Determining the parameters of a Wheatstone bridge

Question

I am trying to use a Wheatstone bridge to condition a thermistor. The arrangement I have is shown below:

^{simulate this circuit – Schematic created using CircuitLab}

I have the following specifications:

1. 10 V output max. from the bridge at maximum temperature (50°C)
2. 0 V output max. from the bridge at minimum temperature (0°C)

I have determined the resistance of the thermistor at the minimum and maximum temperature to be as follows:

1. At max. temperature R = 174 Ω
2. At min. temperature R = 1.1 kΩ

I know that for the Wheatstone bridge to be balanced I need the following for my arrangement:

R1/R3 = R2/Rt

So I have set R3 = 1.1 kΩ and chose R1=R2=1500 kΩ (arbitrary value).

With this I do get 0 V output at 0°C.

My problem is: how do I ensure that the output voltage I get is 10 V at 50°C, as I currently get 8.7 V output with the source voltage set to 24 V.

Answer 1

I am going to assume this is a homework problem, since just by looking at it I can see that the power dissipation in the sensing element is excessive for practical purposes (as Andy has justified numerically in his answer).

Note that for an NTC thermistor the bridge circuit you show is incorrect- the thermistor and R3 need to trade places.

As you may realize this problem is underconstrained since the bridge voltage is not given- so there may be more than one solution if a solution exists. So, let's try to constrain it more.. clearly (although it will never be acceptable most likely) we should like to minimize the power dissipation in the thermistor. In order to do that we can reduce the value of the excitation voltage until there exists only a single solution (modulo some scale factor on the resistor arm).

This is because, as you vary the value of R3 there will be a value that gives you a maximum output change for a thermistor resistance change from 174 ohms to 1100 ohms. Clearly if R3 is open or shorted you get zero change, the maximum is somewhere between.

It can be shown (exercise left for the student- how would you approach it?) that the minimum bridge excitation is 23.20V, and thus R3 should have a value of 437.5\$\Omega\$.

Now we need to solve for the divider to get 0V at 0°C (1100 ohms). Being lazy, I'll set R2 to 437.5\$\Omega\$ and R1 to 1100\$\Omega\$, which will obviously give 0V. If I were more environmentally conscious I could be almost as lazy and pick 4375\$\Omega\$ and 11K\$\Omega\$ (thus almost halving the power consumption of the bridge), but then I'd have to order another resistor value so..

Let's check at 100°C: 6.60V on the passive leg and 16.60V on the thermistor leg. Gosh, it works (for homework purposes).

In reality, you'd use a much lower excitation and an amplifier as Andy suggests.

Answer 2

With a 24 volt supply, the current thru the limb of the bridge containing the thermistor is: -

• Nearly 11mA at 0C
• Nearly 19mA at 50C

At 0C this produces a power in the thermistor of 132mW and 63mW at 50C. To me this is a problem. The self-heating of the thermistor will create a significant measurement error at 0C compared to 50C. I'd use an instrumentation amplifier and keep the excitation levels much lower to prevent significant self-heating errors: -

The picture above shows the AD620 powered from +5V and 0V but a better power regime (for the OP) is +/- 15V with 0V connected to pin 5 (the reference input). The excitation voltage can be much lower (say 2.5 volts) and this means the self-heating effects are 1.5mW at 0C and 0.7mW at 50C i.e. trivial.

However, there is another potential problem of a single thermistor in a Wheatstone bridge and that is linearity. You may design a circuit that produces the correct voltages at the extremes of temperature you require but, between these two points, there is a non-linear mapping between voltage and temperature over and above the basic non-linearity of the thermistor's resistance change with temperature.

It might seem intuitive to place 2 thermistors at opposite sides of the bridge but, with voltage excitation, this won't improve linearity (but does double sensitivity). Here is a good document by ADI explaining all of this (called Bridge circuits by Walt Kester): -

However, if you adopted a current driven bridge configuration then you could use two thermistors and get perfect linearity: -

So, rounding up, try to avoid self-heating effects and, for greater linearity, use two elements in opposing faces of the bridge with the bridge excitation being a constant current like the one below: -

The important voltage to keep constant in the above diagram is that across R1 and, for high accuracy a precision shunt voltage reference would be used in place of R1.