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schematic

simulate this circuit – Schematic created using CircuitLab

I've built this schematic from this link (parallel tuned colpitts VCO), and changed it up a bit for my frequencies/components at hand. I temporarily replaced the varactor with a trimmer cap for tuning purposes and to check if the schematic actually works. I had to increase C3 value dramatically, since otherwise the oscillations wouldn't start at all.

The tank circuit values right now are such that it should produce frequencies somewhere around 30 MHz. (If a trimmer is at 20pF than the combined capacitance is 22 pf, and the frequency is 33.9 MHz, according to this calculator). However when probed with the oscilloscope the output frequency is always close to 12 MHz (from 11.5 to 13 MHz). I say always because the frequency doesn't change much neither when I change the capacitance of the trimmer, nor when I solder in different value capacitors entirely, which I did several times.

I am not very familiar with high frequency circuits and it seems to me that something in the circuit acts like a capacitance in a tank circuit which I can not change. Changing C3 value doesn't change the frequency. Changing C1,C2 does neither however at some point the oscillations just stop. Could the internal capacitance of the transistor cause this behavior?

By the way the sine wave itself is not very clean either it is a bit slanted to the right and looks like the bottom part is cut off at the very end.

Can someone explain to me what might cause this strange behavior, and what can I do to solve the problem?

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  • \$\begingroup\$ "a bit slanted to the " what? An image wouldn't be that bad... also where do you probe and with what input impedance? Measurements always influence the DUT, could that be an issue? Did you try a simulation, how does that behave? \$\endgroup\$
    – PlasmaHH
    Aug 22 '17 at 15:30
  • \$\begingroup\$ Edited and added a description, sorry can't take a picture now. I am probing at the output of C7 cap. The probe is 1 MOhm with 18pF capacitance. Would the probe influence the frequency so dramatically, if at all? \$\endgroup\$ Aug 22 '17 at 15:34
  • \$\begingroup\$ I am not good at these maths, but I would expect ~15MHzish from that one, maybe you can tell us how you came up with that number, so if there is a mistake in the calculation someone can point it out? \$\endgroup\$
    – PlasmaHH
    Aug 22 '17 at 15:36
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    \$\begingroup\$ L1 should not form a tank as drawn ... R1 should absolutely kill the Q (but work out the Q of R1/L1-C2 to be sure). However, it may well couple to L2 unless one of them is screened or you took great care in their mutual orientation. You have decoupled the 5V supply, right? Now, what was C3 originally? 10pF or so? With its current value, C2 is effectively in parallel with the 22pFish tuning capacitasce (C4,6,8), giving 122pF, and a resonant frequency about 14MHz. Add parasitic capacitance and what you're seeing is plausible. \$\endgroup\$ Aug 22 '17 at 16:41
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    \$\begingroup\$ @Andyaka I think you're talking about L2 as the tank, though, surely? In which case you're absolutely right about rE. \$\endgroup\$ Aug 22 '17 at 17:52
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The tank circuit values right now are such that it should produce frequencies somewhere around 30 MHz

Not with the values shown - with the values shown (and a perfect BJT) it is more like 20 MHz.

The "basic" colpitts oscillator produces a frequency that is \$\dfrac{1}{2\pi\sqrt{LC}}\$ where C is the net combination of C1 and C2. This would mean that C = 50 pF hence the oscillation frequency will be 22.5 MHz.

The combination of C4, C6 and C8 will lower this frequency possibly sub 20 MHz.

Next is the BJT - it has a transition frequency of only 250 MHz and could easily add 2 or 3 ns delay to the emitter signal. This type of circuit is basically a phase shift oscillator - rarely does the oscillation frequency closely approach the peaking frequency defined by L2 and C1 (not C2). So, given that it is phase shift oscillator, the 2 or 3 ns added around the loop directly equates to a phase angle "add-in" of about 18 degrees and the oscillator will be forced to run significantly slower than 20 MHz.

A quick estimate is \$\dfrac{1}{1/250 + 1/20}\$ = about 18 MHz.

Throw in a few ten percent tolerances (capacitors and inductor) and the loading effect of the o-scope and you might be down at 12 MHz. Yes, the loading of the scope at the emitter does lower the operating frequency.

EDIT - the math to show what the resonant frequency is: -

enter image description here

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    \$\begingroup\$ Q1 b-e forward biassed across C1 pretty much eliminates C1 from the equation, lowering the resonant freq further. \$\endgroup\$ Aug 22 '17 at 17:36
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    \$\begingroup\$ @BrianDrummond you cannot eliminate C1 from the maths because L2, C1 and C2 add voltage gain of about 6 dB (when C1 = C2) and, importantly, when you look at the influence of rE, make it oscillate as per my answer i.e. C = C1 in series with C2. There is more voltage gain than is needed hence the sine wave shape is never perfect. It can be improved by connecting emitter to the junction of C1 and C2 via 10 to 100 ohms but, as you increase resistance you are dropping the loop gain gradually towards the point where it refuses to cold-start. \$\endgroup\$
    – Andy aka
    Aug 22 '17 at 17:45
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    \$\begingroup\$ @AnthropomorphousDodecahedron use a resistor as mentioned in my comment above and use a faster transistor - you want to oscillation frequency to be largely defined by the inductor and capacitors, not the silicon. In a 10 MHz colpitts oscillator that I did recently, I wanted minimal influence from the BJT so I picked one with an fT of 5 GHz. \$\endgroup\$
    – Andy aka
    Aug 22 '17 at 17:49
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    \$\begingroup\$ A C3 value of anything over 10 nF is potentially going to add parasitic inductance in series with your oscillating inductance. For my 10 MHz oscillator I used 1 nF and if yours doesn't start then the problem lies elsewhere in my opinion. \$\endgroup\$
    – Andy aka
    Aug 22 '17 at 17:55
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    \$\begingroup\$ You can get that info on the internet such as: i.stack.imgur.com/zmMpK.gif - I searched using "self resonance of capacitors". See also this Q and A: electronics.stackexchange.com/questions/172447/… \$\endgroup\$
    – Andy aka
    Aug 22 '17 at 18:14
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You have a Colpitts with 9 reactive LC components.

This config. may be suboptimal for output to meet your requirements of linearity and 2Vpp 27MHz into 50 ohms. It uses an emitter choke to raise the Q and source impedance into the "pi" filter feedback but may be somewhat spurious with reactive loads.

I changed it to a common emitter with R's added to linearize the base impedance and regulate the gain with R ratios to get a somewhat balanced sinusoidal output for the load and V+ used.

Here is my interactive simulation. It is a bit power hungry, as this class is low efficiency and indicated by colour mapping in red.

enter image description here

The AC coupled collector load must always be slightly larger than the Rc value to V+ in order to keep collector current from being starved.

C1 or C2 can be tuned with a varicap with various configs such as L or T divider is a different topic.

In theory with neglecting Miller capacitance, \$LC=\dfrac{1}{\omega ^2}\$ so 27MHz, 1uH, Ceff=34.7pF and in this simulation 70//72pF=35.5pF shows good agreement.

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    \$\begingroup\$ Note that asymmetry is present which can be computed directly from my scope peak outputs; max + min (/ mean*100%) for ac coupled 50 ohm output or 1.17Vmax, -1.10Vmin or +-3% distortion. \$\endgroup\$ Aug 22 '17 at 20:44
  • \$\begingroup\$ Thanks, this really helps. So as far as I understand the emitter choke prevented the signal from basically going into ground in the original design. What serves the same function (if it is even necessary) in your design? \$\endgroup\$ Aug 22 '17 at 21:28
  • \$\begingroup\$ Colpitts comes in all 3 common pins; CB base, CC collector and CE emitter. Output impedances for CC with output at emitter are lower and tend to reduce Q. Mine is CE with current source out. So L is necessary to raise source Z(f) in emitter but not for CE. Did you try my sim? \$\endgroup\$ Aug 22 '17 at 22:07
  • \$\begingroup\$ Don't know where "Colpitts" ends and "Pierce" begins (I think it hinges on where you apply ground). Seems a good arrangement to drive a low-impedance load. It still requires a good local bypass capacitor from DC supply to ground. Upping the power like this may make the design somewhat squirrely. \$\endgroup\$
    – glen_geek
    Aug 22 '17 at 22:11
  • \$\begingroup\$ good call @glen_geek it is more a Pierce which has emitter as AC gnd, but works better IMHO. So improved stability can add small Re. but DC current standby must exceed AC current peak in this with C coupled load if loop is opened \$\endgroup\$ Aug 22 '17 at 22:21

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