Well, both can be explained using voltage divider rule. Assuming the opamp input is connected to a voltage source having a source resistance of Rs and the opamp input impedance is Ri, the voltage going into the opamp is Ri/(Ri+Rs). As can be seen, we would like to have Ri as large as possible. The low output impedance of opamp could be explained similarly be connecting a load having a impedance of RL.
The first stage or the differential input stage of the amplifier must have very high input impedance. This will cause the op-amp to draw very negligible amounts of input current. The very small input current enables the user to utilize the ideal op-amp equations for circuit analysis purposes.
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We also see that the op-amp must have very low output impedance. This minimizes the loading of the output ofthe op-amp by output stage of the op-amp. The outputstage delivers current to the op-amp’s load and it may ormay not have short circuit protection.
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Output impedance of an amplifier: Amplifiers should have very low output impedance and usually it’s assumed to be zero and in this way op-amps will behave like a voltage source and it will be capable of driving a very wide range of loads.
The input resistance of OPA is more bigger,the input bias current of OPA is more smaller,the OPA doesn't produce little load effect for input source!
The output resistance of OPA is more smaller,the OPA just more like a ideal voltage source,so the OPA can deliver big current to load.
low output impedence is need to drive big loading and it's gain will not be affected and have strong fanout capbility. High input impedence is needed to be drived by other low current output stage.
High input impedence and low output impedence are for maximum signal transmission. this is not the same as power transmission. The best configuration for maximum power transmission is a matched input and output.