What you actually end up with is a flyback converter. You may as well get some isolation at the same time.You could use a blocking osc. 1 or 2 transistor circuit designed for a 30% on time. The transistor will be a low current 200V high gain type, or even a small fet. Frequency should be as low as your inductor allows for good efficiency. Start by selecting the core, probably a RM10, the wire will be the smallest you can work with, probably 0.05mm Dia. Then work out how many turns you can fit on it, divide it between primary and sec, then you end up with primary turns. work out what gap you need to stop it saturating at the 60uA Peak current you will have for a 10uA average. that will give you an inductance that will decide your operating frequency, and be as low as possible for that core. ----- Original Message ----- From: "Dave Tweed" To: Sent: Wednesday, September 24, 2003 5:47 PM Subject: Re: [PIC:] Stealing power from the phone line update > Thanks for all the responses! > > gtyler wrote: > > Surely you want a self resonant freq. as high as possible, this means > > the lowest parasitic cap possiblle? > > Yes, that was exactly my point. I probably didn't express it well, but all > the 100mH inductors I've found so far have self-resonant frequencies less > than 100kHz. > > gtyler wrote: > > > You actually need 2 windings, i.e. a transformer so you can keep the duty > > cycle reasonable. > > How does that help relative to the usual simple buck regulator topoology? > > I thought about just driving a step-down transformer with a 100 kHz square > wave oscillator, but the magnetizing (reactive) current for the transformer > will blow my input current budget unless the primary inductance is very > high (>1H @ 100kHz!). > > This leads to the idea of using a capacitor to store energy on the primary > side, leading to some sort of resonant design, which is what I was alluding > to. I don't know much about the design of resonant converters. > > Olin Lathrop wrote: > > I assume you were trying to keep the maximum inductor current low. > > No, just the average current <10uA and the repetition frequency on the > order of 100 kHz. With my 100mH inductor, the peak current is 200 uA. > > > For example, with a 1mH inductor (those are quite available), a 10uS pulse > > at 50V builds up the current to 500mA. That's an average current of 250mA > > for 10uS from the reservoir cap. This would only cause a 25mV drop on a > > 100uF cap. > > Yes, but if I'm keeping the overall average current from the line at 10uA, > then I can't do another pulse for 0.25 seconds. > > Hmm. I guess that works! I was thinking in terms of operating in a (nearly) > continuous-current mode and keeping the switching frequency above the audio > band. However working in an (extremely!) discontinuous mode works as well, > and doesn't require really huge capacitors on the output side of the > regulator, either. At an average output current of 100 uA, I just need > 100 uF to keep the ripple down to 250mV p-p. > > > A more serious problem is how to power the control electronics without > > eating into a substantial portion of the power budget. > > Shouldn't be too hard to deal with, with a 100 uA budget on the regulated > side. There may be start-up issues, but a simple bootstrap circuit to get > things going should be sufficient. > > > Again, this is an interesting intellectual exercise, but in the end trying > > to draw any real power from the phone line is a bad idea. > > Not at all! Phones and caller-ID devices do this all of the time in order > to maintain their internal memories. I'm just trying to see how efficiently > I can use the 10uA the phone company allocates for "leakage" in the CPE > (customer premises equipment). > > michael brown wrote: > > I mean it's fine to talk about comparing the output of the supply against > > a reference and setting the pulse width accordingly, but I still haven't > > come to the part where they explain how to obtain the reference > > efficiently? > > Band-gap references can operate at very low current levels. See for example > National Semiconductor's LM385, which achieves 3% accuracy at a current of > 20 uA, or Analog Devices' ADR280, which gets 0.5% accuracy at 16 uA. > > -- Dave Tweed > > -- > http://www.piclist.com hint: PICList Posts must start with ONE topic: > [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads > -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads