I am struggling to figure out why the current in this is capable of going backwards. To my knowledge the negative feedback is going towards the divider in order to create equal value pins. In this photograph they explain the potentials of the left and right side of the resistor and how these symptoms provide current, but isn't current already going the opposite way?


Another question totally unrelated...

Does virtual ground only apply to when one of the pins is connected directly to ground.

Thank you for all the help. I have been struggling to figure out how op amps work for weeks and am just now starting to make headway.

 

but isn't current already going the opposite way?

Yes, they are going backward. But the Author is using electron flow convection. So the "election current" will flow "from negative to positive"
https://www.allaboutcircuits.com/textbook/direct-current/chpt-1/conventional-versus-electron-flow/

Does virtual ground only apply to when one of the pins is connected directly to ground.

Yes, you are right again. So, if the op-amp inputs are not at ground potential (as in your circuit) in this case we call it a "virtual shot" instead.

This post http://forum.allaboutcircuits.com/showpost.php?p=444315&postcount=9 (start read from
"The more accurate explanation").

 

Regarding direction of the current. This could be a mistake. If you have book errata (google the book or see if publisher website has an errata for this book), you could check and see if it is a mistake. If it is not listed in errata, you could submit it to the publisher.

However. There are actually several different ways to teach the direction of the electrical current. I was taught that current is a flow of electrons from high electrical potential to lower electrical potential. However, this representation of electrical current is not singular. There are other representations taught in other countries. They are just as correct. You may have to examine earlier chapters in this book to see if the authors define their definition of current flow.

 

If it is not listed in errata, you could submit it to the publisher.

But this picture is coming from "allaboutcircuits" textbook
https://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/divided-feedback/

 

"Virtual Ground" is a shorthand approximation that is quite useful when an op amp is amplifying the inputs. Is is due to the very high gain of an op amp. Even the lowly LM741 has a worst case gain of 25,000. So for a maximum output of 18V the input would be 18/25,000=.00072 or just .72 millivolts. That's usually small enough to ignore, so if one op amp input is grounded it is safe to say the other one is too.

BUT... it is called a "virtual" ground because the current into or out of "ground" is zero at this point, due to the high input impedance of the amplifier. It is just the voltage that is at ground.

"Virtual ground" only applies when the op amp is not being used as a comparator, or the output is not in saturation.

And the direction of current in the All About Circuits textbook? Yeah, probably best to chuck that whole thing since it is based on the unscientific concept of current being the movement of electrons.

Thus it is backwards to how current is defined as current is based on the movement of charge

 

Okay then

in this photo on the same website, it shows the opposite. Can you explain this?

 

Can you explain this?

Yes, once again the "electron flow" from the most negative side to the most positive side.

Here you have a conventional current flow.

 

The direction of current flow does not change so it does not matter. Simply use Ohm's Law to calculate the current and the voltage at each resistor. Current direction is important to determine the polarity of a capacitor or diode, not an opamp.

Your opamps have no power supply so how can they do anything? Some opamp circuits have a positive supply, ground and negative supply. Other opamps have a positive supply, a virtual ground at half the positive supply voltage and ground. The half supply is the "virtual ground" and in the opamp circuit that has a positive and negative supply the half the supply voltage is ground. Here are how they are biased (they have input and output capacitors so they amplify AC, not DC):

 

Does virtual ground only apply to when one of the pins is connected directly to ground.

No. In your circuit, none of the signal pins are connected directly to GND. Also, to eliminate some of the input offset voltage error in cases of high circuit gain, even if you put a capacitor in series with R1 between its left side and GND, the inverting input still would be a virtual GND.

ak

 

A virtual ground is a DC voltage with no signal that is a reference voltage acting like a ground.
A virtual ground is usually at half the supply voltage so that the signal can swing equally up and down.
A normal ground reference voltage is also at half the supply voltage (0V) when the supply has dual polarities so that the signal can swing equally up and down.

 

Your opamps have no power supply so how can they do anything?

You've made this statement in multiple posts. It's a fairly common practice to not clutter schematics with power connections.

Do you really think that anyone wouldn't know that circuits will not function without a power source??

These schematics came from National Semiconductor; folks who know a thing or two about circuits:


No explicit power connections or decoupling caps.

 

When a noob asks about the basics then I always show the basics. A noob will not know if the unmarked supply in a datasheet is for a single positive supply or for a dual polarity supply.