Temperature compensation is a common problem among coils, or solenoids. These metals exhibit a positive temperature coefficient with rising temperature. Since the NTC typically has temperature coefficient range of –0.29% to –0.51%, a fixed parallel resistor is used to bring the temperature coefficient down to a usable limit. For example copper has a temperature coefficient of 0.5%/°C and Aluminum has a 0.26%/°C.

**(EX)** Suppose we have a copper solenoid with a temperature coefficient of 0.47%/°C over a temperature range of –40°C to +200°C. Suppose the coil resistance = 9.40W ± 0.50W.** **

**(Q)** Select a thermistor and a fixed resistor to minimize this effect. ** **

**(A)** This is a classic automotive application for a solenoid. The object is to find the most suitable network, which negates the positive temperature effect of the solenoid. In that temperature range if solenoid is left alone it will exhibit a resistance from (9.0 W -18.0 W). This is too much fluctuation, the ideal range will be to flatten the curve at its center without so much resistance fluctuation. The mean resistance of the coil is calculated by [(18-9) /2 ] + 9.4 W =14.90 W. This is approximately about 15.0 W with fluctuation of ± 30%.

The circuit in diagram 2 shows a thermistor in parallel with another fixed resistor and in series with the coil. The total resistance is found by the following equation:

Total resistance = R _{coil} +[ (R_{thermistor})(R_{resistor})] / R_{thermistor} + R_{resistor }= 15.0 Ω.

So if the coil represents about 9.40 W in resistance the total resistance of fixed resistor and the thermistor = 15 – 9.40 = 5.60 W.

The linearization equation for flattening the curve of the thermistor is as follows:

Fixed resistor value = R _{esistor }= R_{R}_{}

Thermistor resistance value = R_{thermistor} = R_{T}

Then the linearization equation = R_{R }= (b – 2T/ b + 2T) (R_{T}) **eqn 4**

Where:

b = Beta = material constant = in our example = 3150°K

T = midpoint temperature of the operating range of the network. In our case it will be 200°C – (-40°C) /2 = 120°C therefore the midpoint temperature will be 200°C –120°C = 80°C.

Using **eqn 4** we have R _{R = }0.6337R_{T}

Now the total resistance for the fixed resistor and thermistor = 5.60W.

The parallel equation describing the relationship between them is:

R _{R}.R_{T }/ R_{R} + R_{T} = 5.60 substitute for R_{R }yields R_{T }= 14.43 Ω

Therefore, R_{R} = 9.14 W.

The resistor is specified at 1% tolerance and the thermistor can normally have tolerance of 15%.

Let us examine the property of these parts at various temperatures.

At 200°C R _{T }= 0.239W_{.}

Therefore, the total resistance of the network = R_{coil }+ resistance of the fixed resistor and thermistor which are in parallel. Network resistance = R_{coil} + 0.230 W.

Total Resistance at 200°C = [(200°C-25°C)(0.47%/°C)](9.40 Ω) + 9.40 +0.230 = 17.36 Ω.

R_{T} at -40°C = 222.4 Ω

Total Resistance of network at -40°C = [(-40°C – (25°C))(0.48%/°/C)(9.40)] + 9.40 + (1/9.14 + 1/222.4)^{-1} = 15.25 Ω.

This shows much less deviation than the coil by itself.

Network fluctuation = 16.30Ω ± 6.5%.