I want to build a circuit that calls for a 1M 1/4W 1% resistor. I couldn't find any at my local electronics store so I bought a pack of 5% instead. Would it be ok to test them until I find one that is within 1% of 1M and use it? Does this tolerance come into affect while the circuit is in use (temperature change), or can I assume that the value I read when I measure it on the bench is the value it will hold while in use?
Out of the pack of 250, there has to be some that are within 1%, right??
Question about resistor tolerances
You most probably can but ultimately it depends on the circuit and its application of course.
First, once you get up to 1M and higher, even 1% is a good chunk of ohms to be off by. In most cases, by the time you are using a resistor this large, its often to limit current to very low values but usually not precision values. For example, we use them in series with a JFET Op amp in a buffer configuration. You don't really need the resistor since the input takes essentially no current but in the event of a failure of the op amp, we do not want excessive current on that node.
Secondly, it has become the practice of some designers and manufacturers to specify 1% resistors in all cases because very often the price difference is tiny and it is easier to stock all 1% values than to stock 1%, 5% and 10% of the same value. (its bad enough stocking caps of the same value with different lead spacing)
The moral of the above 2 points is that it may have never had to be that tight in the first place.
More direct to the point, there is another factor in specifying resistors and that would be the temperature coefficient. This is often specified separately but essentially linked to the overall tolerance. If precision is important for some reason, check the tolerance over the expected temperature range it will see as well as at room temp.
Every once in a while, you need a resistor in a high frequency roll. In this case, the inducatance of the part may also be important. Carbon comp has low L while wire wound and metal film can have higher inductance.
First, once you get up to 1M and higher, even 1% is a good chunk of ohms to be off by. In most cases, by the time you are using a resistor this large, its often to limit current to very low values but usually not precision values. For example, we use them in series with a JFET Op amp in a buffer configuration. You don't really need the resistor since the input takes essentially no current but in the event of a failure of the op amp, we do not want excessive current on that node.
Secondly, it has become the practice of some designers and manufacturers to specify 1% resistors in all cases because very often the price difference is tiny and it is easier to stock all 1% values than to stock 1%, 5% and 10% of the same value. (its bad enough stocking caps of the same value with different lead spacing)
The moral of the above 2 points is that it may have never had to be that tight in the first place.
More direct to the point, there is another factor in specifying resistors and that would be the temperature coefficient. This is often specified separately but essentially linked to the overall tolerance. If precision is important for some reason, check the tolerance over the expected temperature range it will see as well as at room temp.
Every once in a while, you need a resistor in a high frequency roll. In this case, the inducatance of the part may also be important. Carbon comp has low L while wire wound and metal film can have higher inductance.
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ecerfoglio
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- Location: Buenos Aires Argentina
- Contact:
Yes, you "should" measure the resistor at all the temperatures it may have when working (or at least at the lowest and highest expected temperatures) to be able to say that it stays within 1% of the nominal value.Does this tolerance come into affect while the circuit is in use (temperature change), .....
And then there are aging issues
....As Philba said, to measure to 1% you must have a more accurate meter (maybe 0.1%)
If the circuit really needs 1% accuracy, you may series (or paralell, but with values like 1M it's better to series) connect 2 or more resistors to total the required value.
Or you may use a 5% one in a prototype while you get the 1% resistor.
It deppends on the application. If you are building a meter's front end you need 1% so the meter stays accurate, or an audio preamp may call for 1% resistors because they have lower noise....
E. Cerfoglio
Buenos Aires
Argentina
Buenos Aires
Argentina
Thanks for the replies!
The applicaiton is for an op-amp for a cold junction compensator for a J type thermocouple. More specifically, it is for Linear Technology cold junction compensater part number LT1025CN8 and Linear Technology op-amp part number LTKA00CN8.
Linear Tech also has a chip that has the CJC and Amp matched together, part number LTK001CN8 (but I don't have that yet). Unfortunately, there is no datasheet just for the LTKA00CN8 amp, just for the CJC and the CJC+Op-Amp combo.
The accuracy of the temperature reading isn't critical, withing +/- 3 degrees F is fine. The temperature range will be from room temp (79F) to about 320F, and should not exceed 350F.
Here are the specs for LTKA00CN8:
Part Number = LTKA00CN8
Manufacturer Name = Linear Technology
Description = Special-Application Op Amp - Thermocouple Cold Junction Amp
Input Bias Current Max. (A) = 600p
V(io) Max. (V) = 35u
I(io) Max. (A) = 500p
Com Mode Inp Range (VICR) = 1.0
Temp Coef. of VIO = 1.5u
Nom. Supp (V) = 40
Package = DIP
Pins = 8
Military = N
The LT1025 Cold Junction Compensator datasheet can be viewed at http://www.linear.com/pc/productDetail. ... 1073,P1232
The LTK001 Cold Junction Compensator and Op-Amp kit datasheet can be viewed at http://www.linear.com/pc/productDetail. ... 1073,P1181
I really appreciate the help!
The applicaiton is for an op-amp for a cold junction compensator for a J type thermocouple. More specifically, it is for Linear Technology cold junction compensater part number LT1025CN8 and Linear Technology op-amp part number LTKA00CN8.
Linear Tech also has a chip that has the CJC and Amp matched together, part number LTK001CN8 (but I don't have that yet). Unfortunately, there is no datasheet just for the LTKA00CN8 amp, just for the CJC and the CJC+Op-Amp combo.
The accuracy of the temperature reading isn't critical, withing +/- 3 degrees F is fine. The temperature range will be from room temp (79F) to about 320F, and should not exceed 350F.
Here are the specs for LTKA00CN8:
Part Number = LTKA00CN8
Manufacturer Name = Linear Technology
Description = Special-Application Op Amp - Thermocouple Cold Junction Amp
Input Bias Current Max. (A) = 600p
V(io) Max. (V) = 35u
I(io) Max. (A) = 500p
Com Mode Inp Range (VICR) = 1.0
Temp Coef. of VIO = 1.5u
Nom. Supp (V) = 40
Package = DIP
Pins = 8
Military = N
The LT1025 Cold Junction Compensator datasheet can be viewed at http://www.linear.com/pc/productDetail. ... 1073,P1232
The LTK001 Cold Junction Compensator and Op-Amp kit datasheet can be viewed at http://www.linear.com/pc/productDetail. ... 1073,P1181
I really appreciate the help!
I would guess that the tolerance was spec'd for the benefit of stability over time and temp not so much for initial absolute value.
As long as there is a means to calibrate the thermocouple, any error in absolute value would be unimportant unless it changed while in use.
The thermocouple wil be subjected to the temp range you specified, hopefully not the electronics themselves.
An easy test is to connect your resistor to the ohmmeter then put the resistor on top of a light bulb. How much does it change. It probably won't experience self heating from the current flowing within it.
Where are you using the 1M in the circuit, the examples show 256K maximum. If it's in the feedback or creates a bias voiltage or current, use 1% if it is in series with the inputs, who cares, no current flows through them anyway (I(io)500pA)
As long as there is a means to calibrate the thermocouple, any error in absolute value would be unimportant unless it changed while in use.
The thermocouple wil be subjected to the temp range you specified, hopefully not the electronics themselves.
An easy test is to connect your resistor to the ohmmeter then put the resistor on top of a light bulb. How much does it change. It probably won't experience self heating from the current flowing within it.
Where are you using the 1M in the circuit, the examples show 256K maximum. If it's in the feedback or creates a bias voiltage or current, use 1% if it is in series with the inputs, who cares, no current flows through them anyway (I(io)500pA)
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Dave Dixon
- Posts: 436
- Joined: Thu May 01, 2003 1:01 am
- Location: Wichita, KS
- Contact:
The feedback resistance can be varied by as much as 2k via R5. That's 0.2% of the value of R3. Furthermore, R6 which is also in the feedback loop is 10%. However a 10% error on R6 is only 84 ohms while the 1% error of R3 would be 10k.
That loop resistance is critical to determining the gain of the amp which in turn is what sets the 10mV/degC slope. Its a little more complicated because R4,5,6 form a voltage divider which sets the max amount of the output which can be fed back. You could change R6 to a rheostat and use that to further compensate for the error in R3. Changing R6+R3 significantly changes the slope of the output thereby changing the range (for example it might become 20mV/degC)
Go ahead and use the 5% and select the best one. Just be aware that any calibration you do today might not be accurate a week later or if the ambient temperature varies too much. Comparing the temp of two objects side by side or the same object over a few hours should be accurate but comparing measurements taken at various days or weeks may not correlate well.
Who knows what series and parallel connecting resistors does to the tolerance of the resulting network? My intuition says that in series the tolerance adds and in parallel it averages but it never gets better. I couldn't confirm this yet though.
That loop resistance is critical to determining the gain of the amp which in turn is what sets the 10mV/degC slope. Its a little more complicated because R4,5,6 form a voltage divider which sets the max amount of the output which can be fed back. You could change R6 to a rheostat and use that to further compensate for the error in R3. Changing R6+R3 significantly changes the slope of the output thereby changing the range (for example it might become 20mV/degC)
Go ahead and use the 5% and select the best one. Just be aware that any calibration you do today might not be accurate a week later or if the ambient temperature varies too much. Comparing the temp of two objects side by side or the same object over a few hours should be accurate but comparing measurements taken at various days or weeks may not correlate well.
Who knows what series and parallel connecting resistors does to the tolerance of the resulting network? My intuition says that in series the tolerance adds and in parallel it averages but it never gets better. I couldn't confirm this yet though.
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Robert Reed
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- Joined: Wed Nov 24, 2004 1:01 am
- Location: ASHTABULA,OHIO
- Contact:
The Op-Amps output at the R3 injection point will be approximately 0.55 the level of the actual Op-Amp output. From this point R1 & R3 set the major closed loop gain of the Op-Amp ( approx. 100). With a dual supply, the output would normally be in the area of 0 VDCwith 0 VDC input. I am not sure what they are trying to accomplish here,as I am not familiar with your chip, but it looks like some sort of minor gain/ level setting adjustment .So any variation in R1 or R3 will alter the gain and any adjustment to R5 can restore that gain. Once these are set, the real culprit now becomes Ambient temperature. If R3 were under 100k, I would say do not worry too much about tight percentages . But from my experience, carbon film resistors of 1meg and above have poor temperature coefficients and at 10 meg. they are terrible. This is where metal film 1% resistors earn their keep - stability. Often in circuit design a wide range of values is possible to use and especially if any overall adjustments are included. But once set the metal film will hold that value and is the reason I choose them for critical circuits and not neccessarily for their 1% precision.
Hi there,
I agree that 1 percent resistors should be used for R1 and R3 because
that's what they spec for this circuit. However, if you must use a
5 percent resistor for R3 then be sure to use a 5 percent resistor
for R1 as well. The reason for this is because the gain of the circuit
depends heavily on the *ratio* of R3 to R1 (as well as other things)
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. Of course if a 1 percent is used for R3 then a 1 percent
should also be used for R1.
The actual numbers for a change in R3 of 1 percent (holding R1 constant)
show that this results in a change of output of 1 percent, but if R1 is
also changed by 1 percent (in the same direction as a temperature
change would cause) then the output only changes by 0.005 percent,
which is a vast improvement! This means that if R3 were to change
by 5 percent and R1 were to change by 1 percent (using different types
of resistors for R3 and R1) then the output would change by a whole
4 percent, but if both R1 and R3 were to change by 5 percent then
the output would change by about 0.02 percent, which isnt bad either.
Thus, make R1 and R3 the same type of resistors.
The temperature tracking ability of this circuit also depends on the
what type of resistors R4 and R6 are. These two should be the same
type too, and it is preferred to make them the same as the composition
of the pot if possible.
Here are the DC equations for this circuit that were used to understand
the behaviour of this circuit better:
(descriptions of each of these follows below)
para(r1,r2)=(r1*r2/(r1+r2))
Vj=1/400 (a typical input voltage)
A=20000
R1=10
R3=1000
R4=10
R6=8.4
R5a=0.2
R5b=2-R5a
rx=para(R1+R3,R4+R5a)
ra=rx/(rx+R6+R5b)
rb=R1/(R1+R3)
Varm=(Vj*ra-A*ra*Vj*rb+A*ra*Vj-Vj)/(1+A*ra*rb)
Vn=(Varm+Vj)*rb-Vj
Vo=-A*Vn
where
'para' is simply a formula to calculate the resistance
of two resistances in parallel
Vj is the voltage of the J type thermocouple junction
A is the approximate gain of the op amp (change as needed)
the resistors, R1 to R6 are those that appear in the circuit,
in K ohms.
R5a is the upper part of the pot after adjustment
R5b is the lower part of the pot after adjustment
rx,ra, and rb are calculated intermediate resistances
Varm is the voltage at the wiper (arm) of the pot after adjustment
Vn is the voltage that appears at the non-inverting terminal of the op amp
Vo is the output voltage (Vout on the schematic)
It is also assumed that the op amp has very high input impedance.
I agree that 1 percent resistors should be used for R1 and R3 because
that's what they spec for this circuit. However, if you must use a
5 percent resistor for R3 then be sure to use a 5 percent resistor
for R1 as well. The reason for this is because the gain of the circuit
depends heavily on the *ratio* of R3 to R1 (as well as other things)
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. Of course if a 1 percent is used for R3 then a 1 percent
should also be used for R1.
The actual numbers for a change in R3 of 1 percent (holding R1 constant)
show that this results in a change of output of 1 percent, but if R1 is
also changed by 1 percent (in the same direction as a temperature
change would cause) then the output only changes by 0.005 percent,
which is a vast improvement! This means that if R3 were to change
by 5 percent and R1 were to change by 1 percent (using different types
of resistors for R3 and R1) then the output would change by a whole
4 percent, but if both R1 and R3 were to change by 5 percent then
the output would change by about 0.02 percent, which isnt bad either.
Thus, make R1 and R3 the same type of resistors.
The temperature tracking ability of this circuit also depends on the
what type of resistors R4 and R6 are. These two should be the same
type too, and it is preferred to make them the same as the composition
of the pot if possible.
Here are the DC equations for this circuit that were used to understand
the behaviour of this circuit better:
(descriptions of each of these follows below)
para(r1,r2)=(r1*r2/(r1+r2))
Vj=1/400 (a typical input voltage)
A=20000
R1=10
R3=1000
R4=10
R6=8.4
R5a=0.2
R5b=2-R5a
rx=para(R1+R3,R4+R5a)
ra=rx/(rx+R6+R5b)
rb=R1/(R1+R3)
Varm=(Vj*ra-A*ra*Vj*rb+A*ra*Vj-Vj)/(1+A*ra*rb)
Vn=(Varm+Vj)*rb-Vj
Vo=-A*Vn
where
'para' is simply a formula to calculate the resistance
of two resistances in parallel
Vj is the voltage of the J type thermocouple junction
A is the approximate gain of the op amp (change as needed)
the resistors, R1 to R6 are those that appear in the circuit,
in K ohms.
R5a is the upper part of the pot after adjustment
R5b is the lower part of the pot after adjustment
rx,ra, and rb are calculated intermediate resistances
Varm is the voltage at the wiper (arm) of the pot after adjustment
Vn is the voltage that appears at the non-inverting terminal of the op amp
Vo is the output voltage (Vout on the schematic)
It is also assumed that the op amp has very high input impedance.
LEDs vs Bulbs, LEDs are winning.
"Out of the pack of 250, there has to be some that are within 1%, right??"
Don't bet on it.
For one project, we needed two .1% resistors. We couldn't get them in a reasonable time. Since they were in series, we decided to select two resistors whose sum was within .1%. We ordered 100 resistors (1% tolerance) and measured each one. We found that we had 100 resistors that were .3% high, plus or minus abuot .02%. Since all were high, we couldn't select a high/low pair.
Production machinery has become amazingly consistent. Most of the tolerance is not variation between individual parts, but in the accuracy of the machine setup.
Temperature tracking may be a problem no matter what kind of resistors you use. The metal film used for a 10K resistor is probably not the same composition as the metal film used in a 1 meg resistor, and will probably have a slightly different temperature characteristic. Using a 10K metal film with a 1 meg carbon film would be even worse.
If your 1 meg resistor is a carbon composition type, the situation may be much worse. I have seen 5% carbon composition resistors change by 20% when they were soldered into a circuit. Carbon film resistors may be better than this, since the ceramic core is probably more dimensionally stablle.
You may also have a problem with the cold junction. You actually have three junctions (copper to iron, iron to nickel/copper, and nickel/copper to copper) in the drawing. The cold junction compensation is probably for an iron to nickel/copper junction. Since you don't have a cold iron to nickel/copper junction, the compensation may not be correct.
Ideally, you would have four junctions (copper to iron, iron to nickel copper, nickel/copper to iron, and iron to copper) . The iron to nickel/copper junction would be the measuring junction. The nickel/copper to iron junction would be the cold junction. This would have to be at the same temperature as the cold junction compensation chip. The copper to iron and iron to copper junctions should be at the same temperature so that their thermoelectric voltages cancel. This will probably force you to use thermocouple wire rather than copper wire for the leads to the remote thermocouple.
Don't bet on it.
For one project, we needed two .1% resistors. We couldn't get them in a reasonable time. Since they were in series, we decided to select two resistors whose sum was within .1%. We ordered 100 resistors (1% tolerance) and measured each one. We found that we had 100 resistors that were .3% high, plus or minus abuot .02%. Since all were high, we couldn't select a high/low pair.
Production machinery has become amazingly consistent. Most of the tolerance is not variation between individual parts, but in the accuracy of the machine setup.
Temperature tracking may be a problem no matter what kind of resistors you use. The metal film used for a 10K resistor is probably not the same composition as the metal film used in a 1 meg resistor, and will probably have a slightly different temperature characteristic. Using a 10K metal film with a 1 meg carbon film would be even worse.
If your 1 meg resistor is a carbon composition type, the situation may be much worse. I have seen 5% carbon composition resistors change by 20% when they were soldered into a circuit. Carbon film resistors may be better than this, since the ceramic core is probably more dimensionally stablle.
You may also have a problem with the cold junction. You actually have three junctions (copper to iron, iron to nickel/copper, and nickel/copper to copper) in the drawing. The cold junction compensation is probably for an iron to nickel/copper junction. Since you don't have a cold iron to nickel/copper junction, the compensation may not be correct.
Ideally, you would have four junctions (copper to iron, iron to nickel copper, nickel/copper to iron, and iron to copper) . The iron to nickel/copper junction would be the measuring junction. The nickel/copper to iron junction would be the cold junction. This would have to be at the same temperature as the cold junction compensation chip. The copper to iron and iron to copper junctions should be at the same temperature so that their thermoelectric voltages cancel. This will probably force you to use thermocouple wire rather than copper wire for the leads to the remote thermocouple.
Hi again,
Oh yes, i forgot to mention the situation where you try to find
two 5 percent resistors that are within 1 percent of their
marked values.
For this circuit, it is not really necessary to get the two resistors
R1 and R3 within 1 percent of their marked values, it's much more
important to get them both to within the *same* percentage of
their marked values. For example, you find a resistor that is
marked 10k but measures 10.2k instead, which is 2 percent higher
than the marked value, then you must also find a resistor that is
2 percent higher than 1M, which comes out to 1.02M ohms.
With this 2 percent error for *both* resistors, the output change
would be only about 0.01 percent, which is still very very good,
and that's even before calibration. Once calibrated this error
goes away. The reason this works is because again the *ratio* of
the two resistors, R1 and R3, are much more important than the
actual absolute values of the two.
BTW, most circuts like this would go through a calibration procedure
followed by at least one temperature run where the output is monitored
and compared against some known accurate reference.
The nice thing is, if the circuit is always used at room temperature
there shouldnt be too much of a problem.
Oh yes, i forgot to mention the situation where you try to find
two 5 percent resistors that are within 1 percent of their
marked values.
For this circuit, it is not really necessary to get the two resistors
R1 and R3 within 1 percent of their marked values, it's much more
important to get them both to within the *same* percentage of
their marked values. For example, you find a resistor that is
marked 10k but measures 10.2k instead, which is 2 percent higher
than the marked value, then you must also find a resistor that is
2 percent higher than 1M, which comes out to 1.02M ohms.
With this 2 percent error for *both* resistors, the output change
would be only about 0.01 percent, which is still very very good,
and that's even before calibration. Once calibrated this error
goes away. The reason this works is because again the *ratio* of
the two resistors, R1 and R3, are much more important than the
actual absolute values of the two.
BTW, most circuts like this would go through a calibration procedure
followed by at least one temperature run where the output is monitored
and compared against some known accurate reference.
The nice thing is, if the circuit is always used at room temperature
there shouldnt be too much of a problem.
LEDs vs Bulbs, LEDs are winning.
Thanks for the info guys! Although I was able to read the schematic, I didn't truly understand, your explanations really helped! I think I will tkae up Dave's offer on those 1% resistors!
I built the circuit last night, but when I tested it I was receiving an output voltage of 0.554V regardless of the temperature of the thermocouple or the adjustment of the 2k pot. At this point I'm not too worried about accuracy, I'd just like to see the voltage rise with the temperature of the thermocouple.
I have attached a (horribly drawn) layout of how I wired this up on the breadboard. Can you guys confirm that I did this correctly? I checked it over and over again, and it seems I got it right (from what I understand).
The +5v is actually 4.98
The actual, room temperature value of R3 is 0.998M (998K)
I hooked my digital multimeter directly to the thermocouple and was able to read 0- ~5 mV from it (heating it up with a soldering iron).
I built the circuit last night, but when I tested it I was receiving an output voltage of 0.554V regardless of the temperature of the thermocouple or the adjustment of the 2k pot. At this point I'm not too worried about accuracy, I'd just like to see the voltage rise with the temperature of the thermocouple.
I have attached a (horribly drawn) layout of how I wired this up on the breadboard. Can you guys confirm that I did this correctly? I checked it over and over again, and it seems I got it right (from what I understand).
The +5v is actually 4.98
The actual, room temperature value of R3 is 0.998M (998K)
I hooked my digital multimeter directly to the thermocouple and was able to read 0- ~5 mV from it (heating it up with a soldering iron).
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