Operational Amplifier

Thread Starter

Juss

Joined Nov 24, 2013
14
Seriously?

Look at the pic in your problem. Work from the output of the op-amp toward the ground.

At the output of the op-amp we have 4k resistor, there will be some voltage across the 4k resistor. Is it Vp?

From the 4k resistor we have another resistor, the 6k. There will be some voltage drop across the 6k resistor. Is it Vp?
The voltage drop across 4k resistor is Vout? Then the voltage drop across 6k resistor is Vp?
 

shteii01

Joined Feb 19, 2010
4,644
The voltage drop across 4k resistor is Vout? Then the voltage drop across 6k resistor is Vp?
Vp is across 6k. So you now can use voltage divider correctly.

On the more general note. Vout is across 4k in series with 6k. Remember, Vout is from output of op-amp to ground. 4k does not touch ground, so Vout could not possibly be across 4k.

Basically you have Vp plus some other voltage, both of these voltages together are equal to Vout. Vp is across 6k. Some other voltage is across 4k. Together they are equal Vout.
 

Thread Starter

Juss

Joined Nov 24, 2013
14
Vp is across 6k. So you now can use voltage divider correctly.

On the more general note. Vout is across 4k in series with 6k. Remember, Vout is from output of op-amp to ground. 4k does not touch ground, so Vout could not possibly be across 4k.

Basically you have Vp plus some other voltage, both of these voltages together are equal to Vout. Vp is across 6k. Some other voltage is across 4k. Together they are equal Vout.
So is Vp = (6*Vout)/10
 

shteii01

Joined Feb 19, 2010
4,644
For part (d), am I doing it right?

(0-Vx)/4k + (Vout-0)/16k = 0
Vx = 6
so Vout = 24V
I am kinda weak at that, but I think the way you are doing it is wrong.

Here is why.

Your op-amp is powered by +15 volts and -15 volts. These are also known as rails. So your upper rail is +15 volts. Your lower rail is -15 volts. These rails set limits on how high and how low your output can go.

Lets examine what you got so far. You say your output is 24 volts. Your upper rail is 15 volts. The upper rail sets the maximum upper output. Therefore, the maximum you could possibly get from op-amp is 15 volts. You got 24 volts. How is that possible? It is not possible. Which means you did something wrong.
 

Thread Starter

Juss

Joined Nov 24, 2013
14
I am kinda weak at that, but I think the way you are doing it is wrong.

Here is why.

Your op-amp is powered by +15 volts and -15 volts. These are also known as rails. So your upper rail is +15 volts. Your lower rail is -15 volts. These rails set limits on how high and how low your output can go.

Lets examine what you got so far. You say your output is 24 volts. Your upper rail is 15 volts. The upper rail sets the maximum upper output. Therefore, the maximum you could possibly get from op-amp is 15 volts. You got 24 volts. How is that possible? It is not possible. Which means you did something wrong.
I see. Thanks a lot for helping me!
 

shteii01

Joined Feb 19, 2010
4,644
I see. Thanks a lot for helping me!
Like I said earlier, I am kinda weak at this type of op-amp circuits. But I think you are dealing with comparator. So.
Vout=Gain(Vn-Vp)

We already found out Vp in terms of Vout.
Vout=Gain(\(\frac{3}{5}Vout\)-Vn)

Find out Gain.
Find out Vn in terms of Vx and Vout.
 
Last edited:

Jony130

Joined Feb 17, 2009
5,487
Juss - All you need to do is to simply write two nodal equation.
First equation for Vn node and the second one for Vp node.
Then if we assume that we have ideal opamp we have Vn = Vp
Can you do this ?
 

Jony130

Joined Feb 17, 2009
5,487
I also want to add that the example circuit post by OP is very bad example.
Because in this circuit positive feedback wins, and then Vn = Vp is no longer true.
So instead of a amplifier we have a Schmitt trigger.
 

shteii01

Joined Feb 19, 2010
4,644
I also want to add that the example circuit post by OP is very bad example.
Because in this circuit positive feedback wins, and then Vn = Vp is no longer true.
So instead of a amplifier we have a Schmitt trigger.
That is what wiki says too:
"Another typical configuration of op-amps is with positive feedback, which takes a fraction of the output signal back to the non-inverting input. An important application of it is the comparator with hysteresis, the Schmitt trigger."

We have positive feedback. It is a fraction of the output, 3/5 of output to be exact.
 

Jony130

Joined Feb 17, 2009
5,487
Yes, but if we change resistor so that Vn = X * Vout is larger then Vp = Y* Vout.
The negative feedback will start to win and circuit will work as a amplifier.

In our case 4/20 = 1/5 = 0.2 is smaller than them 6/10 = 3/5 = 0.6 and this is why positive feedback wins.
 

shteii01

Joined Feb 19, 2010
4,644
Yes, but if we change resistor so that Vn = X * Vout is larger then Vp = Y* Vout.
The negative feedback will start to win and circuit will work as a amplifier.

In our case 4/20 = 1/5 = 0.2 is smaller than them 6/10 = 3/5 = 0.6 and this is why positive feedback wins.
Since I already have the formula for his Vn, I would not set Vn=Vp. I would simply solve:
Vout=Gain(Vp-Vn)
All the parts except Vout are known. One unknown (Vout), one equation. Just algebra.
 

Jony130

Joined Feb 17, 2009
5,487
But do you know the gain (op amp open loop gain) value? And if we assume Gain = ∞ then this means that Vn = Vp if op amp works in linear region.
So we left with a very simple equations
Vp = Vout*6/(4+6)
and
Vn = Vout*4/(4+16) + Vx*16/(4+16)

and if

Vn = Vp

Vout*6/(4+6) = Vout*4/(4+16) + Vx*16/(4+16)

So all we need to do now is to solve for Vout.
 
Last edited:

shteii01

Joined Feb 19, 2010
4,644
But do you know the gain (op amp open loop gain) value? And if we assume Gain = ∞ then this means that Vn = Vp if op amp works in linear region.
So we let with a very simple equation
Vp = Vout*4/(4+6)
and
Vn = Vout*4/(4+16) + Vx*16/(4+16)

and if

Vn = Vp

Vout*4/(4+6) = Vout*4/(4+16) + Vx*16/(4+16)

So all we need to do now is to solve for Vout.
If I got it solved right, your solution is 24 volts, which is not possible because our rail is 15 volts.
 

shteii01

Joined Feb 19, 2010
4,644
But if you build this circuit in real life you will not get 12V at the output.
The output voltage will be close to positive saturation voltage.
It looks like a problem from a textbook. If we did not want to deal with textbook problem, we should have left this thread alone.
 

WBahn

Joined Mar 31, 2012
29,979
It looks like a problem from a textbook. If we did not want to deal with textbook problem, we should have left this thread alone.
Even as a problem from a textbook the output would not be 12V because that is not stable. Assume the output is +15V and determine the differential input voltage. Is that input compatible with a saturated output? Now do the same for an assumed output of -15V. If either assumption is compatible with the resulting differential input, then the output will saturate at the corresponding rail.

The above technique assumes that the circuit, sans the op amp, is linear (which it is in this case). The more general technique it so compute Vd=Vp-Vn (sans the op amp) and then determining if dVd/dVout is positive or negative. If it is negative at a particular operating point, then the circuit will be stable at that point. If it is positive, then it won't. The reason is simple, if the output changes by some small amount and Vd changes in the same direction (has a positive derivative), then the amplifier will drive the output further in that direction.

If you do this, then you get the constraint that, to be stable within the active region,

Rfp/Rp > Rfn/Rn

Where Rfp and Rp are the feedback and input resistors on the non-inverting side while Rfn and Rn are the feedback and input resistors on the inverting side.

In this case we have

Rfp/Rp = 4kΩ/6kΩ = 2/3
Rfn/Rn = 16kΩ/4kΩ = 4

and the stability requirement is not met.

In fact, if you crank the math, you find that in order to be stable, this configuration has to result in a net negative gain.
 
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