AD628 Programmable Gain Difference Amplifier CMRR Testing

[chandlerbing65nm]

Hello folks,

I have a problem with CMRR testing in lab. I simulated the AD628 CMRR test circuit in LTSpice and Multisim and got the same result, but when I try it in the laboratory I get different output.

Here is the test circuit for the CMRR.


Vin1 = 10V, find Vout1?
Vin2 = -10V, find Vout2?

CMRR = (Vout1 - Vout2)/(delta Vin * System Gain)

Here are my results in lab:


In LTSpice and Multisim, Vout1 when Vin = 10V should be negative like this below:


My expected CMRR result whould be within the limits: +/-178uV/V.
Even the known good units (Good 1-3) are failing, meaning there is something wrong.
I checked the connection multiple times and I'm sure there is nothing wrong with the set-up.

Here is the Datasheet of AD628:
https://www.analog.com/media/en/technical-documentation/data-sheets/AD628.pdf

 

[srizbf]

What about the CMRR calculation ,instead of the nominal value
of R1 and R2 , substituting its precise value by measuring it?

 

[KlausST]

Hi,

when I try it in the laboratory I get different output.

you are measuring in the microvolts range. Thus you need to take care about
* GND bounce
* GND wiring
* thermocouple effects, temperature effects

Maybe the problem is caused by wiring, PCB layout, themperature...

Can you send a photo of your DUT, DUT wiring....

Klaus

 

[chandlerbing65nm]

It's Rf = 10k and Rin = 100k. They are inside the device, within the IC. The only resistors that I can manipulate there are the 10Meg and 10k on the 2nd opamp, for programmable gain. The problem that I have is that I can't get the right output when Vin = 10V, as show in my data.

 

[albbg]

As pointed out by srizbf you have to know exactly the value of R1 and R2 since they have strong effect on the system gain.

 

[wwfeldman]

page 13 of the specification shows the CMRR test set up.

since you are not doing that, you should not expect the same results.

 

[chandlerbing65nm]

page 13 of the specification shows the CMRR test set up.

since you are not doing that, you should not expect the same results.

I cannot use that configuration since the Automated test equipment (ATE) used the config that I posted, and I have to follow that config and produce same results (for the known good IC's) and validate the failing unit(rel 1) in lab if it's a genuine failure or a good unit misread by the ATE.

- - - Updated - - -

As pointed out by srizbf you have to know exactly the value of R1 and R2 since they have strong effect on the system gain.

I'll try to precisely use the R1 and R2 values, Thanks. But how did I get the inversion, the expected Vout1 is negative then I'm getting a positive Vout1, making the calculated CMRR goes beyond the passing limits.

 

[crutschow]

Vin = 10V should be negative like this below:

Why?
There's nothing that states what the polarity of the CM output voltage output will be.
It can be either plus or minus for a positive input.
A simulation is just that. It doesn't show exactly what the real device will do.

My expected CMRR result whould be within the limits: +/-178uV/V.

That's for a 1V input.
The minimum CM rejection ratio is -75dB or 177.83e-6.
For a 10V input that's 1.78mV.
What did you measure?

 

[chandlerbing65nm]

That's for a 1V input.
The minimum CM rejection ratio is -75dB or 177.83e-6.
For a 10V input that's 1.78mV.
What did you measure?

The ATE program that I'm following is using Vin = +/-10V, which attained the passing CMRR within the +/-178uV/V limits. Mean value of CMRR in the dataplots of testing is in ~30-40uV/V. Sample units is 125 IC's.
Also, I've noticed that the CMRR in book is CMRR = (Vin1-Vin2)*(System Gain)/(Vout1 - Vout2). Inverse of what the ATE program is using, maybe because the test developer thought that limits in small units (uV or mV) is more accurate than (MegV or kV).

 

[KlausST]

Hi,

Vin = +/-10V, which attained the passing CMRR within the +/-178uV/V

I assume this is not correct.
+/-10V lets me assume to be 20Vpp.
but the CMRR is +/-178uV for 10V input

example:
Starting from common mode input voltage (CMIV) = 0V:
If you move the common mode voltage to +10V then the ouput may move by (delta. referenced to output of CMIV of 0V)) of +/- 178uV
If you move the common mode voltage to -10V then the ouput may move by (delta) of +/- 178uV (again same ouput voltage range)

Klaus

 

[crutschow]

but the CMRR is +/-178uV for 10V input

The CMRR is specified as +/-178uV/V, not +/-178uV for 10V input.
So for a 10V input, the output would be a maximum of +/- 1.78mv.

 

[KlausST]

Hi,

The CMRR is specified as +/-178uV/V, not +/-178uV for 10V input.
So for a 10V input, the output would be a maximum of +/- 1.78mv.

My bad.
For sure you are correct.

Klaus

 

[chandlerbing65nm]

What about the CMRR calculation ,instead of the nominal value
of R1 and R2 , substituting its precise value by measuring it?

I tried to measure the 10Meg resistor R1 on the set-up and it shows 500k on the multimeter, but when I remove the Rg connection the resistor R1 is 10Meg again.

 

[albbg]

I tried to measure the 10Meg resistor R1 on the set-up and it shows 500k on the multimeter, but when I remove the Rg connection the resistor R1 is 10Meg again.

The measurement of the components must be done when disconnnected from the rest of the circuit. Then take note of the exact reading of the multimeter (that however has a given accuracy that should be taken into account) in order to be able to calculate the system gain with more precision.

 

[FvM]

Post #9 is rather unclear. Do you get the same or different results in ATE and lab setup? 30 uV/V sounds expectable, > 200 uV/V beyond specification.

The setup in post #1 seems useable, but you need to measure Vout with two different input voltages and determine the slope to discriminate CMRR from offset error, did you?