# Electronics Temp Controled Electrobus

Discussion in 'Modding' started by Bhozar, 10 Oct 2002.

1. ### BhozarWhat's a Dremel?

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I was reading this article, http://bit-tech.net/article/52/5 It mentions an electrobus which can be used with a temperature sensor to vary the speed of the fans. It says that it will be discussed in a latter article on how to do it. Does anyone know if it was ever writen or if it will ever be posted?

2. ### linearMinimodder

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Hi Bhozar,

No article yet, but it's a pretty straightforward adaptation of that circuit. It's just that the math gets kind of obnoxious.

One of the very first mods I ever did was to design that exact circuit. The writeup is lost in the mists of antiquity, but I'll sum it up for you.

Take a deep breath....

There are two kinds of thermistors, PTC and NTC. That's positive/negative temperature coefficient. With PTC, temp increase makes resistance increase. With NTC, you can guess what happens.

The biggest thing to worry about with a thermistor is self-heating. a thermistor is a resistor, and if you pass a current through it, it gets warm (that's the definition of resistor). Fortunately, the control current for a voltage regulator is really small (microamps range) so this crazy stunt actually works.

The arrangement of resistors R2 and R1 define Vout:
Vout = 1.25V * ( 1 + R2/R1)

the customary arrangement is to vary R2, which is what the diagram you linked shows (actually, R2 in the above equation is represented by the series combination of the pot and the fixed resistor--this sets a "floor" output voltage greater than 0). But varying R1 can change the output as well, that should be evident from the equation above.

So case A is you have a PTC thermistor. It needs to go into the circuit in place of the pot+fixed resistor series combination (thermistors don't go to zero like a pot).

Case B is that you have an NTC thermistor. It goes in place of R1 to get the same positive change in Vout for a positive change in temperature (cool, huh?).

In either case, you arrange your fixed resistances in order to get the setpoints you want.

For instance, lets say at
20&deg;C you want 7V
35&deg;C you want 12V (or as close as you can get--more on that in a minute)

you get 7V output by having the ratio of R2/R1 be approximately 4.6 (That's (7/1.25) - 1 if you're algebra impaired)

you get 12V (theoretical--more in a minute) output by having R2/R1 = 8.6.

You should be able to calculate the resistance of the thermistor at any given temperature, or have a table provided by the supplier

Now we have a problem--there's two equations and only one independent variable. The values of the thermistor at givent temperatures are constants (not variables for our purposes). So we need to introduce a fixed resistance in series with the thermistor and use it as our second independant variable. Then we solve the system of two equations in two unknowns by normal means, and values for R1 and R2 will fall out.

Let's say you got the 10k NTC thermistor off the shelf at Radio S*ack. I don't have the table of values handy, but for illustrative purposes, let's take its resistance at 20&deg;C as 11k and its resistance at 35&deg;C as 5.5k. (Numbers I made up entirely)

we can substitute directly into the Vout expression:
Vout = 1.25 * (1 + R2/R1)
but let's let Rs be the fixed resistor in series with the thermistor--together they make up R1 in the Vout expression since this is the NTC case.

In general:
Vout = 1.25 * (1 +( R2 / (Rs + Rtherm) ) )
Where Rtherm will be a constant at a given temperature.

All right, let's get some numbers in there:
7 = 1.25 * (1 + ( R2 / (Rs + 11 000) )
12 = 1.25 * (1 + ( R2 / (Rs + 5 500) )

/me waves hands at the algebra...

4.6 * (Rs + 11 000) = R2
8.6 * (Rs + 5 500) = R2

You can see from this form that these are proper linear equations, this is a system of 2 equations in two unknowns, readily solved. So let's do it...

by substitution:
4.6 * (Rs + 11 000) = 8.6 * (Rs + 5 500)
removing parens:
4.6*Rs + 50 600 = 8.6*Rs + 47 300
4*Rs = 3 300
Rs = 825

then again by substitution:
R2 = 8.6 * (825 + 5 500) = 54 400

And voila, you have your values. In the PTC case, you just put a series resistance in series with R2 instead of R1 and solve by the same technique.

Hope that helped.

Now, here's the thing about the maximum Vout. It will be the lesser of Vout as given by the expression, or Vin - Vdrop, where Vdrop is dependent on exactly which regulator you use. Vdrop can be in the range from 1.5 to 0.5 V. You will never get 12V out of an adjustable regulator supplied from 12V. It can't happen. That's what my notes above were trying to get to.

When I built htis, I used a LM317 regulator (since that's what I had) and was terribly dissapointed to find out the above realization the hard way. at 10.5V, my crappy 80mm fans weren't moving a hell of a lot of air. I never put this circuit in effect because of that. For you, if you have 120mm fans especially, this may work out well.

3. ### BhozarWhat's a Dremel?

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thanks for the info! Looks like it would have took you ages to type it out.

I had just come back to the forums to say I am going to build the PWM Fan Controller instead. Just ordered 10 MIC502BN's. Found out that they are now a discontinued item from rapid. They only have 17 left after I ordered mine.

To me the PWM with temperature controll would be a better option, correct me if I am wrong as my understanding is quite basic of electronics.

Hopefully the info you have posted will be useful to someone else, possibly even me if I cant get on with the PWM.

4. ### linearMinimodder

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Let me do this, I'll move this int the electronics forum, it fits better there anyhow.

Use the search for "MIC502" in the electronics forum, I think there were some substiture ICs discussed in one thread that are pin-compatible.

5. ### cpemmaEcky thump

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Going on from linear's maths...

The resistance at any known temperature can be forecast from the thermistor specification. Usually, suppliers quote the resistance at room temperature (25degC) and a B value.

Then resistance at temperature T degC is :

R@T = R@25 * e^(B/(T + 273) - B/(25 + 273))

Where e = 2.7183 and temperatures are in Celsius (adding the 273 converts them to degrees Kelvin for the formula to work)

All very complex, but there's a spreadsheet
here that you can plug your own values in and see what happens.

I used a L200 regulator for a thermal control, as it works well at much lower thermistor currents than a 317T. Details here. It has the same 10-10.5v limit as the 317T, so an improvement would be a low-dropout regulator, or just control a marginally more powerful fan than you need.

OTOH, many people use commercial or home-built 317-based fan controllers and live with the output limit.

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