Boost converter, inductor size?
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From: Jon Kirwan on 21 Jul 2010 05:52 On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden <wrongaddress(a)att.net> wrote: >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. 90 turns of #18 has to be longer than 1". .0403" diameter without insulation times 90 is more than 3.6", so I take it you stacked things up? Wheeler's single layer formula, which I gather isn't bad for low frequency use, is: r^2*N^2/(9*r+10*l) where r is the radius, l is the length, and N is the number of turns, with lengths given in inches and the resulting value given in micro Henries. That doesn't seem to cut it for your case, so I then read that the multilayer formula is: 0.8*a^2*N^2/ (6*a+9*l+10*c) where a is the average radius and c is the radius difference (span.) Assuming you did 6 layers by 15 turns, I then estimate .065" for the wire diameter with insulation and thus c=.39" and l=.98" and a=.51". From that, I get about 107uH. Not 250uH. Let's call that 110uH and talk about this later. In any case, Tim's point about radiation may be important. An air inductor has a very widely dispersed magnetic field (it's not constrained much.) I don't know how to calculate the loss to far field radiation here. But maybe someone can discuss that a little bit. >Works fairly well at 84% efficiency. What are you loading the output with? What is the output voltage? >I lowered the frequency to 12KHz to reduce the diode >switching losses. The current ramps from about 1 amp >minimum to 2.25 peak. Given the earlier discussion and what I think I learned from Tim's comments, let's say your output voltage is 14V and your diode circa 1A is running about 0.8V. From this, I gather that the inductor must handle a percentage of the total power by the factor of (14V+0.8V-9.65V+0.1V)/14V, or 37.5%. (The 0.1V is your transistor switch drop and the 0.8V is the diode drop estimate during conduction.) With assumed usable input power near 9.5 watts, and 37.5% of that in the inductor, about 3.56 Joules must be divided out by 12000 pulses. I get close to 300uJ per pulse, here. Assuming you actually had built 250uH into your air core, this would be about 1.56A at the peak. However, using that 110uH I earlier estimated from that multilayer formula and your other figures, I get about 2.3A at 110uH. Which is much closer to your own comments about the observed 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. Can you talk more about how you are loading things and what output voltage you observe? It's interesting thinking about this and perhaps Tim can add a few points I'm missing out on, as well. Jon
From: Bill Bowden on 21 Jul 2010 13:44 On Jul 21, 2:52 am, Jon Kirwan <j...(a)infinitefactors.org> wrote: > On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden > > > > <wrongaddr...(a)att.net> wrote: > >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. > > 90 turns of #18 has to be longer than 1". .0403" diameter > without insulation times 90 is more than 3.6", so I take it > you stacked things up? > The inner diameter is 1 inch, width is 5/8, 15 turns per layer, 6 layers. It resonates at 20KHz with a 0.22uF cap, so L=288uH. > Wheeler's single layer formula, which I gather isn't bad for > low frequency use, is: r^2*N^2/(9*r+10*l) where r is the > radius, l is the length, and N is the number of turns, with > lengths given in inches and the resulting value given in > micro Henries. That doesn't seem to cut it for your case, so > I then read that the multilayer formula is: 0.8*a^2*N^2/ > (6*a+9*l+10*c) where a is the average radius and c is the > radius difference (span.) Assuming you did 6 layers by 15 > turns, I then estimate .065" for the wire diameter with > insulation and thus c=.39" and l=.98" and a=.51". From that, > I get about 107uH. Not 250uH. Let's call that 110uH and > talk about this later. > > In any case, Tim's point about radiation may be important. An > air inductor has a very widely dispersed magnetic field (it's > not constrained much.) I don't know how to calculate the > loss to far field radiation here. But maybe someone can > discuss that a little bit. > > >Works fairly well at 84% efficiency. > > What are you loading the output with? What is the output > voltage? > In normal operation, the load is the battery, which holds the output voltage constant, regardless of input or output current. The duty cycle is fine tuned for a peak in output current, which seems the best match. But as the battery voltage rises, the duty cycle needs a small adjustment. There is no feedback, so I just leave it set optimum at 13 volts. It's just a 555 timer driving a mosfet. But using a 16 ohm load in place of the battery gives me 1.14A in at 9.55 volts and 12.06 volts out, or 9.1 watts out of 10.9, or maybe 83.5% efficiency. But it was hard to get the same numbers twice since the sunlight kept changing slightly. I'll run the test again using a power supply instead of the panel. > >I lowered the frequency to 12KHz to reduce the diode > >switching losses. The current ramps from about 1 amp > >minimum to 2.25 peak. > > Given the earlier discussion and what I think I learned from > Tim's comments, let's say your output voltage is 14V and your > diode circa 1A is running about 0.8V. From this, I gather > that the inductor must handle a percentage of the total power > by the factor of (14V+0.8V-9.65V+0.1V)/14V, or 37.5%. (The > 0.1V is your transistor switch drop and the 0.8V is the diode > drop estimate during conduction.) > > With assumed usable input power near 9.5 watts, and 37.5% of > that in the inductor, about 3.56 Joules must be divided out > by 12000 pulses. I get close to 300uJ per pulse, here. > Assuming you actually had built 250uH into your air core, > this would be about 1.56A at the peak. However, using that > 110uH I earlier estimated from that multilayer formula and > your other figures, I get about 2.3A at 110uH. Which is much > closer to your own comments about the observed 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. > > Can you talk more about how you are loading things and what > output voltage you observe? It's interesting thinking about > this and perhaps Tim can add a few points I'm missing out on, > as well. > I'm not sure of the current ramps, I was using a 0.1 ohm resistor in series with the coil (for scope monitor) which upsets things a bit. The peak to peak ramp appears to be about 1.25 amps offset by 1 amp, or 2.25 peak to 1 minimum. But that doesn't agree with 280uH if e=L di/dt = .00028* (1.25/25uS) = 14 volts. I think it should be 9.5. Probably inaccurate measurements, and uncalibrated scope. Just ball park figures. -Bill > Jon
From: Tim Wescott on 21 Jul 2010 14:00 On 07/21/2010 02:52 AM, Jon Kirwan wrote: > On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden > <wrongaddress(a)att.net> wrote: > >> 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. > > 90 turns of #18 has to be longer than 1". .0403" diameter > without insulation times 90 is more than 3.6", so I take it > you stacked things up? > > Wheeler's single layer formula, which I gather isn't bad for > low frequency use, is: r^2*N^2/(9*r+10*l) where r is the > radius, l is the length, and N is the number of turns, with > lengths given in inches and the resulting value given in > micro Henries. That doesn't seem to cut it for your case, so > I then read that the multilayer formula is: 0.8*a^2*N^2/ > (6*a+9*l+10*c) where a is the average radius and c is the > radius difference (span.) Assuming you did 6 layers by 15 > turns, I then estimate .065" for the wire diameter with > insulation and thus c=.39" and l=.98" and a=.51". From that, > I get about 107uH. Not 250uH. Let's call that 110uH and > talk about this later. > > In any case, Tim's point about radiation may be important. An > air inductor has a very widely dispersed magnetic field (it's > not constrained much.) I don't know how to calculate the > loss to far field radiation here. But maybe someone can > discuss that a little bit. > >> Works fairly well at 84% efficiency. > > What are you loading the output with? What is the output > voltage? > >> I lowered the frequency to 12KHz to reduce the diode >> switching losses. The current ramps from about 1 amp >> minimum to 2.25 peak. > > Given the earlier discussion and what I think I learned from > Tim's comments, let's say your output voltage is 14V and your > diode circa 1A is running about 0.8V. From this, I gather > that the inductor must handle a percentage of the total power > by the factor of (14V+0.8V-9.65V+0.1V)/14V, or 37.5%. (The > 0.1V is your transistor switch drop and the 0.8V is the diode > drop estimate during conduction.) > > With assumed usable input power near 9.5 watts, and 37.5% of > that in the inductor, about 3.56 Joules must be divided out > by 12000 pulses. I get close to 300uJ per pulse, here. > Assuming you actually had built 250uH into your air core, > this would be about 1.56A at the peak. However, using that > 110uH I earlier estimated from that multilayer formula and > your other figures, I get about 2.3A at 110uH. Which is much > closer to your own comments about the observed 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. > > Can you talk more about how you are loading things and what > output voltage you observe? It's interesting thinking about > this and perhaps Tim can add a few points I'm missing out on, > as well. I'd want to know the inter-coil capacitance. What's the self-resonant frequency of the coil? If you know the self-resonant frequency and the resonant frequency with some capacitance attached you can model the coil as an inductance in parallel with a capacitor and assign values. The closer the self-resonant frequency is to the operating frequency the more switching losses you're going to have, unless you go to a fancy resonant-mode controller. -- 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 21 Jul 2010 15:33 On Wed, 21 Jul 2010 10:44:22 -0700 (PDT), Bill Bowden <wrongaddress(a)att.net> wrote: >On Jul 21, 2:52�am, Jon Kirwan <j...(a)infinitefactors.org> wrote: >> On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden >> >><snip> >> >> 90 turns of #18 has to be longer than 1". �.0403" diameter >> without insulation times 90 is more than 3.6", so I take it >> you stacked things up? > > The inner diameter is 1 inch, width is 5/8, 15 turns per layer, 6 > layers. It resonates at 20KHz with a 0.22uF cap, so L=288uH. Okay. I took things the other way. (Thought that the 1" was the length.) I'll go with your measured values. >><snip> >> >> What are you loading the output with? �What is the output >> voltage? >> >In normal operation, the load is the battery, which holds the output >voltage constant, regardless of input or output current. Yes, but what is that voltage it holds? >The duty >cycle is fine tuned for a peak in output current, which seems the best >match. Okay. Which implies a higher output voltage under load, I think, as a higher voltage across your battery will mean more current into it. And all this means, as you already know, that there is more power delivered, which is the goal. >But as the battery voltage rises, the duty cycle needs a small >adjustment. There is no feedback, so I just leave it set optimum at 13 >volts. It's just a 555 timer driving a mosfet. Okay. Got it. >But using a 16 ohm load in place of the battery gives me 1.14A in at >9.55 volts and 12.06 volts out, or 9.1 watts out of 10.9, or maybe >83.5% efficiency. Thanks. I'm enjoying thinking about the problem and reading about your observations. It's teaching me to apply theory and think better, generally. Tim's knock in the head about the inductor's role in the work involved helped a lot and now I'd like to see if paper and pencil match experimental result, as it should! >But it was hard to get the same numbers twice since the sunlight kept >changing slightly. I'll run the test again using a power supply instead >of the panel. Makes sense!! >><snip> >> Bill: >> >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. >> >> Can you talk more about how you are loading things and what >> output voltage you observe? �It's interesting thinking about >> this and perhaps Tim can add a few points I'm missing out on, >> as well. > >I'm not sure of the current ramps, I was using a 0.1 ohm resistor in >series with the coil (for scope monitor) which upsets things a bit. Given .2 ohms for the inductor itself, it's a 50% increase in the DC resistance. But you should loose an eighth watt or so there, so I don't think it's all that terrible. I might have tried the same thing, since scoping makes observation of the details clear. >The peak to peak ramp appears to be about 1.25 amps offset by 1 amp, >or 2.25 peak to 1 minimum. Doesn't that bother you? It just about cannot be. >But that doesn't agree with 280uH if e=L di/dt = .00028* (1.25/25uS) = >14 volts. From this, I gather you are setting your 555 timer's pulse width so that the mosfet is on for 25us, yes? >I think it should be 9.5. I agree. Assume you are right that L=288uH. (I'm convinced you've tested this.) Assume you are right about 25us. (I'm convinced you can easily see this on the scope, by simply looking.) You and I both know the voltage should be 9.5V -- it's a given, considering the source you have. So from that the dI=V*dt/L, or .825A. Since the others are solid givens, this MUST be true. It cannot be false. So it is clear the 1.25A delta is off in some way. >Probably inaccurate measurements, and >uncalibrated scope. Just ball park figures. Okay. It still bothers me about the 1A minimum. I know you can easily establish where "zero" is at on your scope by simply connecting the probe tip and ground and adjusting that line to a graticule line for reference, so I don't think you can mistake the fact that the pulse voltage bottom is above it. This bothers me. But I believe you are seeing a baseline there, too. Not that you have an exact figure for it. But the fact that it is present. What is missing in my mind is WHY. What would cause it? Oh! Obvious. dt=(L/V)*dI, with V selected for the relaxation time, which will be output+diode-9.5V, or so. At 13V output, this is maybe 13.7-9.65 or close to 4V. This suggests a dt of about 60us. Added to 25us, this is 85us and exceeds what is happening at 12kHz. There just isn't enough time. Try lowering the frequency a little? Jon
From: Jon Kirwan on 21 Jul 2010 16:44
On Wed, 21 Jul 2010 12:33:56 -0700, Jon Kirwan <jonk(a)infinitefactors.org> wrote: >Try lowering the frequency a little? Actually, a lot to kill something near 1A baseline. Worse, though, is that with a lower frequency you need MORE energy per pulse. And that means MORE ON time. So you need to adjust that, too. Let's see: ON dV = 9.5V OFF dV = 4V (or so) V_out = 13V (or so) V_in = 9.65V V_sw = 0.1V (roughly) V_diode= 0.7V (roughly) P_in = 11W (very roughly) L = 288uH V_Lon = V_in - V_sw V_Loff = V_out + V_diode - (V_in - V_sw) Percent of power needed in the inductor's magnetic field is roughly: PCT = V_Loff / V_out In this case, (13V+0.7V-(9.65V-0.1V))/13V, or 31.9%. So, PCT = 0.319 Energy per pulse is then: E_ind = (P_in * PCT / f) = (1/2)*I_peak^2*L Assuming a starting point of zero inductor current: I_peak = V_Lon * t_on / L Also, assuming enough time to relax back to zero: t_off = L * I_peak / V_Loff and now we know, f = 1 / (t_on + t_off) So, t_on+t_off = I_peak^2 * L / (2 * P_in * PCT) Substituting like crazy and flipping around to remove t_off and I_peak in the equation and solve for t_on, I get this: t_on = (1 + V_Lon/V_Loff) * 2 * L * P_in * PCT / V_Lon^2 Dimensional analysis, as a quick check, says: seconds = Henries * Watts / Volts^2 Which is right. Watt is Joules/second, Volt is Joules/ Coulomb, and Henry is Joule-second^2/Coulomb^2. So it seems that I didn't make that kind of gross error. Computing using your L=288uH, PCT=.319, P_in=11W, V_Lon=9.55V, and V_Loff=4.15V, I get t_on = 73.16us. This is a lot longer than you've been using. From this, I also get I_peak=2.426A. This then says t_off=168.36us. So from all of this, I get t_on+t_off should be about 240us long. You might try that. Set t_on+t_off to 240us and t_on to somewhere in the range of 73-74us and see what happens. I'm hoping you see a zero-amp baseline, then. Jon |