Capacitor charging lineair when using it as an integrator in an opamp setup.

[pivu0]

Dear people of Electro Tech,

I study Engineering in Belgium, and I'm studying for my Electronic exam for in a few weeks.
While I was reading about the integrator setup using opamps, I found something that I have trouble to understand.
This is the setup:

**broken link removed**

I know when you have a capacitor in serie with a resistor, you have the differential equation : V = q/C + R*dq/dt, when solved yields Vc = V*(1-e^-t/RC).

However, the capacitor charges up lineair when a constant Voltage is applied at Vi in the setup above. Does it not stand in series with the resistor, as there should flow no current into the opamp?

Thank for reading,

pivu0

 

[MrAl]

Hello there,

The capacitor is not really in series with the resistor. The capacitor is in the feedback path and the resistor is in the input path.
Thus we end up with i=Vin/R, and Vo=-(1/C)*Intg i dt. The same i flows in input and feedback. Thus we integrate Vin/RC and that's just a a constant.

The solution works out not to a capacitor charged by a resistor, but a capacitor charged by a constant current who's current is Vin/R. We end up with the solution Vo= -Vin*t/(R*C). That's why it's a linear charge curve (ramp).

Try this again and see how you make out.

 

[pivu0]

Thanks for the reply!

Hm, but in my textbook, they say, you have to assume no current flows to the opamp (assume its a ∞ resistor), wouldn't this make the resistor and capacitor in series?

And, if the the opamp saturates, wouldn't that remove the feedback?
If you power the opamp with 5V, and you give an input of 20v, Vo wouldn't give feedback right?

 

[ericgibbs]

Thanks for the reply!

Hm, but in my textbook, they say, you have to assume no current flows to the opamp (assume its a ∞ resistor), wouldn't this make the resistor and capacitor in series?

And, if the the opamp saturates, wouldn't that remove the feedback?
If you power the opamp with 5V, and you give an input of 20v, Vo wouldn't give feedback right?

hi,
While Al is sleeping, look up 'Virtual Earth' as explained in OPA's..

https://books.google.com/books?id=A...Q6AEwAA#v=onepage&q=OPA Virtual Earth&f=false

 

[ronsimpson]

1.)"No current flows to the opamp" The inputs (+) and (-) have no (very little) current flow in them. Do not think about current in the inputs.
2.)If the opamp saturates then there is no feedback.
3.)If the power of the opamp is 5V and the input if 20 volts it will work fine.

The opamp wants input+ and input- to have the same voltage. If input+ is larger then the output will move positive. If input- is larger the output will move negative. If input+ equals input- then the output voltage will stay where it is.

Example: the input = 0 volts, the opamp is not saturated, input+ is at ground (0v), input- is 0v. The voltage across R is 0v. The current in R=0. The output of opamp will hold. (no change).
Example: the input =20 volts, the opamp is not saturated, input+ is at ground (0v), input- is 0v. The voltage across R is 20v. The current in R is 20/R. The output of the opamp will move down at a rate to pull R/20 of current from R. To get current to flow in a C you must have a change in voltage across the C. The opamp will have a ramp voltage downward. If the input is -20V the output is a upward ramp.

If the opamp saturates, it can no longer pull current away from the capacitor. Every thing is different. The output stops ramping down and stops at the saturation voltage (-12V for example). Now the current in R will pull input- up away form ground.

 

[dougy83]

wouldn't this make the resistor and capacitor in series?

Your textbook will also say something about negative feedback causing the opamp to output a voltage to keep both its inputs at the same voltage (in the above case, 0V). Yes, they're in series; but they're not being supplied a constant voltage as is assumed in 'Vc = V*(1-e^-t/RC)'. I won't repeat what MrAl has said, but basically the opamp output changes to keep the voltages at its inputs the same.

And, if the the opamp saturates, wouldn't that remove the feedback?

The integrator output cannot operate outside its supply rails & so the feedback cannot cause the inputs to remain equal.

If you power the opamp with 5V, and you give an input of 20v, Vo wouldn't give feedback right?

The output will approach the lower rail voltage. The input voltage only affects the slope of the ramp - if you connected -0.1V as the input it will still saturate; it will just take longer than -20V.

 

[MrAl]

Thanks for the reply!

Hm, but in my textbook, they say, you have to assume no current flows to the opamp (assume its a ∞ resistor), wouldn't this make the resistor and capacitor in series?

And, if the the opamp saturates, wouldn't that remove the feedback?
If you power the opamp with 5V, and you give an input of 20v, Vo wouldn't give feedback right?

Hello again,

Yes for the ideal op amp you assume no current flow into the inverting terminal as you noted, but that's not the only assumption you apply. You also have to apply at the same time the assumption that the inverting terminal is at virtual ground (Eric was nice enough to provide a link for this). That means you not only have the same current flowing in the input as the output, but you also have the inverting terminal stuck to ground potential. In a non op amp circuit the resistor and capacitor junction would be allowed to change voltage and that changes everything as the response goes to k*(1-e^-at) instead of k*t. Thus the op amp makes the capacitor integrate perfectly.
In the frequency domain the R and C in series looks like k/(s+a) while the true integrator with op amp is just k/s.
Of course there are some non idealities that have to be considered in the real world such as offset voltage and current.

All of this also assumes that the op amp is operating in the linear mode too. If the op amp goes out of the linear mode for any reason, the circuit changes completely. This means we have to consider the maximum output current and things like that. If we apply an input that at any time forces the op amp out of the linear mode then we have to figure some op amp output resistance and also no more virtual ground, so we would end up with a very different response.

Thus one of the things that you would have to do first in the real world is to calculate some threshold that would allow you to first determine what mode of operation the op amp was working in. Is it linear or not? Then apply the calculations for that mode and always check to see if it moved out of that mode where other equations would have to be applied.
This is a typical operation for various kinds of electronic devices, not just op amps. Normally we assume a linear mode, but that's not always necessarily going to be true for all time.
You already noted that if you apply 1v for 10 years at first the output ramps down, but once it reaches the negative supply rail that's it. It bottoms out and can not go any lower. If you apply -1v for a long time at first the output ramps up, but once that reaches the positive supply rail that's it too...it tops out and stays there. To add to the problem, many op amps can not even reach as far as their plus and minus rails as there is some other limiting factor such as a transistor saturation voltage drop plus the drivers drop so we may end up with something that can only reach as high as the plus voltage minus 1.5 volts. You'd have to check the op amp specs to find this information out.