Thank you both for your replies.
Sorry for not adding more explanations to my circuit - I've been working on it off-and-on for over a month, so everything's very obvious to
me... The "Accessory Connector" in the top left section is supposed to be connected up to the actual connector for the radio that you see in the insert picture. So, the MIC net is the Microphone (TX)
input to the radio, and the SPK net is the speaker (RX)
output from the radio. So SPK is directly connected to the Headphones and Headset connector (which I just noticed I assigned the wrong net to, SPK should be MIC_EXT) for monitoring the radio RX audio without additional amplification, and to the amplification circuit for driving a loudspeaker. The MIC net is connected to the Microphone biasing circuit, which is directly connected to an electret capsule (i.e. headset / microphone "MIC" connector) at MIC_EXT.
I hope that clarifies your confusion about the MIC biasing, AudioGuru - again, sorry for not explaining that up-front!
External connections often introduce a ground, so, in your case, be cautious if the radio has a DC power jack separate from its accessory connector.
Hi, any external (mains-derived) power to radio or my circuit would come from independent isolated power supplies.
Maybe this is something I need to research / study more. I attached a quick toy circuit of having two voltage sources driving two loads sharing a ground:
Is this a common "nothing can go wrong here" arrangement? Where would the "external connections introduce a ground" problem come into play here? Would I just need to make sure that the external (third) ground is also connected to my common ground to avoid a ground loop? Is a ground loop the specific problem I need to avoid here?
For a class-A circuit to drive a loudspeaker then the currents must be VERY HIGH with VERY LOW resistor values and a transistor that can conduct a high current and be able to get rid of a lot of heat. I simulated your amplifier and it can drive a 10000 ohms load well but has a problem driving 1000 ohms or less.
The circuit you copied and the random 2.2k resistor value you used creates voltage divider. The collector and emitter values are the same so 1/3rd of the 5V can be across the collector resistor in series with the low resistance of the loudspeaker. The maximum peak current is (0.33 x 5V)/2208 ohms= 0.75mA. Then the peak power in an 8 ohm speaker is 0.75mA squared/8 ohms= 0.07 thousandths of a Watt and the average power is half. A cheap clock radio produces 0.5W.
I haven't made any calculations for the actual resistor values I would need to use for this! I'm still at the "pick an amplifier design" stage. It sounds like my simple circuit really is pretty much useless except for demonstration purposes and I should go with an amplifier chip like the LM386 suggested by Ian.M.