An OPAMP with input offset voltage

In summary, the conversation discusses the correct approach for analyzing a circuit using OPAMPs and determining the gain equations. It is mentioned that the resistor values in the circuit may not match those in the prototypical circuit and therefore the equations may need to be adjusted. It is also suggested to analyze the circuit from scratch instead of relying on cookie-cutter formulas. Deriving formulas directly is recommended as it may lead to more accurate results.
  • #1
kl055
26
0

Homework Statement


Capture.jpg

Homework Equations


Superposition
Non-inverting OPAMP: V_out = (1 + R/R)V_in = 2*V_in2
Inverting OPAMP: V_out = -V_in
V_os = 2.5mV

The Attempt at a Solution



1. V_in enabled, V_os disabled

V_out1 = -V_in

2. V_os enabled, V_in disabled

To get V_out2, use the non inv. equation where V_in2 is equal to V_1 + V_os
where V_1 is the node between R and R at the middle bottom of the circuit.
V_out2 = 2*(V_1 + V_os)

3. V_out = V_out1 + V_out2

Is this the correct approach?

Thanks in advance :)
 
Last edited:
Physics news on Phys.org
  • #2
You may want to take a closer look at the gain of the circuit; The resistor values in the network joining Vout and +Vin may not be identical to those of the prototypical circuit from which your gain formula was derived.

The Vos is usually modeled as a fixed voltage source in series with one of the op-amp inputs, typically the + input. The circuit is not particularly difficult to analyze with a bit of nodal analysis, voltage division, etc.

attachment.php?attachmentid=59030&stc=1&d=1369575714.gif
 

Attachments

  • Fig1.gif
    Fig1.gif
    4.5 KB · Views: 772
  • Like
Likes 1 person
  • #3
gneill said:
You may want to take a closer look at the gain of the circuit; The resistor values in the network joining Vout and +Vin may not be identical to those of the prototypical circuit from which your gain formula was derived.
Do you mean that neither the inverting nor non-inverting equations will work in this case and that I need to derive new gain equations using nodal analysis?
 
  • #4
kl055 said:
Do you mean that neither the inverting nor non-inverting equations will work in this case and that I need to derive new gain equations using nodal analysis?

The equations would need adjusting to reflect the actual circuit's values. To adjust the equations, you'll have to know the influence of each resistance on the equation. It would likely be simpler/safer to just analyze the circuit from scratch.
 
  • Like
Likes 1 person
  • #5
Or do you mean that I need to modify my gain equation in step 2 by finding V_3 (by doing voltage division) and plugging that into the equation?

edit: nvm, I need to read that first
 
  • #6
I'm pretty sure that step 2 is correct. The professor did an example in class that was similar enough to that case. But in step 1, I have input voltage at both OPAMP terminals so I do not know what to do. I'd rather take the OPAMP as a black box with a characteristic equation if possible.

In case 1, can't I just use V_3 as a reference (ground) and then modify my equation to become
V_out = -(V_in - V_3)?
The prototypical inverting gain equation assumes that V_+ is ground, so if I take my reference at V_3 the above equation should work?
 
  • #7
The problem is, if you take V3 as the ground reference then all the points that are shown as connected to ground will have to take on some other potential (-V3), and you'll have to adjust Vin accordingly as well (since Vin is a ground-referenced value). Lot's of things are interconnected...

Why don't you try it both ways if you can and compare results?
 
  • Like
Likes 1 person
  • #8
I'll do that. I was just hoping that there was an intuitive way to do it since I'd rather not derive formulas when writing exams unless I have to :S
 
Last edited:
  • #9
kl055 said:
I'll do that. I was just hoping that there was an intuitive way to do it since I'd rather not derive formulas when writing exams :S

I think you're likely to run into more "non-standard" circuits than textbook standard ones. Cookie-cutter formulas are limited in applicability. You're much better off being able to quickly analyze a given circuit on-the-fly.
 
  • Like
Likes 1 person
  • #10
I did the derivation and my "intuitive" formula for part 1 was incorrect (by a factor of 3).
Part 2 was good but I think that I will derive formulas directly from now on.
Thanks for the help
 
  • #11
kl055 said:
I did the derivation and my "intuitive" formula for part 1 was incorrect (by a factor of 3).
Part 2 was good but I think that I will derive formulas directly from now on.
Thanks for the help

Glad to help! Good luck in your studies :smile:
 
  • Like
Likes 1 person

FAQ: An OPAMP with input offset voltage

1. What is an operational amplifier (OPAMP)?

An operational amplifier, or OPAMP, is an electronic device that amplifies an input signal to a higher output voltage. It is commonly used in analog circuits, such as filters, oscillators, and amplifiers.

2. What is input offset voltage in an OPAMP?

Input offset voltage is a small, inherent voltage difference between the two input terminals of an OPAMP. It can cause an error in the output voltage and can be caused by manufacturing imperfections or temperature changes.

3. How does input offset voltage affect the performance of an OPAMP?

Input offset voltage can cause the output voltage of an OPAMP to be slightly different than the expected value, leading to inaccuracies in the circuit's operation. It can also cause the circuit to be less stable, which can result in unwanted oscillations or noise.

4. How can the input offset voltage be minimized in an OPAMP?

There are several techniques that can be used to minimize input offset voltage in an OPAMP, such as using a precision trimming process during manufacturing, using matched components, and implementing circuit techniques like chopper stabilization or auto-zeroing.

5. Can input offset voltage be completely eliminated in an OPAMP?

No, it is impossible to completely eliminate input offset voltage in an OPAMP. However, it can be reduced to a very small value through careful design and component selection.

Back
Top