Electrical – How RF Choke doubles Vcc at collector in BJT Power Amplifier

The BJT power dissipation is minimal - any current it takes comes through R2 which is 15kohms. Supply is 3.3V and if half appeared across the BJT (max power transfer) it would dissipate 182uw!

The easiest way to picture the action of the BJT and D1 is by considering that the transistor cannot turn on too much because if it did, D1 won't have enough voltage across it to conduct current to the BJT's base and turn the transistor on hence, assume that the base is only very slightly forward biased at about 0.4V and therefore D1 must have 2.4V on its cathode.

And here's where you are going to hit problems with the low supply voltage....

On the original example it ran from 9V and you could assume the collector voltage would be at 2.4V + (say) 2V higher (due to R1 volt-drop). With 4.4V on the collector you can calculate collector current (\$\frac{9v - 4.4v}{15k\Omega}\$ = 307uA). And from this you could make a reasonable assumption about base current being 30 times lower\$^1\$ at 10uA. 10uA through a 220kohm resistor produces 2.2V across it and this isn't far away from the 2V I assumed by sticking my finger in the air.

But, on the reduced supply voltage design there isn't the headroom to assume the collector is 2V higher - you can barely assume 0.5V and this is making me think that the base current will be more like 2uA with a collector current of more like 60uA. This 60uA will drop 0.9V across the 15k resistor and clearly this doesn't figure because it would only leave 2.4V on the collector and not enough.

Therefore base current must be even smaller and I have doubts about the noise it will produce if any BUT good luck.

\$^1\$ Quite a few BJTs have poorer current gain at quite small collector currents so assuming 30 for \$H_{FE}\$ isn't unreasonable.

The gain of your transistor circuit depends on what signal load the collector has to fight against. It sees \$R_C\$ which is either 166 ohms or 83 ohms (typo in original question) and it also sees \$R_L\$ which is 50 ohms.

Together they are in parallel and their signal impedance is somewhat less than 40 ohms.

To get gain from this type of circuit, the emitter impedance needs to be lower than the collector impedance (40 ohms max). As \$R_E\$ is (probably) 83 ohms, at low frequency audio the gain will be less than 1 because gain is \$\frac{Z_C}{Z_E}\$ and we've established that \$Z_C\$ is probably less than 40 ohms.

At higher frequencies the 1uF capacitor comes into play and the question is what frequency will it start to lower the emitter impedance and give positive gain. If we ask the question "what frequency does 1uF have an impedance of 83 ohms" the answer is about 1.9kHz.

At frequencies above 1.9kHz you will start to see gain gradually rise. Of course that's useless for your guitar (80Hz low E string).

The problem with this circuit is that you are wanting to drive a 50 ohm load and expect gain. That is unrealistic.

When it comes to using it to amplify an 810kHz signal you will get better results because the 1uF emitter cap has an impedance of less than an ohm and this means you'll have gains of over 50 probably.