# Why won't my 2-transistor oscillator work in the real world?

I am learning about transistors, and I've designed this oscillator circuit.

simulate this circuit – Schematic created using CircuitLab

In the simulator it works perfectly, and in my brain it makes sense, but when I try to implement the circuit in the real world it stabilizes with one side of the capacitor 0.7V above the other. I expect it to repeatedly charge and discharge the capacitor.

My reasoning:

Start w/ the capacitor discharged and all transistors are in their cutoff mode.

If the cap is discharged then it's like a short circuit, so R1/R2 just form a voltage divider, which has a lower voltage than D, so Q1 is not yet forward biased. This means Q2's base is starved for current (which it must get from Q1's collector).

As the cap charges, the voltage at A will eventually exceed the voltage at D, causing Q1 to become forward biased. This will in turn cause current to flow from Q1's collector to Q2's base, forward-biasing Q2 as well.

Now there is a 'connection' (via Q2's collector/emitter) between D and B. This pulls D down to the same voltage as B, keeping Q1 forward biased until the voltages at A and B are equal.

Now the voltages at A, B, and D are equal, reducing the current through the transistors to zero and returning to the original state.

Any information on what I may be missing or how to make my circuit work IRL would be greatly appreciated, thanks!

• You might want to explani why you think it should oscillate so we can point out the mistakes in that reasnong. Feb 2, 2017 at 10:54
• @PlasmaHH, I linked to a simulation showing the expected behaviour. Feb 2, 2017 at 10:58
• Maybe try it with a more useful simulator like ltspice, and you will see that you are not in the initial state, I currently don't have time to show that in detail Feb 2, 2017 at 11:26
• You state: "If the cap is discharged then it's like an open circuit...". Not so. A discharged cap is like a short circuit in the sense that any voltage you apply to it will draw infinite current, or current limited by its own internal impedance, or current limited by the impedance of the voltage source. Feb 2, 2017 at 11:43
• Your circuit can work as a oscillator youtu.be/2a1I1X3RV0g?t=41s
– G36
Feb 2, 2017 at 19:49