# Some questions about drift at the output due to transducer or amplifier

I have instrumentation amplifier box/circuit using a INA128 amplifier. Inside the same box the output of this goes to another amplifier for a larger modifiable gain. All this is used to amplify a force transducer's mV level outputs. The force transducer is using Wheatstone bridges.

I observe large drifts during time and I suspect about the temperature. I cannot verify if the drift is due to the transducer or the amplifier.

So I'm planning to measure the outputs of both amplifier and the force transducer at a constant force/load.

1-) And if I see a change in the force transducer's output it means the transducer is causing this drift and is defect?

2-) If there is no change in the force transducer's output but there is change in the instrumentation amplifier's output it means the amplifier is drifting. How can I calculate the drift of the INA128 after a 10°C increase if its gain is set to 300 for an initial 1mV differential input? I wonder if the drift Im experiencing is normal or not..

• Make a dummy transducer from a stable voltage source and resistive attenuator. Keep that at a stable temperature away from the amplifier, which you can heat/cool as you like (I got some strange looks using a hairdrier!) to test the amplifier on its own.
– user16324
Nov 10, 2017 at 13:21
• What do you mean by "dummy transducer"? Nov 10, 2017 at 16:04
• Something that acts like a transducer' as the amplifier sees it. Produces the same sort of voltage from the same sort of impedance, but allows you to test the electronics independently of a real transducer.
– user16324
Nov 10, 2017 at 17:41
• The temperature coefficient of the external gain resistor RG affects the drift. Page 14 of the INA128's data sheet states "RG’s contribution to gain accuracy and drift can be directly inferred from Equation 1." Nov 11, 2017 at 11:27
• @BrianDrummond Why would I need a dummy load. Cant I just keep the force transducer far away with a constant weight and heat up the amplifier with a hair dryer? Nov 11, 2017 at 14:37

@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 > 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


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