# Thermocouple and Resistors

#### xxxmbgxxx

Joined Aug 15, 2007
3
Could anyone please tell me how I can calculate what resistors (or potentiometer range) is required to reduce the apparent temperature reading from a type K thermocouple to approximately half the actual reading over approximately 10 to 90 degrees (normally 0.5mV to 4.0mv over this range)

Many thanks

#### Dave

Joined Nov 17, 2003
6,969
Wouldn't the most reliable way to achieve this be through software? What DAQ unit are you using?

Dave

#### hgmjr

Joined Jan 28, 2005
9,027
It would be helpful if you could post a schematic of your circuit to aid the members in providing you with assistance.

hgmjr

#### xxxmbgxxx

Joined Aug 15, 2007
3
I am using a datalogger thermometer similar to

http://www.omega.co.uk/Temperature/pdf/HH305_306.pdf

I'm really sorry but I can't go into detail as to why I need this but I can't do it through software .. I need a variable resistor in series with one side of the thermocouple to give an artificial temperature (variable between approximately 25% and 75% of the actual temperature over a temp range of approx 10 to 90 deg C).

Could anyone tell me what resistance is required or how I can calculate this

#### thingmaker3

Joined May 16, 2005
5,083
Type K thermocouples produce voltage in approximate proportion to temperature: http://www.physik.unibas.ch/Praktikum/VPII/SternGerlach/Thermoelement.pdf

Your dataloger will have very high input impedance, so using a series resistor will not be practical. Also, anything in series with a thermocouple, other than thermocouple wire, will adversely affect accuracy.

Perhaps a voltage divider would work, if you are willing to sacrifice accuracy. I would put more trust in a pre-amplifier & attenuator.

#### xxxmbgxxx

Joined Aug 15, 2007
3
thanks for that .. so what effect would a resistor in line have please?

#### beenthere

Joined Apr 20, 2004
15,819
It would limit the current out of the thermocouple. If the circuit reading the thermocouple output has a high impedance, the voltage might not be affected.

#### flip

Joined Aug 20, 2007
1
A resistor in series with a thermocouple like this is not a good idea. As thingmaker3 mentioned, the current that enters the amplifier or signal conditioner is very small, therefore the resistor will have very little effect. This is why the 'measurement circuit' requires a high impedance ... so as to not affect the voltage leaving the thermocouple. This will result in inaccurate readings.
Even if you were successful in reducing the output to the desired value by using a resistor, you still need some method to adjust to the required range or span.
The datalogger thermometer has a high input impedance and is designed to read thermocouples within specific ranges as determined by the logger design.