Question
I have to use an AD8599 and I tried to measure the CMRR with Vcc=+/-5V with
-2V<Vcm
Min). I tried using conditions Vcc=+/-15V with -12.5V<Vcm
same. I carry out these measurements with DC conditions.
Could you help me measure the value given in the datasheet?
I tried 2 bias methods:
In accordance with MIL-STD: Vcc+=+7V; Vcc-=-3V; Vin=+2V and Vcc+=+3V; Vcc-=-7V;
Vin=-2V
With a direct method (instrumentation amplifier between In+ and In-): Vcc+=5V;
Vcc-=-5V; Vin=-2V then Vin=+2V
The Delta of the Vio is about 10µV with -2V<Vcm<2V.
Answer
We use the null opamp in our test set-up. The advantage of this test set up is
that it decouples the DUT output from the input common mode voltage and also
lets you operate the opamp at no load, if needed. This matches the exact
datasheet condition for which the device CMRR is specified.
1.I recommend increasing the closed loop gain to 1000. This will increase
the resolution of measurement. So you can measure Vout and divide it by 1001
to compute the input Vos. CMRR would be 20*log(Delta VCM/Delta Vos).
2.I would not vary VIN. For this test setup, I would do
the measurement the following way:
Condition 1: VCC = +2.5V, VSS = -27.5V, VIN = 0V (this corresponds to a VCM of
12.5V with respect to the part with a +/-15V supply). Measure output voltage of
the opamp.
Condition 2: VCC= +27.5V, VSS = -2.5V, VIN = 0V (this corresponds to a VCM of
-12.5V with respect to the part with a +/-15V supply). Measure output voltage
of the opamp.
Note: VIN is kept fixed at 0V in both cases. But both the supplies are offset
to give the resulting common mode shift. Also, even though the circuit has
about 60 db of closed loop gain, the output does not get saturated because VIN
is fixed at 0V.
CMRR = 20*log (1001*Delta VCM/Delta Vout) where 1001 = Gain.
3.Reduce the gain resistor values to lower the drop due the input bias
current.