Non-linear functions ...
in [Design]
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From: Uwe Hercksen on 26 Nov 2009 10:24 Rich Grise schrieb: > .-----. .-----. .-----. > | | | | | | > problem o---->| ADC |----->| LUT |----->| DAC |-----> answer > | | | | | | > '-----' '-----' '-----' Hello, works well for 8 bit ADCs, also for 10 or 12 bits. With large EPROMs also for 14 or 16 bits. Bye
From: krw on 26 Nov 2009 12:36 On Thu, 26 Nov 2009 16:24:40 +0100, Uwe Hercksen <hercksen(a)mew.uni-erlangen.de> wrote: > > >Rich Grise schrieb: > >> .-----. .-----. .-----. >> | | | | | | >> problem o---->| ADC |----->| LUT |----->| DAC |-----> answer >> | | | | | | >> '-----' '-----' '-----' > >Hello, > >works well for 8 bit ADCs, also for 10 or 12 bits. With large EPROMs >also for 14 or 16 bits. These days it works for any A/D or D/A you can find.
From: Nico Coesel on 26 Nov 2009 14:39 Richard Rasker <spamtrap(a)linetec.nl> wrote: >Hi all, > >I'm currently working with a mass air flow sensor (a Honeywell AWM3100V, see >http://datasheet.octopart.com/AWM3100V-Honeywell-datasheet-57019.pdf), and >I would like to convert the rather non-linear response curve of this device >into a voltage which bears a linear relationship to the actual air flow. >Ideally, I would like to see the air flow converted in millivolts, so that >it can be fed into a 3.5 digit voltmeter directly. > >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. I use Excel for that purpose. If you choose an X-Y graph you can choose to show a math approximation. -- Failure does not prove something is impossible, failure simply indicates you are not using the right tools... "If it doesn't fit, use a bigger hammer!" --------------------------------------------------------------
From: Spehro Pefhany on 26 Nov 2009 15:10 On Thu, 26 Nov 2009 19:39:20 GMT, nico(a)puntnl.niks (Nico Coesel) wrote: >Richard Rasker <spamtrap(a)linetec.nl> wrote: > >>Hi all, >> >>I'm currently working with a mass air flow sensor (a Honeywell AWM3100V, see >>http://datasheet.octopart.com/AWM3100V-Honeywell-datasheet-57019.pdf), and >>I would like to convert the rather non-linear response curve of this device >>into a voltage which bears a linear relationship to the actual air flow. >>Ideally, I would like to see the air flow converted in millivolts, so that >>it can be fed into a 3.5 digit voltmeter directly. >> >>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. > >I use Excel for that purpose. If you choose an X-Y graph you can >choose to show a math approximation. Matlab etc. (if you have it) or Excel can help with this kind of design. I took the solution to the diffeq for capacitor charging from an intial voltage v(t) = Vf * (1- exp(-(t-a)/tau)) and fit that to the data points to find a, Vf so as to minimize the sum of errors squared from each data point. With Excel 2003+ I think solver is not loaded by default, just a less powerful function that will only change a single variable.
From: Fred Bartoli on 26 Nov 2009 15:09
Fred Bartoli a écrit : > Richard Rasker a écrit : >> Fred Bartoli <" "> wrote: >> >>> Spehro Pefhany a écrit : >>>> On Wed, 25 Nov 2009 16:10:31 +0100, Fred Bartoli <" "> wrote: >>>> >>>>> Hmmm, even simpler... >> >> [snip great, simple antilog circuit] >> >>> Thanks for calculating the values I was to lazy to compute. >>> >>> Now I can reveal the whole world the last simplification bit (ahem) : >>> that is, if you make sure the duty cycle is far enough from 100%, which >>> sure would be with a 100Hz clock, then you can just delete the 1V >>> reference and make it a simple resistor (with a small bypass cap) so >>> that the 6.02V reference with the 24K9/Rsource divider just gives you >>> the wanted 1V at the capacitor top. >>> Adjust the source bypass cap to optimize the circuit behavior WRT the >>> mosfet charges. A value from the low tens of pF to maybe 1nF. >>> >>> Also, please applause the effort I made in disclosing the resistor's >>> secret value (just had a cup of coffee :-). >>> >>> That would be: >>> 24K9 10n >>> ___ || >>> 6.02V >--|___|--+--||---. >>> | || | >>> | === >>> Vout >---------. .--------+ GND >>> | | | >>> .-----. | >>> \+ -/ +-|| >>> comparator \ / ->|| optim. >>> V +-||---. .---||-----. >>> >>> | | | | ___ | >>> | '--------+--|___|---+ >>> >>> .--o--. | 4K96 | >>> GND -|D S Q|-------------' === >>> | | GND >>> 100Hz Clk >---|> -| ___ >>> | R Q|---|___|--+-----> Flow >>> '--o--' | 200mV full scale >>> --- >>> --- >>> | >>> === >>> GND >>> >> >> >> Absolutely wonderful, I'm truly amazed how even elegant, simple >> circuits can >> often be optimized and refined even further! >> Now talking about optimization: from this design, it would appear that >> one >> could swap the MOSFET and 4K96 resistor -- which would mean that the >> internal discharge FET (using an ICM7555) can be used, doing away with an >> external MOSFET altogether. >> > > Of course you're absolutely right (I guess one cup of coffee wasn't > enough this morning). > Now you've really squizzed the last remaining bit of optimization out of > this. > > Argh... And also don't forget to shift the 6.02V by one volt too and make it 7.02V, which makes the resistor's secret value 4K14 -- Thanks, Fred. |