How to find Vs and R

[polarKaung]

The circuit can be seen in the following photo, it is just a voltage divider. Rth is a thermistor.
The question is, I want the thermistor to detect from 40 degrees to 80 degrees, and according to the datasheet the resistance decreases from 8k to 4k.
So, the resistance at 40 degrees is 8k, and the resistance at 80 degrees is 4k. I want Vout to vary from 1V to 5V(it is an odd choice).
I need to find appropriate Vs and R for that. I tried, but ended up with negative resistance and voltages less than 5V. So here I am.😂. Thank you

[bob91343]

Watch out for self heating.

[magic]

I tried, but ended up with negative resistance and voltages less than 5V. So here I am.😂. Thank you

Well, rightly so.

You want voltage across R to increase from 1V to 5V and current through R to increase 5 times.
Then Rth current also increases 5 times, its resistance decreases in half, its voltage drop increases 2.5 times.
Both voltages increase - the supply cannot be constant.

With a negative resistor you might be able to pull it off because one voltage drop will decrease while the other increases

[mikerj]

The bottom line is your requirements cannot be met by a simple passive divider, you will need a gain element such as an op-amp.

[magic]

Yep.

With less voltage swing like 2V-3V it's only a 1.5 times increase in R voltage and 25% decrease in Rth voltage (same reasons as before), so 1V is 25% of "cold" Rth voltage and hence Rth voltage is 4V and R is 4kΩ and supply is 6V and everything works. Then it can be amplified to 1V-5V, or the receiving circuit modified to deal with a lower signal.

[bitwelder]

I want Vout to vary from 1V to 5V(it is an odd choice).

As others said, you probably need to use a smaller voltage range than 1..5 V.
Why that specific choice? Do you want to then measure that voltage with an ADC, e.g. with an Arduino?
In that case, how much precision do you want from your measurement?
Looking at the resolution of your ADC you can check whether using a smaller voltage excursion you would be still able to have a measurement sufficiently accurate for your needs.

[Terry Bites]

Here's my handy calculator. Get the beta constant and Ro from the thermistor data sheet and your away. Play with Rs and Vin for best fit. Rsqrd value as near to unity as you can get.
The Rsqrd value is hard to see, thats an excel bug.
This is the equation used for the calculation- it’s all an approximation. Thermistor characteristics are not well defined. Its often better to use the manufacturers look up tables for Rt vs T.
Then just plot Vo= Vin*Rs/(Rt+Rs)
I've added a rescasle feature to the speadsheet-

[golden_labels]

polarKaung:
You have chosen constraints that generate a system of equations having no solutions in the positive quadrant. Here are both cases, charted in the WolframAlpha webapp:⁽¹⁾

• (Rt=4kΩ, Vo=5V) and (Rt=8kΩ, Vo=1V)
• (Rt=8kΩ, Vo=5V) and (Rt=4kΩ, Vo=1V)⁽²⁾

At this point one could start a lecture on what is the valid set of constraints, but there is no need for that. There is one more issue with the approach, which you have chosen. You are using two equations, but have only one degree of freedom in your problem. You can’t arbitrarily choose both upper and lower voltage, because there is only a single variable you can manipulate: the bottom resistance R. It controls both of them at the same time, so they can’t be chosen independently.

But that’s good news too: it makes the problem much easier! You can simply chart all possible upper/lower voltages depending on the resistance. Then choose whatever suits you. For your specific problem the chart is in the attachment (note that he horizontal axis uses logarithmic scale).


⁽¹⁾ Screenshots in the “wa-charts.png” attachment.
⁽²⁾ This one makes no sense physically, but I included it for completeness.

[CaptDon]

Can't really get a 5:1 voltage ratio from a resistance that changes 2:1 without an opamp for gain and an additional couple of resistors or a pot to adjust the offset.

[polarKaung]

Thank you all for the replies, those are really helpful, i will try to think about them.
As for output voltage range, I just picked 1V - 5V randomly  . Actually it is for an assignment and the question doesn't specify the output range. At the start, I was like, i want to choose 0V - 5V but thought it is not possible with just two resistor like that.

Anyway, thank you all.

[Picuino]

A good solution is to choose a 6k resistor (value of the thermistor at midpoint) for better linearity.
 After that, calculate the voltage values.
If you need another range of voltages, then you may consider adding a passive network or op amp.

With this configuration and Vs = 5V

Vmin = 5V * 6k / (8k + 6k) = 2.143V
Vmax =  5V * 6k / (4k + 6k) = 3V

[Terry Bites]

Sorry I messed up some cell locking in the rescaler part.
Here is the corrected sheet LIntherm.xlsx (80.88 kB - downloaded 53 times.)

[ledtester]

If you use a negative voltage instead of ground you can find solutions, e.g.:

R= 1K, Vs = 41, Vg = -4.

at Rth = 8k: Vout = Vs*(R/(R+Rth)) + Vg*(Rth/(R+Rth)) = 41*(1/9) - 4*(8/9) = 1

at Rth = 4k: Vout = 41*(1/5) - 4*(4/5) = 5

[golden_labels]

polarKaung:
Since we basically solved your homework, consider making some challenge for yourself now. I wanted to offer you “what if the divider is not referenced to ground”, but ledtester was quicker by a few minutes and delivered solution for that too. So at least try finding the value of the bottom resistor that will give you the largest voltage swing. That will hold for any voltage, so you may safely assume you are aiming for proportion, not specific voltages. Not necessarily calculating it directly: it’s engineering, so making the right chart is enough. Of course mine already gives some hint, but two lines at an angle may be deceptive to human vision.

[Terry Bites]

Let's chuck in a low duty cycle pulsed supply for a bit more fun.... it needs to run for at least 5 years unattended in a woodland nest box and Tx via GSM. Proper home work!

[golden_labels]

Be serious. The task I have given is just about analyzing the same scenario as given in the original post. If OP arrived to negative resistances, which is a correct  though in practical terms useless solution, or is able to understand the chart I have posted, they have all the knowledge and equations to solve that problem too. And the solution is just one step from that chart; literally just adding a single arithmetic operator.