On Fri, Jan 11, 2002 at 05:03:04AM +1100, Roman Black wrote: > Byron A Jeff wrote: > > > Now we're getting somewhere. This is the issue I want to address. Please > > take a stab at explaining the controller using the 7805. I know that it's > > possible to use the GND pin as an adjust pin. However I just don't see how > > to get over the dropout voltage issue. I just took a look at the National > > Semi LM340 datasheet, and even with an output current of 0A it starts to > > drop out about 1.5V above the regulation voltage. So how can it be used as a > > control element when Vin is in the dropout region? > > Wow! That was a long post. :o) > Here's my views on building a low dropout series > regulator as simple as possible. > > First thing is separate the thing into two parts: > * power (obviously a low sat power device) > * control > > ONLY the power device needs to be low dropout. The > control circuit can be anywhere, and you put it in > the best place, referenced to the zero rail. > > > Low sat PNP > Vin -------------------------E C----------- > Vout > (13.8v) | | B | (Reg 12v) > | R | | > | | | | > | |-----| R1 > | | | > | 7805 C | > |--I O---------B | > G E-----| > | NPN | > | R2 > | | > | | > Gnd > --------------------------------------- > > > Our power stage is just a low sat PNP like a BD204 > or better part. The entire control system is referenced > to Gnd, and a 7805 is used as a cheap easy precision > zener, giving exactly 5v no matter what the input > voltage does. > > R1 and R2 are a voltage divider tied to Vout. They > set the NPN emitter at 5v-0.6 =4.4v whenever the > output is at the correct voltage. Any less than the > correct Vout, and the transistors will turn on as hard > as needed to cause regulation or saturation, whatever > comes first. So that's the "magic" formula!! The R1/R2 ratio is just like the divider on an LM317. So if I calculate R1/R2 so that the midpoint gives 4.4V when the PNP collector is 12V. The transistors will lock in to put the PNP collector to 12V. Young grasshoppa finally understands. Thanks. So I did a quick calc using Vo = (R2/(R1+R2))*Vin with 12.0 for Vin and 4.4 for Vout and got a ratio of R1 = 1.73*R2. > > Make sure the R1/R2 divider passes more than 10x > the max expected PNP base current to keep regulation > fairly accurate. If you keep this current substantially > higher than the PNP base current the regulation will > be close to perfect, at all voltages right down to > saturation. And only at the cost of the base current which is the one item I have no concern over in this circuit. And just to verify the base current will be 10 * (Iout/beta) right? So let's put it all together. I went poking around the Philips NTE/ECG replacements parts site, because there are several Atlanta Distributors. I picked the NTE2314. 15A max, Vce(sat) of 0.25 typ, 0.5 max at 8A draw. Min beta of 30 at 8A. R2 would need to draw 10 * (2/30) -> 0.6A of current @ 4.4V. That works out to 7 ohms. If I use R1=12 ohms and R2=7 ohms it'll set the output voltage to 11.92V. Only one missing piece: what's the purpose of the emitter to base resistor on the PNP. A pullup to somehow force complete cutoff of the transistor? What value might it need to be? Another question: can R1/R2 be fed into a voltage follower? At that low a value the resistors will burn more than a half amp all the time. > -Roman > > PS. It will work just as good with a P type FET > as the power part, if Vin is 12v or higher. Then the Rds determines the dropout. This is a winner! I'll start working on it immediately. Thanks for all your help. BAJ -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listserv@mitvma.mit.edu with SET PICList DIGEST in the body