# Thermocouple bias with LTC2053

This LTC2053 application note shows a thermocouple bias using two 10 MΩ and two 1 MΩ resistors.

I understand the idea that the power supply is divided by two 10 MΩ resistors for good common mode performance.

If the thermocouple is not grounded, no current flows through the thermocouple. If it is, the high resistor values make the current low, causing a low voltage drop across the thermocouple.

What I don't get are the resistor values; why not replace the 10 MΩ with, say, 10 kΩ ones and at the same time replace the 1 MΩ with 10 MΩ (to keep the current low)? What is to be gained by a "weak" 10 MΩ common mode voltage divider?

• The 10M and 1 M Ohm are just providing the necessary current bias for the two inputs, symetrically. The 10K are there for limiting an "overvoltage" eventually. Otherwise, they introduce a little "attenuation" of the thermocouple signal. Jan 18 at 9:02
• Yes, but why 10M and not a lower value (e.g. 10k-100k)? Then the same current could be set by the resistors which are now 1M and could be increased to 10M.
– Hyp
Jan 18 at 9:35
• Of course, one can do it. But current on the 2 10k would be 5/20k=250 uA in place of 5/20 Meg = 0.25 uA ... The goal is to "maintain" a very low current consumption, probably below 1mA. Jan 18 at 9:56
• If you are right and the reason is very low current consumption, then I find it rather strange to have a total consumption of 1.8mA (says app note), most of which is due to the zener diode and then fight for each uA. If the divider used 100k, then it is 5/200k=25uA only. On the other hand, my idea of using a "hard" divider would not bring any benefits either as it would still be "softened" by the 10M (originally 1M) resistors needed for low current connection to the thermocouple.
– Hyp
Jan 18 at 12:02
• Ok for 1.8 mA with some others devices. LTC2053 alone is only 750 uA @2.5V. Jan 18 at 12:10