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Why this oscillator does not work?

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x1 probe must be expected to have a large input capacitance of e.g. 100 pF, considerably affecting the circuit operation. In some cases, it may even stop the oscillation.

Typically, the collector node would be selected as oscillator output, i think. But it should be buffered to drive a low impedance load.
 

x1 probe must be expected to have a large input capacitance of e.g. 100 pF, considerably affecting the circuit operation. In some cases, it may even stop the oscillation.

Typically, the collector node would be selected as oscillator output, i think. But it should be buffered to drive a low impedance load.

But that would not explain the low voltage reading? The frequency reads more or less as expected at 28 mhz, but vpp is still low.
 

Both points are part of the explanation
- the emitter node voltage is already divided down compared to the collector node
- the large load capacitance further reduces the output voltage

In addition, the x1 probe also represents a real load impedance, that can reduce the oscillator amplitude or stop the oscillator. Without the probe load, a high oscillation amplitude can be seen in a simulation, however.

A x10 probe should be used for a realistic measurement.
 

Both points are part of the explanation
- the emitter node voltage is already divided down compared to the collector node
- the large load capacitance further reduces the output voltage

In addition, the x1 probe also represents a real load impedance, that can reduce the oscillator amplitude or stop the oscillator. Without the probe load, a high oscillation amplitude can be seen in a simulation, however.

A x10 probe should be used for a realistic measurement.

Thanks, for the x10 probe I got about 3Vpp on the collecter node although ideally i would want a voltage reading above 6v as per simulation, I am using a pc based DSO.

---------- Post added at 16:05 ---------- Previous post was at 15:12 ----------

I will therefore assume that the low VPP reading (3vpp using the x10 probe) is due to the impedance of the probe, not the actual characteristic of the circuit, but how do i actually simulate this impedance which leads to a low VPP?
 

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