On 10/24/07, PicDude wrote: > I have been searching for a very-low-cost switching replacement regulator for > a PIC-based application (automotive environment), but not coming up with > much, due to cost and space. The goal is to find something smaller but not > be too much more in cost. Efficiency is NOT a priority, except that too much > heat loss results in a heatsinking problem. > > The app requires 5V and uses about 200mA on average, up to 250mA > occassionally. > > Existing setup = SMA rectifier into a 7805 (TO-220) with a stud and standoff > to hold and heatsink the 7805 onto an (alum.) enclosure. At 1.6W (average) > lost to heat, the enclosure does get warm, but not hot. Total parts cost =~ > $0.75 in 1k quantities. > > For switchers, I came up with many arrangements, including this version that I > prototyped quite some time back... > http://www.narwani.net/neil/electronics/78SWxx/index.html > With the parts count, I would definitely need to make this on a separate PCB. > Total parts cost =~ $2.80 in 1k quantities, but labor is higher. Other > lower-cost switchers require more expensive and larger passives. > > So I came up with another idea recently and wondering if I could pull this > off... input voltage (11-14V typical) to the emitter of a PNP transistor, > collector to the input of an LDO 5V regulator (SOT-223 or similar). The > collector would also have a filter capacitor to ground. The base of the > transistor will be driven by a spare PIC pin and that line would also have a > high-valued resistor to ground.. The idea is this... On power up, the base > resistor ensures that the PNP transistor is on, supplies power to the LDO, > and the PIC powers up. After setting up ports etc, the PIC sets up the pin > that drives the PNP transistor to pulse on and off continuously, say 50% > duty-cyle for now. This will not work as described. The PIC cannot turn the transistor off. To turn it off its base voltage needs to go up to the 11-14V input voltage, but PICs operate on 5V. You'll need another common-emittor NPN in there to do the level shifting. That effectively chops the input voltage in half, with > the filter cap selected to ensure that the voltage at the input of the LDO > does not drop below 5V+dropout. This will not work either. The voltage isn't chopped in half. When the transistor is on, it will charge the capacitor to 12V. When it is off, its voltage will decline linearly (since you specified a constant 200mA current.) Basically, you need to add an inductor in the circuit. And a feedback path. Then you'll have a switching regulator. You didn't specify your output voltage tolerance, but the output of a switching regulator will be fine to run digital circuitry like a PIC. No LDO required. In electronics, "chopper" is used to refer to a kind of low-offset amplifier. So that may be a confusing term to use here. Regards, Mark markrages@gmail -- Mark Rages, Engineer Midwest Telecine LLC markrages@midwesttelecine.com -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist