Question about resistor tolerances
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Dave Dixon
- Posts: 436
- Joined: Thu May 01, 2003 1:01 am
- Location: Wichita, KS
- Contact:
Now I'm not sure I get it. It doesn't seem as though R3 is so critical unless the Span adjust R5 is super stable. Wouldn't any changes in temp/etc. affect the 2K adjust pot as much or more than it would the 1 Meg. resistor? It seems as though the potentiometer should be more of a concern in this part of the circuit. Am I missing something? TNX, Dave
Hi again,
Craig, are you sure you can run the V- of the op amp to ground
instead of a negative supply voltage? Try a battery if not just to
see if it helps.
Dave, the pot and two resistors should be ok for temperature
as the pot simply looks like two resistors made of the same
material to this circuit. The upper part of the pot should change
by the same percentage as the lower part of the pot over temperature.
This helps to keep the gain of the circuit constant.
Craig, are you sure you can run the V- of the op amp to ground
instead of a negative supply voltage? Try a battery if not just to
see if it helps.
Dave, the pot and two resistors should be ok for temperature
as the pot simply looks like two resistors made of the same
material to this circuit. The upper part of the pot should change
by the same percentage as the lower part of the pot over temperature.
This helps to keep the gain of the circuit constant.
LEDs vs Bulbs, LEDs are winning.
No, I'm not sure. Isn't "ground" and "a negative supply voltage" the same thing, or do they need to be different voltage sources?are you sure you can run the V- of the op amp to ground
instead of a negative supply voltage? Try a battery if not just to
see if it helps.
If I hook up V- to a battery, I also have to hook up the V+ to the same battery as well, correct? I suppose a 9V battery would work fine? The datasheet says the supply voltage (V+ to V-) is 40V.
Hi again Craig,
You would hook the minus terminal of the 9v battery to the V- of
the chip, and connect the positive (+) terminal of the battery
to ground. Everything else the same as you have it.
Many op amps can not run without a negative supply when the
signal inputs are close to ground. I dont know if your's is or
isnt, but it's easy to try. For the long haul i think i would look
up the specs of the op amp to see if it can work near ground
without a negative supply.
What is the exact part number of the op amp you are using?
You would hook the minus terminal of the 9v battery to the V- of
the chip, and connect the positive (+) terminal of the battery
to ground. Everything else the same as you have it.
Many op amps can not run without a negative supply when the
signal inputs are close to ground. I dont know if your's is or
isnt, but it's easy to try. For the long haul i think i would look
up the specs of the op amp to see if it can work near ground
without a negative supply.
What is the exact part number of the op amp you are using?
LEDs vs Bulbs, LEDs are winning.
The exact model number is LTKA00CN8 made by Linear Technology (http://www.linear.com).
I have look for a datasheet for this part, but I don't think one exists. The only thing I could find is a datasheet for the LTKA00CN8 Op-Amp and the LT1025CN8 Cold Junction Compensator as a combined IC, part number LTK001. The information, specs, and datasheet for LTK001 can be found at http://www.linear.com/pc/productDetail. ... 1073,P1181.
I have look for a datasheet for this part, but I don't think one exists. The only thing I could find is a datasheet for the LTKA00CN8 Op-Amp and the LT1025CN8 Cold Junction Compensator as a combined IC, part number LTK001. The information, specs, and datasheet for LTK001 can be found at http://www.linear.com/pc/productDetail. ... 1073,P1181.
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Robert Reed
- Posts: 2277
- Joined: Wed Nov 24, 2004 1:01 am
- Location: ASHTABULA,OHIO
- Contact:
Quote "and if R3 should change by 5 percent then R1 should also change
by 5 percent to keep the gain pretty much the same for changing
temperatures."
But there is no guarantee that they will change by the same percentage or even in the same direction.
As to carbon film 5% resistors, they are generally quite close to their stated value. My experience- +/- 2%, and not difficult to pick out +/- 1%.
However extreme accuracy is meaningless without the stability to back it up.
Yes you can and do want to run the Op-Amp output string to ground as this is what it is referenced to. The out put swing will be almost rail to rail and centered at grd. potential. This is the beauty of split supplies.(You are using the split supply as shown,aren't you)
by 5 percent to keep the gain pretty much the same for changing
temperatures."
But there is no guarantee that they will change by the same percentage or even in the same direction.
As to carbon film 5% resistors, they are generally quite close to their stated value. My experience- +/- 2%, and not difficult to pick out +/- 1%.
However extreme accuracy is meaningless without the stability to back it up.
Yes you can and do want to run the Op-Amp output string to ground as this is what it is referenced to. The out put swing will be almost rail to rail and centered at grd. potential. This is the beauty of split supplies.(You are using the split supply as shown,aren't you)
Hi again,
Craig, I cant find one either, but i see something on each schematic
that always shows the op amp V- terminal going to a V minus supply,
not to ground. This tells me it needs a negative supply, or at the
very least to test this circuit with one. The 9v battery idea sounds
good to me as a first test.
If you can measure a change in thermocouple output then all that
needs to be done is get the op amp section working.
Robert:
QUOTE
"But there is no guarantee that they will change by the same percentage
or even in the same direction. "
END QUOTE
I've never seen a carbon resistor that does not increase resistance
with temperature. That's like saying "copper sometimes has a negative
temperature coefficient", isnt it, or does my memory serve me wrong?
For that matter, what would make a materials temperature coefficient
reverse? An act of God? <kidding around here he he>
Agreed the 1 percent resistors will in fact track better than the 5 percent.
Thanks for bringing this all up BTW.
Craig, I cant find one either, but i see something on each schematic
that always shows the op amp V- terminal going to a V minus supply,
not to ground. This tells me it needs a negative supply, or at the
very least to test this circuit with one. The 9v battery idea sounds
good to me as a first test.
If you can measure a change in thermocouple output then all that
needs to be done is get the op amp section working.
Robert:
QUOTE
"But there is no guarantee that they will change by the same percentage
or even in the same direction. "
END QUOTE
I've never seen a carbon resistor that does not increase resistance
with temperature. That's like saying "copper sometimes has a negative
temperature coefficient", isnt it, or does my memory serve me wrong?
For that matter, what would make a materials temperature coefficient
reverse? An act of God? <kidding around here he he>
Agreed the 1 percent resistors will in fact track better than the 5 percent.
Thanks for bringing this all up BTW.
LEDs vs Bulbs, LEDs are winning.
Ok, if I understand you correctly, it should be like this:
Is that correct?
Is that correct?
Umm, I don't know what you mean. Well, maybe I do know. When you say "split supply", do you mean two totally different power sources?This is the beauty of split supplies.(You are using the split supply as shown,aren't you)
What exactly do you mean by "close to ground"?Many op amps can not run without a negative supply when the
signal inputs are close to ground.
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ecerfoglio
- Posts: 108
- Joined: Thu Nov 25, 2004 1:01 am
- Location: Buenos Aires Argentina
- Contact:
Yes, you may use a 9V battery as shown in your drawing.
A circuit may draw current between +V and ground, between ground and -V or between +V and -V.
Most op amps must have their inputs and / or outputs between a few volts above -V and a few volts below +V.
If your circuit uses voltages that are "close to ground" (say less than 2V) you must use a split supply, or else a special type of op amp that can swing its output near its -V voltage.
If you expect an output of 10mV/°C (with respect to ground) your reading of 0.554 V should mean 55°C, to measure lower temperatures the op amp must output lower voltages.
Yes, they may be two different power sources (in series) or a single supply with two outputs and three wires:When you say "split supply", do you mean two totally different power sources?
Code: Select all
>> +V (positive with respect to ground),
>> Ground or common (= cero volts reference), and
>> -V (negative with respect to ground)Most op amps must have their inputs and / or outputs between a few volts above -V and a few volts below +V.
If your circuit uses voltages that are "close to ground" (say less than 2V) you must use a split supply, or else a special type of op amp that can swing its output near its -V voltage.
The thermocouple's output voltage is only a few milivolts above (or below) ground.What exactly do you mean by "close to ground"?
If you expect an output of 10mV/°C (with respect to ground) your reading of 0.554 V should mean 55°C, to measure lower temperatures the op amp must output lower voltages.
E. Cerfoglio
Buenos Aires
Argentina
Buenos Aires
Argentina
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Robert Reed
- Posts: 2277
- Joined: Wed Nov 24, 2004 1:01 am
- Location: ASHTABULA,OHIO
- Contact:
Craig
Almost any Op-Amp can work from a single or dual supply(split). The difference is that the inputs have to be biased at one half the power supply voltage when using single supply.This work,but puts constraints on the circuit design . The split supply is always the best choice when possible as it allows for simpler circuit design and less components. Your latest diagram will work ok as shown and will produce a linear output up to its rail limits. The difference here being that it would allow larger negative swings in the output than positive swings(due to -9V and + 5V supply). In your case, you will probably never see swings large enough to approach either rail(assuming input voltages do not exceed certain levels).There may be a problem here if your inputs have to handle large range of swing as the output may exceed the power supply rails. In this case you would have to increase the power supply voltages to accomadate this. You said the chip could handle 40 VDC across its input, so this would put a cap of +/- 20 VDC for a split supply.What you need at this point is a nice test bench power supply for testing these parameters
One thing I am curious about is - what is LT1025. It looks like it may be a voltage source thats already biasing one of the inputs or may be for offsetting the output.
MrAl
Your probably more right and I am more wrong about the carbons temp drift. I think at the moment I was confusing The after solder in temperature effects where some resistors setlled at a slightly different value and direction thean others. I am still not sure that normal temp. coefficients would track for grossly different values as I had run some tests in the past on carbon films and if I recall correctly, there was a larger percentage change in the higher value resistors than the lower ones. However this was a rather limited test for just one project.
Almost any Op-Amp can work from a single or dual supply(split). The difference is that the inputs have to be biased at one half the power supply voltage when using single supply.This work,but puts constraints on the circuit design . The split supply is always the best choice when possible as it allows for simpler circuit design and less components. Your latest diagram will work ok as shown and will produce a linear output up to its rail limits. The difference here being that it would allow larger negative swings in the output than positive swings(due to -9V and + 5V supply). In your case, you will probably never see swings large enough to approach either rail(assuming input voltages do not exceed certain levels).There may be a problem here if your inputs have to handle large range of swing as the output may exceed the power supply rails. In this case you would have to increase the power supply voltages to accomadate this. You said the chip could handle 40 VDC across its input, so this would put a cap of +/- 20 VDC for a split supply.What you need at this point is a nice test bench power supply for testing these parameters
One thing I am curious about is - what is LT1025. It looks like it may be a voltage source thats already biasing one of the inputs or may be for offsetting the output.MrAl
Your probably more right and I am more wrong about the carbons temp drift. I think at the moment I was confusing The after solder in temperature effects where some resistors setlled at a slightly different value and direction thean others. I am still not sure that normal temp. coefficients would track for grossly different values as I had run some tests in the past on carbon films and if I recall correctly, there was a larger percentage change in the higher value resistors than the lower ones. However this was a rather limited test for just one project.
.
With thermocouple output variance already confirmed via your prior direct metering, along with confirming its proper install polarity. Now hows about a quick encompassing system loop test by eyeing your metered output at pin 6 and make a very quick temporary test shorting across of the thermocouple with a 1k resistor and confirm a very healthy voltage swing.
Also was that your circuit, as my referencing was this one as an app note
from the manufacturer....but its using a K type unit ..in a completely different configuration and depending on the 1025's compensatory referencing passing thru the thermocouple:
With the route coming from off pin 7 for K compensated output versus your J compensated correction coming from pin 8.
Plus, that circuitry supply is using a dual supply on the op amp circuitry, as well as the 1025 compensatory IC..
ADDENDA:
I see that the pic / schema below is coming out as an X so I am re-hosting
73's de Edd
[email protected] ...........(Interstellar~~~~Warp~~~~Speed)
[email protected].........(Firewalled*Spam*Cookies*Crumbs)
Why is it that when you're driving and really craning your neck, intently looking for an address, you turn down the volume on the radio?
.
With thermocouple output variance already confirmed via your prior direct metering, along with confirming its proper install polarity. Now hows about a quick encompassing system loop test by eyeing your metered output at pin 6 and make a very quick temporary test shorting across of the thermocouple with a 1k resistor and confirm a very healthy voltage swing.
Also was that your circuit, as my referencing was this one as an app note
from the manufacturer....but its using a K type unit ..in a completely different configuration and depending on the 1025's compensatory referencing passing thru the thermocouple:
With the route coming from off pin 7 for K compensated output versus your J compensated correction coming from pin 8.
Plus, that circuitry supply is using a dual supply on the op amp circuitry, as well as the 1025 compensatory IC..
ADDENDA:
I see that the pic / schema below is coming out as an X so I am re-hosting
73's de Edd
[email protected] ...........(Interstellar~~~~Warp~~~~Speed)
[email protected].........(Firewalled*Spam*Cookies*Crumbs)
Why is it that when you're driving and really craning your neck, intently looking for an address, you turn down the volume on the radio?
.
Hi again,
Craig:
To add to the other replies I'll try to detail what 'close to ground' means.
If either the non-inverting terminal or the inverting terminal is at a voltage
that is close to zero volts (ground) it is said to be close to ground. The
actual voltage level (0.1 volts, 0.5 volts, 1 volts, 2 volts, etc.) is
determined by the actual specs for the op amp. For example, the LM358 can
work right down to zero volts (ground) and it says so in the data sheet.
If an op amp can not work like this it is usually mentioned in the data
sheet somewhere such as under common mode input range. Note that many
op amps can not work correctly without an input that is at least 2 volts
above its V(-) terminal. I dont recall the part number, but one of the
more common FET input op amps is exactly like this...and if you tie the
V(-) terminal to ground and try to run a 0.1v input the op amp will simply
not work at all. On the other hand, if you supply a -9v signal (as you
intend to do soon) the op amp works just fine. The main reason for this
property of an op amp is (as mentioned) the internal circuit input has to be
'biased' properly, and if the voltage isnt high enough it simply does
not get biased enough. If i remember right, the LM358 has PNP emitter
inputs so they work down low (anything more negative than the base causes
the PNP to conduct so it's well biased down to zero volts...the FET input
op amps on the other hand all have some threshold voltage that must be overcome).
Since we dont know the specs of your op amp, we must assume it requires
at least a 2v input above that of the V(-) terminal. Using a 9v battery
to bias the V(-) terminal to approximately -9 volts should work fine.
If the battery is run down (5v instead of 9v) it will probably still
be enough to make the op amp work ok because it probably only requires
about -2 volts or so.
To make sure *both* inputs are not "close to ground" after you hook up
the 9v battery, measure the dc voltage from the non inverting terminal
to V(-) on the op amp and make sure it reads some voltage such as 4 to
9 volts dc. Do the same with the inverting terminal.
Note that the signal inputs are referenced to zero volts (ground) so
the signals are still going to be in the low volts range (less than 1 volt).
ADDED:
I just checked the specs on an LF355 op amp, and found that the
supply voltage range is +/- 15 volts, but the min input common
mode voltage range is only +/- 11 volts. This means the input
of this op amp should not go lower than 4 volts above the V(-)
terminal, nor higher than 4 volts below the V(+) terminal.
This means the input to V(-) terminal should be at least 4 volts,
and the input to V(+) terminal should be at least 4 volts at all times.
With a supply voltage of +5 volts and -5 volts this means the
op amp inputs can be run from -1 volts to +1 volts, but not above
+1 or below -1 volts (either input). With a supply voltage of +5
volts and -9 volts the input can range from -5 volts to +1 volts
without any problem. I guess +9 and -9 volts would be good,
as this would allow an input range of -5 volts to +5 volts, although
in this app it may not be needed.
Robert:
That chip is to compensate for the 'cold' junction of the thermocouple.
It is actually part temperature sensor that senses the ambient temperature
and puts out a voltage that compensates for the fact that the cold
junction of the thermocouple is not at 0 degrees but rather at room
temperature. Since the various types of common thermocouples require
different compensation (volts per degree C) there are several outputs
on that chip where you use the one for the actual type of thermocouple
being used (J,T, etc). Pretty cool i think, im just wondering now
what the price of this chip and the op amp are...
Craig:
To add to the other replies I'll try to detail what 'close to ground' means.
If either the non-inverting terminal or the inverting terminal is at a voltage
that is close to zero volts (ground) it is said to be close to ground. The
actual voltage level (0.1 volts, 0.5 volts, 1 volts, 2 volts, etc.) is
determined by the actual specs for the op amp. For example, the LM358 can
work right down to zero volts (ground) and it says so in the data sheet.
If an op amp can not work like this it is usually mentioned in the data
sheet somewhere such as under common mode input range. Note that many
op amps can not work correctly without an input that is at least 2 volts
above its V(-) terminal. I dont recall the part number, but one of the
more common FET input op amps is exactly like this...and if you tie the
V(-) terminal to ground and try to run a 0.1v input the op amp will simply
not work at all. On the other hand, if you supply a -9v signal (as you
intend to do soon) the op amp works just fine. The main reason for this
property of an op amp is (as mentioned) the internal circuit input has to be
'biased' properly, and if the voltage isnt high enough it simply does
not get biased enough. If i remember right, the LM358 has PNP emitter
inputs so they work down low (anything more negative than the base causes
the PNP to conduct so it's well biased down to zero volts...the FET input
op amps on the other hand all have some threshold voltage that must be overcome).
Since we dont know the specs of your op amp, we must assume it requires
at least a 2v input above that of the V(-) terminal. Using a 9v battery
to bias the V(-) terminal to approximately -9 volts should work fine.
If the battery is run down (5v instead of 9v) it will probably still
be enough to make the op amp work ok because it probably only requires
about -2 volts or so.
To make sure *both* inputs are not "close to ground" after you hook up
the 9v battery, measure the dc voltage from the non inverting terminal
to V(-) on the op amp and make sure it reads some voltage such as 4 to
9 volts dc. Do the same with the inverting terminal.
Note that the signal inputs are referenced to zero volts (ground) so
the signals are still going to be in the low volts range (less than 1 volt).
ADDED:
I just checked the specs on an LF355 op amp, and found that the
supply voltage range is +/- 15 volts, but the min input common
mode voltage range is only +/- 11 volts. This means the input
of this op amp should not go lower than 4 volts above the V(-)
terminal, nor higher than 4 volts below the V(+) terminal.
This means the input to V(-) terminal should be at least 4 volts,
and the input to V(+) terminal should be at least 4 volts at all times.
With a supply voltage of +5 volts and -5 volts this means the
op amp inputs can be run from -1 volts to +1 volts, but not above
+1 or below -1 volts (either input). With a supply voltage of +5
volts and -9 volts the input can range from -5 volts to +1 volts
without any problem. I guess +9 and -9 volts would be good,
as this would allow an input range of -5 volts to +5 volts, although
in this app it may not be needed.
Robert:
That chip is to compensate for the 'cold' junction of the thermocouple.
It is actually part temperature sensor that senses the ambient temperature
and puts out a voltage that compensates for the fact that the cold
junction of the thermocouple is not at 0 degrees but rather at room
temperature. Since the various types of common thermocouples require
different compensation (volts per degree C) there are several outputs
on that chip where you use the one for the actual type of thermocouple
being used (J,T, etc). Pretty cool i think, im just wondering now
what the price of this chip and the op amp are...
LEDs vs Bulbs, LEDs are winning.
First, I want to say thanks! I was able to get it to work! I have a 13.8V Regulated DC power supply (left over from my R/C car racing days) which I built a 5V regulater for.
So, I powered the LT1025 voltage cold junction compensator using that, and powered the Op-Amp using the 5 volt lead from an old AT power supply (in place of the 9V battery).
Notes on the build:
- Project was built on a standard breadboard
- LT1025 Cold Junction Compensator was powered by a regulated 5V (7805) supply; actuall output is 4.98V
- LTKA00 Op-Amp was powered by the 5V lead on an AT switched power supply; actually voltage is 5.18V
- The 1M 1% resistor called for in the schematic was replaced by a 1M 5% resistor with a measured resistance of 0.998M
Here are my observations:
- At room temperature (20-22 C), the voltage output was generally around 0.018 mV, but it would move all over the place +/- 1 mV.
- With the end of the thermocouple sitting about 2 CM above my soldering iron, the reading went up to about 340 mV (34 C). This seemed somewhat accurate (the reading was varying by +/- 5%)
- With the end of the thermocouple touching the head of the soldering iron the reading went up to about 2V (the reading was still varying by +/- 5%)
- I rested a meat thermomeater on the end of the soldering iron and it got up to just over 200 degrees C.
From the observations above, I would say that the setup is working well.
Issues that need to be resolved:
1 - The accuracy at low temperature (under about 70 C) are way off
2 - I would like to get a reading that is not constantly changing +/- 5% or so
In an effort to resolve the above issues, I moved the setup form the breadbaord to a prototyping board with all items soldered in place except for the IC's which are in sockets. This did not resolve any issues.
What to try next:
- Replace the 1M 5% resistor with a 1M 1%
- Try different values of resistors for R6 to see if I can get a more accurate reading at room temperature
- Built another 5V regulator for use with the AT power supply in an effort to more closely match the voltage from the first 5V regulater.
So, I powered the LT1025 voltage cold junction compensator using that, and powered the Op-Amp using the 5 volt lead from an old AT power supply (in place of the 9V battery).
Notes on the build:
- Project was built on a standard breadboard
- LT1025 Cold Junction Compensator was powered by a regulated 5V (7805) supply; actuall output is 4.98V
- LTKA00 Op-Amp was powered by the 5V lead on an AT switched power supply; actually voltage is 5.18V
- The 1M 1% resistor called for in the schematic was replaced by a 1M 5% resistor with a measured resistance of 0.998M
Here are my observations:
- At room temperature (20-22 C), the voltage output was generally around 0.018 mV, but it would move all over the place +/- 1 mV.
- With the end of the thermocouple sitting about 2 CM above my soldering iron, the reading went up to about 340 mV (34 C). This seemed somewhat accurate (the reading was varying by +/- 5%)
- With the end of the thermocouple touching the head of the soldering iron the reading went up to about 2V (the reading was still varying by +/- 5%)
- I rested a meat thermomeater on the end of the soldering iron and it got up to just over 200 degrees C.
From the observations above, I would say that the setup is working well.
Issues that need to be resolved:
1 - The accuracy at low temperature (under about 70 C) are way off
2 - I would like to get a reading that is not constantly changing +/- 5% or so
In an effort to resolve the above issues, I moved the setup form the breadbaord to a prototyping board with all items soldered in place except for the IC's which are in sockets. This did not resolve any issues.
What to try next:
- Replace the 1M 5% resistor with a 1M 1%
- Try different values of resistors for R6 to see if I can get a more accurate reading at room temperature
- Built another 5V regulator for use with the AT power supply in an effort to more closely match the voltage from the first 5V regulater.
Robert:
Edd:
The max temperature range I expect to see is around 350 degrees C. This would produce a voltage of 3.5V at 10mV per degree C. The rail voltage I am using is 5V, so I guess I'd be ok? I could easily bump that up to 13V.In your case, you will probably never see swings large enough to approach either rail(assuming input voltages do not exceed certain levels).There may be a problem here if your inputs have to handle large range of swing as the output may exceed the power supply rails.
It is for offsetting the voltage from the thermocouple. The thermocouple will output a voltage depending on the temperature. The LT1025 is supposed to output a voltage opposite to that of the thermocouple at room temperature. So, if the thermocouple is outputting 5uV at 25 degrees C, the LT1025 will output -5uV, to bring the combined voltage to 0. I am not yet 100% up to speed on all the technical details, but that is pretty much how it works.One thing I am curious about is - what is LT1025. It looks like it may be a voltage source thats already biasing one of the inputs or may be for offsetting the output.
They are actually pretty cheap. Each chip is just over $3. There is also a higher accuarcy LT1025 chip which is about $6. You can get them as an all-in-one chip also for $6.60 or $12.60 for the more accuarate version. You can get free samples of them from Linear Technology http://www.linear.com. Thermocouples are also dirt cheap (if you have the machine to fuse the end, which I do) as they are simply a wire with the ends welded.Pretty cool i think, im just wondering now what the price of this chip and the op amp are...
Edd:
The schematic I am basing my design on is from the manufacture. It is on the last page of the datasheet for the LT1025. The output voltage differs for the various kinds of thermocouples, so their are different pins on the LT1025, as well as different setups. We use j-type thermocouple here at work. J-Type thermocouples consist of one copper wire and on iron wire.Also was that your circuit, as my referencing was this one as an app note from the manufacturer
Hi again Craig,
I would like to see you test it with two batteries, like 9v cells to
see if the noise goes away. The switching power supply could
easily be a problem with this kind of circuit.
If it works with the batteries then the power supply leads will
have to be filtered better.
Oh yeah, and we all forgot to mention thermal noise in the resistors.
The 1 percent film resistors will be better for noise than the carbon
resistors, especially the 1 meg resistor.
I would still try the batteries first though.
One way to calibrate the low end of the temperature range is to
melt some ice in a foam cup, where some of the ice melts into
water and you can dip the end into the water. You may have to
paint the end to water proof it. The reading for this setup should
be zero degrees C. I guess you would adjust the pot to get
a reading that equates to zero degrees C.
To help calibrate the higher end, dip the end in some boiling water
and look for 100 degrees C output.
You may have to find a pot setting that is 'close' for both these
tests and maybe not exact.
BTW how much did you pay for the two ic chips?
Also my understanding of the thermocouple goes like this...
The thermocouple consists of two metals, one 'hot' and one 'cold'.
The cold metal voltage automatically subtracts from the hot
metal voltage because the wires are connected at the ends and
act like two small batteries back to back, where both batteries are
sensitive (one maybe more than the other) to temperature.
If the hot metal is used to measure temperature and say it puts
out 1.5mv at 20 deg C with the cold metal end held at 0 deg C.
This would give the correct output without any compensation, 1.5mv.
However the cold metal end is not at 0 deg C, it is at maybe 20 deg C
too, and say at 20 deg C this metal develops 0.2mv. This 0.2mv
subtracts automatically from the 1.5mv so at the end of the thermocouple
wires you would only see 1.3mv (1.5 minus 0.2mv). Now this
voltage would be incorrect because we would want to see 1.5mv,
so the compensator would have to develop 0.2mv at 20 deg C
and that would be added to the 1.3mv to bring the total up to
the correct value of 1.5mv. Of course this would be amplified to
read perhaps 200mv output.
In other words, the compensator adds the same amount to the
total that is subtracted by the 'cold' metal because it is not
really at 0 degrees (the total would be correct if it was and we
would not need a compensator).
Here's a diagram:
The cold junctions are kept at 0 degrees, while the hot junction is
used to measure the temperature. The two metals used together
like this produce a voltage across the two cold junctions of known
amount for the particular two metals. The problem comes up
where it is not convenient to keep the two cold junctions at
0 degrees, so they are allowed to go to room temperature.
Because of this the voltage across the two cold junctions is
not correct for the type of metals used, so a compensator is
used. The compensator measures the room temperature and
produces a voltage that will add to the voltage across the
two cold junctions to make the sum voltage again correct.
The reason for the two wires is mostly because the voltage
across the one wire has to be measured somehow, and this
requires another wire to allow a voltmeter to sense the voltage
at the hot junction, and of course that wire also develops
a voltage of it's own between it's hot and cold junctions.
If it where not for the need of that second wire, all this would
be so much easier!
I would like to see you test it with two batteries, like 9v cells to
see if the noise goes away. The switching power supply could
easily be a problem with this kind of circuit.
If it works with the batteries then the power supply leads will
have to be filtered better.
Oh yeah, and we all forgot to mention thermal noise in the resistors.
The 1 percent film resistors will be better for noise than the carbon
resistors, especially the 1 meg resistor.
I would still try the batteries first though.
One way to calibrate the low end of the temperature range is to
melt some ice in a foam cup, where some of the ice melts into
water and you can dip the end into the water. You may have to
paint the end to water proof it. The reading for this setup should
be zero degrees C. I guess you would adjust the pot to get
a reading that equates to zero degrees C.
To help calibrate the higher end, dip the end in some boiling water
and look for 100 degrees C output.
You may have to find a pot setting that is 'close' for both these
tests and maybe not exact.
BTW how much did you pay for the two ic chips?
Also my understanding of the thermocouple goes like this...
The thermocouple consists of two metals, one 'hot' and one 'cold'.
The cold metal voltage automatically subtracts from the hot
metal voltage because the wires are connected at the ends and
act like two small batteries back to back, where both batteries are
sensitive (one maybe more than the other) to temperature.
If the hot metal is used to measure temperature and say it puts
out 1.5mv at 20 deg C with the cold metal end held at 0 deg C.
This would give the correct output without any compensation, 1.5mv.
However the cold metal end is not at 0 deg C, it is at maybe 20 deg C
too, and say at 20 deg C this metal develops 0.2mv. This 0.2mv
subtracts automatically from the 1.5mv so at the end of the thermocouple
wires you would only see 1.3mv (1.5 minus 0.2mv). Now this
voltage would be incorrect because we would want to see 1.5mv,
so the compensator would have to develop 0.2mv at 20 deg C
and that would be added to the 1.3mv to bring the total up to
the correct value of 1.5mv. Of course this would be amplified to
read perhaps 200mv output.
In other words, the compensator adds the same amount to the
total that is subtracted by the 'cold' metal because it is not
really at 0 degrees (the total would be correct if it was and we
would not need a compensator).
Here's a diagram:
Code: Select all
<------1.5mv----------->
/-------wire------------o cold \
hot junction O net 1.0mv difference at ends
\-------wire------------o cold /
<------0.5mv----------->
used to measure the temperature. The two metals used together
like this produce a voltage across the two cold junctions of known
amount for the particular two metals. The problem comes up
where it is not convenient to keep the two cold junctions at
0 degrees, so they are allowed to go to room temperature.
Because of this the voltage across the two cold junctions is
not correct for the type of metals used, so a compensator is
used. The compensator measures the room temperature and
produces a voltage that will add to the voltage across the
two cold junctions to make the sum voltage again correct.
The reason for the two wires is mostly because the voltage
across the one wire has to be measured somehow, and this
requires another wire to allow a voltmeter to sense the voltage
at the hot junction, and of course that wire also develops
a voltage of it's own between it's hot and cold junctions.
If it where not for the need of that second wire, all this would
be so much easier!
LEDs vs Bulbs, LEDs are winning.
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