Post by thread:[PICLIST] equations

Are there other equations which can be used to linearize the output of a thermistor? Or are there other ways? If there are, how do Steinhart-Hart equation becomes superior over them? Any website on this? Please help... Thank you very much. Best regards. ____________________________________________________________________ Get free email and a permanent address at http://www.netaddress.com/?N=1 -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email spam_OUTlistservTakeThisOuTmitvma.mit.edu with SET PICList DIGEST in the body

"Jose S. Samonte Jr." <.....dyoweeKILLspam@spam@USA.NET> wrote: Are there other equations which can be used to linearize the output of a thermistor? Or are there other ways? If there are, how do Steinhart-Hart equation becomes superior over them? Any website on this? Please help... Thank you very much. Best regards. ____________________________________________________________________ Get free email and a permanent address at http://www.netaddress.com/?N=1 -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listservKILLspammitvma.mit.edu with SET PICList DIGEST in the body ____________________________________________________________________ Get free email and a permanent address at http://www.netaddress.com/?N=1 -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

At 04:31 PM 2/13/02 +0200, you wrote: >hello everybody , i want to learn how can i solve quadratic equations >by using 16f84. i have 4 equations with 4 unknowns . Which is it? A "quadratic" equation is one in the form a*x^2 + b*x + c = 0, there are two solutions which may be real or complex. You can get the solution from any book x = (- b +/- sqrt( b^2 - 4*a*c))/(2*a) But it sounds like you just have 4 linear equations in 4 unknowns, in which case you can simply use Gaussian elimination or some other textbook method of solving a system of linear equations. Implementing it can be done in C or in assembly with math routines, there's nothing here that isn't straightforward except for the details which will get you every time. ;-) In particular poorly conditioned linear systems can require much more precision in the calculations than what you need in the result. See any textbook on Numerical Analysis for the details, which take a chapter or two. You can imagine the matrix defining your linear system to describe a 4-dimensional elliptical form, the closer to a sphere it looks, (matrix condition number of 1) the better your results will be for a give precision of calculations and coefficients, the more elliptical it looks, the more trouble you are in. The condition number is just the ratio of the largest to smallest eigenvalues. Some useful algorithms are available in the book "Numerical Recipes in C", which you can find on the web. Best regards, Spehro Pefhany --"it's the network..." "The Journey is the reward" .....speffKILLspam.....interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com 9/11 United we Stand -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

hello , for my project i should somehow solve 4 equations with 4 unknowns (something like r^2 -x^2 = (r+A1)^2 - (200-x)^2 eqn.1 r^2 -y^2 = (r+A2)^2 - (200-y)^2 eqn.2 r^2 -z^2 = (r+A3)^2 - (100-z)^2 eqn.3 y^2 + z^2 = r^2 - x^2 eqn.4 i have the values for A1 A2 A3 and asked to find the r,x,y and z values.) i am using 16f84, is 64 bytes ram area is enough and if it is so how to do? i consider using MATLAB but i dont know how to implement MATLAB toolbox to the pic? thanks, -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email EraseMElistservspam_OUTTakeThisOuTmitvma.mit.edu with SET PICList DIGEST in the body

On Sun, Feb 24, 2002 at 11:00:02AM +0200, hasan volkan guducu wrote: > hello , for my project i should somehow solve 4 equations with 4 > unknowns (something like > r^2 -x^2 = (r+A1)^2 - (200-x)^2 eqn.1 > r^2 -y^2 = (r+A2)^2 - (200-y)^2 eqn.2 > r^2 -z^2 = (r+A3)^2 - (100-z)^2 eqn.3 > y^2 + z^2 = r^2 - x^2 eqn.4 > > i have the values for A1 A2 A3 and asked to find the r,x,y and z > values.) > > i am using 16f84, is 64 bytes ram area is enough and if it is so how > to do? > i consider using MATLAB but i dont know how to implement MATLAB > toolbox to the pic? One of the most important computing skills is knowing how to pick the correct tool for a job. A 16F84 has insufficient memory, no floating point, and too limited precision to handle this task. It's like asking how can you complete a 3000 mile trip in 6 hours using a bicycle. The correct answer is that you put you can the bicycle on an airplane and fly to the destination. In this instance the pic is the bicycle. It simply isn't the correct tool for this job. MATLAB is your airplane. It's the right tool. Use it. Good Luck, BAJ -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listservspam_OUTmitvma.mit.edu with SET PICList DIGEST in the body

Hasan Volkan Guducu wrote: >hello , for my project i should somehow solve 4 equations with 4 >unknowns (something like > r^2 -x^2 = (r+A1)^2 - (200-x)^2 eqn.1 >r^2 -y^2 = (r+A2)^2 - (200-y)^2 eqn.2 >r^2 -z^2 = (r+A3)^2 - (100-z)^2 eqn.3 >y^2 + z^2 = r^2 - x^2 eqn.4 > > i have the values for A1 A2 A3 and asked to find the r,x,y and z >values.) > >i am using 16f84, is 64 bytes ram area is enough and if it is so how >to do? > i consider using MATLAB but i dont know how to implement MATLAB >toolbox to the pic? >thanks, Someone else already replied that the pic may not be the right tool for the job, which is certainly good advice as far as it goes. If however you need to solve these equations in real time as part of a pic-based application, then you may be able to do something. As a rule, the more manipulation and simplification you can do by hand, the better off you will be when it comes time to write code. In your case, your equations can be simplified to a single quadratic equation for r, which in general will have two solutions. Are your coefficients (A1 etc.) integers? If so, you may be able to do some further simplifications. If not, you'll need to use some kind of fixed or floating point (or LNS?) representation, and write an optimized routine for the square root. If you know something about the relative sizes of the expected values of the coefficients, you may be able to use Newton's method or similar tricks to further simplify. My main point is this: do as much as you can by hand, to take as much of the computational burden off the pic as you possibly can. If you know anything about the coefficients beforehand, use that information to further simplify. If you take this approach, you may well be able to do it on a pic. Michael V Thank you for reading my little posting. _________________________________________________________________ Chat with friends online, try MSN Messenger: http://messenger.msn.com -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

Hi everyone. Today I've been trying to get an expression for the output voltage for an schematic I've found for a thermistor interface. You can look at the schematic at http://www.olimex.cl/pdf/INEX/ZX-Thermometer.pdf The fact is that for the expression I get, the output voltage is negative. I have used the interface in the real world an the outputs are fine, but for calculations it's not working. I show you the math i'm doing. V(+) = Vdd *10k/(100k+10k) = V(-) // Under the assumption of infinity gain of the opamp V(-) = Vdd - I(R1k)* 1k ==> I(R1k) = (Vdd-V(-))/1k I(R1k) = (V(-) - Vout)/ (12k || RThermistor) // Under the assumption of infinite input impedance After a little algebra i get to Vout = Vdd { 1+ (10k/(100k+10k) -1) * (12k * Rthermistor /(1k * (12k+ Rthermistor))) } But the fact is that this term is negative below 99 celsius degrees, so Vout is negative, and that does not work for a 5V powered opamp, and neither agrees with the datasheet. Is there something wrong with my math or something wrong with the schematic? Any help will be appreciated. -- Lorenzo Luengo Contreras Ingeniero Civil ElectrÃ³nico Laboratorio MIDGEO (LF-106) Facultad de Ciencias FÃsicas y MatemÃ¡ticas Universidad de ConcepciÃ³n ConcepciÃ³n - Chile +56-41-2207400 http://midgeo.udec.cl

> Today I've been trying to get an expression for the output voltage for > an schematic I've found for a thermistor interface. You can look at the > schematic at http://www.olimex.cl/pdf/INEX/ZX-Thermometer.pdf Rushing so incomplete ... . That's going to be easy (as you know) - just a sign error somewhere (as you know). This has been complicated over the normal way of looking at thinbgs by hanging the thermistor reference off Vdd rather than ground. To make it easier on the brain try calulating the drop BELOW Vdd in both cases. V+ = 100/110 of Vdd below Vdd V- = (Vdd-Vout) x 1k/(Rx+1k) below Vdd Rx = Thermistor + 12k in parallel. Solve ... Russell McMaho

Lorenzo Luengo wrote: > Today I've been trying to get an expression for the output voltage > for an schematic I've found for a thermistor interface. You can look > at the schematic at http://www.olimex.cl/pdf/INEX/ZX-Thermometer.pdf After application of a little common sense and consideration of the kinds of things humans are likely to screw up, I think the 100K and 10K resistors are flipped. With 1:10 ratio top to bottom (instead of 10:1 as drawn), the circuit makes sense. The parallel combination of the the 12K resistor and the thermistor causes the output to go to 0 when it is 10K, and higher voltage when it's less. The thermistor would need to go to 60K to make the output negative. Given the values shown (10K and 3.4K for example temperatures), this isn't going to happen over the intended range. That means the 12K resistor could be larger, but it is probably there to shape the curve. The 1K resistor could be made smaller to use more of the available voltage range. ******************************************************************** Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products (978) 742-9014. Gold level PIC consultants since 2000

Thanks Olin!! Flipping the resistors worked great!! It has an offset above the values listed on the datasheet, maybe these are wrong too... :S This datasheet sucks! ;) Well now with modelling done, i just need to make some fitting to convert the A/D readout to a sensible temperature reading. Any advice on this? Which is better?? Solve for T replacing the thermistor equation on this Vout expression? or just a polynomial fit? El 28-09-2010 18:12, Olin Lathrop escribió: {Quote hidden}> Lorenzo Luengo wrote: >> Today I've been trying to get an expression for the output voltage >> for an schematic I've found for a thermistor interface. You can look >> at the schematic at www.olimex.cl/pdf/INEX/ZX-Thermometer.pdf > After application of a little common sense and consideration of the kinds of > things humans are likely to screw up, I think the 100K and 10K resistors are > flipped. With 1:10 ratio top to bottom (instead of 10:1 as drawn), the > circuit makes sense. The parallel combination of the the 12K resistor and > the thermistor causes the output to go to 0 when it is 10K, and higher > voltage when it's less. The thermistor would need to go to 60K to make the > output negative. Given the values shown (10K and 3.4K for example > temperatures), this isn't going to happen over the intended range. That > means the 12K resistor could be larger, but it is probably there to shape > the curve. The 1K resistor could be made smaller to use more of the > available voltage range. > > > ******************************************************************** > Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products > (978) 742-9014. Gold level PIC consultants since 2000. -- Lorenzo Luengo Contreras Ingeniero Civil Electrónico Laboratorio MIDGEO (LF-106) Facultad de Ciencias Físicas y Matemáticas Universidad de Concepción Concepción - Chile +56-41-2207400 http://midgeo.udec.cl

On Wed, 29 Sep 2010 19:12:38 -0400, "Lorenzo Luengo" said: > Thanks Olin!! > Well now with modelling done, i just need to make some fitting to > convert the A/D readout to a sensible temperature reading. > > Any advice on this? Which is better?? Solve for T replacing the > thermistor equation on this Vout expression? or just a polynomial fit? I have always used a lookup table, and then linear interpolation for in-between values. That way you can do the math in a spreadsheet ahead of time and easily change it if you change thermistors or swap other parts around. Cheerful regards, Bob -- http://www.fastmail.fm - Does exactly what it says on the tin

I programmed the chip to output actual ADC readings, either to a PC via, serial port or to a LCD display. I only needed accuracy to a degree or 2. Used freezing and boiling water, and/or next to a known accurate thermometer and the sensor immersed in water over a range of temperatures. The inputted the values in to an Excel spreadsheet, used chart wizard to make a XY chart, add a linear trendline, and use the trendline formula. On 9/29/2010 7:12 PM, Lorenzo Luengo wrote: {Quote hidden}> Thanks Olin!! > > Flipping the resistors worked great!! It has an offset above the values > listed on the datasheet, maybe these are wrong too... :S > > This datasheet sucks! ;) > > Well now with modelling done, i just need to make some fitting to > convert the A/D readout to a sensible temperature reading. > > Any advice on this? Which is better?? Solve for T replacing the > thermistor equation on this Vout expression? or just a polynomial fit? > > El 28-09-2010 18:12, Olin Lathrop escribió: > >> Lorenzo Luengo wrote: >> >>> Today I've been trying to get an expression for the output voltage >>> for an schematic I've found for a thermistor interface. You can look >>> at the schematic at www.olimex.cl/pdf/INEX/ZX-Thermometer.pdf >>> >> After application of a little common sense and consideration of the kinds of >> things humans are likely to screw up, I think the 100K and 10K resistors are >> flipped. With 1:10 ratio top to bottom (instead of 10:1 as drawn), the >> circuit makes sense. The parallel combination of the the 12K resistor and >> the thermistor causes the output to go to 0 when it is 10K, and higher >> voltage when it's less. The thermistor would need to go to 60K to make the >> output negative. Given the values shown (10K and 3.4K for example >> temperatures), this isn't going to happen over the intended range. That >> means the 12K resistor could be larger, but it is probably there to shape >> the curve. The 1K resistor could be made smaller to use more of the >> available voltage range. >> >> >> ******************************************************************** >> Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products >> (978) 742-9014. Gold level PIC consultants since 2000. >> > >

> -----Original Message----- > From: @spam@piclist-bouncesKILLspammit.edu [KILLspampiclist-bouncesKILLspammit.edu] On Behalf > Of Lorenzo Luengo > Sent: 30 September 2010 00:13 > To: RemoveMEpiclistTakeThisOuTmit.edu > Subject: Re: [EE] Help with opamp equations. > > > Thanks Olin!! > > Flipping the resistors worked great!! It has an offset above the values > listed on the datasheet, maybe these are wrong too... :S > > This datasheet sucks! ;) > > Well now with modelling done, i just need to make some fitting to > convert the A/D readout to a sensible temperature reading. > > Any advice on this? Which is better?? Solve for T replacing the > thermistor equation on this Vout expression? or just a polynomial fit? Lookup table with linear interpolation as Bob mentioned. It's faster than calculating polynomials and you can include any arbitrary curve shape. Fill the table with temperature values for ADC counts with a 2^N interval, e.g. for a 10 bit ADC maybe use an ADC interval of 64 giving 16 table values (this is a trade off between table size and maximum error you can tolerate). This makes the linear interpolation part faster since the divide can be performed with shifts. To be honest that circuit is overkill if you already have a micro to perform linearisation; you could simply use a fixed resistor and the thermistor connected as a potential divider (with appropriate decoupling). As long as the impedance is lower than that specified for the A/D input you don't even need to buffer it. By making the fixed resistor value the same as the thermistor resistance in the middle of your required temperature range you already make the curve more linear (it becomes a gentle S shape). Regards Mike ======================================================================= This e-mail is intended for the person it is addressed to only. The information contained in it may be confidential and/or protected by law. If you are not the intended recipient of this message, you must not make any use of this information, or copy or show it to any person. Please contact us immediately to tell us that you have received this e-mail, and return the original to us. Any use, forwarding, printing or copying of this message is strictly prohibited. No part of this message can be considered a request for goods or services. =======================================================================