# Feedback become open circuit, deciding output voltage

#### beggi9

Joined Mar 3, 2016
42

#### MrAl

Joined Jun 17, 2014
9,347
Hello,

Think about this. If the feedback becomes open, the device acts like a comparator, not an op amp.
The inverting input wont be zero then unless all the inputs are zero because what makes that input zero when there are input voltage too is the feedback.

• beggi9

#### beggi9

Joined Mar 3, 2016
42
Hello,

Think about this. If the feedback becomes open, the device acts like a comparator, not an op amp.
The inverting input wont be zero then unless all the inputs are zero because what makes that input zero when there are input voltage too is the feedback.
Ahh okay so the Vref would be zero and inverting input would be higher than that so Vo=v- (-Vcc) ?

#### MrAl

Joined Jun 17, 2014
9,347
Ahh okay so the Vref would be zero and inverting input would be higher than that so Vo=v- (-Vcc) ?
Hi,

I cant be sure what you mean by "Vo=v- (-Vcc)"

Is that v minus a minus Vcc, or is that v minus times minus Vcc, or is that the v minus function of minus Vcc?

In short, since the inverting input has a positive voltage and that is greater than the non inverting input, that means the output will simply be -Vcc (minus Vcc) because the gain causes the output to go to the full negative supply voltage. That's if the op amp type can allow rail to rail operation, which we do assume in many of these theoretical situations.

What you might want to do next is calculate the voltage at the inverting terminal, given the feedback is open.

• beggi9

#### beggi9

Joined Mar 3, 2016
42
Hi,

I cant be sure what you mean by "Vo=v- (-Vcc)"

Is that v minus a minus Vcc, or is that v minus times minus Vcc, or is that the v minus function of minus Vcc?

In short, since the inverting input has a positive voltage and that is greater than the non inverting input, that means the output will simply be -Vcc (minus Vcc) because the gain causes the output to go to the full negative supply voltage. That's if the op amp type can allow rail to rail operation, which we do assume in many of these theoretical situations.

What you might want to do next is calculate the voltage at the inverting terminal, given the feedback is open.
Okay, yes sorry I meant -Vcc (just that in the picture it's marked v- but that could be the marking for the inverting input).
Thank you so much for your help!

#### MrAl

Joined Jun 17, 2014
9,347
Okay, yes sorry I meant -Vcc (just that in the picture it's marked v- but that could be the marking for the inverting input).
Thank you so much for your help!
Hi,

Hey you are welcome, and if you care to you might calculate the voltage that appears at the inverting input with the feedback open. That might not be as important here because all the inputs are positive, but if one of them is negative then that inverting terminal voltage might be negative too or zero, which could change the output voltage. So it is also good to know how to calculate that too. It's up to you though.

• beggi9

#### beggi9

Joined Mar 3, 2016
42
Hi,

Hey you are welcome, and if you care to you might calculate the voltage that appears at the inverting input with the feedback open. That might not be as important here because all the inputs are positive, but if one of them is negative then that inverting terminal voltage might be negative too or zero, which could change the output voltage. So it is also good to know how to calculate that too. It's up to you though.
Hmmm okay how would I go about doing that? V1, V2 and V3 are all parallel so I don't know how I could add them together

#### crutschow

Joined Mar 14, 2008
30,451
Hmmm okay how would I go about doing that? V1, V2 and V3 are all parallel so I don't know how I could add them together
With no feedback the (-) input op amp junction is no longer a virtual ground summing junction, it is basically an open circuit.
Thus you treat the three 10k resistors as just three resistors in parallel to calculate the voltage at their common junction.

#### WBahn

Joined Mar 31, 2012
26,893
How are the three 10kΩ input resistors in parallel? That would only be the case if V1=V2=V3 which they clearly are not.

Since all three input voltages are greater than 0 V, it IS reasonable to conclude that V- > 0 V (when the feedback path is open circuited). That is sufficient to determine the nominal output of the opamp. If we really need V- then simple symmetry tells us that it is equal to 2 V.

• beggi9

#### DGElder

Joined Apr 3, 2016
351
Presumably what Crutschow was eluding to was finding the Thevenin equivalent of the network - where you short the 3 voltage sources which puts the resistors in parallel, so you have a 2V source in series with 3.3K resistor.

#### MrAl

Joined Jun 17, 2014
9,347
Hi,

Yes they are not really in parallel. There is a trick we can use that makes them seem almost like they are, but they are not really in parallel.

The trick when all the resistors are the same value is:
V=(v1+v2+v3)/3

and this defaults to a two source voltage divider when all resistors are the same value and we set v3=0,

and the trick when there are four sources and four equal resistors is:
V=(v1+v2+v3+v4)/4

and note that this last one defaults to an actual three source voltage divider if we set v4=0, but again all the resistors must be equal.

• beggi9

#### WBahn

Joined Mar 31, 2012
26,893
In this case we don't give a rip about the Thevenin equivalent resistance as seen by the inverting input of the amp. We only care about the open circuit voltage at that point and once we have that -- which calculating the equivalent resistance doesn't help find -- we have the information we need.

Note that symmetry answers the question directly as long as two of the source voltages are separated from third source voltage by equal and opposite amounts and that those two sources feed the same resistance. It doesn't matter what resistance the third source is feeding since that resistance has no current flowing in it.

I don't know what crutschow was thinking of -- probably a simple misinterpretation of the circuit or a silly brain fart the likes of which hit all of us from time to time.

• MrAl

#### DGElder

Joined Apr 3, 2016
351
"In this case we don't give a rip about the Thevenin equivalent resistance as seen by the inverting input of the amp"

When you say "we", is that in the imperial sense? The Thev equiv is the first step in calculating the closed loop gain and as such it just may be of interest to someone besides you.

#### WBahn

Joined Mar 31, 2012
26,893
"In this case we don't give a rip about the Thevenin equivalent resistance as seen by the inverting input of the amp"

When you say "we", is that in the imperial sense? The Thev equiv is the first step in calculating the closed loop gain and as such it just may be of interest to someone besides you.
Just "we" as it is commonly used in a conversational sense.

Why would anyone care about calculating the closed loop gain in a circuit where there IS no closed loop to find the gain of?

Did you read the question?

#### crutschow

Joined Mar 14, 2008
30,451
.............
I don't know what crutschow was thinking of -- probably a simple misinterpretation of the circuit or a silly brain fart the likes of which hit all of us from time to time.
That's as good an explanation as any. 