Hartley Oscillator feedback loop

[Quantumplate]

There's a pretty nice description of this circuit on the 'learn about electronics ' website, but there's one bit that still puzzles me. I copied the schematic below. The bit I just don't get: "C2 & R3 form a long time constant to provide an average DC voltage proportional to the amplitude of the positive feedback."

 if C2 & R3 reversed positions that time constant thing might make more sense to me, but of course the emitter needs a resistor in its leg.

So I'm struggling with this...for a couple of years! I hate asking for help that's my problem.
I don't see how the feedback provides a steady DC component that slides the amplifier into class C operation.
Thanks for any help.

[mag_therm]

That doesn't seem sensible to me  either.
Another problem is the loading of the full tank.
Maybe the circuit should be reconnected as a classic Hartley: emitter follower.
Or better, as a source follower.

[Quantumplate]

That doesn't seem sensible to me  either.
Another problem is the loading of the full tank.

Glad it's not just me...
What do you mean by the problem of full tank loading? Lost me there (not difficult)
Here's some more text from the explanation on the site:
When the oscillator is first powered up, the amplifier is working in class A with positive feedback. The LC tank circuit receives pulses of collector current and begins to resonate at its designed frequency. The current magnification provided by the tank circuit is high, which initially makes the output amplitude very large. However, once the first pulses are present and are fed back to the emitter via C2, a DC voltage, dependent to a large extent on the time constant of C2 and R3, which is much longer than the periodic time of the oscillator wave, builds up across R3.

How can a cap in series provide DC?

[mag_therm]

It is a grounded base oscillator with low resistance loading on both the full tank and the tap.
Maybe  it was used where strong output is required  without buffer, with thermal and Vcc stability not so important.
The positive feedback is via C2,   to a divider of  re  ( ~ 26 Ohm) and R3.
I suppose R3 is adjusted for consistent starting and oscillating.

[Quantumplate]

It is a grounded base oscillator with low resistance loading on both the full tank and the tap.
Maybe  it was used where strong output is required  without buffer, with thermal and Vcc stability not so important.
The positive feedback is via C2,   to a divider of  re  ( ~ 26 Ohm) and R3.
I suppose R3 is adjusted for consistent starting and oscillating.

Thanks. Two questions from a beginner's brain:
Why do you say low res. loading? It's high impedance at resonance. You mean the lack of any additional resistance?
And what's re? An emitter resistor not shown?
Thanks

[mag_therm]

I suggest reading about parallel tuned circuit, Q factor and bandwidth.
Read about transistor equivalent circuits (models)
Check the equivalence of tube, bjt and fet and the common emitter, common base, and common collector modes.
Then look at Hartley original tube circuit. And the  Hartley circuits used with bjt and fet today.

[mag_therm]

If you don't have a reference text for all this, maybe someone from UK on here can suggest a low cost book.
I have a good reference book from UK
Fraser "Telecommunications"
 but old and out of print I think.

[Quantumplate]

If you don't have a reference text for all this, maybe someone from UK on here can suggest a low cost book.
I have a good reference book from UK
Fraser "Telecommunications"
 but old and out of print I think.

My GE transistor manual might have something on this. Ok, I've remembered that re refers to the small signal base-emitter resistance. I don't see how that can form part of a potential divider in this mode. The input bypasses it?

[Quantumplate]

Well I found a vacuum tube Hartley in an old book. Ignore the writing on the schematic page. Description of it on the other page. This book uses electron flow, but do the arrows and explanation make sense to you!?

[Quantumplate]

Schematic

[mag_therm]

The lower transistor circuit is cut off.
Can you redo the scan to confirm that is a "real" transistor Hartley?

[Quantumplate]

Sure. Attached some of the text for the transistor Hartley too.  Definitely one mistake in the arrow direction on the schematic compared with the text...and possibly more...the usual infuriating errors rife in technical books.

[mag_therm]

Ok, Note how the full tank circuit is connected to the relatively high impedance base bias circuit. That minimizes the reduction of Q factor of the tank, keeping bandwidth as low as possible
The relatively low impedance drive from collector is fed to the low impedance tap with same purpose of keeping Q high.
A n type fet would have much less loading, being more similar to the original tube grid.
Anyway if you want to build one I suggest to initially use that last circuit you posted, rather than the circuit of the hand drawn common base connection

[Quantumplate]

Interesting, thanks for that. Done some research on why loading of the tank reduces its Q, but no articles are very forthcoming. It makes sense intuitively that draining power off would alter the qualities of the circuit...but then again, pulses of current are fed in to replenish the energy in the tank, so lots of beard stroking at the moment..

Still none the wiser on my original question. Perhaps I should contact the author, although as runs one of the world's most popular educational sites, I won't hold my breath.

[mag_therm]

The best way is to get a reference text at your level which has whole chapters about resonant circuits and the basic oscillators.
A lot of the hits on internet search on these types of subject are incomplete, or oriented to a particular application, so not general.
I think it would be tedious for you to cover your queries by Q&A on a forum, while the basics of Q factor and oscillator  loop gain etc are not covered.

Hope you find your answers.

[Quantumplate]

I think I have the basics covered after reading about 15 books - much prefer them to online articles, although some of those have had good insights.
My problem is 1) holding all the information in my head at my age, and 2) wanting to know and understand everything - at least conceptually, not so much mathematically.

So if you can hit me with a one-liner on why loading reduces Q, then that would be appreciated. I don't think any of my books covered that. I find these circuits rather fascinating as they include just a handful of all the basic components to produce a complex result.

[Quantumplate]

Would the answer be along the lines of:
The current draw reduces the current in the tank which imitates the insertion of a resistor thus lowering the Q?

[mag_therm]

Sorry only text as I am mobile on a Shakey Amtrak.
Parallel RLC resonant tank.
If no external loading, R is the loss resistance of the coil L
Q = 1/bandwidth
Q= R/jWL
So if R is reduced by addition of an external R_ext, then R_effective < R
So Q is less
And Bandwidth is higher.
Normally lower bandwidth makes a more stable oscillator.
Edit R_ext is added in parallel.

[Quantumplate]

If R is the loss resistance of the coil then Q=jwL/R for the inductor surely?
but
Q also = Z/jwL where Z is the impedance of the parallel LC circuit at resonance, so your reasoning still works....I think.

[mag_therm]

Not really,
For R in parallel with L
Q=Rp/jWL
For R in series with L
Q=jWL/Rs
Either will work because we can simply transform
Rp to Rs
In a practical coil,
Q approx 20 say
So you can then calculate both Rp and Rs  for given L and W (at resonance for eg)
Q was defined by engineer at Bell in 1930s I recall.
You can search for him.
The above were originally the only way to define Q which is dimensionless and j term is omitted.
These days Q is sometimes calculated for capacitor.
But capacitor losses << coil losses, usually
    edit correct inventor:
K.S Johnson of Western Electric

[Quantumplate]

Right, think I've got it. In parallel it's the reciprocal formula as you state. Found a line in my notes that was hiding - "Q is reduced by increasing R(L) , but by decreasing shunt resistance".
I think the key learning for me is realising the base-emitter tap loop is effectively a shunt resistance on the tank.
Thanks for your input magtherm.

[mawyatt]

Another way of thinking about Q is the ratio of total energy stored within the resonate tank circuit and the energy lost/removed per cycle from the "tank". Energy is "stored" in L and C and energy is "lost" in R and radiation/coupling.

The water analogy is where the "tank" term comes from I believe. Where a tank of water is stored (L & C) and some water lost to draining (R) and evaporation (radiation), which must be replenished or the water level (stored) will drop. Adding too much water will cause the level to rise and eventually overflow (limit), while adding the right amount (same as lost) will maintain a constant water level.

This gets into an interesting topic regarding oscillator loop gain, which must be identically unity to maintain a constant "tank" level. However, to get to "unity" the oscillator must start with a loop gain >1 and then non-linear "limiting" forces the gain back to unity.

Best,

[Quantumplate]

Another way of thinking about Q is the ratio of total energy stored within the resonate tank circuit and the energy lost/removed per cycle from the "tank". Energy is "stored" in L and C and energy is "lost" in R and radiation/coupling.

The water analogy is where the "tank" term comes from I believe. Where a tank of water is stored (L & C) and some water lost to draining (R) and evaporation (radiation), which must be replenished or the water level (stored) will drop. Adding too much water will cause the level to rise and eventually overflow (limit), while adding the right amount (same as lost) will maintain a constant water level.

This gets into an interesting topic regarding oscillator loop gain, which must be identically unity to maintain a constant "tank" level. However, to get to "unity" the oscillator must start with a loop gain >1 and then non-linear "limiting" forces the gain back to unity.

Best,

Nice info thanks. That's going in my notes.
Your last paragraph relates to my original enquiry. The circuit starts off in class A to give the tank a good thwack, and then 'slides' down to class C for maintenance. No-one seems to know quite why though. At least I'm not alone!

[mawyatt]

Nice info thanks. That's going in my notes.
Your last paragraph relates to my original enquiry. The circuit starts off in class A to give the tank a good thwack, and then 'slides' down to class C for maintenance. No-one seems to know quite why though. At least I'm not alone!

Most electronic oscillators start up with an initial total loop gain >>1, as the waveform envelope increases the oscillator core amplifier begins to enter a non-linear gain limiting region (class C for example) where the gain become "limited". This effect is natural otherwise the envelope would continue to increase unlimited to infinity, which obviously can't happen. When the oscillator achieves Steady State, meaning no variation in amplitude between cycles, the overall loop gain is identically unity (think of this as water in equals water out of the tank, so the "tank" level is constant).

In high "Q" oscillators this can take many cycles to achieve Steady State (think of this as filling up the "tank" with water/energy from empty to a specific level), and thus a relatively slow startup. In simulations this slow startup (many cycles for a high "Q" oscillator) requires significant simulation time, some work arounds were developed long ago but these are more involved and likely out of the scope of this thread.

Anyway, hope this helps.

Best,