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Fully Differential Amplifier closed loop gain (superposition assumption) (Read 2795 times)
monglebest
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Fully Differential Amplifier closed loop gain (superposition assumption)
Nov 10th, 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?
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buddypoor
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Re: Fully Differential Amplifier closed loop gain (superposition assumption)
Reply #1 - Nov 11th, 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!
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LvW (buddypoor: In memory of the great late Buddy Rich)
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monglebest
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Re: Fully Differential Amplifier closed loop gain (superposition assumption)
Reply #2 - Nov 13th, 2017, 12:56pm
 
buddypoor wrote on Nov 11th, 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
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Frank_Heart
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Re: Fully Differential Amplifier closed loop gain (superposition assumption)
Reply #3 - Nov 17th, 2017, 10:50am
 
monglebest wrote on Nov 10th, 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?

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monglebest
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Re: Fully Differential Amplifier closed loop gain (superposition assumption)
Reply #4 - Nov 29th, 2017, 10:08am
 
Frank_Heart wrote on Nov 17th, 2017, 10:50am:
monglebest wrote on Nov 10th, 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
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