An opamp question I can't make sense of

[circuit975]

Greetings to everyone,

I found all values in an opamp question, but I couldn't find the Vout output value. How do we do the formulation?
I mean, with which formula will we calculate vout?

 

[Tony Stewart]

And, changing the question is a cop-out and helps nobody.

That comment was directed to the Op and your comment is irrelevant noise.

 

[ZipZapOuch]

That comment was directed to the Op and your comment is irrelevant noise.

Nice try deflecting -but the comment was directed at you.

 

[MrAl]

Greetings to everyone,

I found all values in an opamp question, but I couldn't find the Vout output value. How do we do the formulation?
I mean, with which formula will we calculate vout?


View attachment 146364

Hi,

There are a couple ways to analyze this circuit. Although you may be able to calculate the output the way the inputs are right now, it's good to know what happens over the entire range of inputs. That's where circuit analysis comes in.

To start, since you are really just looking at voltages and not paying too much attention to time or current, the best model to use is going to be a voltage oriented model for the op amp. We have two inputs:
vn: the inverting input, and
vp: the non inverting input.
We also have a gain called the open loop gain, which is usually 100000 or more but could vary quite a bit. We will just call this AOL for now and then look at what happens when that gain is 100000.

The op amp will amplify the difference between vp and vn by the factor of the open loop gain AOL. That means we have to start:
Vout=(vp-vn)*AOL

See how simple that can be?

There is a limitation though, and that is that if Vout becomes greater than Vcc (+14v) then Vout=Vcc, and if Vout becomes less than -Vee (-14v) then Vout becomes -Vee. So we then end up with:
Vout=(vp-vn)*AOL, with Vout limited as: Vee<Vout<Vcc.

So this is still rather simple, we just have to remember to clamp Vout if it becomes too large or too small.
Now to look at the case in your schematic where vp=3v and vn=2v.
Using that formula, we end up with:
Vout=(3-2)*AOL
which is:
Vout=1*AOL
and with AOL=100000, that means:
Vout=100000 volts.
Since Vout is greater than +14 volts, that means Vout becomes +14 volts.
That's the theoretical limit, but because these devices sometimes have a voltage drop it could be a little lower, from around +12v to +13v. Comparators with a pullup resistor could make it up to nearly +14 volts though, as well as a rail to rail op amp. An LM358 op amp might make it up to 12.5 volts or so.

So there you have it. An op amp in open loop mode amplifies the voltage difference at the input by a large amount.

Now how about small differences between vp and vn?

If we had vp=3 and vn=3, that would be a difference of zero. Using the formula:
Vout=(3-3)*AOL
Vout=0*AOL
Vout=0 volts
This is actually a possibility, although it would be rare to see this in real life. More typically we would see a small non zero difference.

If vp=3 and vn=2.9, we would have:
Vout=(3-2.9)*AOL=0.1*AOL=10000 volts, so Vout is clamped to +14 volts again.

If vp=3 and vn=2.999, we would have:
Vout=(3-2.999)*AOL=(0.001)*AOL=100 volts, so again Vout is clamped to +14 volts again.

Now it gets interesting.
If vp=3 and vn=2.9999, we would have:
Vout=(3-2.9999)*AOL=10 volts, so now we see a voltage in range which means Vout=10 volts.

If we go one more step with vp=3 and vn=2.99999 we see:
Vout=(3-2.99999)*AOL=0.00001*100000=1 volt, which means Vout is now just +1 volt.

If we allow vp to go lower than vn, then we see negative outputs:
If vp=3 and vn=3.1, we would have:
Vout=(3-3.1)*AOL=-0.1*AOL=-0.1*100000=-10000 volts which means Vout is clamped to -14 volts.
If there is also a drop at the bottom of the output range then it may only make it to -13 volts though, just like what happens with the positive output range.

Understanding these ideas also helps to understand what happens in the closed loop cases with feedback.