~analoger~
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Hello,
In the following circuit, the open-circuit voltage at node "a" should be Vs, when connected to a voltage-follower, the open circuit voltage at Vo = Va = Vs? Hence ignoring the source resistance Rs, which could be internal resistance. This way the voltage-follower eliminates the internal resistance of the voltage source and giving the voltage of an ideal source. But the OpAmp itself has output resistance which acts as internal resistance when the circuit is closed at the OpAmp output and the voltage is measured across the connected load.
Now my questions, theoretically an ideal OpAmp can be used to convert a non-ideal voltage source to ideal, and practically if we have an OpAmp with a lower output resistance than the source internal resistance, it can make the source more ideal?
How does it sense and copy the voltage of the source as shown in the picture?
Thanks.
In the following circuit, the open-circuit voltage at node "a" should be Vs, when connected to a voltage-follower, the open circuit voltage at Vo = Va = Vs? Hence ignoring the source resistance Rs, which could be internal resistance. This way the voltage-follower eliminates the internal resistance of the voltage source and giving the voltage of an ideal source. But the OpAmp itself has output resistance which acts as internal resistance when the circuit is closed at the OpAmp output and the voltage is measured across the connected load.
Now my questions, theoretically an ideal OpAmp can be used to convert a non-ideal voltage source to ideal, and practically if we have an OpAmp with a lower output resistance than the source internal resistance, it can make the source more ideal?
How does it sense and copy the voltage of the source as shown in the picture?
Thanks.