Op amp: Find Vo in terms of Vg and A

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SUMMARY

This discussion focuses on deriving the output voltage (Vo) of an inverting operational amplifier (op amp) in terms of the input voltage (Vg) and the open-loop gain (A). The correct formula for Vo is established as Vo = -10Vg/2 = -5Vg, taking into account the finite gain of the amplifier. Participants clarified that the open-loop gain refers to the op amp's gain without feedback and that the current into the input terminals remains zero for an ideal op amp. A reference textbook, "Electronic Devices, 9th ed, T. Floyd," was suggested for further reading.

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Homework Statement



eqnk1z.jpg


Find Vo as a function of Vg and the open-loop gain A.

Attempt

I'm not sure what open-loop gain means.

I redrew the circuit so it made more sense to me, but I'm still not getting it. It seems that Vo is the voltage across both RL and the 4k resistor.

The voltage across the 4k resistor is equal the voltages at the input terminals.

Aside from these bits of information I don't really know how this thing operates. Should the current going into the input terminals still be 0? But at the positive terminal the current is 8/50k right? or does this mean there is a -8 voltage source after the 50k resistor?
 
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The diagram shows neither an RL nor a 4k resistor.
 
ElijahRockers said:

Homework Statement



eqnk1z.jpg


Find Vo as a function of Vg and the open-loop gain A.

Attempt

I'm not sure what open-loop gain means.
It probably means the op amp gain by itself (no feedback or input components). Ground one input, apply a signal to the other input, the OL gain = output/input.

Should the current going into the input terminals still be 0?

Yes. The op amp is ideal except for finite gain.
 
Whoops. Sorry guys, wrong picture. But I figured it out with some help, thank you!
 
Vo=5Vg
 
hoangkyem said:
Vo=5Vg

That's incorrect too.
 
rude man said:
That's incorrect too.
Why? Vo=10/2Vg=5Vg
Since it is Inverting amplifier
 
hoangkyem said:
Why? Vo=10/2Vg=5Vg
Since it is Inverting amplifier

Because you have ignored the finite gain of the amplifier (the "open-loop gain A").

Your expression holds only if the amplifier gain is infinite.
 
rude man said:
Because you have ignored the finite gain of the amplifier (the "open-loop gain A").

Your expression holds only if the amplifier gain is infinite.

I think I'm right. It's is inverting amplifier, so the closed-loop gain (Acl) independent of the op amp's internal open-loop gain (Aol)
Acl=-Rf/Ri where Rf is feedback resistor, Ri is input resistor
→Vo=-10Vg/2=-5Vg

I founded on internet a textbook: Electronic Devices, 9th ed, T.Floyd
You can download and refer in chapter 12, page 618.

Ebook Electronic Devices 9th ed, T.Floyd
 

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