Colpitts Oscillator Frustration

Discussion in 'General Electronics Chat' started by Xupack, May 14, 2014.

1. Xupack Thread Starter New Member

Jun 4, 2013
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Hey everyone!

Long time lurker, first time poster. I'm a fourth year electrical engineering student, my major is geared more towards power applications/renewable energy but have always been interesting in radio engineering (it's what actually got me into electrical engineering) and I'm a bit ashamed to ask such a basic question. I've been trying to build a basic wireless switch for a while now and came across some circuits online. While researching I've encountered the Colpitts oscillator (to generate a carrier frequency) a few times but I can't seem to figure out how it works (I hate building circuits without understanding the ins and outs of it).

I'm trying to figure out this one, in particular: http://www.learnabout-electronics.org/Oscillators/images/Colpitts-01.gif

I understand that we need a positive feedback at the required frequency for oscillation to occur, negative feedback to stabilize the output and some form of amplification. However, I really need someone to explain to significance of each component in the circuit (no need to explain the tank circuit, and the fact that the transistor is in essence an amplifier)

And if it's not too much to ask an explanation for this circuit, as well: http://www.pyroelectro.com/projects/pyro_rf_transmitter_27mhz/img/oscillator_theory.png

2. #12 Expert

Nov 30, 2010
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R1 and R2 set a voltage level on the base and provide base current. R3 limits the collector current as (Vbase - .6V) / R3 = collector current.
C4 sends some signal to the emitter resistor as the feedback path.

There is a parasitic capacitance from collector to base, but a good design will use capacitors big enough to make that insignificant or take it into consideration while doing the math. It's called, "Miller" capacitance and the effect is negative feedback.

C1 renders the base void of AC, so the feedback must be through C4.
That's the half I can do.

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3. KL7AJ AAC Fanatic!

Nov 4, 2008
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That is a somewhat deranged Colpitts; most of the time the tank is fed back through the Emitter, not the Collector.

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4. Brownout Well-Known Member

Jan 10, 2012
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It is fed back to the emitter, through C4. Also, R3 provides negative DC feedback, and stabilizes the bias.

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5. t_n_k AAC Fanatic!

Mar 6, 2009
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Since this is grounded base topology, Miller effect plays little or no part.

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6. #12 Expert

Nov 30, 2010
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Good point. In this case, the Miller capacitance is merely a tiny AC leakage to ground.

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7. vk6zgo Active Member

Jul 21, 2012
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I feel the same,Eric,but I've noticed when Googling for Coipltts,the familiar circuit of our youth is conspicuous by its absence!

When I made a similar comment to yours,it was greeted with contempt by the "yoofs" on the forum I was on!

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8. studiot AAC Fanatic!

Nov 9, 2007
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Xupack, Welcome to AllAboutCircuits.

Here is a rush through treatment that may help.
As a fourth year E Eng student you should be able to follow this OK.

Colpitts and their cousin Hartley oscillators are feedback oscillators.
The Colpitts has a split caopacitor and one inductor, the Hartley has a split inductor and one capacitor.
For both the inductor may be part of a transformer so the output may be taken from a transformer secondary that does not play a part in the oscillator action.

Fig1 shows the general circuit diagram for both types, along with a Nyquist diagram.
The impedances are arranged such that the Nyquist diagram includes the point (1,0) as shown. This makes the setup unstable ie able to oscillate.
More of this later.

For a single transistor implementation both types can appear in in one of three guises, common emitter, common base, and common collector.

The Colpitts versions are shown in Fig3 and Hartley in Fig4.

In Fig5 I have taken the standard configuration from Fig1 and redrawn with a standard transistor current source equivalent circuit.
This is then redrawn to convert the current source as a Thevenin equivalent.

This enables the circuit equation to be written down.

Substituting values for the impedances in the colpitts oscillator we arrive at the expression for the frequency and the condition for oscillation.

I will leave it to you to do the maths for the Hartley.

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Aug 1, 2013
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Four stars.

ak

Apr 19, 2012
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Apr 19, 2012
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12. Brownout Well-Known Member

Jan 10, 2012
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Write the transfer function for feedback through the 3 impedances, and chose f(0) to make the imaginary part = 0.

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13. alfacliff Well-Known Member

Dec 13, 2013
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actually that is not the optimum colpits oscilator, feedback from collector to emitter, grounded base, where do you take off the outut without loading the resonant circuit? on the emitter to base colpitts, you could at least take off output from the collector without loading the resonant tank, increasing "Q" and stability.

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14. Xupack Thread Starter New Member

Jun 4, 2013
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Thanks everyone! I really appreciate all of your replies.

I think I understand the basic operation of the oscillator now.

My next challenge is being able to design a common base Colpitts oscillator. I understand that the basic operation involves the amplifier being in a class A configuration then gradually slides into class C and that's where the challenge is, for me. How can I bias the transistor? I'm using the following circuit as a guideline:

First of all, while analyzing the circuit I began to question the values of the voltages given. For example, calculation of Vbase results in a voltage of 2.45 (9V*5.6k/15k=2.446 V) NOT 2.3 V. Calculation of Vemitter results in a voltage of 1.65 (Vb-0.65V= 1.65 V) NOT 2.2 V. Am I making a rookie mistake here?

Even when using LTSpice I get different values:

Vbase Simulation (Transient Stage)

Vemitter Simulation (Transient Stage)

Putting all the possible mistakes of the circuit aside, my question is how can I bias the transistor to achieve this "sliding class A to C operation"?

As I understood from Learnabout-electronics.org, the transistor should be biased to achieve class A amplification and the rest will work itself out, is this correct?

What's the "correct" quiescent point for a 2N3904 transistor?

I'm looking for a step by step design procedure. Let's assume I have a 2N3904 transistor, looking for a collector current of 10mA and a supply voltage of 9 VDC.

Thanks everyone, your help will be greatly appreciated!

15. dougc314 Member

Dec 20, 2013
38
11
The circuit is a bit odd because of C4. Its also drawn confusingly, because normally the top capacitor is shown from the emmiter to ground, not VCC. But VCC is AC ground so it's correct. If you look at the wikipedia article for Collpits oscillators you'll see that the simplest circuit has an inductor in the collector, with a capacitive transformer from the emmiter to collector. Now, don't forget that the top of the inductor is connected to a battery and is AC ground. That makes the inductor in parallel with C1 and C2 (and the transistor) C1 and C2 act as a transformer because they are in a resonant circuit with the inductor. The equation for the oscillator frequency is simply the equation for a L-C resonant circuit F0=1/(2pi*sqrt(L*C)). In this case C is the series connection of C1 and C2. C4 is superfluous, and if made large enough has no effect. In fact the analysis (fig 5) ignores C4. If you are trying to get a high frequency, say 50 MHz it will oscillate low due to stray capacitance and the junction capacitance's of the transistor. The miller capacitance is still in effect, I believe, its actually across B-C, not B-E. The circuit transform in fig 5. changes the parallel Z3 and current source Hfe*ib into a voltage source with a value of Hfe*ib*Z3 in series with Z3 (Norton to Thevinin) Hfe > C2/C1 is the condition for the closed loop gain to be greater than unity. (C2/C1 is the transformer tap ratio)

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16. t_n_k AAC Fanatic!

Mar 6, 2009
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In grounded base mode (as with the OP's schematic), whilst there is certainly parasitic C-B capacitance the Miller Effect is not an issue.

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17. t_n_k AAC Fanatic!

Mar 6, 2009
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Looking at the schematic it would seem the base and emitter voltages shown are mean values for the circuit actually in steady-state oscillation. Your calculated values are presumably pre-oscillation ....

This might verify a trend (slide) from initial Class A (pre-oscillation) bias towards Class B or Class C operation. The actual values one obtains in practice are not necessarily going to be identical to those indicated on the schematic - owing [say] to device to device parameter variations in the transistor.

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18. Xupack Thread Starter New Member

Jun 4, 2013
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Thank you dougc314. I realize the connections seem a bit odd, but I used this particular schematic because it was the simplest and had actual values in it. Also, I don't think C4 is actually necessary. I built the circuit without it and achieved oscillation with no issues!

Thanks for your response t_n_k! However, I still don't understand how the values for the voltage divider network were obtained (biasing) and the value for Remitter, as well.

I don't have a problem with the tank circuit but the rest is a bet confusing, to me. Should I simply bias the transistor for class A operation? (Since, at switch-on the amplifier is in class A configuration).

I tried something yesterday I'm not sure if it's the optimal way but the circuit oscillated in the simulation; I looked at the circuit in DC conditions which looked something similar to this:

(L1 shorted, C1, C2 and C3 opened)

I then simply assumed a load resistance of zero and continued to bias the transistor for class A operation with Vbase-emitter bigger than 0.7 V and an Ic (Iq) of 20 mA.
After that, I chose a value for Cbase that would have very low reactance at my oscillation frequency and then continued to use the correct values for the tank circuit that would resonate at the required frequency.

Everything worked fine but I'm still not sure if this is the optimal or even the correct way to do it?

Thanks everyone!

19. t_n_k AAC Fanatic!

Mar 6, 2009
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When the oscillator runs in simulation what mean base voltage do you observe? I imagine with the oscillator running in class C mode that the base voltage will be a few volts lower than the emitter. In other words the BE emitter junction is biased well beyond cut-off. I wouldn't put too much confidence in the +2.3V indicated on the schematic as the mean base voltage. I'd probably more likely believe -2.3V.

The normal biasing relationships that hold for class A operation no longer have relevance under large signal class C operation.

I doubt there is actually an optimum pre-oscillation bias setup. Near enough is probably good enough. You just have to get the transistor into the active region. However the discussion in the link below gives some general guidelines for the Colpitts biasing goals ...

http://www.ece.ucsb.edu/Faculty/rodwell/Classes/ece218b/notes/Oscillators2.pdf

In the end this is one of those things that you actually have to build and try in practice. Relying on simulation is not the best option.

Last edited: Jun 21, 2014