From: john/lori (lgnjh_at_earthlink.net)
Date: 2007-04-11 20:11:50
Scott McDonnell wrote: > Back to the drawing board. hmm, not sure if I should comment further, but .. When you first described what you were going to do, I thought you were talking about a shunt regulator, ie you'd supply a fixed amount of current and shunt an adjustable part of that around the cap and the cap get's what's left. I don't think a single transistor series regulator is going to work very well. You make a bipolar transistor current source (current sink actually) like so: v+ | | +----0 load +----0 | |/ Vb ---| |\ V | RE _|_ /// You put a voltage on the base, Vb. That voltage minus the .7V or so of the base-emitter junction appears across the emitter resistor, RE. That sets the current, most of which goes to the load. The transistor will maintain that current as the load varies (with in limits) to maintain the voltage across the emitter resistor. Your load is between the transistor and ground so you really do want a current source so you'll have to turn it upside down (use a pnp) You'll want to be able to put 2.5 V on the cap (just in case it meets the spec instead of what Jim measured :) and you'll need a minimum of .5V or so across the transistor to keep it from going totally non-linear. That leaves you 2V for the emitter resistor. You'll want to keep the base current to a fraction of the current in you bias network. 1 count is 1/255 of you range I'd guesstimate half that. Hfe of 70 looks reasonable to me from the data sheet for the 2N3906 (the complement to the 2N3904) I see a couple of problems. 2.5ma/70(Hfe)x500(fudge factor) gives you about 18ma in your bias network that's a lot of current to be yanking around with your input. Your input and the cap are referenced to ground but the transistor will want a voltage between V+ and the base. So you'll have to take in to account noise on, or variations of, V+ bogax Message was sent through the cbm-hackers mailing list
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