> > > 1) You talked about replacing BUKFET with a PNP bipolar. What about an NPN > > > bipolar? Remove QBUK2, RBUK3, ZBUK2, and RBUK4 and tie the bottom of RBUK2 > > > and the collector of QBUK1 to the base of the NPN. When QBUK1 is off then > > > RBUK2 will turn the power NPN on. When QBUK1 conducts, the NPN will turn > > off. > > > I'm sure that the base current from RBUK2 will need to be higher so that > > > the power NPN will saturate. > > > > I've paper designed something similar. The main problem is that the NPN > > needs positive drive and when it is turned on the base needs to be driven by > > a voltage ABOVE the input voltage. This can be accomplished by using an > > extra high side supply. This can be provided by an extra isolated winding > > on the inductor and a single diode and capacitor. > > Confusion sets in on the computer guy ;-( . Everytime I think I actually > understand this semiconductor stuff, someone comes along and throws a curve: > > The statement you make above makes sense for N-channel MOSFETs. But that's > not my understanding of how NPN bipolars work. First off I thought that > bipolars are current amplifiers, and that voltage wasn't really the determining > factor. Secondly my understanding was that a bipolar can be driven to > saturation with the base voltage well below the collector voltage. And that > in fact that once the base voltage gets above the Vbe that it doesn't in fact > continue to rise but that it sucks down more and more current through the base > while channeling more and more current through the CE junction all the way > to the point of saturation, where Ice/Ibe = hfe. Everything you say is essentially correct as it applies to other circuits ! :-) Yyou just need to think through what happens in this case when an NPN in the place of the BUKFET is turned hard on so that its emitter is pulled all the way up to Vin. What will the base voltage now be relative to Vin? Think about it first and then read the following if needs be. [[[ The confusion arises here because of what happens when the main pass transistor (BUKFET replaced by an NPN) turns on. For an ideal transistor the drop across the transistor Vce will be zero. For a real transistor this will be in the 0.1 volt to 0.3 volt range. Much higher than that and you probably need a different transistor for the current levels you are using. The result is that the emitter is now *essentially* pulled all the way up to Vin. To keep the transistor turned on the base needs to be about 0.6 volts above the emitter so it will need to be about 0.6 volts ABOVE Vin. You are correct that a bipolar transistor TENDS to maintain its Vbe at about 0.6 volts once turned on and just "suck increasing current" as attempts are made to increase this voltage. It's actually an exponential relationship but the curve gets noticeably steep at about this pint and by the time you get to 0.7 volts you would normally be overdriving the device beyond its ratings. BUT, if the source which is providing this Vbe is a voltage source we normally turn it into something approximating a current source by placing a resistor between the turn on voltage and the base. This means we need more than 0.6 volts to cause the required current to flow i the resistor. The larger the driving voltage the larger the resistor for the desired current and the closer this approaches a true current source which will provide "about" the desired current as Vbe moves up its exponential curve. Typically we allow at least a few volts to do this. Say we want 1 mA base drive and we have a 5 volt source then the resistor required is R = V/TI= (5-0.6)/0.001 = 4600 ohms. If the ACTUAL base voltage rose to 0.8 volts the current would be V/R = (5-0.8)/4600 = 0.913 mA. Still OK probably. The result of all this is, when the emitter is pulled to Vin the requirted base drive voltage required is several volts above Vin to apply enough voltate to the base resistor to get the 0.6 volts odd Vbe required. For this we need a :"high side supply". The xtra winding that I show in my more complete diagram can be used to supply this at relatively little cost. (One winding, 1 diode, 1 resistor, 1 capacitor). > I guess my only question out of this is whether or not it is possible to > oversaturate the base. You can damage the transistor by providing too much base current. This is seldom a problem in practice in low power circuits. A saturated transistor takes longer to turn off and some high speed switching designs use reverse schottky clamp diodes from C to B to prevent the transistor fully saturating. This is not applicable here. > For now I abandoned the NPN bipolar for a much simpler reason, the 6W of heat > that would have to be dissapated if I maxed out at the 10A I'm shooting for. > That .6V drop is devastating when you're pumping 10A through... You SHOULD be able to get Vsat down around 0.1 volts with a suitable transitor. At those current levels you need to make sure to study the spec sheet carefully. Many devices are specified for current gains at say 1 amp and have markedly less gain at higher currents. It can be informative to put your transistor on a heat sink, apply different base currents (a few volts supply through a variable resistor) and then to apply the desired collector current (second supply and resistor or variable supply). My experience is that Vsat can get worse quite suddenly as base drive is reduced, that actual current gain (beta) can be quite different than that at specified current levels (not surprisingly) and that there can be wide variations for the same type. Intelligent use of published data and an genetraous amount of base overdrive will usually help. The essentially zero DC drive requirement of a FET makes it much easier to drive in high current situations - (but you still need to provide a means of rapidly moving charge in and out of the gate if you want fast switching.) For a FET a drop of 0.1 volts at 10 A corresponds to an Rdson of 0.010 ohm. - easily enough cahieved in lower voltage FETS (and for $ in higher voltage FETS). > My interest in the question is for the junkbox variety supply. My box has > in decreasing frequency: NPN bipolar, Nchannel MOSFET (usually logic level, > PNP bipolar, Pchannel MOSFET). **** NB NB NB NB NB NB NB *** I said previously that - > > The flyback diode DBUK2 MUST be rated for a peak current capability of > > several times the mean output current and MUST be a high speed part (eg NOT > > 1N400X). Olin has suggested that this is NOT the case due to circuit time constants etc. I'm not convinced yet but his assertion needs looking at,. The ability to use a low cost general purpose diode here would be an advantage. > I have two goals in this arena: understanding and simplicity. Both give me the > ability to throw together a switching PS with readily available parts. As I > stated in my first message, I got the theory, but failed to grasp the control > mechanism. Now I get it. > > For the sake of the following discussion I'm going to abrrev BUK with B only. > > This prompts one last question: What exactly is the purpose of RB2 and QB2 > again? I keep seeing QB1 and QB2 in a darlington configuration but I'm not > understanding why that config is required. Consider if we removed RB2 and QB2 > and attached the collector of QB1 to the bottom of RB3. When Vin is applied > the gate of BF will rise causing it to conduct. QB1 is turned off by PB1 > pulling the base of QB1 low. The coil charges until ZB1 conducts. This turns > on QB1, grounding the gate of the FET turning it off. Eventually the coil will > start to lose energy and ZB1 turns off, turning off QB1 and turning on the > FET again. So it'll oscillate right? > > It seems to be the same exact operation. The only difference I could find is > that the MPSA42 (QB2) had a much higher Vce. But even that's confusing because > QB1 is going to subject to a higher Vce via RB2. > > So anyway just wondering why the darlington config and whether or not > RB2 and QB2 are absolutely necessary? Summary: It's NOT a darlington - look again and you'll see ot's a little unusual. and look at the drive polraity required by the pass FET/transistor. Detail. RB2 is used to provide turn on base drive to QB2. At startup there is no output voltage and this is the only way for the rqgulator to start. In my more complex circuit you will see that I have also added RBUKX from Vout to the base of QB2. This allows Vout to provide some/most of thebase drive for QB2 and allows RB2 to be a higher resistance and therefiore limits dissipation for high Vin (recall that in my case this runs up to 200 volts Vin). QB2/QB1 connection: In a darlington the emitter of the first transistor drives the base of the second. Collector current in the first transistor becomes the base current of the second transistor. In this case quite the opposite occurs. QB1 SHUNTS the base drive to QB2. QB2 is normally turned on by resistor RB2 and turning QB1 on shunts this drive current to ground thereby turning QB2 off. Now look at the relative polarities required. I am using a P Channel FET whose gate must be pulled "low" to turn on. For Vin too high, QB1 is on and QB2 is off and FET gate is high and FET is off. If we changed to your QB1 only design the we get Vin too high, QB1 on, FET gate low, FET on and ..... oh dear. HOWEVER if the FET was an N Channel then this would work as desired. However, we would then need to drive thre FET gate ABOVE Vin during FET turnon and so need a high side supply. (as when on FET drain and source are connected so source is at Vin so gate must be at leat Vthreshold above Vin (plus a bit more). Using a second winding would provide such a supply). A note: My application demanded, among other things, very low cost in volume production and it met this requirement superbly. Power level was moderate (up to 1 amp out at 10 volts). For the low power levels I specified in my original description for this "challenge" then the circuit shown is appropriate. For a low volume high current design I would CONSIDER using a more conventional circuit such as the one Olin just published or one using a commercial IC. These usually have a more controlled drive of the FET gate. Adding the extra transistor in my more complex circuit improves FET turn off markedly. Replacing the diode DBUK1 with a PNP transistor (relative to QB3 connect emitter to emitter, base to base, collector to ground ) adds surprisingly little improvement but ends up looking much like Olin's drive circuit. I trialled this arrangement during development and decided it was not needed. . Hope all that makes sense. regards Russell McMahon -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listserv@mitvma.mit.edu with SET PICList DIGEST in the body