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Fan control, indicating, proportional

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Whoops!  {Up too late.  xP }  Yes, the tempco of a silicon junction is -2.1mV per degree C, NOT 20mV as I stated above, however the mid point of the zero and span pots is 0.625V and the RHI and RLO are about 20mV (10 degrees) above and below, giving a range of about 20 degrees, depending on how you set it up.

Yes the LM3914 gets warm but it's rated for over a watt (1.36W strictly), and when it's at its worst case is when the cooling fan is going hardest.

50mV/-2mV/deg = 25 degree span.

It is also quite possible to re-arrange this sensing input to use a transistor set up for a DC gain of (say) ten and use the transistor to amplify its own change of Vbe by taking the signal off at the collector when the tempco will then be 20mV/degC, (or by whatever you set the gain to).  This may need some capacitive bypassing to stop it amplifying stray AC noise as well.

The main problem with using a sensor with a small tempco is that any stray noise turns up as "chatter" at the control output and is normally met by using a Schmitt trigger with some backlash, but this tends to widen the control range.  In the application above there is some chatter apparent as the control moves between steps, but it is minor and hasn't been a problem.

Really the main problem with this circuit is organising a suitable supply where the only available voltages are the output stage supply rails, perhaps 35 volts, which is a bit too high and has to be limited by a three-pin regulator.  Where I have applied this I have used a power resistor between the supply rail and the input of the three-pin regulator to drop the bulk of the supply voltage before it gets to the regulator input (a so-called "economising" resistor).

In the case of a LM7812 12 volt regulator the minimum input voltage needs to be about 4 volts above the output, so the resistor value is chosen to drop the available supply to 12+4=16 volts at the maximum fan current, 120mA.  For a 35 volt rail this is 35-16 = 19 volts, and at 120mA requires R=E/I = 19/0.12 = 158.33 or 150 ohms at P=E*I = 19*0.12 = 2.28 watts, say a 5 watt resistor.

J M Fahey:
Cool and accurate Math, plus *practical* suggestions, which makes this post even more useful.
Thanks  :dbtu:

Very good answers.  I'm thinking about building one of these to control a PC van in a vent that goes from my laundry room to the living room to help move airflow from the wood stove.  Probably something like 22C it turns on at level 1, and level 10 at 27C.  The whole circuit will be turned on/off by a thermostat in the laundry room.  If it is below 18C it will run, otherwise it's off.  That way it is only circulating when the laundry room needs the heat AND when the living room is hot from the stove running.

Well there's an application I never thought of.  :cheesy:  Just keep in mind that when using a brushless fan that they seem to need a bit of current to get moving initially, but once running they will go down real slow before stalling.

A couple of things to note from the LM3914 data sheet are that operation in bar mode may start to get a bit erratic below 200mV RHI to RLO and while this circuit goes well under that no problems have been encountered, however closing the span even more may encounter non-linearity.

The other point is that having low values for the span and zero pots gives a significant improvement in the internal temperature coefficient of the resistive divider, and this will be come progressively more important as the working span is reduced.

Be interesting to hear how it goes.

Interesting about wanting at least 200mv differential for bar graph mode.  I suppose I could use 10 diodes (they're cheap enough) in series, or I could make a small opamp circuit to boost it. 

Even better, I could set it so that level 1 is 20C and level 10 is 60C (really this would be more ideal as I could take a temperature near the stove as opposed to air temp of the room).  This would give me 40C differential, x 2.1mv per degree C = 84mv swing.  I could put 2-3 in series and have plenty.

Some ideas/questions.

First, please update your original post to show the -2.1mv per diode instead of -20mv.

Second, wouldn't it be pushing ~4.2w at full blow?  Most PC fans I see are 0.15 amps.  LEDs are normally 20ma each, so 200ma for 10.  0.15A + 0.2A = 0.35A x 12v = 4.2w.  If the chip is only good for 1.36w, how's this work without blowing it up?

Or is it that it would push a maximum amps of what goes through the LEDs, IE 0.2A x 12v = 2.4w (still over the chip's rating) and then if that is that case, wouldn't that mean the LEDs and fan split the voltage?  So the LEDs would get say 2.4v and the fan only 9.6v?


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