# How to condition a NTC of very low resistance?

I want to use a temperature sensor which changes its resistance. For 0 degrees celsius its resistance is 0.5325 ohms and for 40 degrees celsius it is 3.264 ohms.

Obviously putting it in series with another resistance to produce a variable voltage is not logical since the current that would circulate is high. (the variable voltage must be within the range of 0 to 5 volts).

It occurs to me to use a small voltage (reference voltage) for the voltage divider and then variable voltage should be amplified but when working with very small voltages there may be problems with electrical noise.

Can it be done better and economically?

Thanks

• You've listed your specs, so what's preventing you from looking for a device? It sounds like you're looking for some type of piezoresistive device? – KingDuken Aug 16 '18 at 14:56
• @KingDuken: He has the sensor; he's asking how to deal with its very low source impedance. – Dave Tweed Aug 16 '18 at 15:00
• A Wheatstone bridge should work, since it measures the differential resistance change. See ametherm.com/thermistor/… – DrMoishe Pippik Aug 16 '18 at 15:15
• Title says NTC, but data implies PTC. What P/N? What is your spec for V/'C and tolerance? Low R PTC's are often used for resettable fuses not accurate thermal sensors – Sunnyskyguy EE75 Aug 16 '18 at 15:29
• What power supplies are you using (negative and positive) and what op-amp are you considering? How far is the sensor from the measurement electronics? – Andy aka Aug 16 '18 at 16:51

## 2 Answers

I assume you want to keep power dissipation low in the sensor, say, well below 1 mW.

If you maintain a voltage of 10 mV across the sensor, the current will range between roughly 2.8 and 20 mA, and the power dissipation will range from 28 to 200 µW.

A circuit to do this would look something like this:

simulate this circuit – Schematic created using CircuitLab

The output voltage will be highest when the temperature is lowest, about 3.2V when the sensor is at 0.5Ω and about 0.5V when the sensor is at 3.5Ω. You can deal with this in software after the ADC or add additional conditioning circuitry to get what you want.

The key parameter for the opamp is its input offset voltage. There are specialized devices on the market that have extremely low values for this parameter. Also, if you want to operate from a single power suplly, keep an eye on the input common mode range.

• However, for good accuracy, you need to be bloody sure that the two ground points shown are in very close proximity. Otherwise, stray resistance between the two points will cause major issues. – WhatRoughBeast Aug 16 '18 at 17:02

The details depend a lot on what kind of accuracy you are looking for. Generally with a low resistance you'd prefer to put a lot of current through it and get more voltage so the noise and drift problems are not as bad. A 3.2 ohm sensor with 6mA through it has as much self heating as a Pt100 with 1mA. So 5-10mA is probably sensible.

10mA will give you a voltage between 5.35mV and 32.6mV, which is similar to thermocouple front end requirement, but without the isolation issues.

simulate this circuit – Schematic created using CircuitLab

Note that if the sensor is at all remote the wiring will add significantly to the resistance (and it will vary with temperature). A 3 or 4 wire topology might be better in that case.

For example, a pair of 1" (25mm) of 0.02" (0.5mm) wide traces on a 1oz (35um) board will cause an error of about +0.7°C if they are in series with the sensor.