# [SOLVED]CMRR simulation for fully differential amplifier

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#### Junus2012

Dear friends,

Can you please tell me about a robust method to simulate the CMRR from a closed loop of a fully differential amplifier using cadence.

for example to use the diffstbProbe with stability analyses or that is not possible

Thank you very much

#### Dominik Przyborowski

How you think is a proper way to do this?

#### FvM

##### Super Moderator
Staff member
Think simple. To measure CMRR, you'll apply a common mode signal and measure the respective transfer function.

Junus2012

### Junus2012

points: 2

#### Junus2012

Dear FvM,

If I apply a common signal and measure the respective transfer function as you suggested, this will give me the common mode gain not the CMRR,

That is the first issue, the second thing is that this simulation is held under open loop condition which I want to avoid it,
that is why I would like to have a method to simulate it from closed loop.

Such method I know from Holberg, but he used it for simulating the single ended op-amp, kindly you can see it below.

#### KlausST

##### Super Moderator
Staff member
Hi,

Gain = V_out / V_in = ratio

Thus - in my eyes - common mode gain is the same as CMRR.
But the gain is less than 1 thus one speaks of "rejection".

Klaus

Junus2012

### Junus2012

points: 2

#### frankrose

For closed loop simulation you don't need diffstbProbe, connect simple AC sources to both inputs of the whole amplifier (with the feedback and input resistors) and run conventional AC analysis.
CMRR is ratio of the differential and common mode gain, so you should simulate both at the same time. For that duplicate your circuit on the testbench, at one use in-phase input signals, at the other use inverted signals (with VCVS from analogLib it is easy). Measure output differential voltage and calculate gain of both test circuits, then just divide the differential gain with the common mode gain.

Junus2012

### Junus2012

points: 2

#### FvM

##### Super Moderator
Staff member
In practice, you don't need to measure differential mode gain to determine CMRR because it's defined by the feedback network with sufficient accuracy.

Junus2012

### Junus2012

points: 2

#### frankrose

At higher frequencies the differential gain drops, and feedback network won't define it.

Junus2012

### Junus2012

points: 2

#### Junus2012

Dear friends,

Thank you all for your nice contribution to my post,

Let me say that you all are agree that I must go to measure the common mode gain and the differential gain.

There is one point might I did't make it clear for you, I don't mean by the closed system is that I have an op-amp connected in feedback,

But what I wanted to say that I need to test the CMRR in such a way similar to the method you learned me, you remember in my former post when I was simulating the AC char of the opamp then you suggested me to simulate in closed loop with diffstbprobe and STB analyses,

The closed loop is just a thing I want to add to my simulation to make it stable independent of the offset voltage (VOS), and why I need it independent of the (VOS) is because that I want to simulate the CMRR over the entire range of the ICMR, so if the VOS is changing when I change the ICMR, then the measurement of the CMRR will fail.

- - - Updated - - -

According to your explanation I have found this configuration from analog devices, please see it below

is it what you suggested right ?

I also plotted the configuration based on it but for my fully differential amplifier to refer to it please in your kind talk

Last edited:

#### FvM

##### Super Moderator
Staff member
Setup 1 is suggested in manufacturer datasheets, e.g. from Analog.

Junus2012

### Junus2012

points: 2

#### Junus2012

Dear FvM,

Ok I will go for this setup, and using the equation given from analog Devices data sheet, I get for my differential amplifier this expression

d(VOdiff)=(d(Vin) / CMRR) * (1+Rf/R1)

Since the differential amplifier is an inverting amplifier I will make the gain of the output VCVS = -1 to inverse the signal.

Simplifying the above equation gives CMRR = [d(vin)/d(Vodiff)] * (1+Rf/R1)

According to the above equation there are three points I would like to confirm it with you please :

1. The gain set by the feedback resistor has no effect on measuring the CMRR as the (vin)/d(Vodiff) will change with the gain value accordingly.

2. The expression in the last formula tell us to take the transfer function from the input to the output as it is (vin)/d(Vodiff), it means we will get the inverse of the CMRR if we make the transfer function from the output to the input

3. Since this simulation setup is performed under closed loop condition, the stability of the amplifier is assured by the feedback and I should not be wary about the input offset voltage

Now to go in to simulation, I will run the AC simulation from Cadence as usual and tell the simulator to plot the transfer function from input source to the output.

Please correct me if I am wrong or confirm me if I am right,
Thank you once again

#### frankrose

I have found this tutorial you are talking about, and even it says "it is clear that this circuit is only marginally useful for measuring CMRR". Only the DC and very low frequency CMRR is measureable with this method, extreme accurate resistors are necessary and if it not a problem then use it, I agree. But run Monte Carlo AC analysis, mismatch of devices will dominate in common mode rejection, not only the mismatch of resistors.
Additionally the fully differential circuits used because of high common mode noise immunity, and tipically noise has wide bandwidth, I think DC/low frequency CMRR is tipically not useful or practical value.

Junus2012

### Junus2012

points: 2

#### FvM

##### Super Moderator
Staff member
The basic idea of the "simple" CMRR setups is to use the same configuration as in actual amplifier application. If it's a 4 resistor configuration, resistor mismatch matters also for real amplifier CMRR, you don't necessary need to determine the ideal IC CMRR.

Junus2012

### Junus2012

points: 2

#### Junus2012

Dear FvM and Frank

Thank you very much for your help

I have simulated the CMRR right now according to the method you proposed of analog devices and the one I plotted as setup 1 before, I set the gain to one so I directly read the CMRR without multiplying it with the feedback gain ratio as I assumed in my former post

I run the AC simulation and found the transfer function which looks very strange, kindly you can see it below

#### frankrose

Your circuit is fully symmetric, CMRR won't result in a meaningful value. I mentioned to run Monte Carlo exactly because of it. And it is inverse common mode gain, not CMRR, but nevermind.

#### Junus2012

Dear frank, sure the CMRR I have to simulate it from monticarlo there is no way to say it is my actual one, but at least the graph should be reasonable, i dont understand what are this oscillation,

However I did one thing, in my former result I plotted the (Vo2-Vo1) by connecting VCVS to the both output to get the fully differential expression,

now I repeated the simulation but by only talking the transfer function from each output individually and the graph is so clear as you can see from the image below which is taken from Vo1.

#### frankrose

This is not oscillation, and totally reasonable.
With common mode input signal, so with the same input signal on both inputs you get the same clear curves at each outputs, that is fine because of the symmetry in your circuit.
At the same time the difference between the 2 outputs is nothing, which means you don't get differential output signal, so common mode gain is very low, the rejection is very high, about 200dB.
The reason behind this that your circuit is absolutely symmetric! There is no mismacth!
Add a small difference between the resistors for example, or to the transistors and you will see totally else curve.

#### Junus2012

Thank you frank,

As you said I need to add mismatch by myself or by the Monticarlo,

But any way we agree that the circuit from my last simulation means or indicate one thing, that I have very high CMRR at least for the right moment of schematic simulation.

And one more thing, I can see you still call my graph as it is common mode gain, not the CMRR

Thank you once again

#### Junus2012

Dear frank

Once you suggested me in your first reply to run the common mode gain and the differential gain setup both in the same testbench, then I get the result from the common mode gain and differential gain in db and subtract them in calculator, I just did it now and please you can see the result to have a comparesion from the analog device (or FvM) method

the first wave in my graph is the common mode gain (AC)
the next below is the differential mode gain (AVD), and the last is the CMRR = (AVD (db)-AC (db))