# Ideal operation amplifier

Discussion in 'Homework Help' started by fdsa, Aug 23, 2012.

1. ### fdsa Thread Starter New Member

Aug 16, 2011
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I really don't understand at all why $U_3=R_1 \times I_1$. Isn't I1 divided over the two resistances? Thanks in advance!

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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If the op-amp is ideal, no current flows into or out of the -ve input terminal and the current I1 is set by the ideal source. So I1 must be Io. Again for an ideal amplifier the -ve input terminal will be at virtual ground because the +ve input is grounded. So one end of R1 is at virtual ground and the other end [Node A] is at a potential of Io*R1 or 6V. From there it's easy to deduce the remaining unknown currents, including ix, since KCL must hold true at Node A.

Last edited: Aug 23, 2012
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3. ### #12 Expert

Nov 30, 2010
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Same thing, different author.

The non-inverting input is at zero volts, therefore, by definition of an ideal op-amp, the inverting input must be at zero volts.

Whatever current Io supplies must be nulled by the op-amp throgh R1. E=IR.

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4. ### ramancini8 Member

Jul 18, 2012
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It seems that with ground at both ends of I0 it is a no-op, I0=0. The real ground end of I0 has to return to a negative voltage to get any current flow from the source.
Considering the given data: I0=I1=.5mA, Vout=I1(R1)=6V, currents in R2, R3 can be found using the current divider rule.

5. ### #12 Expert

Nov 30, 2010
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Io is an imaginary ideal current source. If you can imagine it has infinite impedance, you can imagine it has a battery and a resistor in it.

6. ### ramancini8 Member

Jul 18, 2012
447
119
Well then, lets imagine the complete circuit doesn't exist. No current from the current source as drawn.

7. ### #12 Expert

Nov 30, 2010
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That simplifies the equations!

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I'm not sure exactly what the issue is here ...

An ideal current source of value Io will notionally drive that same current into any resistance. If the source is driving a zero resistance load [a short circuit] then the source terminal voltage would be zero. So it is possible in the OP's schematic to have the source Io with the same potential at either terminal whilst delivering current Io. In the case of the active end of the source the potential is virtual ground rather than a direct connection physical ground.

Of course such an ideal entity does not exist in practice. Nor for that matter, does an ideal op-amp exist in practice.

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9. ### ramancini8 Member

Jul 18, 2012
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Op amps that approach the ideal do exist in practice, while ideal current sources do not exist in practice. My warning is meant to inform the newbie that you can't build this circuit. Guess I should have said that.

10. ### #12 Expert

Nov 30, 2010
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I certainly hope the schools teach the students that they can not actually build an imiginary current source!

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Last edited: Aug 28, 2012
11. ### Tesla23 Active Member

May 10, 2009
323
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I don't think your warning will help the OP understand the operation of the circuit, and I don't think it is right anyway.

This is wrong. There are many physical processes that can generate a current into a short circuit. The first thing that came to mind was a photodiode, the fourth hit on google gives some data: http://sales.hamamatsu.com/assets/applications/SSD/photodiode_technical_information.pdf

Look at Fig 2.2, this does approximate an ideal current source at V=0, Fig 2.3(a) shows you can control the current by varying the illumination, and Fig 2.4(b) shows this exact circuit used to measure the short circuit current of the photodiode and convert it into a voltage.