Basic beginners mistake?

[(*steve*)]

the 2 bias resistors are equal, so the trip point will be when the thermistor resistance matches the current setting of the potentiometer.

Since it's a 15k thermistor, the resistance will vary around this point. I would expect that with a 100k pot, you're using only about 10% of its travel making it quite hard to adjust.

I would measure the resistance of the thermistor when at the temperature you wish to trip at (or pretty close to it).

If that turns out to be 10k, then replace the pot with a 4k7 resistor and a 10k pot. This will give you easier adjustment around the point that you want it to trigger.

Note that because this is a Schmitt trigger, if the temperature rises to the point to trigger the relay, it will need to fall substantially to turn it off again. Increasing the 270k resistor will reduce this effect (and you've already been advised to increase this to 1M).

Because this is a very simple circuit, it is difficult to set the spread of temperatures exactly, nor to make it consistent across all temperatures you might set it to. You could experiment with different values to get the spread you require. Again, measuring the temperature of the thermistor at the points you wish it to turn on and off may help us in suggesting a value that will give an approximate temperature range)

Using a 10 turn (or more) pot will also be effective in making the adjustment easier, however making sure the pot has the appropriate value and placing a sensible series resistor in place would still be advisable.

If you can find the datasheet on this thermistor, and tell us what temperature you want as the trip point (or what range of temperatures you want it adjustable over) then we can suggest something for you.

 

[Bruno]

Thanks for all that, there is only one place nearby that might have multi turn potentiometers so was going to try them at lunch time, otherwise ebay looks good (£2 for 4, inc. post).
Thermistor is a 15 K @ 25 C, I would like the circuit to switch on around 100 - 110 C and off again about 10 deg below the on point if poss, it still has about a 50 degree spread with the 1 M ohm resistor, would going higher on ths resistor help?
I have thought about trying other circuits based on different IC's etc. but none of the ones I have found have any info about how close the on/off points can be set to.

 

[Resqueline]

That much spread with that high feedback resistance seems wrong to me, you quite likely still have a lot of leakage currents on the circuit board.
If you're unable to remove the contaminants responsible for this then you should seek to use as low resistances in the circuit as possible.
You can go as low as 220 ohms on the potential divider resistors before power dissipation starts becoming an issue.
If you use no feedback resistor (= infinite resistance) then you should get no spread at all. Lower positive feedback resistance = more spread. Use a pot?

 

[duke37]

You appear to have a contaminated board so increasing the 1M resistor may not do much. It was suggested on a previous post that an alternative would be to reduce the 10k resistors. Leakage through contaminants would then have a smaller effect.

Reduce the 10k resistors to 1k, this should reduce the deadband by a factor of 10.

Whilst a 10 turn potentiometer will ease setting, it will not affect the operation of the circuit.
To get an estimate of the dead band, you need to know the change of resistance of the thermistor with temperature. Have you got this data? How is your calculus? The voltage difference due to the feedback resistor can be easily calculated. Other ICs will not affect the principles. If I were doing this, since the cost of components is low and the circuit is simple, I would start again, using the deadbug technique so that there will be no board leakage

 

[Bruno]

Got a 10K multi-turn pot and put it in series with a 4K7 resistor and - problem! the circuit has reversed it's operation again, it is back to switching off as the thermistor resistance increases, done lots of cleaning the back of the board but it has not fixed that though there were no odd resistance readings and the voltages at pins 2 & 3 look OK to me, very near 6 volts so have not changed any other resistors.
I started to check resistance against temp on the thermistor and I suspect I may have a faulty one, resistance change on temp rise seemed OK but when I took it off the heat source the resistance did not start to increase for quite some time then it shot up, I got a datasheet off the internet and while I haven't done any exact plots it does not appear to follow the graphs given on the sheet so will get another one tomorrow.
Calculus! I had enough trouble with algebra, anything after that was beyond me. I did a search for "deadbug technique" - would prototyping board be OK, I have noticed it in the shop where I have been buying a few things recently and I think I have seen some in my friends dads workshop.

 

[duke37]

The deadbug technique uses a sheet of copper (PCB) or tinplate and glueing the integrated circuit on it with its legs in the air. Other components are soldered in free air. You may wish to make a pad or two out of plain pcb.
The result is not pretty in my case!
I have made a vibrator replacement with a cmos IC and a couple of fets with suppresion components, Will try to attach a picture.

 

[Bruno]

I got these results from the new thermistor, 15.8 K @ 16 C, 900 ohm @ 90 C & 580 ohm @ 100 C and resistance change on cooling is a lot different to the other one I had, it reacts fairly quickly.
Been thinking about the deadbug method, it seems to me it needs even more planning than the stripboard, might have a go at putting the circuit on this prototyping board.

 

[(*steve*)]

The resistance in series with the thermistor needs to be about 740 ohms.

If you get a 1k trimpot then you should be able to set this value fairly easily (measure it with a multimeter).

Then see what the hysteresis is. You at what temperature does it turn on, and then what temperature does it turn off again. If these are not symmetrical around 95C then change the resistance slightly. Then change the resistor providing +ve feedback to get the switching points as close to 90C and 100C as you wish.

By my calculation, if you have 10k resistors in your reference divider, then the +ve feedback resistor should be 50k

 

[duke37]

Two comments.
1. The measuring of the resistance will affect the resistance and the dissipation in service will give an offset.
2. Start without the feedback resistor to give instant switching and then add the resistor and decrease its value to give the deadband required.

I used to service talking books for the blind (tape players). Coffee on the pcb could cause problems but the worst was salty tomato soup. I managed to clean them up but there were no high value resistors and no ICs so the board could be cleaned on both sides. It doesn't matter what system you use but keep salty finger marks off.

 

[Bruno]

Haven't been able to spend much time on it this weekend but have got the circuit, with the suggested modifications set up on this prototype board and it seemed to be working well in a quick test.