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Typo in (i): "if the source has a peak to peak swing of x volts but a dc offset of y then (neglecting the diode drops) vcc = x/2+y and vss = x/2-y." :) Andy On 19 June 2011 11:01, Andy Fierman <[email protected]> wrote: > Rick is spot on. > > However, there are more things you need to consider: > > i) Does your signal source have a mean DC level of zero? Without C1, > if the source has a peak to peak swing of x volts but a dc offset of y > then (neglecting the diode drops) vcc = x+y and vss = x-y. > > If you have to remove a DC offset then you'll have to put a > transformer between C1 and the recirfiers. Then C1 keeps DC out of the > transformer primary and the transformer secondary provides the > necessary dc path for the rectifiers to work as required. > > ii) I note that C1 and L1 form a series resonant circuit with a centre > frequency at about 22.9kHz. So is your source really bandlimited 15kHz > - 28kHz by the time it gets to your circuit or are you trying to do > the bandpass filtering as part of your circuit? If the latter then you > will have to keep C1, L1 but add the transformer as described in (i) > above. > > The you will have to model that transformer to include at least the > leakage inductance to get the bandpass response right. > > Such transformers are not difficult to design and source. > > iii) What is the source impedance? Does the 8 ohms represent all of > your source impedance or is there more hidden in the source itself? > You will need to allow for all of it to see how the rectified outputs > drop and ripple increases with load current. > > iv) Don't forget that a SMPS represents a constant power or negative > resistance load. As the input voltage drops the current it draws from > the source increases. The actual behaviour of a real SMPS is > complicated by any input undervoltage lockout and soft start features. > This may or may not play well with your source. > > I'd like to make a general point here. > > This isn't a criticism but an important observation: when asking a > question about how to do something, it saves everyone a lot of time, > guesswork and blind alleys if the problem that is to be solved is > clearly stated alongside whatever attempt at a solution that the > specific question may be about. > > Essentially, include the design specification in the original question > otherwise no-one knows the whole story so the question doesn't get a > proper answer in a timely manner. > > Clearly in some instances the design spec may not be something that > can be given openly but usually the part relevant to a question can be > reframed so as to not give away too much. However, there has to be > enough information so that the boundaries of the problem in question > can be understood. > > This question is a classic example. Several people have discussed > removing a part of the circuit that I now strongly suspect (C1 and L1) > is an essential (if inappropriately implemented) part of the circuit > because it wasn't clear what the overall function or scope of the > circuit was. > > Cheers, > > Andy. > > signality.co.uk > > > > On 18 June 2011 21:19, rickman <[email protected]> wrote: >> What is the purpose of C1 and L1? If you want to filter anything, it should >> be AFTER you rectify the signal to DC. A series cap is going to remove low >> frequencies... like DC which is attenuated very highly. So much in fact >> that you can't draw a DC signal through a capacitor. That is why your >> circuit is not working. >> >> If you remove C1 and L1 the circuit will work the way you want it to I >> believe. Also, with an input frequency of 15 kHz or higher, you won't be >> needing 100 uF output filter capacitors for a light load. How many mA is >> your load? How much ripple can you allow? Use those two values to >> calculate the value of output filter capacitor you need. Once I fix your >> circuit by removing the input "filter" I measure 19.14 volts out and 38.2 mA >> of current into a 500 ohm load. Is that what you are shooting for? The 100 >> uF cap gives around 10 mV of ripple. With lighter loads or more ripple the >> cap can be smaller. >> >> Rick >> >> >> On 6/17/2011 4:44 AM, myken wrote: >>> >>> Yeap, it should be a very low power power supply. Vx is not important Vcc >>> and Vss are. >>> Vin can be anything from 15Khz to 28Khz so a transformer is not the most >>> desired option. >>> I have designed two SMPS for Vcc and Vss but there load to the rectifier >>> are not the same, with the described result. >>> I will try the options suggested in this list today. >>> Robert. >>> >>> On 17/06/11 04:13, gene glick wrote: >>>> >>>> On 06/16/2011 02:30 PM, myken wrote: >>>>> >>>>> Hello all, >>>>> >>>>> I would appreciate some expert advice. >>>> >>>> Are you trying to make a low current power supply? >>>> >>>> I agree with DJ - the unequal loading on + and - cycle will average to >>>> something other than zero (unequal capacitors, unequal diodes, etc) If Vx >>>> must always be average zero - you'll need to do something else. >>>> >>>> If you can handle a little voltage drop, don't care what happens to Vx, >>>> and don't mind adding a few parts, make a cheapo regulator with a zener and >>>> BJT? (Or maybe use TL31 instead of zener) >>>> >>>> What about a small transformer, one winding on primary, center tapped on >>>> secondary. Add a diode and a cap for each leg - and there you go! >>>> >>>> Anyway, there's lots of ways to do this. If regulated output is what you >>>> want, a little more work is required. >>>> >>>> >>>> gene >>>> >>> >>> >>> >>> _______________________________________________ >>> geda-user mailing list >>> [email protected] >>> http://www.seul.org/cgi-bin/mailman/listinfo/geda-user >>> >> >> >> >> _______________________________________________ >> geda-user mailing list >> [email protected] >> http://www.seul.org/cgi-bin/mailman/listinfo/geda-user >> > _______________________________________________ geda-user mailing list [email protected] http://www.seul.org/cgi-bin/mailman/listinfo/geda-user
