Analysing BJT circuit

Question

Recently, I found this circuit on the Talking Electronics web page:

I was interested in the modified circuit below, where the feedback capacitor is replaced with short circuit and the 8R speaker with resistor of 8 ohms.

I simulated the circuit with falstad in two situations:

1. When the circuit was running without feedback, I added the short circuit.
2. The short circuit was present at the beginning of the simulation.

(link to the simulation)

Without the feedback, the circuit is an amplifier, with current amplification of about 100 * 100 = 10 000 A/A. In the first scenario, when I add the short circuit while the circuit is running, the circuit continues working as before, but the voltage on the 8 ohm resistor has lower value. Maybe the addition of the short circuit removes some of the charge in the base of the NPN bjt, so it turns a little bit off, which in turn increases it's collector voltage, so the base current of the PNP bjt has lower value, affecting the collector current of the PNP and finally this decreases the voltage on the 8 ohm resistor. In the second scenario, both transistors are cutoff. Maybe this has something to do with the parasitic capacitance of the BJTs and their turn-on time.

I am kindly asking if someone can explain me how and why the circuit behaves this way.

Answer 1

It's not clear in the above schematics where the input is, or how it's coupled. The reason for the voltage changes, though, is that the capacitor doesn't interfere with the DC biasing of the circuit. The short, however, does, since it ties the DC level at the NPN base to the DC level at the output.

The biasing of this circuit seems to be dependent on the gain of the individual transistors, an unreliable system since there is a wide (or potentially infinite) range of variation across individual parts and across environmental conditions. Also, the feedback appears to be positive, since increasing the input (assuming it's coupled to the NPN base) increases the output, so it will increase the gain at the risk of instability.

With the short in the feedback path, it appears that a momentary B-E short on the NPN would turn it off permanently. Once the NPN stops conducting, so does the PNP, and the load pulls the NPN base to ground full time. You'd have to have an input signal capable of driving the load to overcome this.