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

Talking Electronics oscillator circuit

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.

falstad circuit

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.

  • 1
    \$\begingroup\$ In the first case, you made a short circuit. So in real life, you will burn Q2 and Q1. Do you see why? In the second case, the voltage at the Q1 base is to low to open the Q1.330k and 8R for a voltage divider, or the speaker resistance is so low that all the current that is flowing via 330k will continue to flow through speaker resistance instead of a base. The speaker steals (takes) the entire current of the resistor 330k due to the low resistance of a speaker thus, now current is left for a Q1 base. \$\endgroup\$
    – G36
    Commented Sep 18, 2020 at 19:06
  • 2
    \$\begingroup\$ Also, look here electronics.stackexchange.com/questions/338128/… and maybe this as well electronics.stackexchange.com/questions/261288/… \$\endgroup\$
    – G36
    Commented Sep 18, 2020 at 19:07
  • \$\begingroup\$ @G36 Thank you very much for your answer. I think that NPN will burn because it's base current will increase enormously because of the short circuit, and its collector current will also increase, thus increasing the base current on the PNP, burning by the same cause. Also, I found the links you posted very helpful. \$\endgroup\$
    – Bojan
    Commented Sep 18, 2020 at 19:57
  • \$\begingroup\$ "the circuit continues working as before" - that seems very unlikely as the capacitor is an essential part of the circuit that determines the oscillation frequency. \$\endgroup\$ Commented Sep 18, 2020 at 20:31
  • 1
    \$\begingroup\$ Although I know that you didn't asked for this kind of advice, I would add that the best way of learning basics about simple BJT circuits is with real components and a tiny protoboard. The smell and smoke of a burned transistor do miracles in terms of fixing the knowledge. \$\endgroup\$
    – mguima
    Commented Sep 18, 2020 at 23:14

1 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.


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