April 2025 StudentZone Quiz Solution

Nice article! It's fun learning about oscillators. My first thought was that if you wired C1 to Gnd instead of V+, the circuit would work the same since V+ is an AC ground, but you could draw the schematic without any wires crossing. (if you also move C2 to the right of Q1) That may help show the signal path and the feedback around Q1, but maybe not. In the article you state, "Emitter resistor R3 is not decoupled..." but I think C1 is effectively in parallel with it.
So, I ask why is C1 connected to V+ instead of ground? In 99.9% of my experience, capacitors connected to AC or DC ground are connected to DC ground. I think the oscillator would startup either way since the inductor current biases the transistor. Eventually I realized the the current draw on the supply is more constant with C1 wired to V+ since the AC current of C1 largely cancels the AC current of L1.
However, there may be other transient issues with C1 wired to V+. When the power turns off, if V+ suddenly goes to Gnd, the stored charge on C1 will temporarily bring the emitter voltage as low as -9V, which could damage the transistor as the base voltage is held positive by C3. If the V+ supply suddenly gates open, it could be even worse since the stored flux in the inductor may pull V+ negative. On the other hand, with C1 wired to Gnd instead, each of the three capacitors hold each of the three transistor terminals at a safe value while the supply transitions.
Therefore if I were designing this circuit, I would wire C1 to Gnd instead of V+. And I would add an additional bypass capacitor between V+ and Gnd to mitigate the AC loading of the supply. (Also the examples on the https://en.wikipedia.org/wiki/Colpitts_oscillator do this.)

Nice article! It's fun learning about oscillators. My first thought was that if you wired C1 to Gnd instead of V+, the circuit would work the same since V+ is an AC ground, but you could draw the schematic without any wires crossing. (if you also move C2 to the right of Q1) That may help show the signal path and the feedback around Q1, but maybe not. In the article you state, "Emitter resistor R3 is not decoupled..." but I think C1 is effectively in parallel with it.
So, I ask why is C1 connected to V+ instead of ground? In 99.9% of my experience, capacitors connected to AC or DC ground are connected to DC ground. I think the oscillator would startup either way since the inductor current biases the transistor. Eventually I realized the the current draw on the supply is more constant with C1 wired to V+ since the AC current of C1 largely cancels the AC current of L1.
However, there may be other transient issues with C1 wired to V+. When the power turns off, if V+ suddenly goes to Gnd, the stored charge on C1 will temporarily bring the emitter voltage as low as -9V, which could damage the transistor as the base voltage is held positive by C3. If the V+ supply suddenly gates open, it could be even worse since the stored flux in the inductor may pull V+ negative. On the other hand, with C1 wired to Gnd instead, each of the three capacitors hold each of the three transistor terminals at a safe value while the supply transitions.
Therefore if I were designing this circuit, I would wire C1 to Gnd instead of V+. And I would add an additional bypass capacitor between V+ and Gnd to mitigate the AC loading of the supply. (Also the examples on the https://en.wikipedia.org/wiki/Colpitts_oscillator do this.)