Question on Opamp homework problem: Find Iout between these two Opamps

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Discussion Overview

The discussion revolves around a homework problem involving operational amplifiers (opamps), specifically focusing on finding the output current (Iout) between two opamps in a given circuit configuration. Participants explore the calculations related to output voltage and current, as well as the implications of circuit assumptions such as virtual ground.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the output voltage (Vo1) for a non-inverting opamp using the formula and arrives at 10V, while also calculating the current (I3) using Ohm's law, resulting in 0.2mA.
  • Another participant questions the polarity of the top opamp and suggests that the voltage at point A is zero due to the virtual ground assumption, leading to a different interpretation of Vo1.
  • Some participants argue that the circuit may not be correctly drawn, raising concerns about the absence of power supply voltages and the implications for the circuit's functionality.
  • There is a discussion about the feedback configuration of the upper opamp, with one participant stating it has positive feedback and is in full saturation, while the lower opamp is described as having negative feedback and functioning as an inverting amplifier.
  • Participants express differing views on the correctness of the equations and assumptions used in the calculations, indicating a lack of consensus on the approach to solving the problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the problem, with multiple competing views on the circuit configuration, assumptions about voltage levels, and the implications of feedback types.

Contextual Notes

Limitations include potential misunderstandings about circuit polarity, assumptions regarding virtual ground, and the absence of specified power supply voltages, which may affect the validity of the calculations presented.

james weaver
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Homework Statement
find Io
Relevant Equations
ohms law
Hi, so my objective is to find Io, and I've attached a picture of my work. Here is my question: At the point Vo1, if I use the equation for finding the output voltage for a non-inverting op amp I end up getting this:

$$V_{0{1}}=\left( 1+\frac {r_{f}} {r_{1}} \right)V_{s1} = \left( 1+\frac {40k} {10k} \right)2 = 10v$$

However, I can calculate the current ##I_{3}## by using ohms law:

$$\frac{V_{s1}-V_{A}} {10k}=\frac {2v} {10k}=0.2mA$$

From there, I start at Vs1 and add the voltage drops until I reach Vo1, which gives me: ##2-50k*I_{3}=-8v##

My instinct is to go with the -8, since ##I_{3}## would have to be reversed if it were 10v. Can you please help explain to me which one is right and why? Thanks.

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First, I think the polarity of the top amp is backwards.

james weaver said:
I start at Vs1 and add the voltage drops until I reach Vo1, which gives me: 2−50k∗I3=−8v
Nope, the voltage at A is zero (virtual ground assumption), So V01 = 0 - 50KΩ⋅I3.
Alternatively, if you want to start from VS1, you have to count both the 2V from VS1 minus the 2V drop across R1, which cancel each other. So, V01 = VS1 - R1⋅I3 - RF⋅I3.
 
DaveE said:
Nope, the voltage at A is zero (virtual ground assumption),
[edit] Or if you assume the circuit is correctly drawn: [/edit]
Since point "A" is fed with a positive battery voltage AND the positive output from its non-inverting configuration, where is the negative source to put it at zero volts?
 
Tom.G said:
the positive output from its non-inverting configuration
Which is an unlikely HW problem without a good solution since we don't have the PS voltages. So...

DaveE said:
First, I think the polarity of the top amp is backwards.
However, you can solve it the other way if you like.
 
However, the equation he wrote is for a positive gain amp. So it's not just the polarity that's messed up here.
 
To make it clear: The upper opamp has positive feedback and will not work as desired. It is in full saturation.
The other opamp (at the bottom) is wired as a positive gain amplifier (for Vs1) with - of course ! - negative feedback. For Vs2 it will work as an inverting amplifer.
Together, it forms an amplifier which amplifies the difference of both signals (with a different gain factor).
 

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