How would I draw this non-inverting amplifier?

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The discussion focuses on designing a non-inverting amplifier using practical operational amplifier characteristics, specifically with a gain of 10 and a feedback resistor of 100K. Participants emphasize the importance of matching source impedance between the inverting and non-inverting inputs to minimize offset voltage due to input currents. Suggestions include using resistor values from the E24 series for better approximation and ensuring that the offset current is accurately represented in the circuit diagram. The conversation highlights the need for clarity in drawing the circuit to avoid confusion with current flow. Overall, the design considerations revolve around achieving accurate performance while addressing practical op-amp limitations.
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Summary: So far, I have only dealt with ideal operational amplifiers, so I am kind of lost trying to visualize how a practical op amp should look like.

A non-inverting amplifier with a gain of 10 uses 100K as its feedback resistor. It gets its input from a signal source whose source resistance is 5K. The amplifier has zero offset voltage. However, it has an input bias current of 2 µA and an input offset current of 0.2 µA.
Draw the op-amp circuit.

Is this right?
Also, is R2 = 11.11k? I used A = 1 + RF/R2 to get this value

circuit.png
 
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Seems no issues with equivalent circuit schematic above.
Regarding G=10 non-inverting opamp, i prefer to use 91k and 10k, these values are closest approximation to G=10 using E24 preferred numbers series.
 
You have significant input current so you need to match the source impedance of the (–) and (+) inputs. The op-amp output can be seen as having a low impedance.
So with Rf = 100k, and R = 11k11, the source impedance on the (–) input will be Rf in parallel with
R, giving 10k0. To avoid offset voltage due to input currents you need to give the (+) input the same source impedance as the (–) input. The driver has Rsource = 5k0 now, so you need to add an external Rspecial = 5k in series to match the (–) input.

Draw your potential divider so it is obviously a potential divider by keeping the current flowing in a straight line.
Don't make a hedgehog with currents radiating out from the op-amp, keep parallel currents parallel.

gainx10.png
 
Post 2 is not showing the offset current.
 
rude man said:
Post 2 is not showing the offset current.
Or post #3? The offset current was left for later. I judged that matching the source resistance was more important and enough for a first step.
I expected @marcadams267 to have noticed the missing offset current from the way it was drawn. Maybe Ios needs to be drawn as a separate current source between the two input terminals.
 
Baluncore said:
Or post #3? The offset current was left for later. I judged that matching the source resistance was more important and enough for a first step.
I expected @marcadams267 to have noticed the missing offset current from the way it was drawn. Maybe Ios needs to be drawn as a separate current source between the two input terminals.
I couldn't see post#2 until now.
Offset current should be depicted exactly as the OP did in post 1, i.e. as part of bias current.
 
rude man said:
Offset current should be depicted exactly as the OP did in post 1, i.e. as part of bias current.
Then how can you know that the inverting input would be 1.9 uA and the non-inv input 2.1 uA? Could it not be the other way about?

A model should generate two identical 2u0 bias currents, with a sinusoidal offset current of 1u0 between the two inputs. With zero amplifier input that offset would show up on the output of the amplifier, revealing the sensitivity to the Ios max = 2u0 A.
Supply voltage and temperature dependence of bias current can be applied to the two identical current sources. Sweeping the offset current of the model is then independent of the bias.
 
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