On Fri, Aug 24, 2001 at 11:14:45AM +1200, Russell McMahon wrote: [BAJ] >>>>1) You talked about replacing BUKFET with a PNP bipolar. What about an NPN >>>>bipolar? [Russell] >>>I've paper designed something similar. The main problem is that the NPN >>>needs positive drive... [BAJ Again] >>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. > > [Russell] >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). Understood. And if we only pull the base up to Vin, then we'll not get the emitter very close to Vin. In fact it'll end up several volts below. > >>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. I was asking this question in the context of the 20 to 1 input voltage Since the base is pulled above Vin and some base resistor will current limit how do you design so that you don't fry the transistor and yet provide enough base current to get to saturation at lower Vins. [Edited bipolar selection process...] >>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). Since I've been dabbling in switching power supplies recently, high current, high power, high speed rectifiers are actually in my junkbox. I'm fortunate to be in Atlanta and there are several places that stock the required parts. > >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. That's true. But I have a properly specified part in my prototype. > >>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. Duh. It took me a while to see that absolutely everything is reversed from the N-channels that I'm used to using. Source is positive, Drain is negative, Gate must be more negative than drain to get the part to conduct. I hooked the P-channel up backwards and I didn't get that QB2 acts as a sort of inverter. > >Detail. RB2 is used to provide turn on base drive to QB2. And since QB2 is an NPN, it needs a pullup resistor to conduct. > 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). But you still need the startup resistor RB2 right? I see. > >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. Got it. And all of this is a prelude to shunting the base voltage generated by the pullup RB4. I also think that I figured out that RB3 is a current limiting resistor for high Vin. I'm testing with 12-15V Vins. I left it off in my prototype. > >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. And the coil is feeding power to the load. This is the buck phase. >If we changed to your QB1 only design the we get Vin too high, QB1 on, FET >gate low, FET on and ..... oh dear. An avalanche. This didn't occur in my testing because I had the FET wired backwards. I added QB2 and RB2 before I realized that I needed to reverse the Source and Drain. Maybe you should label for us clueless ones out here. ;-) >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). The bit more is based on the current passing through the device. I saw an equation like Rds*Ids... >Using a second winding would provide such a supply). I'm happy with the P-channel. > >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. Noted. Now for my report. After reading (and gaining some understanding) I rushed out and bought parts. I have an IRF9540 P-channel (Ids=19A, Vds=100V, rDS=0.2 ohms), a 10 microhenry 11A coil, a 330 uF cap, and a SB580 5A ultra fast schottky catch diode. Since I wasn't concerned with high Vin, I used ordinary 2N2222 for both transitors. I also dropped the values of all of the resistors since I'm not worried about a high Vin. My concern was getting a bit better snap off on the FET. I bought a handful of different types of zeners, but ended up using a 5V one. I'm still trying to get clued in on why the output voltage isn't 5.6V as you specified but 5.0V right on the mark. I'm pleased to report that it was a total success. Once I added RB2/QB2 and got the FET leads straight, Vout is right at the 5V mark. I did some preliminary high current testing. Built a dummy load out of 4 0.7 ohm 25W ceramics, wired in series/parallel to give a 0.7 ohm 100W load. Final result: Vout=4.75V @ 7.8A !!! A total success. Next up is building a similar boost regulator to get 12V out even with a 6.5V input. I may even try wrapping that second coil so that I can get a low current -12V. >. >Hope all that makes sense. It does. I've been trying to get switchers under my belt for a while. Your design and this thread has taken me several light years further than where I was. I really appreciate it. BAJ -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listserv@mitvma.mit.edu with SET PICList DIGEST in the body