Electronic – Why can’t class A amp drive 8 ohm speaker with just one BJT

bjtclass-atransistors

I designed this class A amplifier. It's my first attempt at designing an amplifier with transistors. My objective is to build a guitar amplifier with tubes, but I figured that building an amp with BJTs and a low-voltage power supply first will enable me to understand what's going on before I attempt anything more complicated.

I want to drive an 8-ohm speaker. The problem, of course, is that if I connect RS, the speaker, suddenly the AC signal sees basically 8 ohms to ground and can't develop any voltage at the Q1 collector.

I think that the proper way to do this is to connect the Q1 collector to the base of another transistor, Q2, configured as an emitter follower, and drive the speaker from the emitter of Q2.

But what I'd like to understand is, what is the limiting factor that prevents me from driving the speaker from Q1? I have a fuzzy understanding that I need Q1 to provide voltage amplification and then an emitter follower to provide the current to drive the speaker, but I don't understand why I can't get both in one place, since all the power is ultimately coming from the same power supply. Is there a way to enable this circuit to drive an 8-ohm speaker without adding a second transistor?

schematic

simulate this circuit – Schematic created using CircuitLab

Best Answer

It can be done. Remove Rs and C2. Let your speaker be RL. RE1 and RE2 should be smaller, because they seriously limit the voltage swing over RL.

Prepare to waste continuously about 9W, half of it in the speaker and possible RE and the other half in Q1 if you want the maximum theoretical audio output power. I bet 2N3904 doesn't stand one second, you need something stronger and well cooled.

ADD due the comments:

12V supply allows about 6V peak voltage swing. The DC resistance of the 8 Ohm speaker is near 8 Ohms. Let's assume it is 7 Ohm. Some RE is useful for linearity and thermal stability. Let's have one RE, only 1 Ohm. Now we have 6VDC over 8 Ohms.That means current =750mA. That means 9W DC power outtake from 12V in idle state.

The speaker should stand 750mA DC if it can handle say 10Watts.

Constructions like this were used by hobbyists in the old days (say 1960 - 1970) because they could get a couple of watts acceptable audio signal AC power without complex push-pull stages and the continuous power loss wasn't a problem if one used 12V car battery as his power supply. A 20cm x 20cm x 4mm aluminium heatsink was able to keep the transistor cool enough. The price of the transistors and other components is today so low that it's useless to avoid the normal push-pull output stage. Even class A amp works best as push-pull.

One trick: Put a massive inductor in parallel with the speaker. The DC current does not go through the speaker and you can multiple the output AC power, because the voltage swing can be more than 12Vpp. The inductor should be about 50 mH or more and it's iron core should be large enough to stand the DC magnetization without getting saturated.

In theory you can add a DC saturation compensating winding to the same inductor core if you drive the compensating DC through another speaker to pick the outputted AC. Otherwise it's a shorted transformer.

Another way to compensate is to add a permanent magnet to the magnetic circuit.