I have a question regarding op amp fundamentals that I am struggling with. It is about the input current for an op amp. I am designing a simple current amplifier using the OPA548. My input is 1V pk-pk and I'm outputting 10V pk-pk and pushing around 333mA. I'm also powering the op amp with +- 12V. Without violating the laws of physics, how can it be proven for my circuit that power in = power out. What is the typical input current for an op amp? I understand that the input current is theoretically zero but then how does the power in = power out? The reason I am asking this is because I am using a 12V power supply that outputs around 1.3 amps to power both an NI module and this op amp. I am wondering if the power supply will output enough current for both the NI module and the op amp? Do I have to be concerned with the current flowing from the rails? Is that current theoretically suppose to be zero? I know my understanding of how an op amp works is not very good and I would appreciate your help in clearing this up for me. Thanks in advance.
Consider your op-amp, there are two power sources. The 12V DC source and your 1Vpp input. So you need to consider both this sources when you calculate the input power.
The power out term should include the power delivered to the load + the power loss in the op-amp + power dissipation in other components in the circuit.
Theoretically with a perfect op-amp, your input power from the 1Vpp signal is 0. The power from the 12V rail is not zero. It is equal to the wasted power internal to the op-amp plus the power given to the load (10Vpp with 333mA). I guess in a perfect op-amp no power is wasted internal to it, so then power from the rails = power given to the output.
The op-amp would just be altering the form of the power (voltage and current) based upon the 1Vpp signal given to it.
Input power is NOT (only) the signal power delivered to the signal inputs but the total power (ac and dc) consumed by the amplifying device. The signal input power can be, in most cases, neglected - if compared with the dc power. Hence, we can say that such an amplifier does not really "amplify"; instead the device "transfers" some portions of the dc power into ac power (signal output).