No, it is not, because what matters is NOT that (R1 + R2)/K be small, but rather that it be small COMPARED to R2.The limiting equation I got was
Vo = (KR2Vi)/(R1 + R2 + KR2)
= R2Vi/[R2+(R1+R2)/K]
So is dimension-ally correct. And in the limit the solution.
K is gain, dimensionless. So if K >> R1 + R2 is an accurate statement.
Regards, Dana.
What I should have said is as K > 00 (infinity, very high G). The equation is dimensionally correct,Vo = (KR2Vi)/(R1 + R2 + KR2)
= R2Vi/[R2+(R1+R2)/K]
So is dimension-ally correct. And in the limit the solution.
K is gain, dimensionless. So if K >> R1 + R2 is an accurate statement.
There is no problem at all withActually I do have an error -
What I should have said is as K > 00 (infinity, very high G). The equation is dimensionally correct,
anytime a constant (K) modifies a dimensioned value the dimensions are maintained and expressed.
In your case you use low values of K, so you are correct in your calculated example.
Regards, Dana.
Such a simple circuit, so much confusion. Nothing weird going on here:Now I've really confused myself here or there's something weird happening that I can't see.
Hi,What does it mean for the dimensionless gain to be much, much larger than the sum of two resistances?
I agree.Such a simple circuit, so much confusion. Nothing weird going on here:
There, nothing to it except for some simple, step-by-step logic and application of basic principles.
- The op amp is ideal: no input offset voltage, no input bias current, and gain is infinite.
- The op amp is in a negative feedback circuit, so start by assuming its output is in the linear region.
- The op amp's (+) input is connected to Vi.
- Since we're operating in the linear region, the (-) input must be at the same voltage as the (+) input.
- If the (-) input is at the same voltage as the (+) input, it is therefore at the same voltage as Vi.
- Thus there is no voltage across R2, and therefore no current through it.
- With no current through R2 and no current into or out of the op amp's (-) input, there can be no current through R1.
- If no current is flowing through R1, there is no voltage drop across it.
- Therefore, the voltage at Vo must equal the voltage at the op amp's (-) input, which is equal to Vi.
- Ergo, Vo = Vi.
Hello again,Dimensional analysis worked just fine and revealed that the claim was fundamentally meaningless.
It was dimensionally inaccurate because a mistake was made -- a mistake that the messed up units shouted at full volume and was roundly ignored.Hello again,
The reason i said that was because the claim worked numerically. Why it's not dimensionally accurate i dont know.
Note that 'a' has to have units of inverse resistance (or conductance).Hello,
The claim was (A=Aol):
A>>R1+R2
Now if A has to be greater than the sum, then set:
A=a*(R1+R2)
Is a = 1,000 S big enough to satisfy A >> (R1 + R2)?and now 'a' is a factor that controls how much greater R1+R2 is.
Then the solution reduces to:
vo=(a*vi*R2)/(a*R2+1)
and with a getting higher and higher, the '1' in the denominator is absorbed by the a*R2 there, so we end up with:
vo=a*R2/(a*R2)*vi
which of course reduces to:
vo=vi
What does it mean for a resistance to be much greater than 1?However, if we make a too small, that '1' stays significant and we might end up with:
vo=R2/(R2+1)*vi
and now we have to depend on R2 alone to get near to vo=vi.
Before we only had to depend on R1+R2 getting larger.
So i guess if you want to say that A can be small if R2 is very large that might work too.
So we might also include R2>>1
Of course the gain can never be unity unless k become large -- in fact, the gain can never be unity unless k is identically infinite!The gain expression can be manipulated into a form which helps to see what's going on:
View attachment 181904
The green and red parts can be seen as error terms. Consider the second parenthetical. If R1>>>R2, the red part approaches 1, the second parenthetical approaches zero, and the gain approaches zero. If R2>>>R1, the red part approaches zero, the second parenthetical approaches unity, and the gain is given by the first parenthetical.
The second parenthetical can never be more than 1, so the gain can never be greater than the first parenthetical. Thus the gain can never be unity unless k becomes large (making the green part zero), no matter what R1 and R2 are.
Because you must have edited after I quoted it -- I started my reply yesterday at about 4:30 pm (about half an hour after you posted) but had to leave to run my daughter around and just now got back to finishing the response. I didn't note that you had changed anything. So here's the last part as you have it now:Hello again,
Why did you leave off the end of that last quote?
Can we say that? Let's see.So i guess if you want to say that A can be small if R2 is very large that might work too.
Well, which is it?So we might also include R2>>1 or rather R2>>R1.
This makes sense because 'a' and R2 are mutiplicatively symmetrical.
Close, but not quite.Another idea might be A>>R1/R2.
There's your dimensional correctness for ya
At least the conditions I came up with actually work. If you are happy coming up with conditions that don't work, I just hope that I don't ever have to rely on anything you designed based on them.But sure, no problem whatever you want to do, you can come up with the conditions then that's fine.
There's no need to guess -- just do that math.Hi,
I was only guessing that's it.
If it weren't dimensionless, then the units would be there. If it lacks dimensions, then it is dimensionless (seems rather obvious when put that way, doesn't it?).A >> 1 + R1/R2
is fine with me. But what are the units for that '1' in that?
Is it dimensionless can we assume that
If we can, how do we know we can?
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by Jake Hertz
by Jake Hertz
by Jake Hertz