Colpitts oscillator - role of feedback resistor - Page 1
Forum section: electronics / beginners

author: aneevuser
author_rank: Frequent Contributor
timestamp: February 02, 2019, 04:54:28 pm
content: In the following video:Mr Offset Voltage discusses a Colpitts oscillator. There is a feedback resistor R1. At about 24:00 (https://youtu.be/3c5u_HRp8m8?t=1456), he briefly mentions this resistor having some kind of transconductance effect.To my mind, "transconductance" would involve some kind of I->V or V->I transformation, but I don't see how that applies here. AFAICS, R1 merely forms a potential divider with the tank circuit.I am not sure what he is getting at here - can anyone explain?

author: Audioguru
author_rank: Super Contributor
timestamp: February 02, 2019, 06:46:25 pm
content: The total positive feedback causes the oscillation. Transconductance has nothing to do with it.

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 02, 2019, 07:02:55 pm
content: Quote from: Audioguru on February 02, 2019, 06:46:25 pmTotal positive feedback indeed is required for sure.If the amplifier were an ideal opamp, it would have zero output resistance. With a zero output resistance, C1 is meaningless. To get a total loop phase shift of 0 degrees, you will need to get 180° to add to the 180° of the inverting amplifier. But, since the inverting opamp circuit has actual input resistance less than infinity, you cannot get the 180° you need. By adding the extra feedback resistor, you can introduce an additional phase shift via R and C1. That is enough to scooch you to 180°.Let me know if you think I am Barkhausen up the wrong tree!

author: David Hess
author_rank: Super Contributor
timestamp: February 03, 2019, 04:50:58 am
content: Transconductance is voltage in and current out which is the case for transistors and tubes. But operational amplifiers do not have transconductance and are voltage output devices making them unsuitable to directly drive the tank circuit. If he had used an operational transconductance amplifier without Rf, then the output resistor R1 would not be necessary and instead of adjusting the voltage gain of the operational amplifier with feedback resistor Rf, the transconductance of the operational transconductance amplifier would have been adjusted with a separate input.Incidentally, current to voltage is transimpedance. Transconductance is measured in mhos (reciprocal ohms) and transimpedance is measured in ohms (reciprocal mhos).

author: aneevuser
author_rank: Frequent Contributor
timestamp: February 03, 2019, 01:22:27 pm
content: Quote from: David Hess on February 03, 2019, 04:50:58 amAt the moment, I'm unconvinced that we need to consider transconductance at all.I've been pondering the Colpitt's circuit in the video, and it seems to me that all we have is:a) an inverting voltage amplifier with gain set by RF and RI in the usual wayb) a phase shift network formed by R1/C1 and L1/C2, with R1 variableThe phase shift network is tweaked to give 180 degrees shift via R1, and it is this inverted voltage that the op amp sees, and amplifies.Also, although the phase shift network does have a resonant frequency, this doesn't necessarily align with the frequency for 180 degree shift (AFAICS), so the fact that we have a tank circuit sitting there is somewhat secondary.All IMHO, of course, but I see no need to analyse this circuit via anything other than the voltages at the moment.I'll try and analyse the phase shift network more theoretically when I get a bit of time, to see if I can justify my remarks above.QuoteUseful distinction - I'll note that for future reference.

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 03, 2019, 01:42:53 pm
content: Quote

author: Doctorandus_P
author_rank: Super Contributor
timestamp: February 03, 2019, 01:50:53 pm
content: "transconductance" is mentioned in the context that the tubes of the first collpits oscillators were transconductive devices.Right after that he says that his opamp is a voltage amplifier.Single inductors or capacitors have at most 90 degree phase shift.Without the feedback resistor then C1 is meaningless (for an ideal opamp).With only one L and C left, the utmost phase degree is 180 degree,The feedback resistors forms an extra pole with C1 so the 180 degree phase shift can be achieved without getting into the limit of the LC circuit.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 03, 2019, 02:23:55 pm
content: For an ideal op-amp, we have a circuit with perfect isolation: signal flows from input to output, and none from output to input (within the amp itself). We also have a finite input impedance (because of the gain setting resistors), and a zero output impedance (which is then made nonzero by the series resistor).This tells us that we should apply network filter theory. We have an input port, an output port, and some system impedance associated with each.The CLC network is then a lowpass network with some phase shift and gain property, which is the important part here, and probably with a strong gain peak (its response will be overall lowpass, but the intent for an oscillator is to have particularly high loop gain at resonance).Tim

author: aneevuser
author_rank: Frequent Contributor
timestamp: February 03, 2019, 03:34:46 pm
content: Quote from: Wimberleytech on February 03, 2019, 01:42:53 pmI think that's a "confused" symbol, no? If so, then my point is that one part of the condition for oscillation is that the total phase shift around the loop be 0 degrees - we are not interested in whether or not the feedback network resonates or not - that's irrelevant in the Barkhausen conditions. So the presence of resonance here is secondary to the oscillation of the circuit, AFAICS.

author: aneevuser
author_rank: Frequent Contributor
timestamp: February 03, 2019, 03:41:29 pm
content: Quote from: T3sl4co1l on February 03, 2019, 02:23:55 pmIsn't the RCLC network the thing that we need to analyse? The CLC doesn't seem to be enough.QuoteI don't follow what you're saying here - we can have a phase shift oscillator with no resonance at all, as in one with a CR feedback network.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 03, 2019, 08:06:20 pm
content: Yes, the Rs in network analysis are implicit as the source and sink resistances.Peaking includes the total loop gain of course, so you can have something like a passive RC network and enough gain to fix it up; but the phase noise will be worse than a sharper tuned network.A Colpitts is usually phrased in terms of a rearrangement of any other oscillator prototype, e.g., Hartley which is a strict bandpass type. It follows that the Colpitts equivalent would have a gain peak, even if it cannot have the same overall bandpass behavior because the lowpass asymptote is set by topology.It's not necessary to stick to that format, though, so you can indeed compensate for less gain peaking with more loop gain. It's not an absolute statement.Detailed analysis of noise response, of course, needs losses of the network, and noise of the amp. Also, nonlinearities are relevant to the conversion of thermal and 1/f noise to sidebands / phase noise. (In short, if you need performance quite that good, you'll need some good tools and/or a lot of tweaking to do it.)Tim

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 03, 2019, 11:57:17 pm
content: Quote from: aneevuser on February 03, 2019, 03:34:46 pmThe IEEE dictionary defines "tank" as follows:tank circuit (signal-transmission system) A circuit consisting of inductance and capacitance, capable of storing electric energy over a band of frequencies continuously distributed about a single frequency at which the circuit is said to be resonant, or tuned. Note: The selectivity of the circuit is proportional to the ratio of the energy stored in the circuit to the energy dissipated. The ratio is often called the Q of the circuit.--The structure of a tank is inductance and capacitance. The nature of a tank is that it will resonate at some frequency.The circuit drawn in the video is a tank circuit. And for his implementation of the oscillator is was necessary. Now, granted, you can achieve the needed phase shift without inductance. But in the video, the circuit analyzed...the LC circuit is necessary.

author: emece67
author_rank: Frequent Contributor
timestamp: February 04, 2019, 01:31:04 am
content: .

author: aneevuser
author_rank: Frequent Contributor
timestamp: February 05, 2019, 09:34:38 am
content: Quote from: emece67 on February 04, 2019, 01:31:04 amThis of course relates to my original question. If I understand correctly, it seems to me that the reference to transconductance is completely misleading - wouldn't we need to have a current source present in the circuit driving the oscillator to meaningfully talk about transconductance?QuoteThis is a pretty nice summary. Thanks.However, I've finally had the time to play around with this op amp based Colpitt's circuit, and whereas I've been able to get it to oscillate reliably, I've found that the frequency of oscillation has varied by up to +10% from the resonant frequency of the tank circuit (the frequency of circuit oscillation has always been greater than the resonant frequency, when it's differed). I don't have time to check these results at the moment, but if I haven't screwed up, it suggests to me that your analysis of oscillation frequency is not the whole story.Is it usual for the oscillation frequency of a Colpitt's oscillator to differ significantly from the resonant frequency?

author: David Hess
author_rank: Super Contributor
timestamp: February 05, 2019, 09:45:46 am
content: Quote from: aneevuser on February 05, 2019, 09:34:38 amTransconductance outputs have a parallel load resistance like voltage outputs have a series resistance. Even if the bias is provided by a perfect current source, the collector or plate resistance is in parallel with the transconductance output. 100s of ohms to kilohms to 10s of kilohms is typical although going higher is completely feasible with additional design complexity like an output cascode which is common in precision current sources.

author: emece67
author_rank: Frequent Contributor
timestamp: February 05, 2019, 09:55:17 am
content: .

author: aneevuser
author_rank: Frequent Contributor
timestamp: February 05, 2019, 09:55:37 am
content: Quote from: T3sl4co1l on February 03, 2019, 08:06:20 pmI know nothing about analysis of phase noise but this is an intriguing comment - can you expand on it? I take it that you're saying that "more gain in feedback network (=> less gain is amplifier) => better phase noise".Is that a general principle? In fact, I'm not even sure what the source of phase noise is that you're referring to here - is it some effect inherent to the resonance of the tank circuit, rather than thermal noise, or whatever?QuoteI don't quite follow this either - are you saying that you can, say, choose to design a Colpitt's oscillator to oscillate away from the resonant frequency of the tank circuit by adjusting the amp. gain suitably? If so, I don't see how.

author: aneevuser
author_rank: Frequent Contributor
timestamp: February 05, 2019, 09:59:37 am
content: Quote from: emece67 on February 05, 2019, 09:55:17 amBy "in circuit", do you mean while physically present in the Colpitt's oscillator circuit? If so, no. I measured it as a separate subcircuit, then connected it to the amp.I'm not even sure how I would measure it when it's part of the whole circuit, apart from measuring the oscillation frequency.

author: emece67
author_rank: Frequent Contributor
timestamp: February 05, 2019, 10:31:04 am
content: .

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 05, 2019, 10:48:22 am
content: Quote from: emece67 on February 05, 2019, 10:31:04 amI would think breaking the loop at the output of the opamp ahead of RF and driving into RF. Otherwise, C1 will not see the resistance it normally sees when the loop is closed.The signal source will probably have a 50ohm output resistance, so that will add to the value of RF and give a small error.

author: aneevuser
author_rank: Frequent Contributor
timestamp: February 05, 2019, 12:39:04 pm
content: Quote from: Wimberleytech on February 05, 2019, 10:48:22 amThat makes sense. I'll re-measure the frequency like this when I get the time - got to put this to one side for now though. I'll redo the whole set of experiments at some point - I'm suspicious that I've screwed something up given the discrepancies in some of my measurements.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 05, 2019, 02:35:11 pm
content: Quote from: aneevuser on February 05, 2019, 09:55:37 amNot really, unfortunately; I'm not thoroughly studied/practiced in the arts of low noise, I just know what I've seen others talk about in the space.A lower gain amplifier can have lower noise, though. Technically, an amplifier acts as a refrigerator for the feedback resistor -- the amp consumes power to oppose the noise power of the feedback resistor. (I think I have that right?)Kind of sad that the electrical noise from a resistor is an extremely low power connection (~picowatts?), so you basically can't insulate something well enough to measure this, let alone make it practical. (To put it another way, you have two degrees of freedom by connecting to the resistor through a 1-dimensional transmission line; the resistor to space and surroundings, however, has zillions of degrees of freedom, because of the spacial fields (photons and phonons) that connect to it. Thermo counts degrees of freedom, so ambient heat wins.)QuoteFor a well-damped network, there is no gain peak, and the phase shift at resonance (which isn't really resonating) needn't be exactly 180 at that point; by the point it is, gain is very low, so more gain is needed to make it oscillate.Well, if you define resonance as 180 degrees, then that's resonance by definition; I suppose it's as much a semantic as a practical difference. For a network more complicated than an all-series or all-parallel RLC, the different kinds of resonance need not match up (impedance becomes real / voltage or current or power reaches peak / phase shift equals a magic number, usually a multiple of 90 degrees / etc.), and it matters what you're doing.Tim

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 05, 2019, 05:22:46 pm
content: BTW, ran a quick simulation with LTSPice and my results were as expected from theoretical calculations.