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Design >> Analog Design >> Fully Differential Amplifier closed loop gain (superposition assumption)
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Message started by monglebest on Nov 10^th, 2017, 8:00am

Title: Fully Differential Amplifier closed loop gain (superposition assumption)
Post by monglebest on Nov 10^th, 2017, 8:00am

I don't understand why output is considered ground when calculating Vin's superposition at amplifier input pin.
http://www.ti.com/lit/an/slyt157/slyt157.pdf

From the above TI notes, I learnt that I should apply superposition
Vn=(Vin-)*R2/(R1+R2) + (Vout+)*R1/(R1+R2) (1)
Vp=(Vin+)*R4/(R3+R4) + (Vout-)*R3/(R3+R4) (2)

Then based on the differential amplifier equation (Vp-Vn)*Adiff=(Vout+)-(Vout-), I can calculate the closed loop gain.

My question is about the superposition equations (1). Why can I assume Vout+ node is grounded when considering (Vin-)'s superposition, which give me Vn=(Vin-)*R2/(R1+R2). If the amplifier is OTA, the output node is high impedance.

Can anyone explain to me why in this superposition, I can obtain Vn=(Vin-)*R2/(R1+R2)? I can understand the other way when calculating Vn=(Vout+)*R1/(R1+R2) because Vin- voltage source should be grounded in Vout+'s superposition. What is missing in my understanding?

Title: Re: Fully Differential Amplifier closed loop gain (superposition assumption)
Post by buddypoor on Nov 11^th, 2017, 2:45am

The superposition principle requires that the influence of the various signal sources is considered successively - all sources grounded (set to zero) except one!

Title: Re: Fully Differential Amplifier closed loop gain (superposition assumption)
Post by monglebest on Nov 13^th, 2017, 12:56pm

buddypoor wrote on Nov 11^th, 2017, 2:45am:

The superposition principle requires that the influence of the various signal sources is considered successively - all sources grounded (set to zero) except one!

Hi, buddypoor

I understand what you say when I execute the calculation of superposition. But why can I do that? In normal operation, there is only voltage source at input, the output should be connected with cap load.

Or maybe I am asking the wrong question when trying to analyze the gain calculation of fully differential pair?

Do you know other way to calculate fully differential amplifier differential gain?

Regards
Gaofeng

Title: Re: Fully Differential Amplifier closed loop gain (superposition assumption)
Post by Frank_Heart on Nov 17^th, 2017, 10:50am

monglebest wrote on Nov 10^th, 2017, 8:00am:

I don't understand why output is considered ground when calculating Vin's superposition at amplifier input pin.
http://www.ti.com/lit/an/slyt157/slyt157.pdf

From the above TI notes, I learnt that I should apply superposition
Vn=(Vin-)*R2/(R1+R2) + (Vout+)*R1/(R1+R2) (1)
Vp=(Vin+)*R4/(R3+R4) + (Vout-)*R3/(R3+R4) (2)

Then based on the differential amplifier equation (Vp-Vn)*Adiff=(Vout+)-(Vout-), I can calculate the closed loop gain.

My question is about the superposition equations (1). Why can I assume Vout+ node is grounded when considering (Vin-)'s superposition, which give me Vn=(Vin-)*R2/(R1+R2). If the amplifier is OTA, the output node is high impedance.

It's true when it is open-loop. Here is a closed-loop application.
-Frank

Can anyone explain to me why in this superposition, I can obtain Vn=(Vin-)*R2/(R1+R2)? I can understand the other way when calculating Vn=(Vout+)*R1/(R1+R2) because Vin- voltage source should be grounded in Vout+'s superposition. What is missing in my understanding?

Title: Re: Fully Differential Amplifier closed loop gain (superposition assumption)
Post by monglebest on Nov 29^th, 2017, 10:08am

Frank_Heart wrote on Nov 17^th, 2017, 10:50am:

monglebest wrote on Nov 10^th, 2017, 8:00am:

I don't understand why output is considered ground when calculating Vin's superposition at amplifier input pin.
http://www.ti.com/lit/an/slyt157/slyt157.pdf

From the above TI notes, I learnt that I should apply superposition
Vn=(Vin-)*R2/(R1+R2) + (Vout+)*R1/(R1+R2) (1)
Vp=(Vin+)*R4/(R3+R4) + (Vout-)*R3/(R3+R4) (2)

Then based on the differential amplifier equation (Vp-Vn)*Adiff=(Vout+)-(Vout-), I can calculate the closed loop gain.

My question is about the superposition equations (1). Why can I assume Vout+ node is grounded when considering (Vin-)'s superposition, which give me Vn=(Vin-)*R2/(R1+R2). If the amplifier is OTA, the output node is high impedance.

It's true when it is open-loop. Here is a closed-loop application.
-Frank

Can anyone explain to me why in this superposition, I can obtain Vn=(Vin-)*R2/(R1+R2)? I can understand the other way when calculating Vn=(Vout+)*R1/(R1+R2) because Vin- voltage source should be grounded in Vout+'s superposition. What is missing in my understanding?

Hi, Frank:

Can you explain it more clearly what's the black magic that closed loop does here?

Thank you
Gaofeng