higher. These are the gains I am achieving when applying 1 volt input:
I have tested an AD8231 in isolation, and it still behaves like this.
Is this an problem you have seen before?
Since the AD8231 is an instrumentation amplifier, its output voltage range
depends on the applied input voltage. So in order to get the wanted output
voltage at a specific gain, you need to observe the input voltage range you
For this purpose kindly refer to the www.analog.com/AD8231 datasheet, section
"Instrumentation Amplifier Performance Curves". Kindly refer to Figure 12 on
page 10 of the datasheet. For your reference I am attaching the screenshot
Figure 12 displays the input common-mode voltage over the output voltage for a
reference voltage of 0V. The area within the triangle is the possible output
voltage range (for larger V_ref this is a diamond).
For an input voltage of 1V the common-mode voltage V_cmi amounts to
V_cmi = (V_+INA - V_-INA) / 2 = (1V - 0V) / 2 = 0.5V.
The output voltage V_out is equal to the input voltage V_in mulitplied by the
gain G plus the reference voltage V_ref:
V_out = V_in x G + V_ref.
In case of a 1V input voltage and a reference voltage of 0V, the output voltage
amounts to 1V for a gain of 1 and 2V for a gain of 2. For your reference I
marked these points in the figure and attached the screenshot below:
From this figure you can see, that for a gain = 1 the output voltage is within
the output voltage range, whereas for a gain of 2 or larger it would be outside
the output voltage range. This explains the difference between your measured
and expected gain for gains larger than 1.
A recommendation is to stay a bit away from the boarder due to errors in the
gain and the offset.
In order to use the gain range of the AD8231, kindly choose an appropriate
reference voltage, input voltage and gain.
For example, at reference voltage of 2.5V and a supply voltage of 5V you can
reach the full output swing (0V to 5V) for input common mode voltages between
1.3V and 3.7V, kindly compare to Figure 14 of the datasheet. So, the maximum
voltage across the input terminals is
V_in = (V_out - V_ref)/ G = (5V - 2.5V)/ 1 = 2.5V for G = 1 and
V_in = (V_out - V_ref)/ G = (5V - 2.5V)/ 128 = 19mV for G = 128.