Understandin the tank circuit in Hartley and Colpitts circuits. - Page 1
Forum section: electronics / beginners

author: Dan Moos
author_rank: Frequent Contributor
timestamp: September 05, 2020, 10:48:46 pm
content: So in a Hartley or Colpitts, one of the reactive elements is split in the middle and grounded.Let's take the Hartley for example. The inductor is tapped and grounded at the middle point. Yet, for purposes of the tank circuit, it's one inductor (the two in series). I guess I don't know how to analyze this thing, but intuitively it seems weird that grounding the midpoint of the inductor doesn't change the resonant point from what it would have been if it were just one continuous inductor.If I re-draw the usual schematic representation, its basically an inductor to ground in parallel with a series cap and inductor to ground. I feel the key to understanding is seeing it this way, but so far I'm stuck.

author: ledtester
author_rank: Super Contributor
timestamp: September 06, 2020, 12:20:47 am
content: Don't make too much of where the grounds are. Here is a Hartley oscillator where the feedback amplifier is capacitively coupled to the tank which allows the center tap of the inductor to be at a completely different potential:The left-hand side of this circuit is a standard class A amplifier implemented with an NPN transistor. Input comes through C2 and the amplified output (phase shifted 180 degrees) comes out C3. The point is that the amplifier provides regenerative feedback to keep the tank circuit going.

author: Dan Moos
author_rank: Frequent Contributor
timestamp: September 06, 2020, 12:32:17 am
content: Quote from: ledtester on September 06, 2020, 12:20:47 amI know i can build these things not knowing why it works. I'm looking for a deeper understanding.Your posted circuit seems to put the supply at the center-tap, which for AC is no different than ground. My questions still remain.I hate "black boxes". I'll admit this has made learning slightly slower, but I do tend to really "know what I know" once I get there since I try to leave no stone unturned.

author: ledtester
author_rank: Super Contributor
timestamp: September 06, 2020, 12:55:38 am
content: QuoteNot sure exactly what you're saying. The center tap is at -6 V. The point is that the center tap can be at whatever potential you want.My explanation of the Hartley is:- first understand how a LC-tank naturally rings- that the top of the tank is 180 phase shifted from the bottom of the tank- you take the signal from the bottom of the tank, send it through a 180-degree phase shifting amplifier and inject it into the top of the tankand that keeps the tank ringing

author: Dan Moos
author_rank: Frequent Contributor
timestamp: September 06, 2020, 01:20:04 am
content: I do understand ALL of that. What I don't understand is how the tank circuit isn't affected by the centertap.

author: ledtester
author_rank: Super Contributor
timestamp: September 06, 2020, 01:42:44 am
content: Ok, you're right that the AC ground isn't affected by V2 in my circuit.Not sure if this will help but... the AC ground needs to be placed in the "middle" of the tank in order for the top and bottom to be of opposite signs in the AC analysis. The feedback becomes a lot less effective the more it is placed toward one end or the other.

author: Dan Moos
author_rank: Frequent Contributor
timestamp: September 06, 2020, 01:50:32 am
content: I think you are still missing my question. I FULLY understand that the tap gets you two opposite phases. What I don't understand is how grounding part of the inductor doesn't change the resonant frequency of the tank.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 06, 2020, 02:34:58 am
content: How would it matter? You can join two masses with a spring and set them bouncing. If the masses are equal, their amplitudes will be equal (and opposite), symmetrical around the center of mass. Or if one is twice the mass of the other, their amplitudes will have that same ratio; and so on. Or if you put the reduced mass* on end end and fix the other end of the spring to "ground"**, and get identical force, amplitude, frequency and damping, with all the amplitude all on the one mass ("ground" is assumed not to move / has infinite mass).*"Reduced mass" is what it says on the tin: a mass-spring-mass system is equivalent to a mass-spring-ground system with mass M = (M1 M2) / (M1 + M2).**"Ground" being a very large mass, Mg >> M, so its motion is negligible. Note that an infinite mass has no effect when the above formula is applied.This still might be hard to intuit, at least without diagrams or animation, or a working understanding of Newtonian mechanics (let alone dynamics; this is a problem of mechanics in motion, after all). I'm guessing that a mechanical analogy will be more effective than solving the electrical system directly -- but for what it's worth, exactly the same relationships hold true!Tim

author: Dan Moos
author_rank: Frequent Contributor
timestamp: September 06, 2020, 03:03:42 am
content: Quote from: T3sl4co1l on September 06, 2020, 02:34:58 amOk, so just so I'm following, in your analogy, the "mass" is the current. In a perfectly split inductor, current is equally resisted at a given freq. If I move the tap, the current is the same, just ratioed differently.Since my two springs still add up, the amount of resistance to current taking place is the same.Am I clumsily getting it right?

author: fourfathom
author_rank: Super Contributor
timestamp: September 06, 2020, 06:48:10 am
content: Or just consider an LC tank floating in free space -- no connections to anything. If you magically give it a kick it somehow, it will ring at the resonant frequency.Now attach a single ground connection to any point in the tank circuit. It doesn't matter if it's at the top, bottom, center-tap, capacitor tap -- it's still going to ring at the resonant frequency. That single-point connection has no effect.It's when you attach impedances at two different points on the tank circuit that you start affecting the resonant frequency and Q.

author: VooDust
author_rank: Regular Contributor
timestamp: September 06, 2020, 08:05:47 am
content: Yes, the center tap is weird. Especially since Ground has this connotation of a black hole that simply swallows everything. Though conceptually it's just a common reference point.See, a ringing LC means the capacitor and inductor are passing charge from and to each other. So if the cap is charged with 5V, it releases its charge in the form of current through the inductor. The inductor would not care about any of its taps, because current flows between points of different potential, the two points being the capacitor's poles.

author: emece67
author_rank: Frequent Contributor
timestamp: September 06, 2020, 09:52:51 am
content: .

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 06, 2020, 01:56:41 pm
content: Quote from: Dan Moos on September 06, 2020, 03:03:42 amTo be perfectly exact, you have your choice of two analogies (which are related by a series-parallel transformation; in a sense, the equivalence between Hartley and Colpitts circuits).The inductor equation is: V = L dI/dtCapacitor: I = C dV/dtIf we let the spring be the inductor, then the masses are the capacitor. The spring has the equation F = kx (Hooke's law), but this is missing something -- rather than position, we should use velocity. Then,F = k int v dtdF/dt = k vv = 1/k dF/dtSo the spring relates velocity to rate of force, in the same way that the inductor relates voltage to rate of current. Likewise the mass:F = m a = m dv/dtMass relates force to rate of velocity, in the same way that the capacitor relates current to rate of voltage.So velocity becomes voltage, force becomes current, amplitude is peak velocity, impedance (Z = V/I) is the ratio of velocity to force, etc.Whereas if we let the masses be the inductors, then the spring is the capacitor. F = m dv/dt means force is voltage, and velocity is current.We can distribute the system's mass into two pieces of any ratio, and they will give the corresponding ratio of amplitudes (with respect to the center of mass).We can also secure the masses with respect to ground -- in any ratio, without much care for the individual masses -- by linking them together. Say we use a pair of geared levers (like a compass and divider might use). The levers themselves will have some spring rate and distributed mass, but as long as those modes can be pushed to high enough frequencies (using stiff enough levers and gearing), that's okay.We can't so easily do the same for the spring. In principle, we could anchor the midpoint of the spring when it happens to coincide with the center of mass. But this is not a good idea: if the anchor is an ideal ground point, then the force in either half of the spring will match only by coincidence, not by equilibrium. Which means, in general, we shouldn't expect this to give a useful phase shift.We could make our own center-tap with two additional springs, one from each mass to ground. This triangle of springs and masses makes a potential divider, keeping the masses centered (at whatever amplitude ratio we like). These act in parallel with the mass-mass spring, so we do need to distribute spring constants appropriately to keep the same resonant frequency and impedance. Downside: notice the two masses act through the new springs, with respect to ground, giving us an even-mode resonance, which is probably undesirable, and comparable in frequency to our intended (odd) mode.But that's just circuit design.We do use large inductors or capacitors to apply bias to circuits, and they contribute resonant modes that can be above or below the design frequency, it's fine as long as they aren't competing with the active oscillation mode.QuoteSort of. What I'd like to get you to see, I think, is that the tap is the center of mass -- a natural zero-motion location, defined by the distribution of mass. In the electrical circuit, it's defined by the ratio of capacitances in the identical way (Colpitts), or inductors (Hartley, no coupling), or turns ratio (Hartley, well coupled; equivalent to masses with linkage).It's not so easy to vary a mechanical system, as it is an electrical one; in the electrical case I can just touch a ground wire anywhere along an inductor, and have that be my ground reference. In the mechanical system, I need to change the ratio of motion for the two levers, at the very least swapping out sets of gears, or pegging different linkages between them. But these are equivalent operations, no matter if one is easier than the other!Tim