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Is it a hartely oscillator??


walid

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Is this circuit using a hartely osc.?

Hi Walid.
Yes, because its coil is tapped and provides positive feedback to the base of the transistor.

1- is the emitter resistor affect the frequency of oscillation?

No, it affects the current in the transistor which changes the output power.

2- this is a transmitter circuit, is it FM or AM and why?

It is simply a continuous carrier frequency without AM nor FM modulation.
If something modulates the amplitude of the carrier then it is AM. If something modulates the frequency of the carrier then it is FM. Some simple modulators produce both at the same time.
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The hartely oscillator has many drawbacks as do ther other simple oscillators. It could have to do with the resonant frequency of the LC and the ability to get high enough gain.

Please i didn't understand.
It may also have to do with any change in load conditions that could prevent or stop oscillation.

what if we use a buffer circuit between?

thank you KevinIV
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  • 3 months later...

Is there any reason for the center-tapped inductor being standard to the Hartley Oscillator? Using two inductors should allow for a better design because this part of the circuit has less tolerance. Or is there a mathematical advantage to using the center-tapped inductor?

A center-tapped single inductor is used because it inverts the signal. The collector of the oscillator transistor produces a signal at one end and the base receives the signal at the other end. The signal is inverted to produce positive feedback to sustain oscillation. The transistor can have a voltage gain of only slightly more than 1 and the oscillator still works.
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The circuit layout can produce 270 degrees of feedback.

Not if the circuit is assembled correctly.

Is C1 and the emitter resistor used for any of the phase lag, or is C1 just a coupling capacitor?

C1 AC couples the feedback from the inductor to the transistor.

I'd be surprised if this circuit works at all with R1 at 1M. The proper way is to bias the transistor with a potential divider.
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Then C1 has a low impedance compared to the transistor input impedance? Where is the collector voltage compared to C1 input voltage?

A simple calculation shows that the 1000pf of C1 has a reactance of only 5.9 ohms at 27MHz so it is almost a dead short with no phase shift.

The center-tapped coil has 45 turns on the collector side and the same 45 turns on the base side so the base signal is the same amplitude as the collector signal.
The center-tap makes the coil like a teeter-totter. When one side goes high then the other side goes low which is 180 degrees phase shift. The transistor inverts so it also has 180 degrees phase shift. The total phase shift of 360 degrees results in oscillation.
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The variable capacitor has to be a higher impedance than the series inductor so that the collector voltage is inverted compared to the base voltage. Unless the feedback capacitor and transistor base add to the phase shift, the amplitude is reduced by at least 75%.

No.
I explained how the center-tapped inductor inverts the signal with 180 degrees phase shift so the transistor provides the other 180 degrees phase shift for it to oscillate.
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