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'[EE]: Are there other equations...?'
2001\04\20@163722 by Jose S. Samonte Jr.

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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.

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2001\04\23@184959 by Jose S. Samonte Jr.

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"Jose S. Samonte Jr." <.....dyoweeKILLspamspam@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.

____________________________________________________________________
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'[PICLIST] solving equations'
2002\02\13@094334 by hasan volkan guducu
picon face
hello everybody , i want to learn how can i solve quadratic equations
by using 16f84. i have 4 equations with 4 unknowns .

anyone having information, please let me know.
thanks.

volkan

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2002\02\13@103313 by Mahmood Elnasser

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face
Hello
Are you using c compiler for the pic or assembly ?
If You are using assembly language get a c compiler.
If u are using c compiler then the problem is solved.


{Original Message removed}

2002\02\13@113714 by Spehro Pefhany

picon face
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"
.....speffKILLspamspam.....interlog.com             Info for manufacturers: http://www.trexon.com
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2002\02\13@161831 by uter van ooijen & floortje hanneman

picon face
> hello everybody , i want to learn how can i solve quadratic equations
> by using 16f84. i have 4 equations with 4 unknowns .

By hand and put the solution in a RETLW table.

Are you serious? Wanna use floating point math?

Wouter

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'[PICLIST] equations'
2002\02\24@040128 by hasan volkan guducu

picon face
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,

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2002\02\24@084618 by Byron A Jeff

face picon face
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

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2002\02\26@131644 by Michael Vinson

picon face
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.


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'[EE] Help with opamp equations.'
2010\09\28@163934 by Lorenzo Luengo
flavicon
face

 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 Electrónico
Laboratorio MIDGEO (LF-106)
Facultad de Ciencias Físicas y Matemáticas
Universidad de Concepción
Concepción - Chile
+56-41-2207400
http://midgeo.udec.cl

2010\09\28@180338 by RussellMc

face picon face
> 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

2010\09\28@181208 by Olin Lathrop
face picon face
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

2010\09\29@191248 by Lorenzo Luengo

flavicon
face
 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 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

2010\09\29@192626 by Bob Blick

face
flavicon
face

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

2010\09\29@193639 by Carl Denk

flavicon
face
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}

>

2010\09\30@062422 by Michael Rigby-Jones

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face


> -----Original Message-----
> From: @spam@piclist-bouncesKILLspamspammit.edu [KILLspampiclist-bouncesKILLspamspammit.edu] On
Behalf
> Of Lorenzo Luengo
> Sent: 30 September 2010 00:13
> To: RemoveMEpiclistTakeThisOuTspammit.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

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