Some questions about drift at the output due to transducer or amplifier
@JimFischer Where is the temperature in that equation? That's the gain equation. 


Like all resistors, the INA128's "gain programming resistor" \$R_G\$ has a temperature coefficient--i.e., \$R_G\$'s resistance value depends upon (is a function of) \$R_G\$'s temperature.

The data sheet for the resistor part you are using will define that part's temperature coefficient of resistance (TCR). A common formula for calculating TCR for parts that have a positive temperature coefficient is shown in equation (1):

$$
TCR=\frac{R-R_0}{R_0\,(T-T_0)}\;\;\;\;\;\;\;\;\;\;(1)
$$

where,

T0 = Reference temperature (e.g., 25 °C)
T  = New temperature (commonly for T &gt; T0)
R0 = Nominal resistance value at the reference temperature T0
R  = Resistance value at temperature T


Solving equation (1) for resistance \$R\$ at temperature \$T\$ gives,

$$
R = R_0\,\left \{ TCR\, (T-T_0) + 1\right \}\;\;\;\;\;\;\;\;\;\;(2)
$$

(n.b. By inspection one sees that for the special case \$T=T_0\$ we have \$R=R_0\$.)

The INA128's voltage gain equation, as specified in its data sheet, is shown in equation (3):

$$
G = 1+\frac{50\,k\Omega}{R_G}\;\left ( \frac{V}{V} \right )\;\;\;\;\;\;\;\;\;\;(3)
$$

To determine \$R_G\$'s effect on gain \$G\$ as a function of temperature \$T\$, i.e., \$G(R_G(T))\$, replace \$R_G\$ in equation (3) with the right-hand side (RHS) of equation (2), and then substitute \$R_G\$ for \$R_0\$:

$$
G(R_G(T)) = 1+\frac{50\,k\Omega}{R_G\,\left \{ TCR\, (T-T_0) + 1\right \}}\;\left ( \frac{V}{V} \right )\;\;\;\;\;\;\;\;\;\;(4)
$$

EXAMPLES

Given,

T0  = 25 °C
TCR = 25 ppm/°C  (25 ppm = 25E-6)
RG  = 12K4 ohms
G   = Gain at temperature T


Example 1)

T = T0 = 25 °C


$$
G(R_G(T=25\,°C)) = 1+\frac{50\,k\Omega}{R_G\,\left \{ TCR\,(T-T_0) + 1) \right \}}\\[0.35in]
=1+\frac{50\,k\Omega}{12.4\,k\Omega\,\left \{ (25\,ppm/°C)\,(25\,°C-25\,°C) + 1 \right \}}\\[0.35in]
=1+\frac{50\,k\Omega}{12.4\,k\Omega}\\[0.35in]
=5.0323\,V/V
$$

Example 2)

T = 40 °C


$$
G(R_G(T=40\,°C)) = 1+\frac{50\,k\Omega}{12.4\,k\Omega\,\left \{(25\,ppm/°C)\,(40\,°C-25\,°C) + 1)\right \}}\\[0.35in]
=5.0307\,V/V
$$