Voltage across the output resistor for this opamp circuit

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Discussion Overview

The discussion centers on determining the voltage across a resistor (R3) in an op-amp circuit. Participants explore various approaches to analyze the circuit, including the application of superposition and the concept of virtual ground. The conversation includes technical reasoning and attempts to clarify the behavior of currents and voltages in the circuit.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that no current flows into or out of the op-amp inputs, but current can flow through the op-amp output, affecting the voltage across R3.
  • There is a discussion about whether the op-amp is passive, with some suggesting that this leads to a specific voltage relationship between the terminals.
  • One participant proposes using the superposition principle to analyze the contributions of different sources to the voltage at the inverting terminal.
  • Another participant expresses confusion about the concept of superposition, questioning how the output voltage can be considered an independent source when it is a result of the input current.
  • Several participants discuss the implications of the virtual ground concept, suggesting that it leads to a specific calculation for the voltage at the inverting terminal.
  • There are conflicting interpretations regarding the sign of the voltage at the inverting terminal, with some stating it is 10V and others indicating it is -10V.
  • One participant asks about the path of current entering or exiting the op-amp output, leading to clarification that the op-amp connects to power supplies that manage the output current.

Areas of Agreement / Disagreement

Participants express differing views on the application of superposition and the behavior of currents in the circuit. There is no consensus on the correct interpretation of the voltage at the inverting terminal, with conflicting claims about its value and sign.

Contextual Notes

Some participants highlight the importance of understanding the assumptions behind the ideal op-amp model, such as the implications of negative feedback and the behavior of currents in the circuit. There are unresolved questions regarding the calculations and the interpretation of voltage signs.

jaus tail
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Homework Statement
Find the voltage across R3
Relevant Equations
No current flows inside opamp
244319

Here is question: Find voltage across R3
options:
10V
11V
5V
2V

Attempt: No current flows through Opamp, so 1mA flows through R3, this gives voltage across R3 = 1ma * 1K = 1V.
But that is not in option.
Where am I wrong? No current flows inside or outside opamp, right?
 
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jaus tail said:
No current flows inside or outside opamp, right?

From - to + ! (The output currect can be considerable) !
I.e. the 1 mA flows through R2 in opposite direction wrt Vb - Va
 
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1 mA will flow left to right through R1, then left to right through R2 and then top to bottom through R3. Right?
 
What about the opamp ? Is it passive ?
jaus tail said:
Right?
That would make Va = 11 V ? and Vb = ##-\infty## ? After all, Vb = ##-## N ##\times ## Va with N a big number...
 
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What does passive mean? This is the only data that is given. The diagram and the question to find Voltage across R3.
 
The key for solving the problem is the following:
For an ideal opamp the voltage between the inv. terminal and ground is zero. This is the result of negative feedback.
Hence, find the expression for this voltage (depends on the input current as well as the opamps output voltage Vout, use the superposition rule) and set it to zero - and solve for Vout.
It is a simple calculation...
 
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jaus tail said:
Where am I wrong? No current flows inside or outside opamp, right?
No current flows into or out of the op amp inputs, but current can and does flow into or out of the op amp output. So the current through R3 is not the same as the current through R2.
 
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LvW said:
The key for solving the problem is the following:
For an ideal opamp the voltage between the inv. terminal and ground is zero.
Does this mean Va is 0?
Or
Can i solve it like:
( 0 - Va ) / R1 = 1mA
This gives Va.
Then (Va - Vb)/R2 = 1mA. It's the same current.
This gives Vb ??
 
jaus tail said:
Does this mean Va is 0?
Yes.
Or
Can i solve it like:
( 0 - Va ) / R1 = 1mA
This gives Va.
No! You don't know what the voltage at the top of the current source is.
Edit - A better way to say it is that you can't assume that the voltage at the top of the current source is zero. You can calculate what that voltage is given that Va=0 and knowing R1 and the current.
 
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  • #10
@jaus tail - do you know the superposition rule?
There are two sources contributing to the voltage Va: The input current source and the opamps output voltage Vout. The influence of both sources on the voltage Va can be (and must be ) calculated separately!
 
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  • #11
Yeah I know superposition. The output voltage is like a series voltage source / battery to right or to left of b?
 
  • #12
I'm confused about superposition. The opamp's output Vout is because of the input current source. So how are there two independent sources? If i consider only Vout and open circuit the current source, then Vout will be zero, cause Vout is because of Iinput.
 
  • #13
is this the explanation:
using concept of virtual ground, we get that inverting terminal of opamp is at ground. 0V
so current through R2 is 1mA. No current flows into opamo.
so (Va-Vb)/R2 = (0-Vb)/R2 = 1mA
So -Vb/10k = 1mA.
This gives Vb = 10V, which is there in option.
 
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  • #14
jaus tail said:
is this the explanation:
using concept of virtual ground, we get that inverting terminal of opamp is at ground. 0V
so current through R2 is 1mA. No current flows into opamo.
so (Va-Vb)/R2 = (0-Vb)/R2 = 1mA
So -Vb/10k = 1mA.
This gives Vb = 10V, which is there in option.
Yes. That's all there is to it.
 
  • #15
Ok. Thanks. Just for knowledge: if current is going into the output of the opamp, then where does it go from there? Like current needs a closed path.
So if current enters opamp from output, where does it come out from?
Or if current is coming out from output of opamp, how does it enter the opamp?
 
  • #16
jaus tail said:
Ok. Thanks. Just for knowledge: if current is going into the output of the opamp, then where does it go from there? Like current needs a closed path.
So if current enters opamp from output, where does it come out from?
Or if current is coming out from output of opamp, how does it enter the opamp?
The op amp has connections to the power supplies, as shown in this figure. They source or sink the output current.
400px-Op-amp_symbol.svg.png
 
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  • #17
jaus tail said:
is this the explanation:
using concept of virtual ground, we get that inverting terminal of opamp is at ground. 0V
so current through R2 is 1mA. No current flows into opamo.
so (Va-Vb)/R2 = (0-Vb)/R2 = 1mA
So -Vb/10k = 1mA.
This gives Vb = 10V, which is there in option.

Not unimportant( the sign): Vb= - 10V.
 
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  • #18
Thanks for the help. I didn't use superposition though. Is there some other method to arrive at answer using superposition?
 
  • #19
jaus tail said:
Thanks for the help. I didn't use superposition though. Is there some other method to arrive at answer using superposition?
The voltage at the inv. input is Vn=Vn1+Vn2.
(1) Vn1 caused by I1 only (for Vout=Vb=0): Vn1=I1*R2=10V.
(2) Vn2 caused by Vout=Vb only (for I1=0, open circuit): Vn2=Vout=Vb (no voltage drop across R2).
(3) Vn=Vn1+Vn2=10V+Vb=0
(4) therefore, Vb=-10V.
 
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