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Gain and stability of fully differential Amplifiers

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hannover90

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Hello all,

with helping of replys to my last thread, I designed a fully differential amplifier with a gain "G", but i can not understand the gain of the whole circuit with a signal-feedback, as showed in the attached figure.

The gain of the whole circuit, I calculated as: A=Rx/Ry for example A=1000KΩ/500kΩ=2, but how is the role of the gain "G" of the fully differential amplifier, which should be high as possible?

Now, i try to check the stability of the designed FDA using cmcdprobe from analogLib, as seen in the attached figure.

Does the AC simulated gain (called in cadance as loop gain) indicate "A" or "G"?

The transient simulated output signals of my DFA for different A=1, A=2, A=3, have a little swing (please, see the attachment ).
Is the DFA instabil?

I would be thankful for any reply.
 

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Last edited:

Hi,

It seems to be stable.

But may I ask why you showed single ended signals, but using a fully differential amplifier?
I think it makes more sense to show the differential signals.

Klaus
 
Hi Klaus,
thanks for your reply,
the third figure shows both output signals of the fully differential amplifier for different A=Rx/Ry=1, 2 and 3.
The first and 2nd figures are zoomed.
 
Last edited:

Hi,

Yes, but there shouldn't be two single ended signals but a singke differential signal.
One line for input, one line for output.

Like it is specified V_in = (Vin+ - Vin-l
--> show V_in.

Klaus
 

Many thanks. I understand. I will simulate again and attach the result tomorrow.
 

On the fourth picture you connected the upper feedback resistor to the Vout_n. Where is the other pin of that Rx? Are you sure your feedback is negative? Rx's other pin should be on the OPAmp's Vin_p pin. I think.
 

Thanks.
My quastion is, does the ac simulated gain loop indicate the gain (G) of the amplifier or the closed loop gain Gx=G/(1+G*A)?
" A" is the gain of the feedback.
 

With normal AC analysis you can simulate the closed-loop gain, and the open-loop gain, but the second is more complicated.
The cmdmprobe is used by STB analysis (not regular AC) which gives back the open loop gain. Not the closed loop gain.
 

First comment, as mentioned by frankrose in post #6, signal polarity is wrongly sketched in the drawing.

My question is, does the ac simulated gain loop indicate the gain (G) of the amplifier or the closed loop gain Gx=G/(1+G*A)?
" A" is the gain of the feedback.

There are some discrepancies in the terms and formulas used in the discussion.

In the above given formula, G is called open loop gain, A is feedback factor and Gx is closed loop gain. But the previously given expression A = Rx/Ry is not correct. Instead, Rx/Ry is the approximate Gx value for G*A >> 1. Feedback factor calculates correctly as

A = Ry/(Rx + Ry)

To derive the exact transfer function, it's useful to refer to an extended feedback scheme with an additional forward factor.

feedback_ext.png

The extended scheme has been used in several previous threads, e.g. https://www.edaboard.com/showthread.php?t=273725

For this scheme, you can derive

Acl=Aol*α/(1+Aol*β)

The used formula signs are

Acl closed loop gain
Aol open loop gain
α forward factor
β feedback factor
Aol*β is also designated loop gain (actually -Aol*β, because the negative sign constituting a negative feedback loop is not included in Aol or β in the schematic.

Referring to the differential amplifier circuit, you get
α = -Rx/(Rx+Ry)
β = Ry/(Rx + Ry)

You can derive the factors for a simpler single ended amplifier circuit

300px-Op-Amp_Inverting_Amplifier_svg.png

The approximate closed loop gain for sufficient large loop gain is

Acl ≈ -Rx/Ry (respectively -Rf/Rin)
 
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