Thermistor with op-amp

[pedot]

i`ve a problem that need your help to solve it..
here i attach the circuit..
fro this circuit,can u explain to me how it work..
maybe you can provide some calculation that tell why that we need to use 68k for r1,10k for r2 and etc..
also the relationship with the thermistor value(100k @ 25)
let me  tell you a little bit that i understand from this circuit..
The thermistor has a negative temperature coefficient (NTC) and when the temperature is high its resistance will be low. This causes the voltage at pin 2 of IC1 to be low. If pin 2 is below the set-point of VR1, the output at pin 6 will be high..
the output of the circuit will be connect to aalog to digital converter(ADC)..after the outpu has been converted i will use 8085 to program it to work..
any suggestion of application that i may use?

[MP]

If you swap the 68K resistor and the thermistor, the circuit will react in the opposite manner. After doing this, you will have to re-calculate the 68K to get the correct scaling on your divider.

MP

[pedot]

i dont have plan to swap tha thermistor with the 68k resistor...
all i need is the operational of my circuit...
i take it from internet...
i want to know why all the resistor consist that value and may need some calculation and formula...
i have to explain all that on my report project..

[ante]

Hi pedot,

If you have someone else doing this work, you will not learn anything from it! Under different circumstances it would be correct to help you, but now it’s not! Besides it would be a bit of cheating too, don’t you think? 8)

[Kevin Weddle]

If you ask me, I would say the inputs are biased around 6 or 7 volts and the gain is reduced to some unknown value but not open loop.The gain is still very high so any difference will send the output away from the 6 volt midpoint. The signal is inverted. There is not much positive feedback and the thermistor needs to change resistance enough to generate a signal in the mV range under regular gain, but in this case it will be much smaller and that's okay. You will really need to tweak the 50K to get your output at that high of gain. But you can realize a linear amplification on the output.

[MP]

Hey, you asked how to get around the negative coefficient and I gave you the easiest solution. The other solution is to do this in your micro program, which is how a datalogger would do it. How are you at Math?
In complete agreement with ante, I will also not do your homework for you. However, I do not mind giving you this hint to get you looking in the right direction.

Kevin: You will need to find the datasheet for this thermistor to calculate other values in the voltage divider. We do not know how linear it is and we only know the resistance at 25C. The data sheet will have a table of resistances for different voltages.

Pedot: How about posting the data sheet?

MP

[audioguru]

Kevin,
The opamp's output is saturated most of the time so it doesn't have any gain. The output doesn't change until the thermistor's voltage divider provides a voltage that reaches the Schmitt trigger threshold of the opamp. Then the opamp switches as fast as it can due to positive feedback and saturates in the opposite direction.
The pot adjusts the center-point of the two Schmitt trigger voltage levels and therefore adjusts the temperature at which switching will occur.
Positive feedback is provided by the 1M resistor for the Schmitt trigger action and its value is high to keep the Schmitt trigger threshold voltages close together.
Schmitt trigger action is provided to give "snap action" to the opamp's switching so it doesn't amplify its own noise nor oscillate at its threshold voltage.

[izach]

one thing i noticed about this circuit is the fact that the pot was placed on the usually fixed potential divider there for  no stable reference voltage would appear at the input.

usually one would have a stable reference voltage to work with then adjust the temprature dependant voltage this would cause a more linear under temprature
switch.

the two potential divders ie the temperature_dependant voltage divider and the reference voltage divder are connected as a wheatstone bridge

any coments on the reason for the pot in the reference voltage??

[audioguru]

any coments on the reason for the pot in the reference voltage??

I think it allows the 68k resistor to reduce the current through the thermistor so it doesn't heat-up.
Changing the reference voltage for the opamp with a variable voltage divider from the supply still allows the temperature threshold to be adjusted or to allow for differences between different thermistors. ;D

[MP]

The pot in the voltage divider connected to the non-inverting input has nothing to do with protecting the thermistor in the voltage divider connected to the inverting input from heat, due to excessive current. The pot allows for adjustment of the set point.

MP

[audioguru]

If the pot was in series with the thermister, instead of being as part of the reference voltage, if it was adjusted to a low resistance then the thermistor would heat-up.
It is best to keep the pot away from the thermistor. ;D

[izach]

here is how i see it but still does not explain the pot in the 'wrong' place

the op amp is connected as an open-loop differential amplifier or voltage comparator, and it s action is such that its output is driven to positive saturation if its negative terminal is  more that a few hundred microvolts negative to the positve terminal and is driven to negative saturation if its negative terminal is more than a few hundred microvolts positive to the positve terminal.

with this in mind the thermistor is a negative tempreture-coefficient device so its resistance falls as the tempreture rises. in practice one would see the pot in the voltage divider and not the stable reference voltage. and thus the pot would be adjusted to balance the bridge, in this conditon there would be zero volts difference between the positve and negative terminals of the op-amp.

consequently when the tempreture rises above the trip level the bridge goes out of balance in such a way that the negative terminal of the op-amp goes negative to the positive terminal and the op-amp is then driven to positve saturation.

therefor i cant see the thermistor warming up as there would allways be a resistance in the voltage divder consisting of the 'balance resistance' pot and the ntc.

any comments ?

[MP]

You would never use a pot in the divider with the thermistor. Even the resistor in series with the thermistor must be a 0.01% tolerance or equivalent to get good results from such a circuit.

If a pot were added to the thermistor divider network, the circuit would suffer in accuracy. The pot in this circuit is used to trim the threshold where the user wants to trigger the comparator. It is not in the wrong place.

MP