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Operational Amplifier problem

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v9260019

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The following statemtnt is from texts
"An independet voltage source is often a representation of a "power supply" or a "singal generator." An independent current source is often designed and constructed with other elements; so too,are the dependent sources. As it happens,
there is a complex electronic unit called the operatinoal amplifier that can be used to construct each and every one of these two-terminal active elements~~~~~"

Is it mean that I can get a independent voltage source, a independent current source, a dependent voltage source ,a dependent current source from "operational amplifiers" right ?????
 

VVV

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Yes, you can use opamps to build those sources.
As an example, imagine the simple amplifier buit with an opamp. Its output voltage looks very much like a controlled voltage source, since the output is proportional to the input, by the gain: Vo=A*Vin.

For current source, the circuits are more cimplex, but the current through the feedback resistor in an opamp amplifier is also linear function of the input voltage.
Assume the + input is grounded. Then the current through the input resistor is simply
Iin= Vin/Rin.
Since the input of the opamp draws theoretically no current, it follows that the current through the feedback resistor equal the above current.
Iout=Iin=Vin/Rin
 

v9260019

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Thanks VVV
So if let Vin is a constant, then it becomes a indpendent source~~~ right ???
 

u04f061

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yes operational amplifiers have this quality that u can get all above 4-types of sources from them
 

VVV

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Yes, if Vin is a constant, then the output is constant and since the output impedance of the opamp is relatively low, it behaves like a fair voltage source.
Sometimes this property is used to create a "virtual ground".

Imagine that you need a negative voltage. If you use an opamp in a follower connection, with its input biased to some arbitrary voltage with respect to ground, then the output will be at that voltage, too.
But if in your circuit you consider the output of the opamp to be the ground, or the "virtual ground", then the former ground is negative with respect to the virtual ground. Thus, it's like you have a negative supply on board, with respect to this virtual ground, and a positive one, derived from the normal supply voltage minus the virtual ground potential.
The only catch is that the "ground" and the "virtual ground" are not really connected together. But this method is used sometimes, to avoid a negative supply on board.
 

Santoshalagawadi

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see freind i am not getting your question please send it in short form manner
 

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