Boost converter, inductor size?
in [Basics]
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From: Bill Bowden on 21 Jul 2010 20:03 On Jul 21, 12:33 pm, Jon Kirwan <j...(a)infinitefactors.org> wrote: > On Wed, 21 Jul 2010 10:44:22 -0700 (PDT), Bill Bowden > > <wrongaddr...(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 I got some better figures using a 9.5 volt PS and 16.3 ohm load. Output is 12.7 at 9.9 watts. Input is 9.5 at 1.16 amps, or 11 watts. efficiency is around 90% which is closer to what I added up considering 1/2 watt loss in the diode and another 1/2 watt in the coil. The mosfet resistance is only 28 milliohms, so not much lost there. Frequency is 8.33KHz, duty cycle is 40uS and 80uS = 33% Assuming the inductor is 280uH, then di/dt is 9.5 / .00028 = 33929 amps per second, so the current should ramp up 1.36 amps in 40uS. But the problem is on the down ramp, the voltage should be 12.7 + 0.8 for the diode, or 13.5 minus the supply of 9.5 = 4 volts. So di/dt should be e/L = 4/.00028 = 14286 amps per second, or 1.143 amps in 80uS. But this doesn't agree with the ramp up value of 1.36, so something is missing. How would you determine the minimum current point? I think the idea is to drop the inductance from infinity where the current is constant to a smaller value where the minimum current gets close to zero. So, I might take a few turns off the inductor and save some resistance, and let the current get closer to zero, but I don't think the efficiency will go much past 90% that I already have. -Bill
From: Jon Kirwan on 22 Jul 2010 06:25 On Wed, 21 Jul 2010 17:03:24 -0700 (PDT), Bill Bowden wrote: ><snip> >I got some better figures using a 9.5 volt PS and 16.3 ohm load. > >Output is 12.7 at 9.9 watts. Input is 9.5 at 1.16 amps, or 11 watts. >efficiency is around 90% which is closer to what I added up >considering 1/2 watt loss in the diode and another 1/2 watt in the >coil. The mosfet resistance is only 28 milliohms, so not much lost >there. Frequency is 8.33KHz, duty cycle is 40uS and 80uS = 33% Okay. So about 120us period, 40us ON and 80us OFF. Got it. >Assuming the inductor is 280uH, then di/dt is 9.5 / .00028 = 33929 >amps per second, so the current should ramp up 1.36 amps in 40uS. But >the problem is on the down ramp, the voltage should be 12.7 + 0.8 for >the diode, or 13.5 minus the supply of 9.5 = 4 volts. So di/dt should >be e/L = 4/.00028 = 14286 amps per second, or 1.143 amps in 80uS. But >this doesn't agree with the ramp up value of 1.36, so something is >missing. This is exactly the point. And I think you've got your numbers right. >How would you determine the minimum current point? That's a little trickier. I had to think a moment to get a clue. I might still be wrong, but here is the argument. If I'm lucky, I will derive something quantitative below. You are driving (forcing) the situation using a clock and duty cycle. The inductor doesn't control any of that. It is, instead, driven by it. However, it does respond. Imagine you set your duty cycle to some value that causes a finite dI to be (t_on*V_on/L). The V_on can't be controlled -- it just is. The t_on is set by your clock source. It also just is. The same with L. So dI is fixed. It doesn't matter what happens on the output. The inductor has no choice in the matter. The dI will be some value. What isn't said here is what the baseline I happens to be. Let's call this baseline current through the inductor as I_0. We hope and expect I_0 = 0A. But this isn't necessarily so. What we can compute is dI. So the inductor current will have to rise from I_0 to I_1 and it _will_ be the case that I_1=I_0+dI. There is no choice on that matter. In the steady state case, which we still aren't sure of, I_0 will be stable and so will I_1. We hope I_0 is zero and I_1 is simply dI. But hold that thought for a moment. Now, your clock source also sets t_off. In the steady state case, we know what the dI must be for the t_off period, too. It has to be the same. If not, then it's not steady state. And it turns out that we should think about that V_off value, which will be L*dI/t_off. Since dI=(t_on*V_on/L), this works out to V_off=[V_on*(t_on/t_off)] -- the voltage that _must_ be across L when the switch is off if a steady state exists. Now, this V_off might not actually be what you imagine. In other words, you might think it is simply the output voltage plus a diode drop and less your input voltage because that is what is needed. But it isn't what happens. Suppose t_off is shortened a bit. Look at the equation for V_off. Note that t_off is in the denominator? If t_off gets smaller, V_off gets larger!! This means the output voltage also __increases__. It has to (assume for a moment that the diode voltage drop remains close to the same value as before.) But how can that be? Doesn't that mean that the load (resistive, for now) will use more power? Yes, it does. You might think that is a good thing. You've upped the output voltage purely by reducing t_off!! You could reduce it even more and get a still higher output voltage. But more output power means more input power. And you still have a fixed number of clocks per second. So more power must mean more energy per pulse! How do you get more energy per pulse?? You've already determined that there is a fixed dI, completely determined by L, t_on, and your input source voltage. So that can't change. What can happen to increase the energy per pulse? Well, take note that a fixed change in the magnitude of the inductor current is NOT the same as representing a fixed change in the magnitude of the energy stored in that magnetic field. You have these two circumstances: E_0 = (1/2) * L * I_0^2 E_1 = (1/2) * L * I_1^2 The difference is then: dE = E_1 - E_0 = (1/2) * L * (I_1^2 - I_0^2) But I_1 = I_0+dI. So, dE = (1/2) * L * ((I_0+dI)^2 - I_0^2) = (1/2) * L * (I_0^2 + 2*dI*I_0 + dI^2 - I_0^2) = (1/2) * L * (2*dI*I_0 + dI^2) = (1/2) * L * dI * (2*I_0 + dI) = (1/2) * L * dI^2 + [L * dI * I_0] Note the last term enclosed in brackets? When I_0 is non- zero and the same sign as dI, there is an additional amount of energy in each pulse. Another way of noting this fact is assume dI=1A and then to compare when I_0=0A and I_0=1A. dE when I_0=0A is some number we can call X. dE when I_0=1A will be 4X-1X, or 3X. In other words, more energy is taken up and released in each pulse when I_0=1A, even though there is the same dI in both cases. The point I'm getting to is that the inductor _must_ yield a larger V_off if you shorten t_off. To do that, the output voltage increases. That causes more power to be required. Which, because the frequency is constant, requires more energy per pulse. But to do that, I_0 must rise upwards so that each pulse _can_ deliver more energy to meet the power requirements. So it rises from zero. But this then causes a fixed DC current in the inductor, which means more loss in the resistive parts and the diode, as well. Which is not desired. The computation of exactly what I_0 must level out as depends upon the output load's response to a change in the voltage. for example, let's do some computations using some of your numbers. R_out=16.3, V_Lon=9.5V, t_on=40us, t_off=80us, f=8333Hz, L=288uH, etc. We don't know V_out, but we can compute it from V_Lon*t_on/t_off. This is 4.75V, in short. Assuming your diode voltage is about 0.7V, this means 4.05V added to the 9.6V input. Call it 13.6V. However, there is some resistance you mentioned which at a couple of amps average (grossly assuming a lot) is maybe .5V? So call it about 13.1V on the output. The PCT of power in L should be about 4.75V/13.1V or, say, about 36%. And the power is 13.1V^2/16.3Ohms, times that 36%, or in other words about 3.79W is delivered by the inductor itself. With f=8333, this means almost 455uJ per pulse. We know from t_on=40us that the dI=1.32A. Energy from that part, using (1/2)*L*dI^2, is about 251uJ. Since we need a total of about 455uJ, this leaves about 204uJ that needs to be made up in the L*dI*I_0 term. This suggests I_0=.534A. Which may or may not be about where you are. But that seems to be where theory takes me, right now. >I think the idea is to drop the inductance from infinity where the >current is constant to a smaller value where the minimum current gets >close to zero. >So, I might take a few turns off the inductor and save some >resistance, and let the current get closer to zero, but I don't think >the efficiency will go much past 90% that I already have. I think you are losing some efficiency you could get back, simply because you have a DC current flowing all the time, increasing the mean power dissipation in resistances. Not to mention that this would be a problem if you weren't using an air core inductor. I hope the above isn't too terrible to follow. It took me some moments of thought to realize _why_ the minimum inductor current would rise above zero on its own. But once I realized that dI is determined by t_on and that the inductor off-voltage is determined by dI and t_off and that shortening t_off only means that the off-voltage goes UP, the rest seemed to fall into place for me and make some sense. Jon
From: Don Klipstein on 22 Jul 2010 16:39 In article <l88d46h5bl21v35s21t3lvqh02r4n0jv9h(a)4ax.com>, Jon Kirwan wrote: >On 20 Jul 2010 21:34:07 (PDT), Bill Bowden <wrongaddress(a)att.net> wrote: <SNIP what leads to this to edit for space> >>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. If Bill's spool is 1 inch long by 5/8 inch diameter and his 18 AWG wire is .065 inch in overall diameter, such as hookup wire, then this inductor as described would have inductance around 110 uH. I doubt it will have even .01 microfarad of interlayer capacitance. Looks like I would want to hear more details about this coil. For example, if the spool was 1 inch in diameter and 5/8 inch long and the wire was AWG 18 magnet wire more like .047 inch in diameter, Bill could have wound 6 layers of 13 turns and a 7th layer of 12 turns. The radius difference c becomes .33 inch The average radius a becomes .665 inch The length b is .625 inch The inductance in microhenries at this rate would be 222 microhenries. - Don Klipstein (don(a)misty.com)
From: Bill Bowden on 22 Jul 2010 18:41 On Jul 22, 1:39 pm, d...(a)manx.misty.com (Don Klipstein) wrote: > In article <l88d46h5bl21v35s21t3lvqh02r4n0j...(a)4ax.com>, Jon Kirwan wrote: > >On 20 Jul 2010 21:34:07 (PDT), Bill Bowden <wrongaddr...(a)att.net> wrote: > > <SNIP what leads to this to edit for space> > > > > >>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. > > If Bill's spool is 1 inch long by 5/8 inch diameter and his > 18 AWG wire is .065 inch in overall diameter, such as hookup > wire, then this inductor as described would have inductance > around 110 uH. > > I doubt it will have even .01 microfarad of interlayer > capacitance. > > Looks like I would want to hear more details about this coil. > > For example, if the spool was 1 inch in diameter and 5/8 inch long > and the wire was AWG 18 magnet wire more like .047 inch in diameter, > Bill could have wound 6 layers of 13 turns and a 7th layer of 12 turns. > > The radius difference c becomes .33 inch > The average radius a becomes .665 inch > The length b is .625 inch > > The inductance in microhenries at this rate would be 222 microhenries.. > > - Don Klipstein (d...(a)misty.com) Yes, the spool is 1 inch inner diameter and 5/8 length. 6 layers of 15 turns each, 90 turns total of #18 enamel copper wire. I used this calculator: http://www.pronine.ca/multind.htm -Bill
From: Bill Bowden on 23 Jul 2010 00:58
On Jul 22, 3:25 am, Jon Kirwan <j...(a)infinitefactors.org> wrote: > On Wed, 21 Jul 2010 17:03:24 -0700 (PDT), Bill Bowden wrote: > ><snip> > >I got some better figures using a 9.5 volt PS and 16.3 ohm load. > > >Output is 12.7 at 9.9 watts. Input is 9.5 at 1.16 amps, or 11 watts. > >efficiency is around 90% which is closer to what I added up > >considering 1/2 watt loss in the diode and another 1/2 watt in the > >coil. The mosfet resistance is only 28 milliohms, so not much lost > >there. Frequency is 8.33KHz, duty cycle is 40uS and 80uS = 33% > > Okay. So about 120us period, 40us ON and 80us OFF. Got it. > > >Assuming the inductor is 280uH, then di/dt is 9.5 / .00028 = 33929 > >amps per second, so the current should ramp up 1.36 amps in 40uS. But > >the problem is on the down ramp, the voltage should be 12.7 + 0.8 for > >the diode, or 13.5 minus the supply of 9.5 = 4 volts. So di/dt should > >be e/L = 4/.00028 = 14286 amps per second, or 1.143 amps in 80uS. But > >this doesn't agree with the ramp up value of 1.36, so something is > >missing. > > This is exactly the point. And I think you've got your > numbers right. > > >How would you determine the minimum current point? > > That's a little trickier. I had to think a moment to get a > clue. I might still be wrong, but here is the argument. If > I'm lucky, I will derive something quantitative below. > > You are driving (forcing) the situation using a clock and > duty cycle. The inductor doesn't control any of that. It > is, instead, driven by it. However, it does respond. > > Imagine you set your duty cycle to some value that causes a > finite dI to be (t_on*V_on/L). The V_on can't be controlled > -- it just is. The t_on is set by your clock source. It > also just is. The same with L. So dI is fixed. It doesn't > matter what happens on the output. The inductor has no > choice in the matter. The dI will be some value. What isn't > said here is what the baseline I happens to be. Let's call > this baseline current through the inductor as I_0. We hope > and expect I_0 = 0A. But this isn't necessarily so. What we > can compute is dI. So the inductor current will have to rise > from I_0 to I_1 and it _will_ be the case that I_1=I_0+dI. > There is no choice on that matter. > > In the steady state case, which we still aren't sure of, I_0 > will be stable and so will I_1. We hope I_0 is zero and I_1 > is simply dI. But hold that thought for a moment. > > Now, your clock source also sets t_off. In the steady state > case, we know what the dI must be for the t_off period, too. > It has to be the same. If not, then it's not steady state. > And it turns out that we should think about that V_off value, > which will be L*dI/t_off. Since dI=(t_on*V_on/L), this works > out to V_off=[V_on*(t_on/t_off)] -- the voltage that _must_ > be across L when the switch is off if a steady state exists. > > Now, this V_off might not actually be what you imagine. In > other words, you might think it is simply the output voltage > plus a diode drop and less your input voltage because that is > what is needed. But it isn't what happens. > > Suppose t_off is shortened a bit. Look at the equation for > V_off. Note that t_off is in the denominator? If t_off gets > smaller, V_off gets larger!! This means the output voltage > also __increases__. It has to (assume for a moment that the > diode voltage drop remains close to the same value as > before.) > > But how can that be? Doesn't that mean that the load > (resistive, for now) will use more power? Yes, it does. You > might think that is a good thing. You've upped the output > voltage purely by reducing t_off!! You could reduce it even > more and get a still higher output voltage. But more output > power means more input power. And you still have a fixed > number of clocks per second. So more power must mean more > energy per pulse! > > How do you get more energy per pulse?? You've already > determined that there is a fixed dI, completely determined by > L, t_on, and your input source voltage. So that can't > change. What can happen to increase the energy per pulse? > > Well, take note that a fixed change in the magnitude of the > inductor current is NOT the same as representing a fixed > change in the magnitude of the energy stored in that magnetic > field. > > You have these two circumstances: > > E_0 = (1/2) * L * I_0^2 > E_1 = (1/2) * L * I_1^2 > > The difference is then: > > dE = E_1 - E_0 = (1/2) * L * (I_1^2 - I_0^2) > > But I_1 = I_0+dI. So, > > dE = (1/2) * L * ((I_0+dI)^2 - I_0^2) > = (1/2) * L * (I_0^2 + 2*dI*I_0 + dI^2 - I_0^2) > = (1/2) * L * (2*dI*I_0 + dI^2) > = (1/2) * L * dI * (2*I_0 + dI) > = (1/2) * L * dI^2 + [L * dI * I_0] > > Note the last term enclosed in brackets? When I_0 is non- > zero and the same sign as dI, there is an additional amount > of energy in each pulse. > > Another way of noting this fact is assume dI=1A and then to > compare when I_0=0A and I_0=1A. dE when I_0=0A is some > number we can call X. dE when I_0=1A will be 4X-1X, or 3X. > In other words, more energy is taken up and released in each > pulse when I_0=1A, even though there is the same dI in both > cases. > > The point I'm getting to is that the inductor _must_ yield a > larger V_off if you shorten t_off. To do that, the output > voltage increases. That causes more power to be required. > Which, because the frequency is constant, requires more > energy per pulse. But to do that, I_0 must rise upwards so > that each pulse _can_ deliver more energy to meet the power > requirements. So it rises from zero. But this then causes a > fixed DC current in the inductor, which means more loss in > the resistive parts and the diode, as well. > > Which is not desired. > > The computation of exactly what I_0 must level out as depends > upon the output load's response to a change in the voltage. > > for example, let's do some computations using some of your > numbers. R_out=16.3, V_Lon=9.5V, t_on=40us, t_off=80us, > f=8333Hz, L=288uH, etc. We don't know V_out, but we can > compute it from V_Lon*t_on/t_off. This is 4.75V, in short. > Assuming your diode voltage is about 0.7V, this means 4.05V > added to the 9.6V input. Call it 13.6V. However, there is > some resistance you mentioned which at a couple of amps > average (grossly assuming a lot) is maybe .5V? So call it > about 13.1V on the output. The PCT of power in L should be > about 4.75V/13.1V or, say, about 36%. And the power is > 13.1V^2/16.3Ohms, times that 36%, or in other words about > 3.79W is delivered by the inductor itself. With f=8333, this > means almost 455uJ per pulse. > > We know from t_on=40us that the dI=1.32A. Energy from that > part, using (1/2)*L*dI^2, is about 251uJ. Since we need a > total of about 455uJ, this leaves about 204uJ that needs to > be made up in the L*dI*I_0 term. This suggests I_0=.534A. > > Which may or may not be about where you are. But that seems > to be where theory takes me, right now. > > >I think the idea is to drop the inductance from infinity where the > >current is constant to a smaller value where the minimum current gets > >close to zero. > >So, I might take a few turns off the inductor and save some > >resistance, and let the current get closer to zero, but I don't think > >the efficiency will go much past 90% that I already have. > > I think you are losing some efficiency you could get back, > simply because you have a DC current flowing all the time, > increasing the mean power dissipation in resistances. Not to > mention that this would be a problem if you weren't using an > air core inductor. > > I hope the above isn't too terrible to follow. It took me > some moments of thought to realize _why_ the minimum inductor > current would rise above zero on its own. But once I > realized that dI is determined by t_on and that the inductor > off-voltage is determined by dI and t_off and that shortening > t_off only means that the off-voltage goes UP, the rest > seemed to fall into place for me and make some sense. > > Jon I think I see the error. The scope readings are not exact and the duty cycle may be 38Us to 82 instead of 40/80. Hard to read a couple microseconds out of 120. This makes the current ramps more equal so that 33929 amps per second on the up side for 38uS is about 1.29 amps and 14286 amps per second on the down side for 82uS = 1.17 amps. Pretty much the same. So they look about equal and probably are if I could read the scope accurately. A small error in duty cycle reading upsets the apple cart. Now, as to where the peak and valleys should occur, I think we should consider the case of infinite inductance where the current will be constant, or 779 ma for 68% of the time, or about 1 amp average. So, using the smaller inductor, the current waveform should move above the center by 1.2 amps, or 2.2 amps peak and then back down to 1 amp which is close to what I saw on the scope. Just some thoughts. -Bill |