hi,i want to connect load cell with picolog 1000 series data logger. suggest me how to connect it?

I use an amplifier circuit with the load cell so, the output ranges from 0 to 5 volts. My picolog can accept maximum volt of 2.5 volt.

how do i connect them and store the data ?

If you have any program for conversion of voltage to other values share a link so i can learn.

## picolog 1000 with load cell

### Re: picolog 1000 with load cell

Hi Arun,

I apologise for the delay in answering your question. The run up to Christmas and fall-out after it has kept us very busy!

First of all, just in case you haven't done so, the easiest way to do what you ask may be to adjust the amplifier to a maximum output of 2.5V if that is possible.

If that isn't possible then, I assume that in your comment of "if you have any program.....so I can learn", when you say 'Program' you mean instruction, or tutorial. Unfortunately we don't have any official material that you can learn from (creating an official document was on our list of things to do). However, I can give you an explanation on this post. I will have to assume that you at least have a basic level of knowledge in Electronics, e.g. understanding concepts such as current, voltage, ohms law, resistors in parallel and in series, etc (otherwise this would be an extremely long post).

So, to connect a voltage that can be between 0V and 5V, to an input of one of the PicoLog 1000 data loggers, you need to reduce the voltage down to 2.5V by creating a voltage divider circuit on our 'Small Terminal Board for the PicoLog 1000 data logger' (this is discussed on page 3, of the Small Terminal Board User Guide here: https://www.picotech.com/download/manua ... -guide.pdf). As this needs to divide the voltage in half, the values of the resistors for the potential divider should be the same (so the circuit including the connection to the amplifier will look like the 1st Diagram below where R1 and R2 are the same)

The actual values that you choose for both resistors in the voltage divider must not be too low. The reason for this is that the 2 resistors of the voltage divider, combined, create another potential divider with the output impedance of the amplifier (the first diagram, above). So, if the resistance of the potential divider is not high enough, there will be an error in the calculated values for the resistors in the potential divider (they will be slightly too small due some of the 5V output appearing across the Amplifier resistor on the output).

The values chosen must also not be too high. This is because R2 also forms a current divider with the input resistance (Radc) of the Data logger (as shown in the second diagram, below). If you choose high values for the resistors and the value of R2 is too high relative to Radc, then too much current will be flowing through Radc. This will mean that not enough current will be flowing through R2 to accurately divide the voltage across R1 and R2 in half (using the same resistor values for the voltage divider). So, the voltage being measured at the data logger input will be too small and, and once again, there will be an error in the voltage divider (whether the error will be too large for your measurement goals will depend upon the accuracy that you need for your measurement).

Typically, if you want maximum accuracy in your measurement, then you need to consider the stated accuracy of the PicoLog 1000 device you are using and then make sure that the total error is small enough in comparison to the stated accuracy to be insignificant (ideally < 1/3). However, you would also need to consider the accuracy values of the sensor and amplifier that you are using, and as I don't know what they are, and to avoid making this explanation too complicated (e.g. with uncertainties, offset errors, etc) I will focus on keeping the errors to below the 1% stated accuracy of the data logger.

So, let's assume that the output impedance of your amplifier is 50Ω (which is a typical output impedance used) and that you're using the PicoLog 1012, with an accuracy of 1%. Then the value of the resistors must be greater than 50Ω * 100 , and less than 1MΩ (the value of Radc) / 100, i.e. 5kΩ < R < 10kΩ. Also, Let's assume that this is your first time purchasing discrete resistors to solder onto a board. In this case you need to know that resistors are sold in values that are specified in a table, which relates to the resistors 'Tolerance'. This is because they are manufactured to be close to the value specified in the table, where the maximum + or - percentage error is the Tolerance value (popular tolerance values are 10% 5% and 1%).

Getting back to the actual values to choose for the 2 resistors, if we choose 7.5kΩ for R1 and R2 (midway between 5KΩ and 10KΩ) , and if we look at 1% tolerance values (to match the 1% error of the Data Logger) as shown in this link http://www.brannonelectronics.com/image ... 0VALUE.pdf then it's possible that you could have high positive and negative percentage errors for both resistors (i.e. ≃7.5kΩ -1%, and ≃7.5KΩ + 1%), which means that the total error in voltage at the midpoint of the potential divider would be >-1%, or <+1% (depending upon which way round you place the resistors in the divider). So, you could go ahead and purchase 2 of these to make a suitable voltage divider on the Terminal Board.

Quite often though, what you will find is that there is no value that exactly matches the values you have calculated, e.g. if instead you had chosen a resistor value of 8kΩ for your voltage divider, in the 1% tolerance resistor table the nearest preferred value would be 8.06kΩ. This would add more error (i.e. the worst-case error due to the tolerance + the offset error between your calculated value and the nearest preferred value) which would be an additional worst-case error in the resistance value of +0.75%. So you could end up buying resistor values of ≃ 8kΩ + 1.75% = 8.14kΩ, and ≃ 8kΩ - 0.25% = 7.98kΩ. Inserting these values into the equation for the potential divider would give us a voltage at the input of the data logger as defined below:

Vin = (5 * 7.98)/(8.14 + 7.98) = 2.475V which would be a voltage error of -1%

(notice that we would always put the smaller resistor value in the bottom leg of the voltage divider, to make sure that the input voltage doesn't go above 2.5V and therefore out of range in the data logger)

So these resistor values, if they happened to be the absolute worst case values would be equivalent to our 1% error limit. As the likelihood of exact worst case values in both cases would slim, we could go ahead and purchase 2 of these for a suitable divider.

As a further example, if we assume that the difference between the calculated value and nearest preferred value resulted in a voltage error of >1% (or outside of the accuracy you want) then you have would have 4 options. Option (1) would be that you could either use resistors in parallel or in series, to make up a more accurate value, e.g. you could choose resistor values of 4.01kΩ, and 3.99kΩ in series, which would add up to exactly 8KΩ, giving you two perfect nearest preferred values for the legs of the divider. Option (2) would be that you could look for better or ideal nearest preferred value resistors using a smaller tolerance table e.g. 0.5% tolerance (which would not give you any improvement for the value of 8kΩ). If options 1 and 2 will not reduce the error enough then option (3) would be that you would have to include the Data Logger input impedance, if they are too high (so R1 would become Rout + R1), or the Amplifier output impedance, if they are too low, in the calculation of the voltage divider (as described at the bottom of page 3 in the link that I gave you). Option (4) would be that you would need to use a high input impedance voltage follower (or unity gain buffer) between the voltage divider and input of the Data Logger, or a switch the amplifier you are using for a lower output impedance amplifier in order to match the impedance that you need for the divider.

Note that we started out by making the assumption that your sensor and amplifier error were less than 1%. if these are larger, or if you don't need accuracy as high as 1%, then you would obviously have more freedom to select which resistors you want from which tolerance band.

In case, from your comment "if you have any program...so I can learn", when you said 'Program', instead you meant calculator, then there is one here that can automatically calculate the resistor values in a voltage divider for you: http://www.mantaro.com/resources/impeda ... ulator.htm, and can also help you calculate the nearest preferred value to your calculated resistor value from a resistor tolerance table, along with the percentage error.

If you're interested in learning the electronics behind how to change a + and - voltage to one that just goes from 0 to 2.5V to connect to a PicoLog 1000 input using the Small Terminal Board, i.e. how to apply offset and scaling (as mentioned on page 4 of the Small Terminal Board User Guide) then below is a document that gives the equations that you need for the resistor values and shows you how these equations were found. The document hasn't been made generally available because it is incomplete, but most of the parts that are missing are explained above.

Regards,

Gerry

I apologise for the delay in answering your question. The run up to Christmas and fall-out after it has kept us very busy!

First of all, just in case you haven't done so, the easiest way to do what you ask may be to adjust the amplifier to a maximum output of 2.5V if that is possible.

If that isn't possible then, I assume that in your comment of "if you have any program.....so I can learn", when you say 'Program' you mean instruction, or tutorial. Unfortunately we don't have any official material that you can learn from (creating an official document was on our list of things to do). However, I can give you an explanation on this post. I will have to assume that you at least have a basic level of knowledge in Electronics, e.g. understanding concepts such as current, voltage, ohms law, resistors in parallel and in series, etc (otherwise this would be an extremely long post).

So, to connect a voltage that can be between 0V and 5V, to an input of one of the PicoLog 1000 data loggers, you need to reduce the voltage down to 2.5V by creating a voltage divider circuit on our 'Small Terminal Board for the PicoLog 1000 data logger' (this is discussed on page 3, of the Small Terminal Board User Guide here: https://www.picotech.com/download/manua ... -guide.pdf). As this needs to divide the voltage in half, the values of the resistors for the potential divider should be the same (so the circuit including the connection to the amplifier will look like the 1st Diagram below where R1 and R2 are the same)

The actual values that you choose for both resistors in the voltage divider must not be too low. The reason for this is that the 2 resistors of the voltage divider, combined, create another potential divider with the output impedance of the amplifier (the first diagram, above). So, if the resistance of the potential divider is not high enough, there will be an error in the calculated values for the resistors in the potential divider (they will be slightly too small due some of the 5V output appearing across the Amplifier resistor on the output).

The values chosen must also not be too high. This is because R2 also forms a current divider with the input resistance (Radc) of the Data logger (as shown in the second diagram, below). If you choose high values for the resistors and the value of R2 is too high relative to Radc, then too much current will be flowing through Radc. This will mean that not enough current will be flowing through R2 to accurately divide the voltage across R1 and R2 in half (using the same resistor values for the voltage divider). So, the voltage being measured at the data logger input will be too small and, and once again, there will be an error in the voltage divider (whether the error will be too large for your measurement goals will depend upon the accuracy that you need for your measurement).

Typically, if you want maximum accuracy in your measurement, then you need to consider the stated accuracy of the PicoLog 1000 device you are using and then make sure that the total error is small enough in comparison to the stated accuracy to be insignificant (ideally < 1/3). However, you would also need to consider the accuracy values of the sensor and amplifier that you are using, and as I don't know what they are, and to avoid making this explanation too complicated (e.g. with uncertainties, offset errors, etc) I will focus on keeping the errors to below the 1% stated accuracy of the data logger.

So, let's assume that the output impedance of your amplifier is 50Ω (which is a typical output impedance used) and that you're using the PicoLog 1012, with an accuracy of 1%. Then the value of the resistors must be greater than 50Ω * 100 , and less than 1MΩ (the value of Radc) / 100, i.e. 5kΩ < R < 10kΩ. Also, Let's assume that this is your first time purchasing discrete resistors to solder onto a board. In this case you need to know that resistors are sold in values that are specified in a table, which relates to the resistors 'Tolerance'. This is because they are manufactured to be close to the value specified in the table, where the maximum + or - percentage error is the Tolerance value (popular tolerance values are 10% 5% and 1%).

Getting back to the actual values to choose for the 2 resistors, if we choose 7.5kΩ for R1 and R2 (midway between 5KΩ and 10KΩ) , and if we look at 1% tolerance values (to match the 1% error of the Data Logger) as shown in this link http://www.brannonelectronics.com/image ... 0VALUE.pdf then it's possible that you could have high positive and negative percentage errors for both resistors (i.e. ≃7.5kΩ -1%, and ≃7.5KΩ + 1%), which means that the total error in voltage at the midpoint of the potential divider would be >-1%, or <+1% (depending upon which way round you place the resistors in the divider). So, you could go ahead and purchase 2 of these to make a suitable voltage divider on the Terminal Board.

Quite often though, what you will find is that there is no value that exactly matches the values you have calculated, e.g. if instead you had chosen a resistor value of 8kΩ for your voltage divider, in the 1% tolerance resistor table the nearest preferred value would be 8.06kΩ. This would add more error (i.e. the worst-case error due to the tolerance + the offset error between your calculated value and the nearest preferred value) which would be an additional worst-case error in the resistance value of +0.75%. So you could end up buying resistor values of ≃ 8kΩ + 1.75% = 8.14kΩ, and ≃ 8kΩ - 0.25% = 7.98kΩ. Inserting these values into the equation for the potential divider would give us a voltage at the input of the data logger as defined below:

Vin = (5 * 7.98)/(8.14 + 7.98) = 2.475V which would be a voltage error of -1%

(notice that we would always put the smaller resistor value in the bottom leg of the voltage divider, to make sure that the input voltage doesn't go above 2.5V and therefore out of range in the data logger)

So these resistor values, if they happened to be the absolute worst case values would be equivalent to our 1% error limit. As the likelihood of exact worst case values in both cases would slim, we could go ahead and purchase 2 of these for a suitable divider.

As a further example, if we assume that the difference between the calculated value and nearest preferred value resulted in a voltage error of >1% (or outside of the accuracy you want) then you have would have 4 options. Option (1) would be that you could either use resistors in parallel or in series, to make up a more accurate value, e.g. you could choose resistor values of 4.01kΩ, and 3.99kΩ in series, which would add up to exactly 8KΩ, giving you two perfect nearest preferred values for the legs of the divider. Option (2) would be that you could look for better or ideal nearest preferred value resistors using a smaller tolerance table e.g. 0.5% tolerance (which would not give you any improvement for the value of 8kΩ). If options 1 and 2 will not reduce the error enough then option (3) would be that you would have to include the Data Logger input impedance, if they are too high (so R1 would become Rout + R1), or the Amplifier output impedance, if they are too low, in the calculation of the voltage divider (as described at the bottom of page 3 in the link that I gave you). Option (4) would be that you would need to use a high input impedance voltage follower (or unity gain buffer) between the voltage divider and input of the Data Logger, or a switch the amplifier you are using for a lower output impedance amplifier in order to match the impedance that you need for the divider.

Note that we started out by making the assumption that your sensor and amplifier error were less than 1%. if these are larger, or if you don't need accuracy as high as 1%, then you would obviously have more freedom to select which resistors you want from which tolerance band.

In case, from your comment "if you have any program...so I can learn", when you said 'Program', instead you meant calculator, then there is one here that can automatically calculate the resistor values in a voltage divider for you: http://www.mantaro.com/resources/impeda ... ulator.htm, and can also help you calculate the nearest preferred value to your calculated resistor value from a resistor tolerance table, along with the percentage error.

If you're interested in learning the electronics behind how to change a + and - voltage to one that just goes from 0 to 2.5V to connect to a PicoLog 1000 input using the Small Terminal Board, i.e. how to apply offset and scaling (as mentioned on page 4 of the Small Terminal Board User Guide) then below is a document that gives the equations that you need for the resistor values and shows you how these equations were found. The document hasn't been made generally available because it is incomplete, but most of the parts that are missing are explained above.

Regards,

Gerry

Gerry

Technical Specialist

Technical Specialist