Talk:Buck converter/Archive 1

Archive 1

No Explanation of self-regulation

It is mentioned that buck converters are self-regulating and not prone to allowing variation in input voltage to affect output voltage. This is not obvious given the equation $D={\frac {V_{o}}{V_{i}}}$. Efadae (talk) 03:35, 7 May 2009 (UTC)

Buck regulators are not naturally self-regulating. In general, there must be some sort of control circuit that cycles the switch at the desired frequency and duty cycle. In most practical regulators, this control circuit monitors the output (and possibly also input and switch current) and controls the duty cycle in order to maintain regulation. The quality of the regulation will depend on the details of this control loop. If there is no control loop (ie fixed frequency and duty cycle), the output will be a constant multiple of the input (under DC conditions), with dependence on the output load iff the load is light enough for discontinuous mode operation (and the converter uses a diode instead of synchronous rectification). The article should probably deal with the issue of regulation more directly. Evand (talk) 05:07, 7 May 2009 (UTC)

Voltage Divider is Incorrect

A voltage divider cannot be used as a regulation circuit. On the other hand a voltage divider in conjunction with a transistor can regulate a load although at that point your divider is now a linear regulator.

Stabilization has no bearing on this particular conversation.

Aramoe

Nice Job ! Amfortas 08:49, 19 April 2006 (UTC)

Seconded; but

- Does anyone know where the name "Buck" comes from?

- Is it true that SMPSes were primarily designed for aviation purposes (weight etc.)?

- Why not mention Schottky diodes as useful middle ground between normal diodes and sync.rect.?

(JAB 6 aug 2006, 23:07 CEST)

Circuit Operation Title Non-Conventional

The phrasing for the section "Circuit Operation" seems non-conventional to me. Wouldn't it usually be titled "Theory of Operation"?

Phosphoricx (talk) 18:11, 3 January 2008 (UTC)

Buck switching power supplies have evolved to have very low output impedances for the applications found in telecommunications today.

The assumption that the output capacitance could be "infinite" disregards both power converting theory and the practical applications of that theory.

158.116.144.3 (talk) 20:18, 23 February 2016 (UTC)

Linear Inductor Current Rise

DemoMaster (talk) 06:46, 27 May 2008 (UTC)How come that "The current through the inductor rises linearly" , shouldn't it be exponential rise?!

This is a very fine point which is ignored in analysis of switching regulators. When the inductor is not saturated inductor current is quite linear over the switching time period. The very short switching period causes the exponential charge/discharge effect to be so small as to be insignificant. The effect of changing inductance as its bias changes is much more significant and even that is ignored in most first-order analyses such as this. Jimmyswimmy (talk) 16:43, 7 January 2009 (UTC)

Actually, the linear rise/fall of current is due to the large capacitor in the circuit. It functions as an ideal voltage source and as a result, the inductor does not see a resistance. This means the time constant of the inductor circuit is infinity, which leads to a linear function of current vs. time.

The reason for the capacitor is to simplify analasys and reduce current ripple. Without it, the circuit will still work but the current would be a peice-wise exponential waveform. The average voltage and current values still hold. —Preceding unsigned comment added by 69.196.190.37 (talk) 04:06, 8 February 2010 (UTC)

Expression for duty cycle in presence of non-idealities

I just did the derivation for duty cycle expression in case of the inductor loss alone (assuming switches to be ideal) and I did not get the V_L term in the denominator. Could someone check it out?

EDIT: I've just checked in Erickson's Fundamentals of Power Electronics, 2nd Edition and there is no voltage drop across the inductor term in the denominator, only in the numerator. Naumz (talk) 11:50, 14 May 2009 (UTC)

Theory of Operation - Ideal Behavior

Should the article not point out that the derivations assume that the output capacitor is infinitely large?

If V_o remains constant, the current through the load I_R is also constant. Therefore, the current into the output capacitor I_C must vary (oppositely) as I_L. But since V_o is constant, ${\frac {dV_{o}}{dt}}=0$, hence ${\frac {I_{C}}{C}}=0$. For I_C that is non-zero, C must be infinite. --Oneover0 (talk) 00:24, 14 November 2009 (UTC).

MMM, of course assuming C is infinite is great for analysing the steady state case but a converter with infinite output capacitance could never start as no matter how much current it delivered to the output cap it would never raise the output voltage from zero. I've tried to add a paragraph explaining that what we are about to analyse is an idealised circuit. Plugwash (talk) 01:30, 4 February 2012 (UTC)

Buck?

Why is this converter called "buck"? Maybe someone can add one sentence on the origin of the term? --57.79.167.152 (talk) 10:44, 27 June 2011 (UTC).

The name “Buck Converter” presumably evolves from the fact that the input voltage is bucked/chopped or attenuated in amplitude and a lower amplitude voltage appears at the output.

Source: Design and Implementation of ZCS Buck Converter Project Report for Final EvaluationSubmitted by: Gyana Ranjan Sahu(10602019) Bimal Prasad Behera(10602044) Rohit Dash ( 10602043

MPPT?

I think a boost converter can be used for maximum power point tracking. A boost converter has continuous current flow from the photovoltaic's source, rather the switch is on or off. On the other hand, the buck will chop the PV power off, and only make use of it when on. To me this suggests it is not a good choice for maximizing an available power source as MPPT is aiming for. — Preceding unsigned comment added by 71.223.13.156 (talk) 19:58, 16 August 2012 (UTC)

Anthropomorphic Windings?

I loved the human explanation in terms of an inductor's "reluctance" (sic) to allow a change, followed by the sympathetic resonance of "the inductor doesn't want it to change from 0, so it will attempt to fight the increase" (etc.).

I suppose it can change but it has to be induced to do so. Coercion might have an effect, too, if applied frequently.

Back in the last century we used to put it in such impersonal terms: an inductor tends to oppose current, because with flow it must build a magnetic field (in the reluctant medium of the core).

But this is much more fun. We could have EMO (Electro-motive Onslaught?) instead of EMF ;-)

Let's keep it this way.

One is one and one is one (talk) 21:32, 4 July 2013 (UTC)

Freewheeling or flyback?

"Freewheeling" diode or "flyback" diode? What did your prof call it in your class notes? Or can we arrive at a sacred consensus? Caling the same thing by two different names is confusing. --Wtshymanski (talk) 17:14, 14 November 2013 (UTC)

On the "Concept" Section...

The quoted text: "When the switch is first closed, the current will begin to increase, and the inductor will produce an opposing voltage across its terminals in response to the changing current. This voltage drop counteracts the voltage of the source and therefore reduces the net voltage across the load." Makes it seem like the inductor is generating a voltage of reversed polarity to "counteract" the source voltage. And the phrase "voltage drop" may be miscognited to draw a parallel to the properties of a semiconductor such as a simple diode, where there is a constant, unrecoverable voltage drop, rather than a constantly changing voltage. I request that someone with more background increase the clarity here. 7 June 2014 — Preceding unsigned comment added by 68.228.204.160 (talk) 06:23, 8 June 2014 (UTC)

Error in figure 9

Not sure if I'm reporting this the right way, but there appears to be an error on the labeling of one of the switches in Figure 9. The second switch (going top-to-bottom) is labeled S21, when in fact it should be labeled S12. There are 'n' phases in this circuit, and each one has a switch 1 (Sn1) and a switch 2 (Sn2). The label S21 is already used correctly further down the figure to represent the first switch in phase 2.

On ripple voltage and a quantification thereof also pertaining AoE edn 3 (please discuss) - PhD Physics, Cambridge, UK

For the continuous case, the article states current difference between load and input for on-state, but of course it must be remembered that in the off-state current is being supplied also, with current coming from the capacitor being supplemented from the inductor here, not just from the capacitor.

I was looking to quantify the ripple voltage here, and it's not so easy using the usual Q = CV, and related I = C*dV/dt, due to the current supply from the inductor always. Let's perhaps think deviation from the steady-state here or DC level, in terms of currents, and perhaps (delta-I)/2 as being that flowing into the capacitor ONLY (assume constant load current - Iav), as sinusoidal on top of a DC component etc. Considering the impedance of the capacitor, which tends to zero as C becomes infinite, as being much lower than that of the load, it might be a good idea to assume this to a good first approximation. My best upper-bound here then in terms of ripple voltage would therefore be ((delta-I)/2)*(1/(omega*C)), omega = 2 * pi * ripple frequency, as half delta-Vpp-ripple.

But certainly, the "ripple" section for the continuous case needs overhauling here, as per my first paragraph. If I have further thoughts I will post back. I had a quick look in my Art of Electronics 2nd ed but not much on this here. Perhaps there's better in version 3. May be something relevant in Duffin (Electricity & Magnetism) but I haven't looked yet. Perhaps also in a book by Tooley (electronic circuits; blue cover - 2nd ed?) but I've never seen so many errors in a book in all my life!

Please discuss how best to quantify ripple here. Nice work though from what I've seen so far. — Preceding unsigned comment added by 86.17.132.85 (talk) 14:15, 28 December 2015 (UTC)

Further comments pertaining quantification of output ripple and AoE 3rd edn) PhD Physics, University of Cambridge UK

Please see: https://artofelectronics.net/wp-content/uploads/2016/02/AoE3_chapter9.pdf

AoE (Art of Electronics 3rd edn) Chapter 9 page 645 -> 4) - Buck converter and Voltage Ripple: how to calculate. You'll need edition 3 as there are no equations in edition 2, just qualitative treatment.

I thought and said above that voltage ripple (peak to peak amplitude; I used delta I/2 above so just amplitude there not peak-to-peak) was approx: delta I / (omega * C) i.e. ripple current times impedance. If we assume it's all going to the capacitor - infinite capacitance and all or certainly say 1.7 ohms worth for a 4.7uF capacitor at 20kHz ripple freq so about 0.17 Ohm for a 47uF etc we see this is a good approx'n.

If we do some elementary calculus on integral Imin to Imax of Idt/C (C = Q/V; so dV = dQ/C, and I = dQ/dt) for delta V, we get (Imax-Imin) * delta T / 2C, or delta I * delta T / 2C where delta T is t-on say (t-on from AoE edn 3 pages 642-646). You can easily get it from an areal approach to the integration; it's a triangle, hence the "2". And so I understand approximately AoE 3rd edn's: Vripple = Ipeak * ton / 2Cout for voltage ripple.

What I don't understand is why the current goes from 0 to Ipeak AoE Chapter 9 page 645 at 4). Wikipedia also crudely states voltage ripple as: dV = i dT / C but this talks as if you're not getting any current from the inductor during the cycle and treats as if just simply a capacitor discharging.

It seems systemic that nobody seems to do a proper treatment of output ripple voltage (mine's peak to peak here not amplitude as my first comment). If you compare the two versions above (mine) there's only a pi in it, as the former equates to: delta I * T / (2 * pi * C) where T is 1 / f, the time period of the ripple (not t-on). So it's going to be a multiple of t-on, say 3 * ton which is roughly pi * ton and so the pi and 3 roughly cancel and the approaches are somewhat equivalent. So you'd say we're possibly along the right lines here.

Please can someone explain to me why AoE edn 3 uses 0 to Ipeak (ILpeak) here instead of ILmin to ILmax (= ILpeak). Surely we can consider just the alternating current as all going into the capacitor or reacting off of it if you like due to the very low impedance and thus output ripple is as the above simple treatment (either of the above it would seem) isn't it? Of course AoE 3rd edn's approach would approximate mine as Iout tends to Iout(min), but surely it would be way off for Ipeak (AoE edn 3) >> Iout(min).

I have emailed one of the authors of AoE 3rd edn to explain their approach further (P Horowitz). They may wish to update AoE / Wikipedia if this information is useful, but the above is what I use for my calculations of ripple voltage.

---

Some hours later...

Looks like I may have made a mistake here but it's not too far off from my previous working (I've been looking at this the last few hours). Page 643 AoE edn 3 (see link above or get a copy). Draw the IL vs time on the time axis (move it down) so Iout is along the time axis. Now it looks more like the current into/out of the capacitor vs time. As per integral Idt, we see that the areas are (Imax-Imin)/2 * t-on/2 * 0.5 + (Imax-Imin)/2 * t-off/2 * 0.5 = (Imax - Imin) / 8 * T (that's height * base * 0.5; please remember to divide before multiplying; and we evidence conservation of charge into/out of the capacitor). Now we see that we have Vripple or delta V = delta I / 8C * T. Compare it with delta I / (omega * C) where omega = 2 * pi * f as approximate ripple voltage of a sinusoid and f = 1/T. Or delta I * T / (2C * pi) or around delta I * T / 6C. These three approximations all look very similar but I prefer this last one and the ripple current times impedance for the rough guess (6 is still pretty close to 8). I'm not 100% completely sure on this but it looks right to me. And confidence comes around as all these approximations are very similar.

And so thus far please consider ripple voltage (peak-to-peak) as [delta V = delta I * T / 8C] (where T is the total time period of ripple) a very good approximation, or substitute the 8 for 2 * pi as you like from our initial example to begin with.

NB. Looks like I have only considered current flowing into the capacitor here and not out of it when calculating the integral Idt below the graph of Iout (average output current) along time axis as described above (a graph showing current into/out of the capacitor vs time; the rest - Iout - flowing to the load). The peak-to-peak ripple voltage is hence delta V = delta I * T / 4C (twice that derived above, where delta I is the peak-to-peak ripple current). This is what should be compared with delta V = delta I / (omega * C) = delta I * T / (2 * pi * C) from simple AC circuit theory: impedance times AC current. We see a factor of 4 vs factor of 2 * pi ~ 6.3

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Looks like it is in fact a factor of 8 NOT 4

See: http://www.onmyphd.com/?p=voltage.regulators.buck.step.down.converter#h3_outputripple

We note that as per the integration I was alluding to above, that Vc-min starts at t-on/2 (just after charge has flowed out of the capacitor) and reaches a maximum (Vc-max) at t-off/2. This would have been clear if I had drawn the Vc vs time waveform chart. What this means is that the above treatment of just current flowing into the capacitor is all that is needed and hence Vppripple = delta V = delta I * T /8C. Winfield Hill kindly did a SPICE analysis at 100kHz square wave C = 25uF L = 25uH Iout = 0.5A & 1A Iripple using just this formula and observed the correct ripple voltage using this formula. Got there in the end! — Preceding unsigned comment added by 80.6.69.144 (talk) 12:55, 14 June 2018 (UTC)