Boost converter, inductor size?
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From: Jon Kirwan on 16 Jul 2010 08:00 On Fri, 16 Jul 2010 04:53:01 -0700, Jon Kirwan <jonk(a)infinitefactors.org> wrote: >f=t_on I meant f=1/t_on. Jon
From: Tim Wescott on 16 Jul 2010 11:23 On 07/16/2010 04:53 AM, Jon Kirwan wrote: > On Thu, 15 Jul 2010 20:09:35 -0700, Tim Wescott > <tim(a)seemywebsite.com> wrote: > >> <snip> >> If the inductor current is just kissing zero on each cycle then it needs >> to ramp up to twice average, or 2A (ignoring losses) peak. >> <snip> > > Hi, Tim. That would ONLY seem correct to me in the extreme > case where f=t_on. (An impossible case.) Not in the case > where, let's say, t_on equals t_off so that D=50%, for > example, where it would seem to be pushed to 4X, not 2X. > > In any case, very fundamental considerations would suggest > (and these completely ignore some other relationships that > are necessary, such as nearly fixed limitations on t_off due > to the allowable voltage across the L when transferring > energy to the cap): > > I_peak = SQRT[ (2*P) / (f*L) ] > > This is simple to observe, since it is nothing more than > figuring out the (1/2)*I^2*L energy at I_peak, times the > number of such pulses allowed in a second, which must match > the input power available, I'd think. (Power is energy per > unit time, after all.) Well... Note my constraint: if things are arranged (i.e. if frequency and inductance is selected such that the inductor current goes to zero just as the transistor switches on). When an inductor has a constant voltage across it, the current ramps at a constant rate (assuming constant inductance with current, which isn't always valid in power electronics -- close your eyes to that). If the current ramps up from zero to 2A, then 2A to zero, then _immediately_ goes from zero to 2A again, the average current is 1A. > 9.65W (the 9.65V time 1A) available on input (and all this > assumes such a switcher even makes sense at all for a solar > panel, which I'm not convinced of because it wants to supply > a constant current and not a wildly swinging one) That's what input caps are for -- to smooth out the current seen by the panel, by letting the capacitor supply the AC portion. > divided by > the OP's 25kHz yields 386uJ as the energy storage required in > the inductor. From that, I get 3.93A for 50uH and 3.78A for > 54uH. Neither of which are 2A and both of which are much > closer to the 4X factor, using D=50%. > > Even then, it's probably higher still because there are other > considerations I can imagine. For one example, D will be > more like 30%, as the OP mentioned, and the t_off time will > likely be longer than the t_on, at steady state when the V > across the inductor during t_off will be 4V or so as opposed > to the t_on V of about 9.5V. > > But very basic energy-in/energy-out considerations without > any duty cycle or other limitation considerations will > suggest that at 25kHz and 50uH (assuming those are even > tenable with each other, which is yet another question not > answered) the current needs to be higher than 2A here just to > meet the energy transfer per unit time requirement. > > Or maybe I'm missing something. I am a modest hobbyist and > have had zero electronics training, after all. Not so -- the inductor is not storing all of the energy that gets transferred. Think about it -- for the 28us that the forward diode is conducting you are still getting current flow from the input side -- the inductor is only making up the difference between the 9.65V and the 13.8 (or whatever). Thus, you are overestimating the amount of energy that must be stored in the inductor. I missed the OP's frequency spec. You know that if conduction is continuous then the duty cycle has to be 30% just from the voltage ratio. So you know that the FET is on for 12us, and that it reaches 2A peak. To do this with that voltage in, L = (9.65)(12us)/(2A) = 58uH. If you want to do the energy balance equations go ahead, but take into account that during forward conduction of the diode some of the energy is going directly into the output. In fact, it'll be 70% of the input power. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
From: Tim Wescott on 16 Jul 2010 12:38 On 07/15/2010 08:09 PM, Tim Wescott wrote: > On 07/15/2010 07:33 PM, Bill Bowden wrote: >> I have a DIY solar panel that delivers 9.65 volts (24 cells) at about >> 1 amp and need a boost converter (12-14) to charge a SLA battery. I >> want to use a air core inductor (30-60uH) to avoid special ferrite >> cores. I think a small spool of copper wire on a plastic core will be >> about 50uH. >> >> According to this boost converter calculator, with 9.65 volts in, 13 >> out, diode drop of 0.7, transistor drop of 0.1 (IRFZ44 mosfet), freq >> of 25KHz, output current of 700mA, and inductor ripple current of >> 300%, the inductor is 54uH. But it also indicates the peak inductor >> current is 1.75 amps with a duty cycle of 30%. I'm not sure what the >> inductor Ipp (2.1 amps) is? >> >> Anyway, I don't see how the peak current can be only 1.75 amps with a >> duty cycle of 30%. If the input current is a constant 1 amp, then it >> must be around 3 amps during the 30% time the transistor is on. I can >> add a input cap to supply 3 amps during the 30% on time (1 amp >> average). And if the current ramps from 0 to a peak and averages 3 >> amps, the peak would seem to be around 6 amps? >> >> What am I missing? >> >> http://www.daycounter.com/Calculators/Switching-Converter-Calculator.phtml >> > > A good calculator? > > The 30% duty cycle is the on time of the transistor -- the 13.8V side is > getting current about 70% of the time. > > The 30% duty cycle sounds about right -- 9.6V is about 70% of 13.8V, so > for 13.8V side to match the 9.6V side it needs to be pulled to zero for > about 30% of the time. (that makes sense, really). > > If the inductor current is just kissing zero on each cycle then it needs > to ramp up to twice average, or 2A (ignoring losses) peak. The more > inductance the more the inductor peak (and trough) will approach the > average 1A. So 1.75A is possible. > I forgot to mention -- there's no reason to make an air-core inductor for this, or to obtain a core for a custom inductor. Switching supplies have gotten popular enough that you can almost always buy the inductor you need from the likes of DigiKey or Mouser. I suspect that an air core inductor would have problems with parasitics, as well as being huge and wanting to radiate a lot. There's a _reason_ people use cores. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
From: Tim Wescott on 16 Jul 2010 12:44 On 07/16/2010 04:59 AM, Jon Kirwan wrote: > On Thu, 15 Jul 2010 19:33:31 -0700 (PDT), Bill Bowden > <wrongaddress(a)att.net> wrote: > >> I have a DIY solar panel that delivers 9.65 volts (24 cells) at about >> 1 amp and need a boost converter (12-14) to charge a SLA battery. I >> want to use a air core inductor (30-60uH) to avoid special ferrite >> cores. I think a small spool of copper wire on a plastic core will be >> about 50uH. >> >> According to this boost converter calculator, with 9.65 volts in, 13 >> out, diode drop of 0.7, transistor drop of 0.1 (IRFZ44 mosfet), freq >> of 25KHz, output current of 700mA, and inductor ripple current of >> 300%, the inductor is 54uH. But it also indicates the peak inductor >> current is 1.75 amps with a duty cycle of 30%. I'm not sure what the >> inductor Ipp (2.1 amps) is? >> >> Anyway, I don't see how the peak current can be only 1.75 amps with a >> duty cycle of 30%. If the input current is a constant 1 amp, then it >> must be around 3 amps during the 30% time the transistor is on. I can >> add a input cap to supply 3 amps during the 30% on time (1 amp >> average). And if the current ramps from 0 to a peak and averages 3 >> amps, the peak would seem to be around 6 amps? >> >> What am I missing? >> >> http://www.daycounter.com/Calculators/Switching-Converter-Calculator.phtml > > Are you even sure that is the right approach? Does this > solar panel work well driving an inductor along a sloped ramp > from I=0 to I=some_peak_value and then nothing at all while > the switcher goes into its t_off period? As mentioned elsewhere, you use smoothing caps for this. > And how do things vary with solar motion across the sky? In the obvious way: power drops with falling light. The less light, or the more obliquely light falls on the panel, the less power. > Is the 9.65V(a)1A simply a peak? Or? It's probably the peak. You can model a solar cell pretty well as a current source (from photon impingement) in parallel with a silicon diode. As the voltage across the cell rises the internal current across the junction increases, ultimately until the cell has no external current at all. As the voltage across the cell falls the power lost to ohmic heating in the cell rises and the amount of power you can extract goes down. > And what kind of load should it "see?" Just the right one. Tracking the maximum power point of the array is not trivial, particularly when you take the variation of the diode characteristics with temperature into account. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
From: Jon Kirwan on 16 Jul 2010 14:40
On Fri, 16 Jul 2010 08:23:19 -0700, Tim Wescott <tim(a)seemywebsite.com> wrote: >On 07/16/2010 04:53 AM, Jon Kirwan wrote: >> On Thu, 15 Jul 2010 20:09:35 -0700, Tim Wescott >> <tim(a)seemywebsite.com> wrote: >> >>> <snip> >>> If the inductor current is just kissing zero on each cycle then it needs >>> to ramp up to twice average, or 2A (ignoring losses) peak. >>> <snip> >> >> Hi, Tim. That would ONLY seem correct to me in the extreme >> case where f=t_on. (An impossible case.) Not in the case >> where, let's say, t_on equals t_off so that D=50%, for >> example, where it would seem to be pushed to 4X, not 2X. >> >> In any case, very fundamental considerations would suggest >> (and these completely ignore some other relationships that >> are necessary, such as nearly fixed limitations on t_off due >> to the allowable voltage across the L when transferring >> energy to the cap): >> >> I_peak = SQRT[ (2*P) / (f*L) ] >> >> This is simple to observe, since it is nothing more than >> figuring out the (1/2)*I^2*L energy at I_peak, times the >> number of such pulses allowed in a second, which must match >> the input power available, I'd think. (Power is energy per >> unit time, after all.) > >Well... > >Note my constraint: if things are arranged (i.e. if frequency and >inductance is selected such that the inductor current goes to zero just >as the transistor switches on). I made two points before your comment here, so I'm not sure which of the two you are addressing. The first was about 2X vs 4X and evolves from V = L*dI/dt and depends exactly the point you make where the inductor current goes to zero. Without intuition, here's the algebra: V = L*dI/dt Assuming inductor current goes to zero and dt=t_on: V_in = L * I_peak / t_on Now assuming that f=1/t_on (which we know isn't true): V_in = L * I_peak * f And therefore, f = V_in/(L*I_peak) But, with P=V_in*I_in (power): I_peak = SQRT[ (2*P) / (f*L) ] So, substituting for f from above: I_peak = SQRT[ (2*P) / ([V_in/(L*I_peak)]*L) ] I_peak = SQRT[ (2*P) / (V_in/I_peak) ] I_peak^2 = (2*P) / (V_in/I_peak) I_peak = (2*P) / V_in But P=V_in*I_in, with 100% efficiency, so: I_peak = (2*V_in*I_in) / V_in Which works out conveniently to: I_peak = 2*I_in Of course, the problem in the above flow is that we assumed f=1/t_on. Clearly, it isn't, as t_off is non-zero. If you use the above and use f=1(2*t_on), assuming D=50%, then you will get: I_peak = 4*I_in The second point I'd made above was about the basic idea of multiplying power by time to get energy and then observing what that _must_ mean regarding inductor current, once again also assuming that it goes to zero each cycle. So this also does not violate the assumption you called out, again. In fact, I took it as a given. >When an inductor has a constant voltage across it, the current ramps at >a constant rate (assuming constant inductance with current, which isn't >always valid in power electronics -- close your eyes to that). Agreed. >If the >current ramps up from zero to 2A, then 2A to zero, then _immediately_ >goes from zero to 2A again, the average current is 1A. I'm not sure I read this correctly. There is the highlighted word _immediately_ that suggests 0-2A, but I know that _time_ is required to do that. So _immediately_ cannot mean zero time. Which leaves me confused. In any case, what I think needs to be focused upon is energy and time, not average current. However, without that intuition, the algebra above arrives at >> 9.65W (the 9.65V time 1A) available on input (and all this >> assumes such a switcher even makes sense at all for a solar >> panel, which I'm not convinced of because it wants to supply >> a constant current and not a wildly swinging one) > >That's what input caps are for -- to smooth out the current seen by the >panel, by letting the capacitor supply the AC portion. I guess I failed to note the OP writing that part. I thought about it, though. Just didn't see it. In any case, I wonder about __proper__ design for solar panels here. Not the simple stuff used in "photovore" robots, for example. But real, meaningful design with a solar panel and something more on the order of 10 watts and more varying downwards as the sun moves. >> divided by >> the OP's 25kHz yields 386uJ as the energy storage required in >> the inductor. From that, I get 3.93A for 50uH and 3.78A for >> 54uH. Neither of which are 2A and both of which are much >> closer to the 4X factor, using D=50%. > > >> Even then, it's probably higher still because there are other >> considerations I can imagine. For one example, D will be >> more like 30%, as the OP mentioned, and the t_off time will >> likely be longer than the t_on, at steady state when the V >> across the inductor during t_off will be 4V or so as opposed >> to the t_on V of about 9.5V. >> >> But very basic energy-in/energy-out considerations without >> any duty cycle or other limitation considerations will >> suggest that at 25kHz and 50uH (assuming those are even >> tenable with each other, which is yet another question not >> answered) the current needs to be higher than 2A here just to >> meet the energy transfer per unit time requirement. >> >> Or maybe I'm missing something. I am a modest hobbyist and >> have had zero electronics training, after all. > >Not so -- the inductor is not storing all of the energy that gets >transferred. Think about it -- for the 28us that the forward diode is >conducting you are still getting current flow from the input side -- the >inductor is only making up the difference between the 9.65V and the 13.8 >(or whatever). Thus, you are overestimating the amount of energy that >must be stored in the inductor. I'm thinking exclusively about the steady state case -- _after_ the output capacitor has reached it's design voltage. There is allowable droop. But not in my wildest imagination did I guess that the diode would be forward conducting every cycle!! Only in the early startup time. Which I set aside for thinking purposes. Once the output, within ripple considerations, is established, the ONLY way energy gets transferred is when that diode is forward conducting. And if the output voltage doesn't droop back down to 9V (which I can't accept it does do), then this only happens when the inductor dumps. And, I have to believe, the OP intends this to be just about all 10 watts at this steady state. Yes, I would be over-estimating in the case of "startup." Obviously. But once the output voltage is established and if the entire panel output of nearly 10W is supposed to continue to supply energy to the output side at 100% efficiency (I think we both take this as a given, for now) then the inductor is fully involved in achieving that, I think. Or? >I missed the OP's frequency spec. You know that if conduction is >continuous then the duty cycle has to be 30% just from the voltage >ratio. Yes. >So you know that the FET is on for 12us, and that it reaches 2A >peak. To do this with that voltage in, >L = (9.65)(12us)/(2A) = 58uH. > >If you want to do the energy balance equations go ahead, but take into >account that during forward conduction of the diode some of the energy >is going directly into the output. In fact, it'll be 70% of the input >power. 10W power-in needs to be transferred to 10W power-out, at 100% efficiency. Assuming that the output capacitor is at or near it's rated output voltage of 12-14V, this seems to suggest that the 10W must be divided out into (1/2)*L*I^2 pulses, since it cannot be directly flowing via a reverse-biased freewheeling diode as the V_in is less than 10V and the V_out is decidedly higher. As I had pointed out, this means almost 400uJ at 25kHz. Could you help me by a more thorough description explaining by what mechanism power might be transferred -- once steady state is achieved? I'm not seeing it and that must be my fault. Jon |