^{simulate this circuit – Schematic created using CircuitLab}

The objective is to drive 1 watt into the speaker. R1 represents the output impedance of the previous common collector stage.

Here's what I think I know:

- 1 watt into 4 ohms means 2V RMS or 2.8V peak.
- If the beta of Q1 is something like 70 then the input impedance of the emitter follower is something like 700–pretty low.
- In the quiescent state C1 is charged and the current through the speaker is zero.
- When the voltage rises to +2.8V, 700mA (2.8V/4 ohms) goes to SPKR1. (Confused about whether it all goes to SPKR1, seems like some must go through R2.)
- When the voltage falls to -2.8V, 700mA gets pulled back through SPKR1 and must be sunk through R2.

The problem in a sentence is that R2 is too high to sink the current coming back from SPKR1 when the voltage is low. If the emitter voltage is about 6V then when it swings -2.8V to 3.2V, R2 will sink only 320mV. So the returning current causes the emitter voltage to rise, turning off Q1 and creating a flat spot at the bottom of the waveform.

If I decrease R2, then the input impedance drops and it's hard to maintain the voltage swing. Also, the quiescent current increases and I start needing a big heat sink for Q1.

My question is: is there a process for determining a value for R2 that can meet the objective of driving 1 watt into the speaker, or is this just a waste of time?

## Best Answer

As you know, the peak AC input signal for delivering any particular power into any particular speaker load is:

$$V_\text{P}=\sqrt{2\:R\:P}$$

In your case, with \$1\:\text{W}\$ and \$R_\text{SPKR}=4\:\Omega\$, this works out to \$V_\text{P}=2.83\:\text{V}\$ and \$I_\text{P}=710\:\text{mA}\$.

Also, your capacitor needs to have a low impedance relative to the speaker, so let's say about \$400\:\text{m}\Omega\$ for the capacitor. If you decide that \$f=200\:\text{Hz}\$ is the lowest frequency you want to handle, this suggests a capacitance of \$2\:\text{mF}\$. A bit larger than you show. Let's also assume you set the capacitance, appropriately.

In general, you want the BJT to be operating with a collector current dynamic range of less than 5:1. (That's about \$42\:\text{mV}\$ variation of \$V_\text{BE}\$.) This means that your minimum emitter current should be \$I_{\text{E}_\text{MIN}}\ge \frac{1}{2}\left[I_\text{P}+\frac{V_\text{P}}{R_\text{E}}\right]=\frac{1}{2}\frac{V_\text{P}}{R_\text{E}\vert\vert R_\text{SPKR}}\$. In your circuit's case, this means \$I_{\text{E}_\text{MIN}}\ge 495\:\text{mA}\$. Whatever you decide for \$I_{\text{E}_\text{MIN}}\$ is your choice, though. That's just a recommendation.

Once you have that value, and I'm picking \$I_{\text{E}_\text{MIN}}=500\:\text{mA}\$, you can work out that the quiescent emitter voltage needs to be \$V_{\text{E}_\text{Q}}\ge R_\text{E}\left(I_\text{P}+I_{\text{E}_\text{MIN}}\right)+V_\text{P}\$. In your case, this means \$V_{\text{E}_\text{Q}}\ge 14.83\:\text{V}\$. I'm picking \$V_{\text{E}_\text{Q}}= 15\:\text{V}\$. (Your BJT's base will have to be driven with a quiescent voltage that is one \$V_\text{BE}\$ higher.)

(You could skip the steps shown above and simply compute \$V_{\text{E}_\text{Q}}\ge \frac{3}{2}V_\text{P}\left(1+\frac{R_\text{E}}{R_\text{SPKR}}\right)\$ and get the same answer. Note that the larger the ratio of \$\frac{R_\text{E}}{R_\text{SPKR}}\$ the higher the quiescent emitter voltage needs to be.)

The BJT's collector voltage should be at least \$2\:\text{V}\$ higher than the quiescent voltage plus the peak. So that tells you the rail you'll need, at minimum.

Just did a quick run on the basis of above analysis, except that I didn't bother working out the \$V_\text{BE}\$ for the BJT. I just set the DC bias for the input signal as shown. Close enough.

Spice reports the power in the speaker at \$1.0043\:\text{W}\$ and the RMS voltage across the speaker as \$V_\text{SPKR}=2.0043\:\text{V}_\text{RMS}\$. Not bad for shooting from the hip, so to speak.

Sadly, Spice also reports that the power for the BJT here is about \$3.9\:\text{W}\$ and for \$R_\text{E}\$ is \$20.7\:\text{W}\$. This ignores all the driver stuff that would normally have to back up this output section and would only increase the power dissipation. But in round numbers this means burning off about \$25\:\text{W}\$ in a circuit that will deliver \$1\:\text{W}\$ to the speaker. (Which is a good thing, I suppose, if you own a lot of stock with your local power company. You'll get some of your cost back in dividends.)

You can work this backwards. Given that your \$V_\text{CC}=12\:\text{V}\$, then find that \$R_\text{E}\le 2.5\:\Omega\$. (From: \$R_\text{E}\le R_\text{SPKR}\left(\frac{2}{3}\frac{V_\text{CC}-V_\text{P}-2\:\text{V}}{V_\text{P}}-1\right)\$.)

## FINAL NOTES

Here is an LTspice schematic that can be used to program up and test out any particular crazy-minded class-A amplifier of this kind. I've used a particular BJT in it that can handle some current -- but feel free to change it. Other than that, the specs are hopefully clear. Make some changes and run it.

Aside from setting up the desired power and load values, only one of the two -- either \$R_\text{E}\$ or else V_\text{CC}\$ -- can be set, as the other one is then determined by that choice. So set exactly one of them, not both. The other must be set to zero.

The .ASC file follows here to save time having to write up the above schematic, by hand.