From: Jasen Betts on 18 Jul 2010 05:44 On 2010-07-17, Jon Kirwan <jonk(a)infinitefactors.org> wrote: > 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: >>> >>> 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? The control input is from the switchers' output, that's whay it's called closed-loop --- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---
From: Jon Kirwan on 18 Jul 2010 10:03 On 18 Jul 2010 09:44:32 GMT, Jasen Betts <jasen(a)xnet.co.nz> wrote: >On 2010-07-17, Jon Kirwan <jonk(a)infinitefactors.org> wrote: >> 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: >>>> >>>> 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? > >The control input is from the switchers' output, that's whay it's >called closed-loop That much, I know. I'm just wondering about the addition of more C on that end of the control loop. Jon
From: Jon Kirwan on 18 Jul 2010 10:48 I suppose it depends, though, on what the closed control loop observes to perform its function. It just seems to me that a capacitor is not a panacea for all designs. It may actually be a problem, for some. This is a question, not a claim. Jon
From: Tim Wescott on 18 Jul 2010 12:05 On 07/18/2010 02:44 AM, Jasen Betts wrote: > On 2010-07-17, Jon Kirwan<jonk(a)infinitefactors.org> wrote: >> 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: >>>> >>>> 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? > > The control input is from the switchers' output, that's whay it's > called closed-loop In a photovoltaic array the control system needs to take input from both sides of the switcher if it's going to suck the maximum possible power from the array. And then (to answer Jon's question) you have to take the input capacitance into account -- but it's not that difficult. -- 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: Bill Bowden on 21 Jul 2010 00:34
On Jul 16, 9:38 am, Tim Wescott <t...(a)seemywebsite.com> wrote: > 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..... > > > 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. > Well, I wanted to experiment with air core inductors, so I made a 250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns of #18 wire and 200 milliohms resistance. Works fairly well at 84% efficiency. I lowered the frequency to 12KHz to reduce the diode switching losses. The current ramps from about 1 amp minimum to 2.25 peak. I tried a fast recovery diode against a regular rectifier diode and only got a 1.5 % difference, so I guess diode recovery time doesn't matter much at 12KHz. The inductor current appears to ramp from 1 amp minimum to 2.25 amps peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt. Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%, so I don't where the other 6% went, but it works ok. The battery is fully charged. -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 |