I'm actually curious because when I tried playing around with some
circuit simulation I had accidentally interchanged the inverting and
non-inverting input in an inverting configuration. I knew my output
was funky so I tried doing hand calculations and my equations came out
the same.

??
Thanks!

"MRW" (m...@gmail.com) writes:
> I'm actually curious because when I tried playing around with some
> circuit simulation I had accidentally interchanged the inverting and
> non-inverting input in an inverting configuration. I knew my output
> was funky so I tried doing hand calculations and my equations came out
> the same.
> 
I hope you aren't saying you connected the feedback resistor to the
non-inverting input.  Because that won't work.

Inverting has a series resistor from the signal input into the inverting
input, with a feedback resistor from the output to the inverting
input.  The non-inverting input goes to ground, either directly or
through a resistor that exists to keep things balanced DC wise.

Non-inverting has the same resistor from the output to the inverting
input.  Then a resistor from the inverting input to ground.  Then
the signal is fed into the non-inverting input.

There always has to be feedback from the output to the invering input.

  Michael

On Feb 1, 11:50 am, e...@FreeNet.Carleton.CA (Michael Black) wrote:
> I hope you aren't saying you connected the feedback resistor to the
> non-inverting input.  Because that won't work.
>
> Inverting has a series resistor from the signal input into the inverting
> input, with a feedback resistor from the output to the inverting
> input.  The non-inverting input goes to ground, either directly or
> through a resistor that exists to keep things balanced DC wise.
>
> Non-inverting has the same resistor from the output to the inverting
> input.  Then a resistor from the inverting input to ground.  Then
> the signal is fed into the non-inverting input.
>
> There always has to be feedback from the output to the invering input.
>
>   Michael

Hello Michael:

I'm actually creating a what-if scenario. What if in an inverting
opamp configuration the opamp inputs are flipped? How would I prove
via calculations that it will not work? That was just my curiosity.

"MRW" <m...@gmail.com> wrote in message 
news:1...@l53g2000cwa.googlegroups.com...
> On Feb 1, 11:50 am, e...@FreeNet.Carleton.CA (Michael Black) wrote:
>> I hope you aren't saying you connected the feedback resistor to the
>> non-inverting input.  Because that won't work.
>>
>> Inverting has a series resistor from the signal input into the inverting
>> input, with a feedback resistor from the output to the inverting
>> input.  The non-inverting input goes to ground, either directly or
>> through a resistor that exists to keep things balanced DC wise.
>>
>> Non-inverting has the same resistor from the output to the inverting
>> input.  Then a resistor from the inverting input to ground.  Then
>> the signal is fed into the non-inverting input.
>>
>> There always has to be feedback from the output to the invering input.
>>
>>   Michael
>
> Hello Michael:
>
> I'm actually creating a what-if scenario. What if in an inverting
> opamp configuration the opamp inputs are flipped? How would I prove
> via calculations that it will not work? That was just my curiosity.
>

An ideal opamp is just a comparator and does not depend on which which input 
pin is used but only on the difference in the signal.

When you add a feeback loop then you need to make sure that you feed back to 
the right input else you increase your amplifciation differential instead of 
decreasing it(which an be useful but( isn't what an op amps about).

Your basic ideal op amp is



----- I1 ---|\
            | \
            R  >-- O = G*(I1 - I2)
            | /
----- I2 ---+/


I1,I2 are your input's, O is your output and R is infinite resistance.


In this case I1 and I2 are identical. You flip them and it doesn't matter 
except on your gain(you will flip the O in sign).

If you use some feedback then fliping which input it connects too flips it 
in the equation but now your feedback gain differential changes in sign and 
now adds instead of subtracts.  So you get infinite output. That is the 
reason why the feeback is connected to the inverting pin. If not then its 
kinda useless to use feedback(Maybe there are some uses for it though).

Jon

On Feb 1, 12:22 pm, "Jon Slaughter" <Jon_Slaugh...@Hotmail.com> wrote:
> Jon

Hi Jon:

Ideally, just using the opamp golden rules, if I were to calculate the
close loop gain for an inverting opamp configuration, I would get Vo/
Vi = - Rf / Ri. I've been told that the "minus sign is due to the fact
that the op-amp is inverting." But if my inputs are flipped for the
same inverting configuration (+ becomes -, - becomes +), then my
calculation steps would still be using the same methods and
assumptions, so I would still get Vo/Vi = -Rf / Ri (?), right?

Would it be valid to say that since the feedback network is still the
same as the correct inverting configuration, but the opamp inputs are
just flipped, we can just flip the sign on the close loop gain?
There's no mathematical step for this?

On 1 Feb 2007 09:53:32 -0800, "MRW" <m...@gmail.com> wrote:

>On Feb 1, 12:22 pm, "Jon Slaughter" <Jon_Slaugh...@Hotmail.com> wrote:
>> Jon
>
>Hi Jon:
>
>Ideally, just using the opamp golden rules, if I were to calculate the
>close loop gain for an inverting opamp configuration, I would get Vo/
>Vi = - Rf / Ri. I've been told that the "minus sign is due to the fact
>that the op-amp is inverting." But if my inputs are flipped for the
>same inverting configuration (+ becomes -, - becomes +), then my
>calculation steps would still be using the same methods and
>assumptions, so I would still get Vo/Vi = -Rf / Ri (?), right?
>
>Would it be valid to say that since the feedback network is still the
>same as the correct inverting configuration, but the opamp inputs are
>just flipped, we can just flip the sign on the close loop gain?
>There's no mathematical step for this?
>

No. The feedback becomes positive, the gain becomes infinite (or
more!) and the opamp slams to one of the power rails.

Interestingly, with an ideal opamp model, some simulators will show a
zero-input, zero-output state here, which doesn't happen in real life.

John

"MRW" <m...@gmail.com> wrote in message 
news:1...@v33g2000cwv.googlegroups.com...
> On Feb 1, 12:22 pm, "Jon Slaughter" <Jon_Slaugh...@Hotmail.com> wrote:
>> Jon
>
> Hi Jon:
>
> Ideally, just using the opamp golden rules, if I were to calculate the
> close loop gain for an inverting opamp configuration, I would get Vo/
> Vi = - Rf / Ri. I've been told that the "minus sign is due to the fact
> that the op-amp is inverting." But if my inputs are flipped for the
> same inverting configuration (+ becomes -, - becomes +), then my
> calculation steps would still be using the same methods and
> assumptions, so I would still get Vo/Vi = -Rf / Ri (?), right?
>
> Would it be valid to say that since the feedback network is still the
> same as the correct inverting configuration, but the opamp inputs are
> just flipped, we can just flip the sign on the close loop gain?
> There's no mathematical step for this?
>
>


Heres a very simple analysis:

A feedback of the output voltage Vo into the positive or negative input can 
be represented as x*Vo where x is the fractional amount of Vo that we are 
feeding back. 0 <= x <= 1. We cannot stick any less unless we have some 
other means of amplification(such as another op amp).

Since you know that for open loop op amp one has

Vo = A*(V2 - V1)

where V2 is the non-inverting input and V1 is the inverting input.

If we add a feedback path to the non-inverting then we have V2 = V2 + x*Vo 
or

Vo = A*(V2 + x*Vo - V1)

solving for Vo gives

Vo = A/(1 - x*A)*(V2 - V1)

The same on the inverting pin gives

Vo = A/(1 + x*A)*(V2 - V1)

So here is the main difference, for the negative feedback we have the 
amplification factor

A/(1 + x*A) = 1/(1/A + x)

Notice that this factor is always between A/(1 + A) and A.

For the positive feedback we have

A/(1 - x*A) = -A/(x*A - 1)

With positive feedback we invert the output for certain x but more 
importantly we get infinite gain when x = 1/A and its very unstable around 
this point. If we are slightly below 1/A then we are non-inverting but 
slightly above we are inverting. Also there is no way to reduce the below a 
factor of 1. So here it is always amplification(which is key).


The equations themselfs don't look much different but the - sign is a big 
deal.

Op amps are made with large A and not small A. For a real op amp we have

1/(1/A + x) ~= 1/x

or

1/(1/A - x) ~= -1/x

but in the second case we have unstability because x might be around 1/A 
(1/A is close to 0).


Now if you understand all that its not hard to see that a real op amp 
doesn't like positive feedback. We could use it on an ideal opamp and get 
gain for anything above 1. But in the real world the op amps will surely it 
1/A when fed back into. You cannot make x so large to feed back all of the 
input and because it actually does this in a continuous way(the op amp has 
to "gradually" turn on) it will end up "hitting" 1/A and just loose control.

So while its true that they look like they are inverses of each other. i.e. 
1/x vs -1/x, In reality the second case is very unstable and is only a 
theoretical possibility.

In any case the negative feedback does everything the positive feedback can 
and more.  (we can get gains from 1 to infinity. (very similar to positive 
feedback case)


Hope this is more clear,
Jon

MRW wrote:
> On Feb 1, 12:22 pm, "Jon Slaughter" <Jon_Slaugh...@Hotmail.com> wrote:
>> Jon
>
> Hi Jon:
>
> Ideally, just using the opamp golden rules, if I were to calculate the
> close loop gain for an inverting opamp configuration, I would get Vo/
> Vi = - Rf / Ri. I've been told that the "minus sign is due to the fact
> that the op-amp is inverting." But if my inputs are flipped for the
> same inverting configuration (+ becomes -, - becomes +), then my
> calculation steps would still be using the same methods and
> assumptions, so I would still get Vo/Vi = -Rf / Ri (?), right?
>
> Would it be valid to say that since the feedback network is still the
> same as the correct inverting configuration, but the opamp inputs are
> just flipped, we can just flip the sign on the close loop gain?
> There's no mathematical step for this?


Your so called mathematical model is only valid for a linear circuit, and a 
comparator with hysteresis(this is what you get) is a *non-linear* circuit 
element, which nevertheless can also be descibed with a set of equations, 
but completely different ones. You can calculate the threshold values going 
from negative to positive and vice versa. just assume one of the stable 
stages and calculate which input is needed to change the output sign.
-- 
ciao Ban
Apricale, Italy 

On Thu, 1 Feb 2007 13:42:17 -0600, "Jon Slaughter"
<J...@Hotmail.com> wrote:

>
>"MRW" <m...@gmail.com> wrote in message 
>news:1...@v33g2000cwv.googlegroups.com...
>> On Feb 1, 12:22 pm, "Jon Slaughter" <Jon_Slaugh...@Hotmail.com> wrote:
>>> Jon
>>
>> Hi Jon:
>>
>> Ideally, just using the opamp golden rules, if I were to calculate the
>> close loop gain for an inverting opamp configuration, I would get Vo/
>> Vi = - Rf / Ri. I've been told that the "minus sign is due to the fact
>> that the op-amp is inverting." But if my inputs are flipped for the
>> same inverting configuration (+ becomes -, - becomes +), then my
>> calculation steps would still be using the same methods and
>> assumptions, so I would still get Vo/Vi = -Rf / Ri (?), right?
>>
>> Would it be valid to say that since the feedback network is still the
>> same as the correct inverting configuration, but the opamp inputs are
>> just flipped, we can just flip the sign on the close loop gain?
>> There's no mathematical step for this?
>>
>>
>
>
>Heres a very simple analysis:
>
>A feedback of the output voltage Vo into the positive or negative input can 
>be represented as x*Vo where x is the fractional amount of Vo that we are 
>feeding back. 0 <= x <= 1. We cannot stick any less unless we have some 
>other means of amplification(such as another op amp).
>
>Since you know that for open loop op amp one has
>
>Vo = A*(V2 - V1)
>
>where V2 is the non-inverting input and V1 is the inverting input.
>
>If we add a feedback path to the non-inverting then we have V2 = V2 + x*Vo 
>or
>
>Vo = A*(V2 + x*Vo - V1)
>
>solving for Vo gives
>
>Vo = A/(1 - x*A)*(V2 - V1)
>
>The same on the inverting pin gives
>
>Vo = A/(1 + x*A)*(V2 - V1)
>
>So here is the main difference, for the negative feedback we have the 
>amplification factor
>
>A/(1 + x*A) = 1/(1/A + x)
>
>Notice that this factor is always between A/(1 + A) and A.
>
>For the positive feedback we have
>
>A/(1 - x*A) = -A/(x*A - 1)
>
>With positive feedback we invert the output for certain x but more 
>importantly we get infinite gain when x = 1/A and its very unstable around 
>this point. If we are slightly below 1/A then we are non-inverting but 
>slightly above we are inverting. Also there is no way to reduce the below a 
>factor of 1. So here it is always amplification(which is key).
>
>
>The equations themselfs don't look much different but the - sign is a big 
>deal.
>
>Op amps are made with large A and not small A. For a real op amp we have
>
>1/(1/A + x) ~= 1/x
>
>or
>
>1/(1/A - x) ~= -1/x
>
>but in the second case we have unstability because x might be around 1/A 
>(1/A is close to 0).
>
>
>Now if you understand all that its not hard to see that a real op amp 
>doesn't like positive feedback. We could use it on an ideal opamp and get 
>gain for anything above 1. But in the real world the op amps will surely it 
>1/A when fed back into. You cannot make x so large to feed back all of the 
>input and because it actually does this in a continuous way(the op amp has 
>to "gradually" turn on) it will end up "hitting" 1/A and just loose control.
>
>So while its true that they look like they are inverses of each other. i.e. 
>1/x vs -1/x, In reality the second case is very unstable and is only a 
>theoretical possibility.
>
>In any case the negative feedback does everything the positive feedback can 
>and more.  (we can get gains from 1 to infinity. (very similar to positive 
>feedback case)

---
Very nice, but it seems you've left out the effect of the delay
through the opamp.


-- 
JF

On Feb 1, 2:42 pm, "Jon Slaughter" <Jon_Slaugh...@Hotmail.com> wrote:
>...<snip>...
> Hope this is more clear,
> Jon

Thanks, Jon! :-) Much clearer.