# Oscillator questions

Discussion in 'General Electronics Chat' started by Piggins, Dec 5, 2014.

1. ### Piggins Thread Starter New Member

May 5, 2014
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First question is abot the tapped inductor. What part of the voltage over the tapped coil goes to the emitter of the transistor? Is it the voltage over L2 or the whole voltage over the inductor? Does it work by raising the emitter voltage and closing the transistor when the voltage over the coil is enough compared to the base.

Also it says that the input waveform is at the emitter. I dont understand that at all, where does that wave come from? Or is that the voltage formed over R3 when C2 couples the coil voltage there?

It says in the explanation that when the top part of L1 conducts current a current is also induced in the lower part of the circuit. Does that mean that the lower part is at negative DC level then?

2. ### MikeML AAC Fanatic!

Oct 2, 2009
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The transistor in the Hartley Osc you linked to is operated as a common-base amplifier.
Where is the input to a common-base amplifier?
Where is the output from a common-base amplifier?
Is a common-base amplifier an inverting or non-inverting amplifier?
Study up on Barkhausen Criterion

On the second question: Study up on center-tapped transformers

Last edited: Dec 5, 2014
3. ### alfacliff Well-Known Member

Dec 13, 2013
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the emitter current flowing through the coil provides the feedback to the base tap. the emitter feedback is in phase, an oscilator with collector feedback has 180 degree phase shift .

4. ### Piggins Thread Starter New Member

May 5, 2014
26
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Thanks for the replies. I read the article about the center-tapped transformers and it was very informative, it did clarify the bit about the common emitter version of the hartley oscillator.
But unfortunately I still cant grasp the common base one , the explanation says:
Because in Fig. 2.1.6 (and Fig. 2.1.1) the top of L1 is connected to +Vcc, it is, as far as AC signals are concerned, connected to ground via the very low impedance of C5. Therefore waveform X across L1, and waveform Y across the whole circuit are in phase.

These are the figures referenced.
I cant understand why the whole signal is in phase.
My best guess is if I was to think the L1, L2 coil as a 3 tap autotransformer with the center being the connection going to the emitter and with the top grounded. So L2 would be in anti phase then?
Thanks.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I have uploaded a conceptual equivalent circuit which may clarify the circuit behaviour.

6. ### Piggins Thread Starter New Member

May 5, 2014
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Thanks for the reply. So the current goes thru L2 in phase with the oscillating signal in the tank?

7. ### MikeML AAC Fanatic!

Oct 2, 2009
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I have prepared the following LTSpice simulation of a typical grounded-base Hartley as it oscillates.

Look at the circuit and plots and answer the following questions:

Why is there no AC signal present at V(supply)? At V(base)? (Can these be considered "ground" for AC signals?)

Why is the DC level of V(base) where it is?

Why does V(collector) swing to a voltage higher than V(supply)?

Why is the peak-to-peak value of V(tap) approximately 0.2 of the peak-to-peak value of V(collector)? (look at the turns ratio)

Why are V(tap) and V(collector) centered on V(supply)? (What is the resistance of an inductor?)

Where is the AC signal ground point for the autotransformer?

Why is V(tap) in phase with V(collector)? (Your question about how an autotransformer works)

Why is V(emitter) so small compared to V(collector)?

Why is V(emitter) more-or-less in phase with V(collector)?

Which capacitor primarily determines the frequency of oscillation?

What is the function of capacitor C2? C3?

When you can answer all these questions, you will have a good understanding of this oscillator works.

Last edited: Dec 9, 2014
8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Terminology can become a little fuzzy sometimes so it's not clear what you mean (for instance) by "the oscillating signal in the tank"

There will certainly be an AC voltage across the tank which could be measured at the collector - ignoring the DC offset. See Mike's simulation.
Presumably, you might concede that to be the tank oscillating signal. Ignoring the loading effect of the emitter connection on the tank inductor winding tap point, the tap will produce a voltage which is a specific fraction of the total tank AC signal voltage. The fractional value would (ideally) be directly proportional to the ratio between the turns to the tap point (from AC ground) and the overall tank winding turns from AC ground to collector. In practice the tap point will be loaded by whatever AC impedance is seen "looking into" the emitter terminal. So there will be some discrepency between ideal and actual tap point AC voltage. Because the tap point voltage is simply ratiometric there is no phase inversion - as would the case with the typical relationship between the input and output voltage phase relationships with an auto-transformer. One might have to allow for some phase displacement in the case of the practical oscillator depending upon the nature of the various non-ideal circuit components.