion of 1V) has an average value of zero volts. It is said to have a *peak* value of 1V and a *peak-to-peak* value of 2V. If the wave were squashed down to occupy a rectangle, the height of the rectangle would be 1 divided by the square root of 2. 1/1.414 approx. or 0.707V. This is called the RMS (or *root mean square*) value.

Now we can calculate the voltages or currents required to produce (say) 10 Watts in an 8 ohm speaker. The formula for voltage would be W=V^2/R. Substituting the values we have, we get 10=(V^2)/8. Rearranging we get V^2=10*8=80. Hence V=sqrt(80)=8.944V. This however is the RMS value. In order to obtain the *peak* value we must multiply by sqrt(2) or 1.414 with the result 12.65V. We must then double this to get the *peak-to-peak* value of 25.3V. This is the voltage swing required to produce 10W (RMS) power in a load of 8R. All other considerations aside the amplifier must be able to produce such a swing across the load, be it between 0 and 25.3V or -12.65V and +12.65V.

It's obviously not impossible to put a transformer in the collector circuit of a BJT, but this is rarely done in an audio amplifier (it's not uncommon in RF output stages). Unless the quiescent voltage is 0V (which means a push-pull amplifier) the amplifier must be AC coupled to prevent DC flowing in the speaker. Transistor output stages may be arranged as common emitter or emitter follower.

Taking the case of the common emitter, for greatest efficiency the output impedance (collector resistor) must match that of the speaker. This means that half the available voltage swing appears across the output impedance and the voltage available to the amplifier must be double that appearing across the load.

We can similarly calculate the *peak* current required in the load. Since in class A the current varies from zero to twice the *peak* level, the quiescent current in the stage can be set to the *peak* level or half the *peak-to-peak* value. Often the actual value used will be 110% of the *peak* level and the supply voltages increased by a similar factor to account for losses.

In your case we want 0.8W in 8 Ohms. This gives us V^2 = 6.4V and VRMS = 2.53V. Vpk=3.58V and Vpk-pk = 7.16V so your supply voltage needs to be ~15V (for a common emitter) and the quiescent current ~0.5A for class A.

Unfortunately you will find that designing a common emitter amp with 8 Ohms in the collector circuit has its own problems. It's easier in many respects to take an opamp with rail-to-rail capability and drive an emitter follower or effectively design a discrete rail-rail opamp and follow that with an emitter follower.

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