One reason is that the transistor gain is degraded at high frequencies. To pick a specific example, the ON semiconductor BC546 has a gain-bandwidth product (GBP) of 100MHz at 1mA collector current (see figure 6 in the linked datasheet). This means that at a frequency of 27MHz, the current gain (beta) is about 100MHz/27MHz = 3.7, not 100.
At 27MHz, stray capacitances in the transistor (amplified by the Miller effect) may well also be playing a role in reducing the gain.
Simply replacing the transistor with one more suited to high frequencies may be sufficient to fix the problem. You may get away with just choosing a different general-purpose transistor: the 2N3904, for example, is a little better with a typical GBP of 300MHz. A better solution is probably to choose one of the many transistors designed for high frequency applications. To pick one at random, the PN5179 from Fairchild has a typical GBP of 2000MHz.
Because of the Miller effect, the common collector amplifier is not especially well suited for high frequency amplification, and topologies such as the common base amplifier are often used for signals at several tens or hundreds of MHz. However, at 27MHz I suspect you will be OK with a common emitter amplifier.
An additional factor limiting the gain is that the impedance of C4 || R6 needs to be added to r_e when calculating the emitter resistance at signal frequencies. Usually C4 is chosen to have negligible impedance at signal frequencies compared to the r_e of the transistor, but at 27MHz the impedance of your R6 || C4 is about 55Ω (dominated by the 59Ω impedance of C4). Switching C4 to a 1nF or 10nF capacitor should increase the gain by more than a factor of two.
In an amplifier like this, your objective in selecting R1 and R2 is to bias the amplifier halfway between two extremes. The two extremes are:
- The transistor is fully on, and the collector current is limited only Rc and Re.
- The transistor is fully off, and the collector current is zero.
If you hit either of these extremes, the output is clipped. So if we can bias the amplifier to be halfway between these extremes, then we have maximized the input signal amplitude that can be amplified without clipping either the positive or negative side.
We can make a couple simplifying assumptions:
- Because the transistor has high current gain (β > 75, most likely), we can consider that the current into the collector is equal to the current out of the emitter. It also follows that the current through Rc must equal the current through Re.
- Because we are only interested in biasing at DC, we can ignore all capacitors as if they are open circuits.
- Because the saturation voltage for a BJT transistor is small (0.2V) relative to the supply voltage (9V), we can assume this saturation voltage is 0.
- This amplifier's output will be connected to a high impedance, so we consider this current to be zero. Notably, a speaker is not a high impedance (8Ω is typical). If you want to connect this circuit to a speaker, you need a buffer amplifier.
So the first question is this: in the first extreme, when the collector current is limited only by Rc and Re, what is that current?
Since Rc and Re are in series, we can add those resistances together and calculate the current with Ohm's law:
$$ I_c
= {9\mathrm V \over R_c + R_e}
= {9\mathrm V \over 2.2\:\mathrm{k\Omega} + 1.2\:\mathrm{k\Omega}}
\approx 2.65\:\mathrm{mA} $$
Remember, this is the current through the collector of the transistor at one extreme of clipping. The other extreme is no current at all. So halfway between these points is just half of \$I_c\$, or about 1.32mA.
So what voltage needs to be at the base to make \$I_c = 1.32\:\mathrm{mA}\$ ?
The base emitter junction of a BJT transistor is effectively a diode. And we've already established that the current through Rc, the collector, and Re are equal (see simplifying assumption #1). So we can simplify this problem a bit:
simulate this circuit – Schematic created using CircuitLab
Given that we know the current through Re (1.32mA), we can calculate the voltage across it with Ohm's law:
$$ 1.32\:\mathrm{mA} \cdot 1.2\:\mathrm{k\Omega} = 1.58\:\mathrm V $$
We also know that the forward voltage of any silicon diode is about 0.6V. Added to the 1.58V above, that means if we want the current to be 1.32ma, then V1 will need to be:
$$ 1.58\:\mathrm V + 0.6\:\mathrm V = 2.18\:\mathrm V $$
So now you just need to come up with a pair of resistors for R1 and R2 that make a voltage divider with an output of 2.18V.
You must also keep in mind that the current into the transistor base will introduce some error into your voltage divider. We can estimate that this current will be the collector current, divided by the transistor's gain. If you then pick your voltage divider values such that the current through the divider is at least 10 times the base current, then this error will be negligibly small.
To keep the math simple it's reasonable to guess that transistor gain (β) is 100. So the base current will be something like 0.0265 mA. You want the current through the voltage divider to be at least 10 times this, or 0.265 mA. By Ohm's law:
$$ R_1 + R_2 < 34\mathrm{k\Omega} $$
Finally, you will want to adjust your simulator to input a much smaller amplitude signal. The output signal can't be possibly more than 9V peak-to-peak, and actually less than that because the transistor can't drive the output all the way to the supply rails. Since this is an amplifier, that means the signal will need to be very much less than 9V peak-to-peak, otherwise you will see clipping and attenuation.
Best Answer
There is a bit of redundancy in your design - both Q1 and Q2 do exactly the same thing - you can get rid of one of them and have R3 and R6 connected to the same point such as the collector of Q1.
The other main problem is that the largest peak-to-peak voltage you'd be able to get on the output is about 1.5 volts because the output transistors are wired as emitter followers in push-pull - to make either transistor conduct, folk tend to use the 0.7 V base-emitter rule - i.e. the base needs to be 0.7 volts greater than the emitter (NPN) and 0.7 volts less than the emitter (PNP).
Given the speaker is 8 ohms and the biggest sinewave from the amp (on a 3V3 supply) is 0.53 volts RMS, the power into the speaker is 35 mW (before things start to clip/distort).
I'd throw it away and get a chip that does this sort of thing - even a H bridge amplifier suitable for motors is going to be better and you'll get a lot more power.