Gain in inverting and noninverting opamps

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Discussion Overview

The discussion revolves around the gain calculations in inverting and non-inverting operational amplifiers (op-amps), focusing on the derivation of voltage differences and the principles of negative feedback. Participants seek clarification on specific derivations and the underlying concepts of op-amp behavior.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants inquire about the derivation of voltage differences in op-amp circuits, specifically which voltages to subtract and how to understand the calculations presented in the images shared.
  • Warren emphasizes the importance of understanding the virtual short and virtual open principles of op-amps to derive currents and voltages easily.
  • Bob S provides equations related to the gain of a non-inverting op-amp configuration and discusses the relationship between Vout and V-, suggesting it is derived from a voltage divider circuit.
  • There is a question regarding the necessity of subtraction in the voltage calculations, with some participants expressing confusion about the clarity of the derivations.
  • Warren argues that the use of subtraction is explicit and necessary for finding the current through a resistor, countering claims of confusion.
  • Participants reference external resources and textbooks that explain the concepts of virtual short, virtual open, and virtual ground in detail.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the derivations and principles involved, with some agreeing on the importance of the virtual short concept while others remain uncertain about specific calculations. The discussion does not reach a consensus on the clarity of the derivations or the necessity of certain steps.

Contextual Notes

Some participants express confusion over the derivation steps and the application of voltage subtraction, indicating potential gaps in understanding the foundational concepts of op-amps. The discussion includes references to specific figures and equations that may not be fully resolved within the thread.

Who May Find This Useful

Readers interested in operational amplifier theory, circuit analysis, and those seeking clarification on the principles of negative feedback and voltage calculations in op-amp configurations may find this discussion beneficial.

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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