# Investigating a Thermistor.

INVESTIGATING A THERMISTOR

After deciding to investigate the properties of a thermistor, I chose to be more specific, and to look at repeatability, accuracy, and sensitivity. I also thought of many different ideas as to what the experiment could be used for: for fridge…??? controlled???. The proceeded by looking at different circuits suitable for exploring sensitivity. I looked into the ‘whetstone bridge’ circuit:

The wheatstone bridge circuit enables more accurate readings,

However, I decided on another circuit that acted as a potential divider only using one fixed resistor, which seemed equally suitable for detecting small changes of volts. The reason for this being that I thought it would be interesting to see the effect(s) of changing the fixed resistor (R):

After recognizing the importance of (R), I decided on the equipment I was going to use, and then tried using some algebra to tackle the problem of finding the value of R that would give me the biggest change DVo.

Equipment:

1 power generator set at 2volts

1 thermistor (RS 0.47Kohms 232-4538)

an assortment of fixed resistors

some crocodile clips

a digital voltmeter

Here are my jottings:

Vo/Vs=R/R+RTh

Vo=Vs(R/R+RTh)

Vo=Vo t2 – Vo t1

Vo=VsR[1/ R+RTh t2 – 1/ R+RTh t1]

At this point I thought it appropriate to take a numerical approach, and use Excel to plot a graph of the equation as I was having difficulty simplifying or manipulating it further. For RTh t1 and 2 I put in 5 and 105, as an RS data sheet showed these to be the values at 100 and 0 degrees.

This graph suggests to me that using a fixed resistance of about 25ohms would produce the highest value of DVo, and therefore would produce the most sensitive readings. However, I suspected that the resistance of the fixed resistor should be roughly equal to the thermistors resistance (0.47Kohms), so I used the internet, and found that the ???? website stated:??????????????????. Due to these conflicting ideas, I decided to perform the experiment using a range of thermistors.

The problem of self heating then occurred to me, which is extremely important as the thermistor responds to temperature. So to address the problem I used some more algebra:

V=IR

Which can be written as I=V/R

P=I2R

Therefore P=(V/R) 2R,

so subbing in V and R (or Rth) I get P=(2/470) 2x470=8.51E-3W

Because the power is in micro watts, there is no reason why my experiments should be noticeably effected by the thermistor’s self heating.

I thought of any other ways that may lead to an unreliable experiment, which lead me to thinking how I was going to accurately control the ...