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I just wired up a test harness to see what voltage the smoke machine would cut off current to the heating coil when I stopped(pretty unusual) and thought(don't know where this came from...) I should probably test it without anything connected to the PCB just to see what happens. So I did and lo and behold, it thought it was up to temperature... Check the voltage accross the pins... 2.5-3V there... Resistance of the temperature probe? 3 ohms... Hmmm... might not be a thermocouple... time for some more tests...

Attach the probe to PCB... heater turns on... remove it heater turns off... I can hear the relay ticking so that's not what caused it to overheat... some more research needed methinks... Wikipedia here I come... Write comment (0 Comments)
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Rather than just getting a high resistance to drop the voltage I realized that a higher current would also. By putting a resistor in parallel with the potentiometer I can increase the current flowing through the first resistor and use a smaller, more available resistor there to drop the same voltage.
This is the schematic I did up in KTechLab to test my theory. the 1M resistor represents the smoke machine input. the max. voltage possible with this arrangement is about 30mV(50mV with a 5V supply). This means I should be able to supply the required 25mV for my max. temperature.
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Looking at the Wikipedia page on thermocouples I found this table:
TypeTemperature range °C (continuous)Temperature range °C (short term)Tolerance class one (°C)Tolerance class two (°C)IEC Color codeBS Color codeANSI Color code
K0 to +1100−180 to +1300±1.5 between −40 °C and 375 °C
±0.004×T between 375 °C and 1000 °C
±2.5 between −40 °C and 333 °C
±0.0075×T between 333 °C and 1200 °C
IEC Type K Thermocouple.svg
BS Type K Thermocouple.svg
MC 96.1 K Thermocouple Grade Color Code.svg
J0 to +750−180 to +800±1.5 between −40 °C and 375 °C
±0.004×T between 375 °C and 750 °C
±2.5 between −40 °C and 333 °C
±0.0075×T between 333 °C and 750 °C
IEC Type J Thermocouple.svg
BS Type J Thermocouple.svg
MC 96.1 J Thermocouple Grade Color Code.svg
N0 to +1100−270 to +1300±1.5 between −40 °C and 375 °C
±0.004×T between 375 °C and 1000 °C
±2.5 between −40 °C and 333 °C
±0.0075×T between 333 °C and 1200 °C
IEC Type N Thermocouple.svg
BS Type N Thermocouple.svg
MC 96.1 N Thermocouple Grade Color Code.svg
R0 to +1600−50 to +1700±1.0 between 0 °C and 1100 °C
±[1 + 0.003×(T − 1100)] between 1100 °C and 1600 °C
±1.5 between 0 °C and 600 °C
±0.0025×T between 600 °C and 1600 °C
BS Type N Thermocouple.svg
BS Type R Thermocouple.svg
Not defined.
S0 to 1600−50 to +1750±1.0 between 0 °C and 1100 °C
±[1 + 0.003×(T − 1100)] between 1100 °C and 1600 °C
±1.5 between 0 °C and 600 °C
±0.0025×T between 600 °C and 1600 °C
BS Type R Thermocouple.svg
Not defined.
B+200 to +17000 to +1820Not Available±0.0025×T between 600 °C and 1700 °CNo standard use copper wireNo standard use copper wireNot defined.
T−185 to +300−250 to +400±0.5 between −40 °C and 125 °C
±0.004×T between 125 °C and 350 °C
±1.0 between −40 °C and 133 °C
±0.0075×T between 133 °C and 350 °C
IEC Type T Thermocouple.svg
BS Type T Thermocouple.svg
MC 96.1 T Thermocouple Grade Color Code.svg
E0 to +800−40 to +900±1.5 between −40 °C and 375 °C
±0.004×T between 375 °C and 800 °C
±2.5 between −40 °C and 333 °C
±0.0075×T between 333 °C and 900 °C
IEC Type E Thermocouple.svg
BS Type E Thermocouple.svg
MC 96.1 E Thermocouple Grade Color Code.svg
Chromel/AuFe−272 to +300n/aReproducibility 0.2% of the voltage; each sensor needs individual calibration.
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To test my hypothesis about the temperature sensor being a thermocouple I realized that a thermocouple would have something like an op-amp as a buffer amp and to get the low voltage up to a logic level. An op-amp has an (ideally) infinite (actually) really high input impedance. This high impedance means that the amp does not load down the input circuit with high current draw.

I=V/R
I=V/infinity
I=0

This is why an op-amp has a high input impedance.

The temperature probe connections have a resistance of around 1000k. That's pretty high, which when coupled with the polarity markings indicates that there is an op-amp on the board and the temperature probe is a low current voltage source. Also known as a thermocouple.

This means I can now simulate the presence of the temperature probe by supplying an appropriate voltage(not sure how big yet though). By simulating the presence of the thermocouple I can test if the electronics will turn off the heater current at some point which means that it was just the thermocouple that failed. The outer braid of the thermocouple has failed where it enters the remains of the aluminium so that makes me think that something went badly wrong with the thermocouple.

While I had the multimeter out I also measured the heater coil's resistance- this came out at about 700k. So with a little math we can calculate the current flowing through the coil and also the wattage(not really that important but might be interesting to see how much power the electronics use).

I=V/R
I=230/700k  (Pretty sure 230V isn't RMS but I know it isn't Peak to Peak so it shouldn't give us a number that is too high)
I=230/700 000 (Now in SI units)
I=0.003288571A    (Seems very very low ???)

Now for power consumption:

P=IV
P=230*0.003288571
P=0.76W

This is clearly wrong so I decided to check again... It turns out that my fingers have a resistance of around 700k ohms and it was this resistance I measured. The heating coil appears as an open circuit so I think I'm going to have to melt out the remains of the Aluminium block and get a closer look. Write comment (0 Comments)
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2019-07-19