Boost converter, inductor size?
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From: Bill Bowden on 16 Jul 2010 22:32 On Jul 16, 5:41 pm, Tim Wescott <t...(a)seemywebsite.com> wrote: > On 07/16/2010 04:18 PM, Bill Bowden wrote: > > > > > On Jul 15, 8:09 pm, Tim Wescott<t...(a)seemywebsite.com> 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..... > > >> 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. > > > This is where I get confused. If the inductor current just kisses > > zero, then it has released all the energy into the load and must > > recharge during the next 30% time frame. If the power in is equal to > > the power out, then the inductor must ramp from 0 to 6 amps during the > > 30% time to average 3 amps for 30% of the time, or a 1 amp average > > continuous input all the time. > > I think that where you get confused is when you forget that the inductor > is permanently attached to the source. > > The inductor current goes from 0 to 2A during the transistor on time, > and from 2A to zero during the transistor off time. The source is > _always_ delivering current to the inductor, but the inductor is only > delivering current to the load 70% of the time. > > > Now, if a very large inductance brings the current peaks and valleys > > closer together so they are much the same, then it seems the inductor > > current would be 3 amps, so the transistor can switch on for 30% of > > the time and supply 3 amps, or 1 amp average. So, I don't see how > > increasing the inductance can ever reduce the current below 3 amps. > > See above. With an infinite inductance flowing 1A the input current > will _always_ be 1A exactly, and the output will be 0 30% of the time > and 1A 70% of the time, for 700mA. > > HTH. > Yes, that makes it clearer. So, if the transistor were removed, the load would still get current through the diode all the time, and the output voltage would be around 9 volts. The extra 5 volts is obtained by the transistor pulling the inductor to ground for 30% of the time and adding current to the inductor. But in the case of the infinite inductance, the current doesn't change. I was thinking all the power was switched by the transistor, but it's actually only 30%. Good explanation. -Bill > -- > > Tim Wescott > Wescott Design Serviceshttp://www.wescottdesign.com > > Do you need to implement control loops in software? > "Applied Control Theory for Embedded Systems" was written for you. > See details athttp://www.wescottdesign.com/actfes/actfes.html
From: Jon Kirwan on 17 Jul 2010 10:25 On Fri, 16 Jul 2010 13:53:13 -0700, Tim Wescott <tim(a)seemywebsite.com> wrote: >On 07/16/2010 11:46 AM, Jon Kirwan wrote: >> On Fri, 16 Jul 2010 09:44:57 -0700, Tim Wescott >> <tim(a)seemywebsite.com> wrote: >> >>> 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. >> >> That does NOT seem optimal to me. I do think it works great >> on tiny photovore robots because it is cheap and easy to do >> -- but that isn't a case where we are talking watts of power >> and getting the most out of it. And at 10W, I wonder at the >> sizes required anyway. >> >>>> 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. >> >> I kind of figured. But it would help to be sure. >> >>> You can model a solar cell pretty well as a >>> current source (from photon impingement) in parallel with a silicon >>> diode. >> >> As a current source and diode, you'd probably NOT want the >> voltage to reach enough to cause significant leakage. In >> short, you might want something like a transimpedance amp. >> Except for the fact that a transimpendance amp would seek >> close to zero voltage, and that times lots of current isn't >> "much power." So you really want to allow some middling >> point where voltage and current (product) is at a peak. I >> would guess some kind of odd shape here with a peak power >> position that is difficult to track. >> >> A capacitor isn't my first idea of a good tracking circuit. >> >>> 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. >> >> I think I gather. >> >>>> 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. >> >> Yes. Hence why I don't think slapping a cap across the panel >> is optimal. > >You're not using the cap for tracking for heaven's sake! You're using >it to smooth out the current draw from the panel. It's all or part of >the input filter that any wise designer always puts on the front end of >a switcher. Thanks, Tim. But would you really want to paste capacitors across a solar panel? Slapped across the input side of a closed loop control? Jon
From: Jon Kirwan on 17 Jul 2010 10:40 On Fri, 16 Jul 2010 13:51:39 -0700, Tim Wescott <tim(a)seemywebsite.com> wrote: >On 07/16/2010 11:40 AM, Jon Kirwan wrote: >> 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. > >That's all well and good, but it simply cannot be right. If the diode >conducts for the entire time that the transistor is off, then the output >side of the inductor is always either connected to the load (13.8V) or >ground (0V). The input side is constant, so the inductor current must >be a sawtooth. The average current is going to be the mean of the peak >and trough of this sawtooth. So if the average is 1A and the trough is >0A, then there's absolutely nothing left for the peak to be but 2A. > >If your fancy math comes up with an answer that doesn't match, it means >that either your math or your assumptions are incorrect. > >Perhaps you should simulate this on LTSpice and see what it says... I need to do that, thanks. >>> 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. > > > > L D > ___ > .-------UUU----o---->|-------. > | | | > | | | > Vin /+\ ||-+ /+\ Vout > ( ) ||<- ( ) > \-/ -||-+ \-/ > | | | > | | | > === === === > GND GND GND >(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) > >Please explain how it is possible for the diode to _not_ conduct when >there is forward current in the coil and the transistor is off, as will >happen every single cycle at the moment that the transistor turns off. It will forward conduct when the transistor is off, Tim. Because of the inductor. That's never been at issue. The point here is that the forward conduction is DUE to energy stored in a magnetic field in the inductor. And _that_ is the fact at issue in our discussion. This energy _must_ be factored into any discussion. Since the input side is at a lower voltage than the output, at steady state, there is no other method by with power is transferred, except by way of pulsed magnetic field energy storage. As such, it's a very simple calculation to know how much energy must be present in each pulse if there are exactly 25000 of them per second. (25kHz.) I'm not sure what is wrong about our dialog, yet. There is something "cross purposes" about it. And I have to take the blame here. You are the trained expert, not me. >> 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. > >And if you're at all smart, you size the inductor so that it takes the >entire off-time of the transistor to "dump". Not just during startup -- >all the time. I'm not all that smart, Tim. I'm just a bystander looking in. >> 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. > >So, you've got 9.65V on the input side, 4.15V across the inductor, and >you flow a charge of (1A)(28us) = 28uC, for a total energy of >(13.8V)(28uC) = 386uJ. Yes, per pulse at 25kHz. >So how much of the energy is delivered by the >inductor, and how much is delivered by the input voltage source? Okay. I think I'm following, a little. >Keep in mind that the only connection to the input source is the coil, Okay, yes. >so any >current into the coil comes from the input source, which constrains the >input source current to that of the coil. Okay. I think I follow the argument you are making here. In my own words, what you are saying here is that the inductor provides only "part" of the net voltage and that both the inductor and the panel together provides the current so that the effect is a "sharing" of net power. The result of that is that only a portion of the total transfer of energy per unit time needs to be present, stored temporarily as energy in the magnetic field each pulse. So I'm mistaken to imagine that approximately 10J of energy needs to divided down by 25000 to get each pulse's energy because some of the energy arrives directly by way of the source voltage and current. Jon
From: Tim Wescott on 17 Jul 2010 12:47 On 07/17/2010 07:40 AM, Jon Kirwan wrote: > On Fri, 16 Jul 2010 13:51:39 -0700, Tim Wescott > <tim(a)seemywebsite.com> wrote: > >> On 07/16/2010 11:40 AM, Jon Kirwan wrote: >>> 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. >> >> That's all well and good, but it simply cannot be right. If the diode >> conducts for the entire time that the transistor is off, then the output >> side of the inductor is always either connected to the load (13.8V) or >> ground (0V). The input side is constant, so the inductor current must >> be a sawtooth. The average current is going to be the mean of the peak >> and trough of this sawtooth. So if the average is 1A and the trough is >> 0A, then there's absolutely nothing left for the peak to be but 2A. >> >> If your fancy math comes up with an answer that doesn't match, it means >> that either your math or your assumptions are incorrect. >> >> Perhaps you should simulate this on LTSpice and see what it says... > > I need to do that, thanks. > >>>> 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. >> >> >> >> L D >> ___ >> .-------UUU----o---->|-------. >> | | | >> | | | >> Vin /+\ ||-+ /+\ Vout >> ( ) ||<- ( ) >> \-/ -||-+ \-/ >> | | | >> | | | >> === === === >> GND GND GND >> (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) >> >> Please explain how it is possible for the diode to _not_ conduct when >> there is forward current in the coil and the transistor is off, as will >> happen every single cycle at the moment that the transistor turns off. > > It will forward conduct when the transistor is off, Tim. > Because of the inductor. That's never been at issue. > > The point here is that the forward conduction is DUE to > energy stored in a magnetic field in the inductor. And > _that_ is the fact at issue in our discussion. This energy > _must_ be factored into any discussion. > > Since the input side is at a lower voltage than the output, > at steady state, there is no other method by with power is > transferred, except by way of pulsed magnetic field energy > storage. As such, it's a very simple calculation to know how > much energy must be present in each pulse if there are > exactly 25000 of them per second. (25kHz.) > > I'm not sure what is wrong about our dialog, yet. There is > something "cross purposes" about it. And I have to take the > blame here. You are the trained expert, not me. > >>> 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. >> >> And if you're at all smart, you size the inductor so that it takes the >> entire off-time of the transistor to "dump". Not just during startup -- >> all the time. > > I'm not all that smart, Tim. I'm just a bystander looking > in. > >>> 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. >> >> So, you've got 9.65V on the input side, 4.15V across the inductor, and >> you flow a charge of (1A)(28us) = 28uC, for a total energy of >> (13.8V)(28uC) = 386uJ. > > Yes, per pulse at 25kHz. > >> So how much of the energy is delivered by the >> inductor, and how much is delivered by the input voltage source? > > Okay. I think I'm following, a little. > >> Keep in mind that the only connection to the input source is the coil, > > Okay, yes. > >> so any >> current into the coil comes from the input source, which constrains the >> input source current to that of the coil. > > Okay. I think I follow the argument you are making here. > > In my own words, what you are saying here is that the > inductor provides only "part" of the net voltage and that > both the inductor and the panel together provides the current > so that the effect is a "sharing" of net power. The result > of that is that only a portion of the total transfer of > energy per unit time needs to be present, stored temporarily > as energy in the magnetic field each pulse. > > So I'm mistaken to imagine that approximately 10J of energy > needs to divided down by 25000 to get each pulse's energy > because some of the energy arrives directly by way of the > source voltage and current. Yup. -- 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 17 Jul 2010 14:16
On Sat, 17 Jul 2010 09:47:49 -0700, Tim Wescott <tim(a)seemywebsite.com> wrote: >On 07/17/2010 07:40 AM, Jon Kirwan wrote: ><snip> >> Okay. I think I follow the argument you are making here. >> >> In my own words, what you are saying here is that the >> inductor provides only "part" of the net voltage and that >> both the inductor and the panel together provides the current >> so that the effect is a "sharing" of net power. The result >> of that is that only a portion of the total transfer of >> energy per unit time needs to be present, stored temporarily >> as energy in the magnetic field each pulse. >> >> So I'm mistaken to imagine that approximately 10J of energy >> needs to divided down by 25000 to get each pulse's energy >> because some of the energy arrives directly by way of the >> source voltage and current. > >Yup. Okay. So I think I follow, now. I had completely neglected the 'pressure' that exists on the other side of the inductor in figuring things. Thanks for the kick in the side. I just ran a quick behavioral simulation (near perfect switch, etc) and used a "realistic" schottky diode that burned a little power and tweaked things so that the output was a steady 13V at 0.7A. The MBRS1100 I used here presented about 0.69V when forward-conducting. Input voltage was set to 9.65V. (Used .IC to init the output node.) The result is: Input power: 9584 mW Output power: 9101 mW Schottky power: 482 mW Misc power: 1 mW Those figures include numerical rounding errors by LTSpice. Peak inductor current is 2.13A and with 50uH is about 113.5uJ per pulse. At 25kHz, this works out to about 2836 mJ/s and it runs nicely with 11us ON time and 29us OFF time. The inductor current has fallen to zero after a little more than 27us, so there is a tiny period of zero inductor current before the next pulse starts. In short, to deliver 9100mJ to the load about 2840mJ of it appears in pulsed magnetic storage in the same period. About 31% or so. Working the other way, from theory to practice, I get 13V plus the schottky's 0.69V for 13.69V, then minus the 9.65V source voltage, for 4.04V net. This, compared against the output voltage of 13V gives 31.08%... which is very close to the required estimate for the inductor energy factor. I think I get it, now. Thanks, Jon |