Why is the output voltage of the Buck Converter independent of load resistance? - Page 1
Forum section: electronics / projects

author: hkBattousai
author_rank: Regular Contributor
timestamp: March 17, 2015, 01:21:14 pm
content: A lot of online sources (ex:Wikipedia) give the input-output voltage relation of the Buck Converter asVout = D . Vin;where, D is the duty cycle of the driving PWM signal.A similar formula is given for each SMPS topology. In each of them, the output voltage is only dependent on the duty cycle and several other parameters (e.g.; transformer turns ratio, etc) depending on the topology, but it is always independent of the load resistance.Consider the circuit above. If the load resistance is infinite, the capacitor voltage will be equal to the input voltage after enough switch cycles. And, if the load resistance is zero, the capacitor will never fill up, so the output voltage will always stay zero. Thus, I can intuitively say that the output voltage must be dependent on both the duty cycle and the load resistance.Why do the formulas discard load resistance?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 17, 2015, 01:33:31 pm
content: They assume continuous current mode operation, zero resistive (or other) loss in all components, infinite capacitors and inductors (and, accordingly, the conditions only occur at t --> infty, since it takes infinite time for such components to stabilize), and infinitesimal switching speed.This is the easiest case to analyze, because the inductor is connected to* a source which is an ideal voltage source at all times (always zero ohms impedance). In DCM, the "source" is effectively open circuit for part of the waveform, and the relation does depend on other variables.*Or connected to through a transformer, or is a transformer.Tim

author: nixfu
author_rank: Unknown
timestamp: March 17, 2015, 01:35:14 pm
content: >Why is the output voltage of the Buck Converter independent of load resistance?African or European?

author: hkBattousai
author_rank: Regular Contributor
timestamp: March 17, 2015, 02:07:17 pm
content: Quote from: T3sl4co1l on March 17, 2015, 01:33:31 pmIf we assume that the output capacitance is infinite, the load resistance value will indeed be unimportant. Because the capacitor will take forever to charge or discharge, so it will preserve its voltage level. Do we assume that the capacitor is practically very large (instead of being mathematically infinite - so large that again it makes the value of the load resistance unimportant for at least one switching period which also makes the system controllable) and we are choosing the best duty cycle (and it will equal to Vout/Vin) which will keep the capacitor voltage fixed at its current level?Quote from: nixfu on March 17, 2015, 01:35:14 pmI am Asian. Did I make a grammatical mistake?

author: krish2487
author_rank: Frequent Contributor
timestamp: March 17, 2015, 02:17:02 pm
content: Quote from: hkBattousai on March 17, 2015, 02:07:17 pmAny SMPS filter cap will NOT be inifnite and will be the closest standard available.Any SMPS primarily functions by correcting any line and load variations several thousand times per second.It corrects the line/load variations by a feedback mechanism, a mechanism which will be upset by a theoritically inifnite capacitance (or in your terms very large).To answer your original question, any smps is designed to work within a given limited set of constraints - like load (0 - 100%), line (x volts to y volts), temperature ( -15 C to 55 C) etc.Any practical SMPS cannot be designed to work under any inifite spectrum. Having said that, the reason why you see a smsp design equations being independant of load resistance is this - the equations define mathematically, the function of the SMPS while operating under a given range of conditions, within which the output voltage is largely independant of the aforementioned constraints.Try operatin beyond this ranges, you ll see it fails drastically, if not spectacularly.Teslacoil has mentioned continuous current mode, try operating an smps in discontinous mode, you ll see as he has mentioned.

author: Marco
author_rank: Super Contributor
timestamp: March 17, 2015, 02:31:22 pm
content: Quote from: hkBattousai on March 17, 2015, 01:21:14 pmCause the math works out that way.What you should be asking is "How can it regulate the voltage at fixed duty cycle even though the load resistance changes?". To which the answer is, by changing the frequency (only DCM can be done with fixed frequency).

author: Richard Head
author_rank: Frequent Contributor
timestamp: March 17, 2015, 04:45:17 pm
content: T3sl4co1l summed it up perfectly. That's the reason.In fact, the control-output transfer function actually changes between discontinuous current mode and continuous current mode. The optimum control loop compensation is also different for the two modes and it's usually compensated for the worst case which is CCM.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 17, 2015, 05:01:02 pm
content: Quote from: hkBattousai on March 17, 2015, 02:07:17 pmNononono, that's why I specified time approaching infinity. If you were to set up the analysis, you can do it in such a way that time goes to infinity "faster" than the capacitance and inductance, therefore an infinity of time constants have elapsed and the circuit is in equilibrium.It would just as well suffice to set finite component values and infinite frequency and time, assuming your analysis is not so nuanced that you'll get into trouble with an infinite frequency (e.g., the waveform ceases to be a waveform at all, because the discontinuities become "dense": V(t-->a) might be VCC or GND at any arbitrary point a).Without invoking time or equilibrium at all, you can also set up the problem so that the initial values of the capacitors (which, if infinite, will indeed be very stable voltages..) are at the equilibrium voltages, and solve for the voltage where current change goes to zero.All equivalent, so it's all fine.QuoteIt's required literature/film in the English speaking world I think...Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 17, 2015, 05:07:22 pm
content: Quote from: krish2487 on March 17, 2015, 02:17:02 pmIt was a theoretical question, about a theoretical drawing, from an equally theoretical source. It's also open loop (D is a variable in the equation, not defined by an error amplifier). None of this "practical" or "real" business!Tim

author: DutchGert
author_rank: Frequent Contributor
timestamp: March 17, 2015, 05:56:09 pm
content: Quote from: hkBattousai on March 17, 2015, 01:21:14 pmBeause it is based on the assumption that the output C is big enough to hold the same voltage regardless of the load.This formula tells u something about the unloaded regulator. When u start doing the math to find component values to make a actual converter u start taking into acount the actual output load.

author: krish2487
author_rank: Frequent Contributor
timestamp: March 17, 2015, 06:13:56 pm
content: Correct!! But any practical SMPS will not be open loop. It can also be argued that the D, for a practical SMPS, is a function of the error amplifier.:-)Quote from: T3sl4co1l onTodayat 04:07:22 AM

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: March 18, 2015, 12:56:27 am
content: Hi,The output voltage is only independent of load resistance if the current in the inductor is continuous. This can happen if:1) The average current is high enough so that the minimum value of inductor current is greater than 0.2) The Buck regulator is a synchronous buck, the diode is replaced by a MOSFET and current can flow in both directions.I like this explanation:Assume all components are loss free.That is the switch is ideal, the diode is ideal, the inductor has no resistance, the capacitor has no loss.We can then make the following observations:The average output voltage is Vout.There is no dc voltage across the inductor, it has no resistance.No dc current flows in the capacitor.The average voltage at the left hand side of the inductor is the Vin x Ton /Ton + ToffTon / (Ton + Toff) = duty-cycleSo Vout = duty_cycle x VinIf you introduce losses this is no longer true.Jay_Diddy_B

author: P_Doped
author_rank: Contributor
timestamp: March 19, 2015, 04:30:09 am
content: The conditions for Vout = D * Vin are:1) lossless switches and L-C filter2) Vout = constant3) volt-sec balance across inductor is in effect; namely, if D*Ts = duty cycle * period & D'*Ts = (1 - duty cycle) * period:(Vin-Vout)*D*Ts = Vout*D'*Tswhich you can rearrange (realizing that D + D' = 1) to get Vout = D * VinThe volt-sec balance means that the switcher is in equilibrium and the instantaneous inductor current at time Ts is the same at 2*Ts.All these "ideal" Vin -> Vout relationships for switchers give you the CCM duty cycle. Losses will always cause the duty cycle to increase in order to draw more power out of the input source to account for these losses. One way to compare 2 implementations is to see which one gives you the close to ideal D for the correct regulated output voltage. The closer one is more efficient.