Non-linear functions ...
in [Design]
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From: Tim Williams on 25 Nov 2009 16:35 "Fred Bartoli" <" "> wrote in message news:4b0d6bec$0$17807$426a74cc(a)news.free.fr... >> .-----. .-----. .-----. >> | | | | | | >> problem o---->| ADC |----->| LUT |----->| DAC |-----> answer >> | | | | | | >> '-----' '-----' '-----' > > .-----. .-----. .-----. .-----. .-----. > | | | OPE | | ABA | | OPE | | | slow > problem o---->| ADC |-->| RA |-->| CUS |-->| RA |-->| DAC |-->answer > | | | TOR | | | | TOR | | | > '-----' '-----' '-----' '-----' '-----' > problem o---- .--------------------. .--------------------. | | | | | 10 kilomonkeys |--->| 10 kilotypewriters |---> answer | | | | (eventually) '--------------------' '--------------------' Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
From: Spehro Pefhany on 25 Nov 2009 18:31 On Wed, 25 Nov 2009 16:10:31 +0100, Fred Bartoli <" "> wrote: > >Hmmm, even simpler... >You could use the 555 with an external mos switch with it's source >shifted, say 1V (guess why) and omit the constant subtraction from your >Vout input. > > > R C RC = C2 > ___ || > C1 >--|___|--+--||---. > | || | > | === > Vout >----------. .--------+ GND > | | | > .-----. | > \+ -/ +-|| > comparator \ / ->|| > V +-||---. > | | | > | +1V | > .--o--. | > GND -|D S Q|-------------' > | | > Clk >---|> -| ___ > | R Q|---|___|--+-----> Flow > '--o--' | > --- > --- > | > === > GND Very nice, Fred! These should be very close to practical values: 24K9 10n ___ || 6.02V >--|___|--+--||---. | || | | === Vout >---------. .--------+ GND | | | .-----. | \+ -/ +-|| comparator \ / ->|| V +-||---. | | | | 1V | .--o--. | GND -|D S Q|-------------' | | 100Hz Clk >---|> -| ___ | R Q|---|___|--+-----> Flow '--o--' | 200mV full scale --- --- | === GND
From: Tim Williams on 25 Nov 2009 18:32 "Bill Sloman" <bill.sloman(a)ieee.org> wrote in message news:1dd7bac6-d23b-4ba7-819c-820f8d336a19(a)w19g2000yqk.googlegroups.com... > In a related circuit I used a thick film platinum resistance sensor to > compensate for the change in the therma voltage - kT/h which is > proportional to absolute temperature and about 26mV at 300C (room > temperature). Er, KbT/q is volts. And it's 300K. Lead melts at 300C. Just how long have you been unemployed, Bill? Ordinarily you're just a sourpuss, but when you start reporting incorrect facts, you're actively sabotaging things. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
From: Bill Sloman on 25 Nov 2009 20:40 On Nov 26, 12:32 am, "Tim Williams" <tmoran...(a)charter.net> wrote: > "Bill Sloman" <bill.slo...(a)ieee.org> wrote in message > > news:1dd7bac6-d23b-4ba7-819c-820f8d336a19(a)w19g2000yqk.googlegroups.com... > > > In a related circuit I used a thick film platinum resistance sensor to > > compensate for the change in the therma voltage - kT/h which is > > proportional to absolute temperature and about 26mV at 300C (room > > temperature). > > Er, KbT/q is volts. And it's 300K. Lead melts at 300C. > > Just how long have you been unemployed, Bill? Ordinarily you're just a > sourpuss, but when you start reporting incorrect facts, you're actively > sabotaging things. 26 mV is reporting a voltage in volts. Sorry about the 300C. It was meant to be 300K - which is 26.85 Celcius, and a bit warmer that European room temperature - 20C - or US room temperature - 25C - but the value traditionally associated with declaring kT/h to be 26mV. By claiming that 300C was room temperature I did provide enough extra information to avoid confusing all but the terminally stupid kT/h was given as 30mV in the applications I read when I was young, but we've got pickier since then. As for "reporting incorrect facts" you've just told us that 26 mV isn't a voltage, as if millivolts weren't an an absolutely standard unit with the dimension "volt". Sorry to be a sourpuss about this, but your tone hasn't given me much alternative. Spehro's crack about melting lead was distinctly more civilised. -- Bill Sloman, Nijmegen
From: Przemek Klosowski on 25 Nov 2009 22:16
On Tue, 24 Nov 2009 22:19:26 +0100, Richard Rasker wrote: > These are the values (F=flow): > F (ccm) Vout (V) > 0 1.00 > 25 1.90 > 50 2.67 > 75 3.27 > 100 3.75 > 125 4.17 > 150 4.50 > 175 4.80 > 200 5.00 > > The first problem was simple: finding a suitable mathematical function > which fits the curve; I looked at something along the lines of > Vout=c1*(1-e^(-F/c2))+1, and it turns out that c1=5 and c2=125 provides > a near-perfect fit. The second problem was to find an inverse function > -- no problem there either: F=-c2*ln(1-(Vout-1)/c1) -- leading to the > third and rather trickier problem, which of course is to implement that > inverse function in an actual circuit. > A second degree polynomial fit is not too bad either. V = -7.6087e-05*F^2 + 3.4724e-02 *F + 1.0566 BTW, your c1/c2 didn't work for me; shouldn't it be 128,125 rather than 5,125? |