operational-amplifierthermistor
Given the circuit layout here with a 9V input and the 10k ohm thermistor specs given, what at what temp would the green LED light and at what temp would the red LED light?
I am expecting green to come on at about 17.5C and then red to come on at about 37C. Green and red should never be on at the same time and nothing should be on at under about 17.5C.
Is this correct?
Would it work without change for a 3V input? If not, what would need to change?
The resistance of a thermistor changes with temperature. In the datasheet for the thermistor there will be a graph that shows the resistance over some temperature range. If you cannot get the datasheet for the termistor, don't buy that thermistor. Digikey has datasheets for everything they sell, but eBay can be hit or miss.
Sometimes the resistance goes up as the temp increases. This is called a Positive Temperature Coefficient, or PTC thermistor. Other times the resistance decreases as the temp increases. This is a Negative Temperature Coefficient, or NTC thermistor.
When a thermistor is advertised as being a 100K thermistor, that means that it has a 100K ohm resistance at some "normal" temperature-- usually room temperature. The datasheet will tell you what the "normal" temp is. Another reason why you should not buy one without the datasheet.
Next is the problem of figuring out how to control the RGB LED with the thermistor. Judging from your question, I am assuming that you're new to electronics. This is unfortunate, as the problem of controlling the LED can be difficult. Especially if you want to control the color based on the temp.
One thing that you likely do NOT want to do is to use the thermistor as the current limiting resistor that is normally found with LEDs. There are several reasons for this. #1 is that the resistance of most thermistors is too high to be practical as a current limiting resistor. Not impossible, just not usually practical. #2 is that thermistors often cannot handle a lot of current flowing through them that is required for an LED. Again, not impossible but you must carefully choose the thermistor. #3 The change in resistance over the required temperature range is probably not right to get the desired look you seek. And the big one is: #4 If you run a lot of current through the thermistor you will actually heat up the thermistor, thus corrupting your temperature measurement.
There are several things you might want to control based on the temp. The main ones are the intensity of the LED and the color of the RGB led. Controlling the color is probably what you want, but it is the most difficult. The reason why it is the most difficult is because the conversion from temp to R, G, and B is not linear (or even monotonic).
The easiest way to do it is with a microcontroller. Use the thermistor as one resistor in a voltage divider and feed that signal to the ADC input of an MCU. Then some software in the MCU reads the temperature and figures out the proper duty cycle for PWM-ing the R, G, and B LED's. I know, this isn't easy-- but it is the easiest way. There are other methods to do this but they are much harder and/or do not provide the best visual appearance.
Here is what I was thinking before:
This is a basic inverting opamp circuit. L1 and C2 are only there to filter the power supply to the opamp a little. They are not strictly required, but might reduce noise depending on what else is going on around this circuit and how clean the 3.3 V supply is.
R3 and R4 form a voltage divider that holds the + input at a fixed 500 mV. C1 removes noise that might be coming from the 3.3 V supply.
The real meat of this circuit is R2, the feedback resistor, and R1, the thermistor. The way the feedback is arranged, the opamp will do whatever it takes to keep its - input the same as the 500 mV on its + input. That means there will be a constant 500 mV accross the thermistor. The current thru the thermistor is then a function of its temperature. This current only comes from R2, so the voltage accross R2 is inversely proportional to the resistance of the thermistor.
This circuit won't utilize the full A/D input range, but will do significantly better than just a bare resistive divider as was discussed in your previous question. At 140 Ω, the current will be 3.57 mA, which produces 1.68 V accross R2. This voltage on R2 is added to the 500 mV bias, so the minimum OUT value is 2.18 V. At 98 Ω, OUT is 2.90 V, for a total range of 719 mV. That would be 223 counts of your A/D, which is more than typical thermistor accuracy can support.
You can get a wider output range by using a lower bias voltage and making R2 bigger accordingly. The value of R2 is directly proportional to the gain of this circuit. I showed 500 mV as a example because it seemed like the maximum sufficient value, but 250 mV would give you more than twice the A/D range. I wouldn't go much lower than that since other errors and sources of noise would start to get significant.
One advantage of this circuit is that it keeps a low voltage on the thermistor, which makes self-heating negligeable. At the worst case, the 500 mV is applied to 98 Ω, which causes a dissipation of only 2.6 mW. If you use 250 mV bias it goes down by a factor of 4 to 640 µW. Unless you have a very unusual situation, that amount of self-heating should be irrelevant.
One issue to keep in mind is that the output is dependent on the 3.3 V supply level. However, since you specifically mentioned 3.3 V, it sounds like it is produced by a regulator, so should be fine.
If you only have 3.3 V power available, you need a rail-to-rail opamp as I show. A TL081, for example, would not work here without both higher and lower supply voltages.
Best Answer
The idea with a schematic is to show the schema of the design. It should read from left to right and, where possible, with the higher voltages at the top. This way signals will generally progress from left to right and current from top to bottom.
simulate this circuit – Schematic created using CircuitLab
Figure 1. The OP's schematic redrawn.
It's also clear that R2 and R8 form a second potential divider. The voltage at B will be given by \$ V_B = \frac {R8}{R8+R2} \$. Therefore the switching point will be when \$ V_A = V_B \$ which occurs when \$ \frac {R8}{R8+R2} = \frac 1 2\$. You can work that out.
You have a second reference point for C so you can work out the switching threshold for that too.
I think you are correct for your 37°C calculation. I didn't work out the other.
The circuit is a little unusual in that D2 gets its current from OA2 and sinks it into OA1. This has the problem that when OA1 is output high and OA2 is output low that D2 is reverse biased. It's a good idea to keep the LED reverse voltage to less than 5 V or it may be destroyed. Try grounding it in the same way as D1 instead.
simulate this circuit
Figure 2. Using transistor drivers to work on 3 V supply.