Oscillator buffer - unexpected behavior

Thread Starter

atferrari

Joined Jan 6, 2004
4,095
I've been testing this 1,8 MHz oscillator (circuit attached) followed by a buffer. Managed to complete the adjustments probing always on point A (buffer's transistor not soldered yet).

After soldering the buffer's transistor, surprise!! found no signal in points B or C. I recalculated the bias divider with current values shown to no avail.

Finally, after replacing the transistor twice, not sure why I probed A, B & C using both points and found that as long one of the points was probing point A, I would get output signal in B (and in C as well). I swapped the points, tested both channels with independent trigger, changed to X10 and back to X1. No change.

After lifting point D in C14, I applied a 400 mV pp sine from my signal generator and got the expected output at B (and C as well).

My questions:

a) Why I get signal at the output only if a probe is on point A?

b) Accepting that the oscillator's Ic is not enough, how to actually calculate the necessary to drive the buffer appropriately? Every time I read that the input impedance is related to the parallel of R13//R14, I get lost. Sorry but the little I knew of Thevenin, Norton and God knows who, is gone. It seems that my 73 years are claiming their share on me. Cannot help much.

Could anyone explain in simple terms the most basic so I can go ahead with my calculations? Gracias for that.

Clapp 06 flawed.png
 

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upand_at_them

Joined May 15, 2010
519
And since the probe will essentially be in parallel with the capacitor you should find out the capacitance of your probe and adjust C15 so that the total (caps in parallel get added) is the pre-adjusted value.
 

MrChips

Joined Oct 2, 2009
21,890
The input impedance of Q4 circuit is too low.
Try using a JFET instead. JFETs take very low currents on the gate input.
 

Thread Starter

atferrari

Joined Jan 6, 2004
4,095
The value of C15 is critical. Putting a scope probe at A or D change C15. Try changing the value of C15 slightly.
And since the probe will essentially be in parallel with the capacitor you should find out the capacitance of your probe and adjust C15 so that the total (caps in parallel get added) is the pre-adjusted value.
Dead-on suggestions, gentlemen. Thanks for that!

The Owon manual says 10 pF +/- 5 pF so I added 8 pF in parallel: output coming and going. Adding 22 pF instead, got a steady output. :) C15 now = 77 pF.

@ronsimpson

I repeated part of the circuit provided here by member @Bordodynov for my first Colpitts.

Ron, could you explain what justifies the necessity of C15? Is that so evident when designing an oscillator like this from scratch?

BTW, LTSpice seems not to notice changes in value or even if C15 is not there.

Gracias again.
 

oliverb

Joined Feb 23, 2020
6
I'm going to speculate that R16 may be too large, and that this is "starving" the oscillator. Adding probe tip capacitance lowers the impedance at the collector, making the feedback path from the emitter more effective. The theoretical implementation appears to tie the collector to the supply and run the transistor as an emitter follower.
 

peterdeco

Joined Oct 8, 2019
225
I did this a while back. I used a small signal NPN, common emitter configuration, and biased it like a class A audio amplifier with a resistor on the collector. It did amplify the source but you can avoid that with some emitter resistance.
 

Thread Starter

atferrari

Joined Jan 6, 2004
4,095
@ronsimpson
@Danko
@Bordodynov

Sorry for tagging you, gentlemen; I am bumping this thread because I am really interested to get some answers. Just in case, please check post #7 and my circuit in OP.

The actual circuit working quite well in the breadboard.

I am repeating my questions here:

How do you recognize the need to include C15 in the circuit? IOW what is the actual function of C15? Kind of a decoupling for RF?

Why in the tens of schematics you find around in the Web, nobody seems aware of the need?

How do you calculate / estimate the required value? At the current frequency (1,8 MHz) Xc = 1.400 ohms app.

LTSpice seems not very "interested" in C15. Even if you eliminate it, nothing seems to change at the output.

Thanks to anyone who could reply.
 

Danko

Joined Nov 22, 2017
991
How do you recognize the need to include C15 in the circuit? IOW what is the actual function of C15? Kind of a decoupling for RF?
Capacitor C15 decreases Zout @ point D, so generator is less affected by load, but Vout decreases too.
You can reduce R16 to 9.1k or 8.2k, then probably generator will work as in LTspice (even without capacitor C15).

BTW, try simulation with R16=10.7k (No oscillation).

ADDED: FFT of Vout
With C15 = 56p:
1598826057852.png

Without C15:
1598827460299.png
ADDED:
1598836971844.png
 

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Thread Starter

atferrari

Joined Jan 6, 2004
4,095
Capacitor C15 decreases Zout @ point D, so generator is less affected by load, but Vout decreases too.
You can reduce R16 to 9.1k or 8.2k, then probably generator will work as in LTspice (even without capacitor C15).

BTW, try simulation with R16=10.7k (No oscillation).

ADDED: FFT of Vout
With C15 = 56p:
View attachment 216003

Without C15:
View attachment 216005
Thanks for replying, Danko.

While I still do not understand clearly why, I am aware that C15 is necessary in the circuit.

Thinking of recalculating the oscillator with a higher Ic, maybe around 4 mA.
 

Thread Starter

atferrari

Joined Jan 6, 2004
4,095
Looks promising and simpler but sorry @Danko ; I am a stupid.

When looking at your circuit I realized that since the beginning I obviated what I should have had considered upfront: with Vout I want to drive a frequency counter implemented with a PIC micro. That's why I thought of a buffer since the first attempt but failed to think of the pk-pk values.

The micro's Vcc is 5V. To measure my signal with no comparators in between but directly to one input, Vout should be a sinusoid of 4V pp riding an offset of +2,5V, IOW going from 4,5V to 0,5V.

That's the sole application intended for this oscillator. Gracias.
 
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