# Inverted op amp

Discussion in 'Homework Help' started by fleur_, Jan 5, 2016.

1. ### atferrariWell-Known Member

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I would never open any .doc document posted here which is a nice way to get a stupid virus if you are not careful. Why not an humble .pdf?

2. ### fleur_New Member

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Okay no problem.pdf it is

3. ### fleur_New Member

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This is my report,I have also attached a rubric to give you a rough idea of what is expected.
I would really appreciate it if you could check it out.

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5. ### audioguruWell-Known MemberMost Helpful Member

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The actual gain IS NOT -20dB which is a loss, not a gain. Instead the actual gain is 20dB and is -10 times. The dB number will not say if the signal is inverted or not. I think if you say a signal has a gain of 10 then we also do not know if it is +10 times which is non-inverting or if it is -10 times which is inverting.

Use simple arithmetic to determine the output impedance of the signal generator. If the 10k feedback resistor value is actually exactly 10k ohms and the actual gain is -9.65 then the input resistance is 10k/9.65= 1036.3 ohms. If the 1k resistor is actually exactly 1k ohms then the generator impedance is 36.3 ohms.

6. ### fleur_New Member

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got it.Thank you !

7. ### fleur_New Member

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I am trying to plot the theoretical graph of the capacitor voltage against time for the circuit components values.I have plotted the x axis on excel,however I am struggling with plotting the y axis.I am aware that that I have to use this equation Vcap=Vs(1-e^-t/RC) Vs being 2V R=1000 and C=0.020
Can't upload an excel so I uploaded a PDF to show the values for the x axis.

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8. ### audioguruWell-Known MemberMost Helpful Member

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Your circuit on page 1 does not have a capacitor. Did you add a 20nF (0.02uF) series input capacitor? If the input is 2VDC then this capacitor charges through the 1k input resistor exponentially. But since the gain is -10 and the supplies are only 15V then the output of the opamp will saturate wren it reaches 13.5V to 14V, then the capacitor will continue charging much slower.

9. ### fleur_New Member

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Oh I should have mentioned that this is a different question and is not linked to the circuit on page 1. I made a homemade capacitor using foil and paper and used a 1k resistor.Then I calculated the RC time constant T=RC (this is probably where i went wrong) 0.632x2v=1.264v R=1000 T=20μs =20x10^-6 (not sure about this ) so in order to get C I rearranged the equation to C=T/R thus 20μs/1000=0.02 so now that I know what C is I need to use the equation for the voltage across a capacitor being charged through a resistor in order to plot the graph on excel as i was saying earlier but I have no clue on how to use this equation
Vs=2v,R=1000,C=20μs

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10. ### audioguruWell-Known MemberMost Helpful Member

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Why did you make a Mickey Mouse capacitor and guess about its value when you can buy an accurate capacitor for almost nothing?
If the time constant of a capacitor charged with 1k resistor is 20us then its value is 20nF which is 0.02uF. Buy a 20nF or 22nF capacitor, measure it and see if it is the same.

11. ### fleur_New Member

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because that's what I have been told to make in class. I still don't know how to plot this graph...

12. ### audioguruWell-Known MemberMost Helpful Member

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I have never plotted a graph of the voltage of a charging capacitor, I simply looked at the graph already in a book or on the internet. If you do not know how to do it then ask your teacher to teach you how to do it.

13. ### gophertActive Member

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fleur_

The gain of -10 is in the first (input signal is 1v peak-to-peak) but the sine wave goes up as output wave goes down (output peak-to-peak is 10 V).

Gain in second graph is 10, note they are in phase.

Graph 2 (gain of +10)

14. ### audioguruWell-Known MemberMost Helpful Member

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It might be confusing to the student to see that your scope has 7/10ths of a volt for each line instead of 1.0V.

• Agree x 1
15. ### gophertActive Member

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It confuses me as well but, since it is an iPad app, the number of controls for the user (me) are very limited.