Gain in inverting and noninverting opamps

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The discussion focuses on understanding the derivation of gain in inverting and non-inverting operational amplifiers (op-amps), specifically the calculations involving voltages and currents through resistors. Key concepts include the "virtual short" principle, which states that an ideal op-amp forces its inputs to have equal voltage, and the importance of voltage differences in calculating current through resistors. Participants seek clarification on why certain voltage subtractions are used in the equations and how these relate to the overall gain calculation. References to textbooks that explain these principles in detail are also provided. The conversation emphasizes the necessity of grasping these foundational concepts to effectively analyze op-amp circuits.
uzair_ha91
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Can you explain the yellow highlighted portions (the derivation part)? How do we get to know which voltages to subtract? Is there also a good link which provides a good explanation of negative feedback mechanism?
Correction: (V2 in fig 18.28 is actually Vin)

http://img89.imageshack.us/img89/7360/new1ed.jpg
http://img199.imageshack.us/img199/5502/new2ob.jpg
 
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uzair_ha91 said:
Can you explain the yellow highlighted portions (the derivation part)? How do we get to know which voltages to subtract? Is there also a good link which provides a good explanation of negative feedback mechanism?
Correction: (V2 in fig 18.28 is actually Vin)


Not sure I understand the question. It's a Differential amplifier. What voltages do you think should be subtracted?
 
Do you understand the virtual short and virtual open principles of op-amps? The currents and voltages are easy to derive once you understand those two principles.

- Warren
 
berkeman said:
Not sure I understand the question. It's a Differential amplifier. What voltages do you think should be subtracted?

I'm talking about the calculation of currents through resistors in op amp (inverting and non-inverting) which is given in the images I posted. I need help in those (highlighted part),, how did the author do it?

chroot said:
Do you understand the virtual short and virtual open principles of op-amps? The currents and voltages are easy to derive once you understand those two principles.

- Warren

Nope.
 
An ideal op-amp behaves as a "virtual short," because it attempts to force its inputs to have equal voltage on them.

In your Figure 18.28, for example, one of the inputs is tied directly to ground, or zero volts. The other input will then be driven to the same voltage. Since resistor R1 has Vin on one side and ground on the other, the current through it is Vin/R1.

- Warren
 
chroot said:
An ideal op-amp behaves as a "virtual short," because it attempts to force its inputs to have equal voltage on them.

In your Figure 18.28, for example, one of the inputs is tied directly to ground, or zero volts. The other input will then be driven to the same voltage. Since resistor R1 has Vin on one side and ground on the other, the current through it is Vin/R1.

- Warren

Exactly, but why confuse the reader by writing Vin-V-/R1 and then equalizing it to Vin/R1? Why the subtraction? And similarly for I2 and the currents in the 2nd figure of noninverting amplifier..
Are we trying to find net voltage or something?

Bob S said:
Redbelly98's illustration in post #2 is an excellent example of a voltage follower with gain.
The equations are:

1) Vout = A(V+ - V-) [ where opamp gain A is arbitrarily large]
2) Vin = V+
3) V- = R1*Vout/(R1 + R2)
4) Vout/A = Vin - R1*Vout/(R1+R2)
5) Vin = Vout(1/A + R1/(R1+R2))
Let A ==> infinity
6) Vout = [(R1+R2)/R1]*Vin

Bob S

I don't understand how you got that value for V- in the 3rd step.
 
uzair_ha91 said:
Exactly, but why confuse the reader by writing Vin-V-/R? Why the subtraction?

They're not confusing the reader. They're being explicit, which is usually the opposite of confusing.

To find the current through a resistor, you take the difference in voltage across it, and divide that by the resistance. To find a difference, you use subtraction.

- Warren
 
okaaay...well that was quite simple...I guess all the new stuff is interfering with my old concepts
Thanks!
 
  • #10
From Bob S
Redbelly98's illustration in post #2 is an excellent example of a voltage follower with gain.
The equations are:

1) Vout = A(V+ - V-) [ where opamp gain A is arbitrarily large]
2) Vin = V+
3) V- = R1*Vout/(R1 + R2)
4) Vout/A = Vin - R1*Vout/(R1+R2)
5) Vin = Vout(1/A + R1/(R1+R2))
Let A ==> infinity
6) Vout = [(R1+R2)/R1]*Vin


uzair_ha91 said:
I don't understand how you got that value for V- in the 3rd step.
The equation relating Vout and V- in line 3 is just a voltage divider circuit, since the ideal negative opamp input has infinite impedance and has no input bias current.
Bob S
 
  • #11
Dear uzair ha91
The virtual short, virtual open and virtual ground concepts are explained well in the following two reference books (these concepts are explained in the chapter devoted for opamp) :

1) Thomas Floyd, Electronic Devices [Exellent book for those who like practical work]

2) Adel Sedra, Microelectronic Circuits [A universal reference book for Electronic and Microelectronic]

You can download these books together with lots of other books from:

Admin: link deleted
 
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