And you really think that's more accurate than using the processors A/D converter to measure an analog voltage, such as from the bridge circuit?

I really don't know unless one analyzes the errors involved in the two approaches.
What I do know is I can extend the resolution (number of bits) by counting for longer.
By counting over integral numbers of 50/60Hz AC mains cycles, one can reduce line frequency noise, in effect taking averages over a given period.

 

What I do know is I can extend the resolution (number of bits) by counting for longer.

True.
But I was talking about accuracy, not resolution.
You can also increase the resolution of an A/D converter by averaging the readings.

 

I think temperature to frequency conversion has several error sources that other techniques do not have, such as the self-heating caused by the capacitor discharging through the thermistor.

The thing I like about thermistors is that with a decent 0.1% bias resistor (and a higher 25C resistance), they really are calibration-less in many applications. I worked up a variation of Wally's circuit back in the mid 80's. over a +25 to +70 range the peak error was <0.6% calculated, and under 2% with real-world resistors. Not bad for no cal.

ak

 

Let us take a popular MCU platform, Atmel ATmega328 in Arduino.
It has an internal 10-bit ADC. Hence resolution is 1:1024 which is 0.1%.

If you use a crystal based timer module, you are starting out with a basic uncertainty of better than 10 ppm or 0.001%. That is two orders of magnitude improvement. I understand that you do have to account for errors in supply voltage, capacitance temperature coefficient, logic level transitions etc. That analysis is too complex for me to perform. The alternative is to test the two options and collect actual comparative data.

I have designed and built meteorological stations using this technique and have achieved 0.1°C accuracy over full operating range.

 

If you use a crystal based timer module...

Oh...you need a crystal, too? You failed to mention that in your BOM above.

 

Hi all,
I want to design a circuit that at 20 degrees Celsius the output voltage is 0V.
I then need to determine the voltage for temperatures 21-30 degrees Celsius which have given resistances as below:

172234

To get 0V you for a particular temperature you have to use a bridge circuit as Mr Chips says (or have a negative voltage rail). The simplest readout however is as you showed in your own schematic, but then you have to live with a voltage offset at 20C (but you can simply subtract this from all the readings).

 

Hi All,

What I'm struggling to understand and feel there is something incorrect is that, I have simulated a voltage divider circuit with a resistor and thermistor @ 2814 ohms to give a temp of 20 degrees and a voltage of 0V.
When the temperature rises in increments of 1 up to 30 degrees Celsius my output voltage goes negative which does not seem correct. I am specifically using the Omega 44004 thermistor.

 

Hi NMC,
Well your thermistor resistance drops with increase in temperature, so in your potential divider circuit, the voltage across it will also decrease. So if you artificially set your 0V reference at 20C, then yes, you will appear to get a negative voltage.. but this is relative.
Your potential divider divides up your battery voltage, so you can only ever measure a fraction of this, it can never be zero (unless your thermistor is a perfect conductor.)

 

Hi All,

What I'm struggling to understand and feel there is something incorrect is that, I have simulated a voltage divider circuit with a resistor and thermistor @ 2814 ohms to give a temp of 20 degrees and a voltage of 0V.
When the temperature rises in increments of 1 up to 30 degrees Celsius my output voltage goes negative which does not seem correct. I am specifically using the Omega 44004 thermistor.

The behavior you describe is inherent in a device with a negative temperature coefficient. Increasing temperature implies decreasing resistance. The other huge disadvantage of the voltage divider is that the load impedance will change the behavior of the divider in significant and possibly unexpected ways. You might want to simulate that while you're at it. Then you should go back and try harder to understand the bridge circuit.

 

When the temperature rises in increments of 1 up to 30 degrees Celsius my output voltage goes negative which does not seem correct.

Reverse the positions of the thermistor and its bias resistor.

This is easier to discuss with a schematic that has reference designators for each component.

ak

 

To get 0V you for a particular temperature you have to use a bridge circuit as Mr Chips says (or have a negative voltage rail). The simplest readout however is as you showed in your own schematic, but then you have to live with a voltage offset at 20C (but you can simply subtract this from all the readings).

Your attachment is going to error when I try to open it pmd34?

 

Sorry I pasted a quote in and it attached an unrelated image into it!

 

Still cant get 0V at 20 degrees Celsius with the thermistor resistance 2814 ohms.

 

you have to measure your voltage between point 2 and point 3

 

The circuit in post #10 is about the simplest and cheapest that will provide the really accurate results that you want. For a whole lot more money and MUCH MORE EFFORT you can do it other ways and probably be as accurate. But with the information that you gave us that post from Crutchow is quite adequate. It is very close to what I was going to suggest.

 

The circuit in post #10 will provide what you are asking for and it is the simplest and cheapest approach. There are other more complex and expensivew ways to get the results but why waste effort and money that way. Crutschow gave a very good circuit in post #10.

 

Hi all,

Thanks for the replies.
Apologies it is an Omega 44004.

I want to design a circuit that at 20 degrees Celsius the output voltage is 0V.
I then need to determine the voltage for temperatures 21-30 degrees Celsius which have given resistances as below:

20 degrees C = 2814 Ohms
21 = 2690
22 = 2572
23 = 2460
24 = 2354
25 = 2252
26 = 2156
27 = 2064
28 = 1977
29 = 1894
30 = 1815

Below is a circuit which I tried to simulate -

View attachment 172234

Is this what your trying to do?

eT

 

Here's the circuit redrawn to make the bridge aspect of it more appaent:

Notice that, instead of reading the bridge's unbalanced voltage across the center of the bridge, it keeps the bridge voltage at zero by changing the voltage at the bottom of one of the legs.
This gives an output voltage that's double what the difference would be across the unbalanced bridge.

 

Here's the circuit redrawn to make the bridge aspect of it more appaent:

Notice that, instead of reading the bridge's unbalanced voltage across the center of the bridge, it keeps the bridge voltage at zero by changing the voltage at the bottom of one of the legs.
This gives an output voltage that's double what the difference would be across the unbalanced bridge.

View attachment 172472

Notice an even more important thing, which is that by keeping the sensor current constant the output versus temperature becomes linear, not just "close to linear" like some other circuits do. I am impressed by that, by the way. My method of providing a constant current uses fewer parts, but it is harder to adjust, and a bit less efficient.

 

Notice an even more important thing, which is that by keeping the sensor current constant the output versus temperature becomes linear, not just "close to linear" like some other circuits do.

Disagree. The blue line in post #38 is not straight.

Since the thermistor temp-vs-ohms relationship is more complex than a simple logarithm or exponential, no circuit based on only gain and offset functions can produce a linear volts vs temp output. Granted, some circuits are better than others, and over such a small temperature span several of them look linear, but they are not. Here is why:

https://en.wikipedia.org/wiki/Steinhart–Hart_equation

ak