OPAMP question

When I get the same answer by two methods, either I am right or I am twice wrong.

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Heidi said:

Please forgive my poor English ability. What does it mean?

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When I go from question to answer using one method, that is good.
When I go from question to answer using method (1) and then again using method (2) and I get the same answer the is very good.
OR
Maybe I am two times wrong.

Heidi said:

I have to get some sleep now, it is almost 2:30 am in Taipei.

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I have walked many Km in Taipei. After work there is nothing to do. (hotel room) I take a map and some money and go walking for three hours every night. Taipei is a very safe place to walk at night.

Heidi said:

The voltage at node p is Vp=0.5V, so is the voltage at node n, Vp=Vn=0.5V.

The current through R3 (let's call it I3) flowing to the left is I3=0.5/2k=0.25mA.
The current through R4 to the right is I4=0.5/1k=0.5mA.
These two currents must come from the current through R5, that is from node out to node n, hence there's a voltage drop of 0.75m*4k=3V, so Vout=0.5+3=3.5V.
The short circuit current Is=Vout/R6 + Vn/R4 =3.5/4k + 0.5/1k =1.667mA.

The open circuit voltage V0 can be found to be Vo=0.625V by Ron's method, so the output resistance is equal to Vo/Is=0.625/1.667m=375Ω.

Is it correct?

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Yes, it is. The output impedance is the same as the Thevenin or Norton impedance, which is obtained by the method of open voltage/short current.

As you can see, if R=1000, then your answer is correct. If R=10000, then the output impedance will be 3750 ohms. Any load you put on Vo will share the voltage proportionately with the output impedance.

Ratch

Ratchit said:

Yes, it is. The output impedance is the same as the Thevenin or Norton impedance, which is obtained by the method of open voltage/short current.
View attachment 92855
As you can see, if R=1000, then your answer is correct. If R=10000, then the output impedance will be 3750 ohms. Any load you put on Vo will share the voltage proportionately with the output impedance.

Ratch

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Thank you, Ratch. I think I've got the point.

The Thevenin equivalent circuit for the original one in #1 has a Thevenin voltage Vth=Vopen=5*Vin/8, a Thevenin impedance Zth=Zout=3*R/8. Therefore, for any load resistors, the voltage across and current through the load can be easily found by voltage divider rule.

Thank you, JoeIester.
Thank you, Ron, and hope you enjoy your work and walking in Taiwan!

BTW, in the following circuits, how do we vary the values of V4 and V5 at the same time contineously, with the relationship, say, V5=0.625*V4 in PSpice or LTspice?

One could simply write the KCL equations (nodal analysis). Numbering the nodes like this:

Node 1 is Vin
Node 2 is the "+" input of the opamp
Node 3 is the "-" input of the opamp.
Node 4 is the opamp output (not Vo)
Node 5 is Vo, the output of the circuit.

This problem needs 4 KCL equations and one constraint equation (for the opamp ouput, Node 4).

The equations can be put in the form of a matrix, an admittance (Y) matrix.

The Y matrix is inverted to give a impedance (Z) matrix.

From the elements of the Z matrix we can find anything we want.

For example, the diagonal of the Z matrix gives the impedance at all 5 nodes at once, except that the value of element 4 of the diagonal is not an impedance, because the 4th equation is a constraint equation, not a KCL equation.