0
\$\begingroup\$

I have not got a lot of experience with PCBs, but I have moved a breadboard circuit to a PCB and I am seeing an issue.

The objective of the board is to sense temperature. When the circuit is on the breadboard it works as intended, but when I get the corresponding PCB from the manufacturer, the temperature values are wrong.

I am looking for feedback or suggestions that might help pinpoint this issue. Below are the schematic and the PCB layout from Eagle.

schematic

PCB

\$\endgroup\$
7
  • 3
    \$\begingroup\$ Please edit your question to specify both the exact temperature sensors in use, and full details of the error. What happens if you remove the temperature sensors and substitute jumper wires to the old ones on the breadboard? What if you solder the ones from the breadboard onto the PCB instead? Touching it? Stood off from it? Could it be a self heating issue? Measuring something like room temperature in an integrated product can be harder than you might initially assume. \$\endgroup\$ Oct 20, 2020 at 1:50
  • \$\begingroup\$ Are those 100k pull-ups R4 & R5 for I2C lines? If so then that's a really high value unless you're running your I2C clock rreeelllyyy sslllooooooww. \$\endgroup\$
    – brhans
    Oct 20, 2020 at 2:31
  • \$\begingroup\$ The 100K resistors bias the temperature sensors. Can't say more than that; it's a big secret. \$\endgroup\$
    – AnalogKid
    Oct 20, 2020 at 2:56
  • \$\begingroup\$ Damaged during reflow. BOM error. Schematic error (netlist of breadboard does not match netlist of PCB). PCB has other components that are getting warm and conducting heat to the sensor. It is anyone's guess what may be wrong. Please provide a lot more detail as per Chris Stratton's comment. \$\endgroup\$
    – mkeith
    Oct 20, 2020 at 3:00
  • 1
    \$\begingroup\$ The 100k ohm resistors are used to calculate the resistance of each probe and ultimately used to find the corresponding temperature for the probe. The probe/thermistor has a resistance of 100k ohms at room temp. This online guide was used as a reference: create.arduino.cc/projecthub/iasonas-christoulakis/… \$\endgroup\$ Oct 20, 2020 at 9:58

1 Answer 1

0
\$\begingroup\$

Years ago I was approached by an Application Engineer with "I have 5 degree Centigrade error in this circuit, using a type-K thermocouple. What to do?"

Over the next few days, we uncovered several principles for such circuits and such PCBs.

Such as

  • place the local (cold junction) temperature sensor in between the 2 thick pins of the special type-K socket

  • create a thermal short around that type-K socket, so both pins have the same temperature

  • have NO HEAT dissipation near the type-K connector, so thermal flux GRADIENTS can be taken to near_zero

  • do not use a high_power MCU to perform the math.

  • do create a thermal open between the type-K connector(and cold junction sensor) and the MCU and other heat dissipating components (such as USB interface or LDOs)

  • be aware of the HEAT from you own forehead, flowing into the copper of the PCB and causing thermal gradients.

How to do this?

Use 4 layers of PCB; around the type-K connector you must thermally connect 3 of the layers with vias every 1cm, so those vias and layers as ensemble operate as near_isothermal region, ensuring the 2 type-K pins are at same temperature and that the cold_junction sensor is indeed sensing only the type-K (thick copper lugs 2mm thick by 6mm wide) temperature.

Why does this work? Copper foil of the standard thickness is 70 degree C thermal gradient per square of foil per watt of heat flow.

A typical via between layers is about 1:1 aspect ratio: 1/48" diameter, 1/16th" depth if 2_layer PCB. Such a via in 4 layer (1/16 total stack height) has 3x shorter aspect ratio (1/48 diameter, 1/48 depth ==> 1/16 periphery, 1/48 depth), thus have 70/3 = 23 degree C thermal gradient per watt of heat flow.

If you have 20 such vias spread out over a 2" by 2" region where the type-K connector (and the cold junction sensor) are placed, and your heat comes from only your FACE, assuming the sensor is installed symmetrically between the 2 pins of type-K), and you have THREE layers (the 4rth for the cold junction sensor wiring), and you have ONE WATT heat intrusion from your face/forehead, then the thermal error is

23 degrees / ( 20 vias * 3 un_slit copper foil planes 1.4 mils thick)

or

23/60 = 0.4 degree error at most

What success did the Application Engineer have? his 5 degree error immediately dropped to less than 1 degree, and since then he has created his own business line of thermocouple PCBs.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.