What is the curent of a unity gain differential op amp?

[24Volts]

Hello,

In the following op amp configuration, can someone explain how much current is going through ra and rf?

If the ra voltage drop is 0 Vdc, how could there be current going through ra??

thanks!

 

[(*steve*)]

The same current flows through RA and RF.

This is one of the fundamental principles.

In this case, if your voltages are all correct (I say IF) then the current is zero because all points are at the same potential.

 

[24Volts]

yes that's what I thought also....

Okay, so the inverting input as seen by the op amp is 0volts right?
And therefore, since we have unity gain, the output results by what the
non inverting input sees (which is 3Vdc) minus what the inverting input sees
(Which is 0vdc) resulting in Vout of 3vdc... I guess !!! right?

thanks (*Steve*)

 

[(*steve*)]

Okay, so the inverting input as seen by the op amp is 0volts right?

The differential voltage is zero -- they are the same voltage -- Is that what you mean? Because that's theoretically correct.

And therefore, since we have unity gain, the output results by what the
non inverting input sees (which is 3Vdc) minus what the inverting input sees
(Which is 0vdc) resulting in Vout of 3vdc... I guess !!! right?

No, the output is what the non-inverting input sees less what the inverting input sees, times the gain of the op-amp (which may be 1,000,000, but is theoretically infinite). [edit: mathematically Vo = A(Vi+ - Vi-)]

Remember that I said above that the the inverting input and the non-inverting input are THEORETICALLY at the same voltage. Practically, they're not. Ignoring other imperfections, in this case the non-inverting input (assuming a gain of 1,000,000) is 3 millionths of a volt (3uV) more positive than the inverting input.

Practically that's going to mean that not all of those 3V readings will be 3V. They will differ by a few microvolts, and the amplifier's gain will be some small fraction of a percent different to 1.000000.

Practically it's also unimportant.

Practically we can assume the gain is infinite. So the output voltage divided by infinity is zero, so the difference between the inputs is zero. And zero times infinity is the number we want... (does that last step sound like magic or religion?)

The reason why zero times infinity is the number that we want is because the number that we want results in the difference in the inputs being zero.

And the number we want is determined by the feedback resistors, and the input voltage.

In this case we want the output voltage to equal the input voltage so we pick a ratio of resistors which will cause the inputs to be equal when the overall gain is some desired value.

To be perfect, an op amp must have infinite input impedance, zero output impedance, infinite gain, zero phase shift, infinite bandwidth, zero offset voltage, infinite supply rail voltage, and probably some other things.

In practice, even allowing for significant deviations to these, in many cases we can just assume perfection. The practical differences are usually negligible as long as we don't have extremely high or low resistances, or demand very high closed loop gain, or want to operate at very high frequencies or high power.

To gain a real understanding of why these formulae for calculating gain work, you need to start from the basic equation Vo = A(Vi+ - Vi-), and then solve for Vo using a practical circuit (So Vo in terms of Vi-, R1, and R2 -- assuming Vi+ is a constant voltage). For this you'll need to be able to do some mathematical analysis of simple linear circuits.

Then you look at the output voltage at differing values of A, finally looking at the limit as A approaches infinity.

The result of the last step is the traditional equation for op-amp closed loop gain.

Hopefully, once you've done the math yourself (and understand it) you will have a significant insight into the standard inverting amplifier configuration.

If you can't do that, you may have to just accept that some of the magic just happens. (or to have faith).

 

[24Volts]

The differential voltage is zero -- they are the same voltage -- Is that what you mean? Because that's theoretically correct.

oooops ! yes that's what I meant!

No, the output is what the non-inverting input sees less what the inverting input sees, times the gain of the op-amp (which may be 1,000,000, but is theoretically infinite). [edit: mathematically Vo = A(Vi+ - Vi-)]

Remember that I said above that the the inverting input and the non-inverting input are THEORETICALLY at the same voltage. Practically, they're not. Ignoring other imperfections, in this case the non-inverting input (assuming a gain of 1,000,000) is 3 millionths of a volt (3uV) more positive than the inverting input.

Ooooh !!! sorry .... let me reword that:

"May I say that the output's results is by what the
non inverting input sees (which is 3Vdc + 3uv) minus what the inverting
input sees (Which is aprox 3vdc) resulting in a very small difference
which is being amplified by a very high gain resulting in a Vout of 3vdc... right? "

To gain a real understanding of why these formulae for calculating gain work, you need to start from the basic equation Vo = A(Vi+ - Vi-), and then solve for Vo using a practical circuit (So Vo in terms of Vi-, R1, and R2 -- assuming Vi+ is a constant voltage). For this you'll need to be able to do some mathematical analysis of simple linear circuits.

Yes... I'm going through that these days!!

Hopefully, once you've done the math yourself (and understand it) you will have a significant insight into the standard inverting amplifier configuration.

yes !!

thanks for your help (*Steve*)

 

[24Volts]

Hello (*Steve*),

I have a few more questions if I may....

No, the output is what the non-inverting input sees less what the inverting input sees, times the gain of the op-amp (which may be 1,000,000, but is theoretically infinite).

Is this what is typically know as the open loop gain?

and the amplifier's gain will be some small fraction of a percent different to 1.000000.

Is this what they call the closed loop gain.. the one determined by rf/ra... ?

thanks!

 

[(*steve*)]

Yes, and yes!

 

[KrisBlueNZ]

Another point: you should remove the marking "virtual ground" from the inverting input in your schematic. The inverting input of an op-amp is only a virtual ground if the non-inverting input is grounded. (Also the op-amp must be connected with negative feedback, and it must be operating within its output voltage and current limits, for the inverting input to be called a "virtual ground"; both of those things are true for your circuit.)

 

[24Volts]

Today, 07:40 AM #8
Another point: you should remove the marking "virtual ground" from the inverting input in your schematic. The inverting input of an op-amp is only a virtual ground if the non-inverting input is grounded. (Also the op-amp must be connected with negative feedback, and it must be operating within its output voltage and current limits, for the inverting input to be called a "virtual ground"; both of those things are true for your circuit.)

understood!

thanks guys