Dave Dilatush wrote: > > Roman Black wrote... > > >Dave Dilatush wrote: > > >> One way of keeping Q1 from being fried by inductive kickback would be > >> to place D1 right across the main inductor L1, in the same manner in > >> which one would place a snubber diode across a relay coil. > >Actually this is not so. :o) > > Really? Build the circuit- physically, with real components, or > virtually, with a simulator- and observe its behavior. :) Ummm, actually I have number of stepper motor drivers using the "standard" slow-decay system and all showing excellent SMPS efficiency... I credit "Jones on Steppers" :o) > >To correct the circuit I would place a schottky > >diode across the coil directly. This IS actually > >efficient, it's similar to the "slow decay" system > >used in large stepper motor drivers, > > Stepper moter drivers != switching regulators. Hmm. Sounds like an argument! :o) So you're saying the switching efficiencies shown in motor drivers don't translate to current supplied from point A to point B?? > Whatever similarity you're seeing here may be visually appealing, but > not particularly relevant. Putting a catch diode directly across an > inductor is a good way to prevent inductive kickback from frying > whatever device is driving the inductor, but it doesn't do a darn > thing for efficiency. > > >...and instead > >of the coil energy being dumped into the input or > >output supply when Q1 turns off, the magnetic field > >of the coil can only decay very slowly, so the > >energy is maintained IN the magnetic field until > >the next time Q1 turns on again. > > If we assume the above statement is true (see next paragraph for an > explanation of why it isn't), then we are left wondering how the > inductor's stored energy ever gets transferred to the output. The > answer is that it doesn't. With Vin > Vout (that is, with pin 2 of L1 > positive with respect to pin 1), the energy stored in L1 during the > next conduction cycle can only increase, not decrease. What ends up > happening is that in between the periods in which Q1 is turned on, the > stored energy in L1/L2 gets dissipated as heat, partly in D1 and > partly in R1 and R3. Now I understand your argument better. :o) > >This should have > >similar efficiencies to other SMPS flyback systems, > > The very essence of a flyback SMPS is that it stores energy from the > power source in an inductor during one part of its operating cycle, > then transfers the inductor's stored energy into the load during the > other part of the cycle. > > Your circuit contains no physical means whatsoever for transferring > that energy: energy is certainly stored in the inductor when Q1 is on, > but the only way it gets out is as heat. OK. Now we get to the crux of the argument... First, I have the greatest of respect for your 25 years SMPS design experience, I have similar years but probably from a different background. :o) Please argue the following (simplified) points: When you apply volts to a series inductor, it takes ENERGY to build the magnetic field in the inductor, at the same time current does now flow until that field is built, hence the XL=2pifL formula. So in most SMPS supplies we rely on total (series) current being limited by XL. So with any series inductor we get the energy we stored in the magnetic field, that we have already paid for, which can THEN be distributed in 1 of 3 ways when the main switcher turns off: 1. Field collapses- energy fed back to psu input cap. 2. Field collapses- energy fed forward to psu output cap. 3. Energy is maintained within the field, by slow decay, (ie, the coil is short-circuited) as in stepper and most pwm DC motor drivers. In a standard buck supply the field is encouraged to collapse, and the energy is fed by the flyback diode to the output cap. Cool. It's efficient. BUT, if we keep the magnetic field WITHIN the coil, as in a motor driver, the initial energy can't go anywhere, hence no waste. The next time Q1 turns on, we don't need to re-establish the field because it's still there. And WHEN Q1 turns on, the current is fed straight to the output, because we don't need to first establish the field. So after the first cycle, each Q1 turnon simply transfers current to the output, giving us current transfer with no resistive losses. Which is the essense of SMPS efficiency, same current in as current out, but the voltage difference is taken care of... :o) Remember that once we have paid the energy cost to "charge" an inductor with field, that energy MUST go somewhere. If we short circuit the inductor with a Vf 0.2v schottky diode, the inductor energy CAN'T collapse, due to simple physics. To make this easy for argument sake, if we charge the inductor with 20vdc, and short it at 0.2vdc, for equal time periods, you get 100:1 efficiency, in terms of energy flow. :o) Please argue, I appreciate anyone who has made SMPS's for 25 years, as I know that you learned X things in those years and I learned Y things in those years and it sure beats arguing the football scores... :o) -Roman -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.