Because an infinite input impedance would be desired if possible. This is because no current would flow into (or out of) the input of the amplifier, no voltage would be dropped across the internal impedance of the signal source and the whole voltage provided by the signal source would appear at the amplifier's input and thus a higher output voltage would be got.
because the gain is infinite and output voltage is at finite level: Uout = Uin x Gain = 0 x ∞ = finite voltage. This basic idea (input voltage at zero) simplifies the OPAMP usage in real applications. Real operational amplifier requires properties close to ideal requirements: • High amplification gain (open loop) – for example: gain ...

Applying these assumptions to the standard op-amp model results in the ideal op-amp model shown in Figure 3. Because R i = ∞ and the voltage difference V p – V n = V i at the input port is finite, the input currents are zero for an ideal op-amp: in = ip = 0 (6) Hence there is no loading effect at the input port of an ideal op-amp: V i =V s (7)
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output resistor RO of the op-amp and the load resistor RL and output in Figure 1.1.2. Here, the signal can be output without being attenuated if the RO is sufficiently smaller than the RL (RO=0) because the second term can be approximated by 1. Such an op-amp is called an ideal op-amp. Usually, op-amps with high input ?

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  • Apr 28, 2013 · While an ideal op amp has infinite gain, a real op amp has very high, but finite DC gain. It also has a pole at a fairly low frequency that causes the op amp’s gain to roll off, eventually reaching unity gain at some much higher frequency. This pole is referred to as the dominant pole.
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Nov 13, 2018 · Because the op amp terminals are open circuit (or, at least, we are pretending they are) you know the current through R1 must also be the current through R2. So we know the voltage across R2 which ...

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