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I want to estimate this industrial amplifier's temperature drift.

There is a section in the data sheet in ratings section as follows:

enter image description here

Basically the amplifier I use above is now set to around total gain of G = 2300. I measured this. And a 10kg nominal force transducer attached to this amplifier has 2mV/V sensitivity. This means for 10V excitation the transducer output increases 20mV for 10kg increase and for a 1kg increase the transducer output increases 2mV.

So to summarize, there is a transducer whose output increases 2mV per 1kg increase. This then goes to the industrial amplifier in question whose total gain i set to 2300 for now.

An example(after setting to around zero volt for no load):

Transducer is loaded 1300 gr and outputs 2.1mV; the amplifier outputs 6.178V

Transducer is loaded 1800 gr and outputs 3.1mV; the amplifier outputs 8.504V

So you can see the total gain is around 2300 from (8.5-6.17)V/1mV.

My question is how can I interpret the data sheet of this amplifier and estimate the change in the amplifier's output voltage for a 10K temperature increase? I don't understand how to do this by data-sheet's way of writing. So as you see above at a constant temperature I obtained around 2.3V increase for a 1kg load increase. What would that increase in voltage be if the temperature would be 10K degrees more? In other words what would be the total drift of the amplifier output for a 10K temperature increase?

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Referring to your previous post titled A question about temperature change of a transducer, the load cell's sensitivity is defined as \$C_n=2\,mV/V\$, and the load cell's temperature effect on sensitivity \$TK_C\$ is specified as \$0.05\,\%\,of\,C_n/10\,K\$:

$$ C_n=C_{n0}+0.05\,\%\;C_{n0}\left ( \frac{T-T_0}{10} \right ) \;\;\;\;\;\;\;\;\;\;(1) \\[0.35in] \rightarrow C_n=C_{n0}\, \left \{ 1 + \left ( \frac{1/20}{100} \right ) \left ( \frac{T-T_0}{10} \right ) \right \} \;\;\;\;\;\;\;\;\;\;(2) \\[0.35in] \rightarrow C_n = C_{n0} \left ( 1 + \frac{T-T_0}{20,000} \right ) \;\;\;\;\;\;\;\;\;\;(3) $$

where

T0   = Load cell's initial (or reference) temperature, e.g., 25 °C
T    = Load cell's final temperature
Cn0  = Load cell's sensitivity when the load cell's temperature is T0
Cn   = Load cell's sensitivity when the load cell's temperature is T

The amplifier module's data sheet claims that for each 10K increase in ambient temperature, the amplifier affects the load cell's sensitivity1 by less than 0.1% of full scale, where "full scale" is \$C_n\$ in equation (4):

$$ C_n^{'} < C_n + 0.1\,\% \; C_{n}\left ( \frac{T-T_0}{10} \right ) \;\;\;\;\;\;\;\;\;\;(4) \\[0.35in] \rightarrow C_n^{'} < C_n\, \left \{ 1 + \left ( \frac{1/10}{100} \right ) \left ( \frac{T-T_0}{10} \right ) \right \} \;\;\;\;\;\;\;\;\;\;(5) \\[0.35in] \rightarrow C_n^{'} < C_n \left ( 1 + \frac{T-T_0}{10,000} \right ) \Bigg\rvert_{C_n=RHS\;of\;equation (3)} \;\;\;\;\;\;\;\;\;\;(6) \\[0.35in] \rightarrow C_n^{'} < C_{n0} \left ( 1 + \frac{T-T_0}{20,000} \right ) \left ( 1 + \frac{T-T_0}{10,000} \right ) \;\;\;\;\;\;\;\;\;\;(7) \\[0.35in] \rightarrow C_n^{'} < C_{n0} \left \{ 1 + \frac{3(T-T_0)}{20,000} + \frac{(T-T0)^2}{200,000,000} \right \} \;\;\;\;\;\;\;(8) $$

Various formulae in the amplifier module's data sheet have a "Sensitivity in mV/V" parameter; substitute \$C_n^{'}\$—i.e., the RHS of inequality (8)—for that sensitivity parameter.

:: DISCLAIMER :: Double-check these equations before using them in any product.


NOTES

1 At the top of page 17 of the amplifier module's data sheet there is a formula (above Example 1) where the word "sensitivity" refers to the load cell's sensitivity. So I'm assuming the "Effect of a 10 K-change in ambient temperature on sensitivity" spec on page 47 of the amplifier's data sheet also refers to the load cell's sensitivity.

2 The equations above are for the special case where the starting temperature \$T_0\$ and ending temperature \$T\$ are the same for both the load cell and the amplifier. I've provided a separate answer below for the more general case where the load cell's starting/ending temperatures can be different from the amplifier's starting/ending temperatures.

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  • \$\begingroup\$ One thing a bit confuses me here. I use 10V excitation voltage. But the sensitivity is defined for 1V excitation i.e mV/V. In my case I want to estimate the total drift. Your Cn' is the total drift voltage for the sensitivity. Should that be 100*Cn' in my case?(for 10V excitation) \$\endgroup\$ – floppy380 Nov 14 '17 at 10:13
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Using MATLAB I cranked out a formula to determine the effect on load cell sensitivity for the general case, where the load cell's starting/ending temperatures can be different from the amplifier module's starting/ending temperatures.

Cn = L_Cn0 + (A_T-A_T0)*A_TKC*L_Cn0/1000 + (L_T-L_T0)*L_Cn0*L_TKC/1000 + (L_T-L_T0)*A_T*A_TKC*L_Cn0*L_TKC/1000000 + (L_T0-L_T)*A_T0*A_TKC*L_Cn0*L_TKC/1000000

where

:: Load cell parameters ::
L_Cn0 = Sensitivity at temperature L_T0 (e.g., 2 mV/V)
L_TKC = Temperature effect on sensitivity (e.g., 0.0 5% of Cn/10K)
L_T0  = Initial temperature
L_T   = Final temperature

:: Amplifier parameters ::
A_TKC = Temperature effect on load cell sensitify (e.g., <0.1% of Cn (Full Scale))
A_T0  = Initial temperature
A_T   = Final temperature
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  • \$\begingroup\$ Is A_TKC in your formula 0.0001 or 0.1 ? Similarly is L_TKC 0.0005 or 0.05? \$\endgroup\$ – floppy380 Nov 16 '17 at 20:35
  • \$\begingroup\$ A_TKC = 0.1 and L_TKC = 0.05 \$\endgroup\$ – Jim Fischer Nov 17 '17 at 0:16
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@DonCarlos> One thing a bit confuses me here. I use 10V excitation voltage. But the sensitivity is defined for 1V excitation i.e mV/V. In my case I want to estimate the total drift. Your Cn' is the total drift voltage for the sensitivity. Should that be 100*Cn' in my case?(for 10V excitation)

I'm not 100% certain I correctly understand the information in the Clip IG datasheet--e.g., the two examples shown on page 17 for the AE101 Measuring Amplifier unit mention S21 switch selections that don't make sense to me given the information in Table 4.1 on page 16.

Looking at the formula at the top of page 17, my best guess as to the actual meanings for the formula's terms is as follows:

  1. Part load := \$L_{applied(max)}\$, Your system's maximum applied load
  2. Nominal load := \$L_{rated(max)}\$, The load cell's max rated load
  3. 10V := \$V_{out(max)}\$, The desired maximum output voltage, where \$V_{out(max)}\le10\,Volts\$
  4. Measuring range in V := \$V_E\$, The bridge's excitation voltage(?)
  5. Sensitivity in mV/V := \$C_n\$, The load cell's specified sensitivity
  6. Nominal measurement range (mV/V) := \$R_n\$, AE101's measurement range sensitivity

EXAMPLE 1

Design requirements

  1. I need to measure masses not to exceed 75 kg, so I'll pick a load cell whose Max. capacity \$E_{max}\$ is 100 kg.
  2. At full load (75 kg applied), the AE101's output voltage \$V_{out(max)}\$should be 9 Volts for my particular application.

Data Sheet Specs

  1. According to the load cell's data sheet, the load cell's reference excitation voltage \$U_{ref}\$ (a.k.a., \$V_E\$) is 5 Volts.
  2. According to the load cell's data sheet, the load cell's sensitivity \$C_{n}\$ is 2 mV/V.

Calculations

$$ R_n = \left ( \frac{L_{applied(max)}}{L_{rated(max)}} \right ) \left ( \frac{V_{out(max)}}{V_E} \right ) C_n \;\;\;\;\;\;\;\;(1) \\[0.35in] = \left ( \frac{75\,kg}{100\,kg} \right ) \left ( \frac{9\,V}{5\,V} \right ) 2\, mV/V \\[0.35in] \rightarrow R_n = 2.7\,mV/V $$

Looking at Table 4.1 in the Clip IG data sheet, for \$V_E=5\,V\$ and \$R_n=2.7\,mV/V\$, range selection S21 switches 5 + 8 are turned ON. After setting S21, an accurate 75 kg load is applied to the load cell, and then calibration potentiometer P22 is adjusted for an output voltage of 9 Volts.

Following calibration, the AE101's output voltage is defined by equation (2) below, where calibration pot P22's setting is the calibration constant 'k', and 'F' is the applied mass in kilograms:

$$ V_{out} = R_n\,k\,F \,\Bigg\rvert_{R_n = RHS\,of\,equation\,(1)} \;\;\;\;\;\;\;\;(2) \\[0.35in] \rightarrow V_{out} = \left ( \frac{L_{applied(max)}}{L_{rated(max)}} \right ) \left ( \frac{V_{out(max)}}{V_E} \right )\,C_n\,k\,F \;\;\;\;\;\;\;\;(3) $$

For EXAMPLE 1 above, k's value is 44.444...V/kg. [Hint: Given Vout(max)=9V, F=75kg, Rn=2.7mV/V, solve for k.] Basically, the calibration constant \$k\$ adjusts the load cell's sensitivity \$C_n\$.

EXAMPLE 2

Assume the calibration procedure from EXAMPLE 1 is performed at temperature \$T_0=25\,°C\$. Now I want to know the maximum limit of output voltage for an applied 75 kg load at temperature \$T=50\,°C\$. (n.b. Both the load cell and the AE101 start at 25 °C and end at 50 °C.) In equation (3) replace \$C_n\$ with the RHS of equation (8) in my previous answer to your original question. This is left as an exercise for the reader. ;-)

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