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I was reading about opamp integrator concept in a text as such:

enter image description here

I marked the terminals of caps as a and b. Above formula assumes the b terminal as being the positive end I guess.

But regarding the direction of I2, can't we say that the positive terminal of the capacitor is a?

But then if it the positive terminal is a, then it is connected to the ground. What is the + and - ends of the capacitors in the above figure?

I know it is basic circuit theory but somehow I'm confused at this point.

Isn't there only one option of the polarity of a cap at a moment for a given current direction across it?

EDIT:

Besides these, I also do not get why in sim for the inverting input (point X) the voltage is not zero for 10 secs of time duration. According to op amp theory Vx must be around zero but it is around 1V here:

enter image description here

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  • \$\begingroup\$ The simulation shows you the steady-state response. Op-amp is in negative saturation. In transient options, you need to check this box "Skip initial operating point solution". And for R1 = 1k and Vin = 1V the capacitor (1uF) will charge to -12V in \$ T= 1\mu F *\frac{12V}{1mA} = 12ms = 0.012s \$, \$\endgroup\$ – G36 Jun 19 '17 at 17:59
  • \$\begingroup\$ But whatever the response type is, isn't it violating one of the opamp golden rules which is Vx = V- = V+ = GND in this case? \$\endgroup\$ – user1245 Jun 19 '17 at 18:18
  • \$\begingroup\$ Could you write this to your answer in a detailed way? i thought "Skip initial operating point solution" would skipping something but in this case it reveals something. \$\endgroup\$ – user1245 Jun 19 '17 at 18:20
  • \$\begingroup\$ No, when the opamp output is in saturation (positive or negative) this "golden rule" do not hold anymore. Because the opamp output voltage cannot be large than supply voltage. \$\endgroup\$ – G36 Jun 19 '17 at 18:20
  • \$\begingroup\$ Try read this e2e.ti.com/blogs_/archives/b/thesignal/archive/2012/05/08/… \$\endgroup\$ – G36 Jun 19 '17 at 18:22
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The picture is worth a thousand words

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Logically interpret it as non-polar but practically when chooosing a part Vin- = Vin+=0V so if an e-cap (polar) was used on a single supply, it's + polarity would be on the output.

Some old drafting std's use this symbol for all caps polar or not so dont be confused. A polarized cap in America shud have the + symbol. EU , ISO and IPC stds may be different. ( I foget which is most current std but your location, company and mileage may vary)

This academic question just shows it backwards to conventional symbols.

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  • \$\begingroup\$ If lets say Vi is positive constant DC, then to me "a"(also the GND) is more positive than "b" for I2 to flow that direction for the capacitor C. So to me Vo must be negative value if Vi is positive. But the integration in the book for the Vo looks positive? Do you think I2 should be negative in that integration? \$\endgroup\$ – user1245 Jun 19 '17 at 16:25
  • \$\begingroup\$ Yes for +V1, Vo=-ve and Io=- and Io=-I1 at all times until saturated output. So the formula remains correct polarity, for cap. As you can see it is always an inverting integrator, so Vo = - integral of Vin \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 19 '17 at 16:31

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