Boost converter, inductor size?
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From: Bill Bowden on 15 Jul 2010 22:33 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 -Bill
From: Tim Wescott on 15 Jul 2010 23:09 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. -- 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: Jasen Betts on 16 Jul 2010 04:40 On 2010-07-16, Bill Bowden <wrongaddress(a)att.net> wrote: > 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. it's an inductor not a resistor. current flows while the transistor is off. that's the whole idea of a boost converter. --- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---
From: Jon Kirwan on 16 Jul 2010 07:53 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.) 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) 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. Jon
From: Jon Kirwan on 16 Jul 2010 07:59
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? And how do things vary with solar motion across the sky? Is the 9.65V(a)1A simply a peak? Or? And what kind of load should it "see?" Jon |