Boost converter, inductor size?
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From: Tim Wescott on 16 Jul 2010 16:54 On 07/16/2010 01:42 PM, Jamie wrote: > 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 >> >> >> -Bill > What's wrong with using a charge pumped converter ? Well, other than the fact that efficiency would absolutely suck, nothing I guess. -- 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 16 Jul 2010 18:54 On Jul 16, 4:59 am, Jon Kirwan <j...(a)infinitefactors.org> wrote: > On Thu, 15 Jul 2010 19:33:31 -0700 (PDT), Bill Bowden > > > > <wrongaddr...(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.... > > 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 There is a large capacitor across the panel, so it can deliver high currents for short times. The output impedance of the panel is close to zero for 50uS or more. The 9.65 at 1 amp is the peak in bright sunlight. I got a bargain on 25 chipped cells on ebay. The seller was smart enough not to give me 36 cells that would be ideal for a 12 volt panel. He wants me to buy 2 sets of 25 and have a few left over. -Bill
From: Bill Bowden on 16 Jul 2010 19:18 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. 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. And another problem is if the inductor current was a constant 3 amps, the output power would be (13X3) or 39 watts for an input of only 9.65 watts? How do we get the extra power? -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: Bill Bowden on 16 Jul 2010 19:23 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. 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. And another problem is if the inductor current was a constant 3 amps, the output power would be (13X3)*.70 or 27 watts for an input of only 9.65 watts? How do we get the extra power? -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: Tim Wescott on 16 Jul 2010 20:41
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. -- 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 |