# Coolant Temperature Sender

### From WikiLEC

## Background

The Coolant Temperature Sender (not to be confused with the Coolant Temperature Sensor used by the ECM) is also a negative temperature coefficient (NTC) thermistor that allows the temperature gauge to display engine temperature. The temperature gauge is provided with a precision 10±0.1V supply and current flows through the gauge and via the sender to ground.

The temperature gauge has caused much confusion over the years as the normal running temperature (~80°C) is approximately 1/4 scale and the temperature at which the radiator fans cut in (~100°C) and it has been postulated that the sender (Isuzu) and gauge (Vauxhall) are not matched.

## Analysis

Empirical data for the effect of various resistors on the temperature gauge is as follows:

Column 1: Sender resistance (Rsender)

Column 2: Sender voltage (Vsender)

Column 3: Gauge position (Pos%)

Column 4: Temperature (Temp)

35.6 | 4.00 | 8/8 |

48.7 | 4.80 |

50.0 | 4.90 | 6/8 | 100°C

60.0 | 5.30 |

70.0 | 5.75 | 4/8 |

85.3 | 6.20 |

100. | 6.60 | 2/8 | 80°C

120. | 7.00 | 1/8 |

153. | 7.50 | 0/8 |

200. | 8.00 |

The reference voltage (Vref) was measured as 10.1V.

Some simple analyses can be performed:

a) The gauge resistance (Rgauge) can be determined as:

I = Vref / (Rsender + Rgauge) = Vsender / Rsender

Or Rgauge = Vref × Rsender / Vsender - Rsender

The average value from the above 10 data points gives Rgauge = 53.4 ohms.

b) The resistance of the sender (thermistor) is R = Ro × exp(β × ( 1 / T - 1 / To)) where

Ro is resistance at To

β = 1/b with c=0 in the Steinhart-Hart equation

T = temperature (Kelvin) = Temp + 273.15

To = reference temperature (Kelvin)

Fitting to the approximate temperatures at 1/4 and 3.4 scale gives values of Ro(+25°C) = 1085 ohms and β = 4560.

c) Temperature gauges are usually linear with current but with an offset current before they start to read. Plotting the gauge position (Pos%) against the current (I = Vref / (Rsensor + Rgauge)) gives a good fit to the curve

Pos% = 15.6 × I - 0.775

The offset current is therefore 0.775 / 15.6 or approximately 50mA.

d) Combining all of the above gives us an equation for the gauge position vs temperature:

Pos% = 15.6 × Vref / (Rgauge + Rsender) - 0.775 = 15.6 × 10.1 / (53.4 + 1085 × exp(4560 × ( 1 / (Temp + 273.15) - 1 / 298.15))) - 0.775

This results in a reasonably linear relationship between temperature (69.3°C to 111°C) and gauge position (0% to 100%) of:

**Pos% = 2.41 × Temp - 167** with R²=0.9998

## Gauge improvement

By adding a fixed resistor in parallel across the sender, it is theoretically possible to increase the gauge output: This will be particularly prevalent at the bottom end of the gauge where the sender resistance is highest and consequently most affected by the parallel resistor. Obviously, this will introduce some additional non-linearity but it should be possible to find a value which gives a beneficial increase without too much non-linearity and without causing full scale deflection at too low a temperature.

If, for example, we want to make 5°C above normal operating temperature (85°C) to read half scale: The sender resistance at 85°C is 83.5 ohms and the sender resistance required for half scale deflection is 70.1 ohms so the parallel resistor required is:

R = 1 / (1 / 70.1 - 1 / 83.5) = 433 ohms.

This results in a reasonably linear relationship between temperature (58.7°C to 108°C) and gauge position (0% to 100%) of:

Pos% = 2.04 × Temp - 122.5 with R²=0.9982

Similarly, to read half scale at 80°C requires the resistance to be reduced from 100 ohms to 83.5 ohms so:

R = 1 / (1 / 70.1 - 1 / 100) = 233 ohms.

This results in quite a bit of non-linearity and the gauge starts to read at 45°C, however it has a reasonably linear relationship between temperature (65.2°C to 106°C) and gauge position (25% to 100%) of:

Pos% = 1.56 × Temp - 71.2 with R²=0.981

In all cases the power dissipated in the resistor is manageable and 300 ohms is less than ¼W and 56 ohms is less than ½W.