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#### arthur442

Joined Jul 10, 2021
18

How can I find V3 and I2?

What i have tried is:

Va = Vb

I2 = ( 5v - V3 ) / ( 5k + 2k ) = ? -> ( How can i find V3 ?)

( 10k / 5k ) + 1 = 3 = gain amp 1

Vo1 = 3 * Va
Io1 = Vo1 / 12.5k
V3 = V2 = Vo1 - ( Io1 * 2.5k )
But I don't know what else to do?

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#### arthur442

Joined Jul 10, 2021
18
View attachment 243379

How can I find V3 and I2?

What i have tried is:

Va = Vb

I2 = ( 5v - V3 ) / ( 5k + 2k ) = ? -> ( How can i find V3 ?)

( 10k / 5k ) + 1 = 3 = gain amp 1

Vo1 = 3 * Va
Io1 = Vo1 / 12.5k
V3 = V2 = Vo1 - ( Io1 * 2.5k )
But I don't know what else to do?

Ignore the attachments, I didn't know they were added

#### ericgibbs

Joined Jan 29, 2010
18,673
hi 442,
One of the attachments has the answer for Vo and V3..?

E

#### Jony130

Joined Feb 17, 2009
5,487
Notice that I2 = (Vin - Vo)/(10kΩ) therefore V3 = Vo + I2*3kΩ

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#### arthur442

Joined Jul 10, 2021
18
hi 442,
One of the attachments has the answer for Vo and V3..?

E
hi 442,
One of the attachments has the answer for Vo and V3..?

E
I know, but I don't know how they get it?

#### ericgibbs

Joined Jan 29, 2010
18,673
hi 442,
@jony in post #4 gives you a starting point.
E

#### arthur442

Joined Jul 10, 2021
18
Notice that I2 = (Vin - Vo)/(10kΩ) therefore V3 = Vo + I2*3kΩ
Okay, but shouldn't you first find Va to get Vo1 and then calculate V2 to get the output Vo? Or am I wrong?

#### Jony130

Joined Feb 17, 2009
5,487
All you need is to write two equations.

The first one is Va = Vb and the second one is V3 = V2. And try to express Va in terms of I2 and Vin and Vb and V2 in terms of Vo1.

#### RBR1317

Joined Nov 13, 2010
712
I don't believe there should be any need to calculate any currents since all voltages are determined by voltage dividers and the constraint conditions: Va=Vb & V3=V2.

#### LvW

Joined Jun 13, 2013
1,749
I don't believe there should be any need to calculate any currents since all voltages are determined by voltage dividers and the constraint conditions: Va=Vb & V3=V2.
Yes - however, this assumes that the circuit would have a stable DC operating point.
But the main (outer) loop provides positive DC feedback. The circuit is unstable.

#### The Electrician

Joined Oct 9, 2007
2,971
Yes - however, this assumes that the circuit would have a stable DC operating point.
But the main (outer) loop provides positive DC feedback. The circuit is unstable.
This is true provided the opamps are assumed to be ideal: zero input current, infinite gain, zero ohms output impedance.

However, if the two opamps both have identical gain, and that gain is less than about 7.051, the circuit will be DC stable.

If the + and - inputs of either opamp are reversed, the circuit will be DC stable.

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#### LvW

Joined Jun 13, 2013
1,749
This is true provided the opamps are assumed to be ideal: zero input current, infinite gain, zero ohms output impedance.

However, if the two opamps both have identical gain, and that gain is less than about 7.051, the circuit will be DC stable.

If the + and - inputs of either opamp are reversed, the circuit will be DC stable.
Do you really think that it matters if the opamps are to be considered as real or ideal?
The DC loop gain for ideal opamps is Aloop=+1.7 and for real devices it perhaps 1% lower ...

I do not understand the statement in the last line of your post....do you recommend to swap both opamp input nodes or only one (which one?).

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#### arthur442

Joined Jul 10, 2021
18
I found it.
I2 = 1.4mA
thanks for helping

#### LvW

Joined Jun 13, 2013
1,749
I found it.
I2 = 1.4mA
thanks for helping
Arthur, are you aware that you have computet something which has no meaning in practice?
You have assumed Va=Vb and V2=V3 - however, this assumption holds only for a stable circuit.
But the whole circuit is unstable - it does not work!

#### The Electrician

Joined Oct 9, 2007
2,971
Do you really think that it matters if the opamps are to be considered as real or ideal?
The DC loop gain for ideal opamps is Aloop=+2.45 and for real devices it perhaps 1% lower ...
I consider an ideal opamp to have infinite gain. Opinions may vary, but to me an otherwise ideal opamp with finite gain is not ideal.

I do not understand the statement in the last line of your post....do you recommend to swap both opamp input nodes or only one (which one?).
"either opamp" means one of them only.

#### The Electrician

Joined Oct 9, 2007
2,971
I found it.
I2 = 1.4mA
thanks for helping
Because of the fact that there is positive feedback, this circuit will latch up (opamp outputs will be clamped to power supply rails, or nearly so, in a real circuit) if the opamp gains are greater than 7.051. The voltages at the nodes will not really be what a calculation assuming linear operation would give you even though the mathematics gives you a number. Your instructor should give you credit for noticing this.

For extra credit, assume the + and - inputs of the rightmost opamp are reversed and then calculate I2 and V3. This calculation will give you a valid result.

By the way, where did this circuit come from? If it's from a textbook, what is that textbook?

#### LvW

Joined Jun 13, 2013
1,749
I consider an ideal opamp to have infinite gain. Opinions may vary, but to me an otherwise ideal opamp with finite gain is not ideal.
I agree, of course.
But the question is not how an ideal opamp is to be defined, but:
Does it matter if we consider an opamp as ideal or real - as far as the stability of the DC operating point is concerned
This involves the loop gain only (and the stability criterion). In both cases, the loop gain > +1.
So the circuit will not assume a stable DC operating point.

"either opamp" means one of them only.
This will not stabilize the circuit.

#### LvW

Joined Jun 13, 2013
1,749
Because of the fact that there is positive feedback, this circuit will latch up (opamp outputs will be clamped to power supply rails, or nearly so, in a real circuit)
Aha - fine. It seems we have come to an agreement.

#### arthur442

Joined Jul 10, 2021
18
Because of the fact that there is positive feedback, this circuit will latch up (opamp outputs will be clamped to power supply rails, or nearly so, in a real circuit) if the opamp gains are greater than 7.051. The voltages at the nodes will not really be what a calculation assuming linear operation would give you even though the mathematics gives you a number. Your instructor should give you credit for noticing this.

For extra credit, assume the + and - inputs of the rightmost opamp are reversed and then calculate I2 and V3. This calculation will give you a valid result.

By the way, where did this circuit come from? If it's from a textbook, what is that textbook?
Sorry but I don't fully understand you as I have now found I2.
And by this I can calculate the rest of the circuit right?

Top right you see the solution for Vo and V3.

#### LvW

Joined Jun 13, 2013
1,749
Top right you see the solution for Vo and V3.
Arthur - as I have mentioned in my post, all these solutions are not correct.
You are not allowed to use the condition (Va=Vb and V2=V3)

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